Green's functions for ordinary differential equations

Construction and properties of Green's functions for boundary-value and initial-value problems.

In this chapter 4

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Green's functions in one dimension

Green's functions in one dimension

Theorem

Green's function for a boundary-value problem

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Defines the Green's function of a second-order two-point boundary-value problem by the homogeneous equation off the source point, continuity, the jump $1/a_2(\xi)$ in the first derivative and the homogeneous boundary conditions; derives the jump from $LG = \delta$ , constructs $G$ explicitly from two homogeneous solutions and their Wronskian, proves that $u(x) = \int_a^b G(x,\xi)f(\xi)\,d\xi$ is the unique solution, and proves the reciprocity relation $G^*(x,\xi) = G(\xi,x)$ and hence the symmetry of $G$ for a self-adjoint problem.

Prerequisites: Adjoint and self-adjoint differential operators, The Dirac delta in calculations, Structure of linear equations and the Wronskian, Existence and uniqueness of solutions

Definition(The boundary-value problem, and its Green's function)AdaptedAdapted from Stakgold & Holst (2011), §3.2, (3.2.1)–(3.2.2) pp. 191–192 and (3.2.7)–(3.2.8) p. 194; Riley, Hobson & Bence (2006), §15.2.5, (15.59)–(15.62) and the summary (i)–(iii), pp. 511–513; Arfken, Weber & Harris (2013), §10.1, Eqs. (10.5)–(10.6), p. 448

Throughout, $a \le x \le b$ is a bounded closed interval and

Lu \;=\; a_2(x)\,u'' + a_1(x)\,u' + a_0(x)\,u ,

where $a_0, a_1, a_2$ are continuous on $a \le x \le b$ and $a_2(x) \ne 0$ there. (Stakgold and Holst's (3.2.1) asks a little less: continuity of the coefficients on the open interval $a < x < b$ , with $a_2 \ne 0$ on the closed one. We follow Hassani in asking for continuity up to the endpoints, because the proof of the jump condition below uses boundedness of $a_1/a_2$ and $a_0/a_2$ on a closed neighbourhood of the source; nothing else in the unit needs the extra strength.) The data are a forcing function $f$ , piecewise continuous on $a \le x \le b$ , and two numbers $\gamma_1,\gamma_2$ prescribed through two boundary functionals

\begin{aligned} B_1u &= \alpha_{11}u(a) + \alpha_{12}u'(a) + \beta_{11}u(b) + \beta_{12}u'(b) ,\\ B_2u &= \alpha_{21}u(a) + \alpha_{22}u'(a) + \beta_{21}u(b) + \beta_{22}u'(b) , \end{aligned}

whose coefficient rows $(\alpha_{11},\alpha_{12},\beta_{11},\beta_{12})$ and $(\alpha_{21},\alpha_{22},\beta_{21},\beta_{22})$ are independent, so that the two conditions are genuinely different. The boundary-value problem (BVP) is

Lu = f \ \ (a < x < b), \qquad B_1u = \gamma_1, \quad B_2u = \gamma_2 .

The conditions are unmixed when each involves only one endpoint, that is $\beta_{11} = \beta_{12} = \alpha_{21} = \alpha_{22} = 0$ ; then $B_1u = \alpha_{11}u(a) + \alpha_{12}u'(a)$ and $B_2u = \beta_{21}u(b) + \beta_{22}u'(b)$ . The completely homogeneous problem is the one with data $\{0;0,0\}$ .

Unlike an initial-value problem, a BVP may have no solution or many. Since the difference of two solutions with the same data solves the completely homogeneous problem, the alternative is sharp: if the completely homogeneous problem has only $u \equiv 0$ , the BVP has at most one solution; if it has a nontrivial solution, the BVP has either none or many. For the rest of this unit we assume the completely homogeneous problem admits no nontrivial solution; the other case is the Fredholm alternative.

Green's function. Write $x_{<} = \min(x,\xi)$ and $x_{>} = \max(x,\xi)$ . The Green's function $G(x,\xi)$ of the problem $(L;B_1,B_2)$ is, for each fixed source point $\xi$ in $a < \xi < b$ , the function of $x$ satisfying

$LG = 0$ on $a < x < \xi$ and on $\xi < x < b$ ;
$B_1G = 0$ and $B_2G = 0$ , the functionals acting on $x$ ;
$G$ is continuous at $x = \xi$ ;
$\displaystyle \left.\frac{\partial G}{\partial x}\right|_{x=\xi^+} - \left.\frac{\partial G}{\partial x}\right|_{x=\xi^-} = \frac{1}{a_2(\xi)}$ .

Conditions 1, 3 and 4 are Stakgold and Holst's (3.2.7), and together they are abbreviated by the single equation

LG(x,\xi) = \delta(x - \xi), \qquad B_1G = B_2G = 0 ,

their (3.2.8), with $\delta$ the Dirac delta, the equation being read in the sense of distributions in $x$ . The next statement goes one way round this equivalence: it derives 3 and 4 from the delta equation, given 1. The converse — that a function with properties 1, 3 and 4 satisfies $LG = \delta$ distributionally — is a matter of running the same computation backwards and is not carried out here. Physically, $G(x,\xi)$ is the response at $x$ , under homogeneous boundary conditions, to a unit source concentrated at $\xi$ .

Note that the boundary conditions must be homogeneous for a Green's function to exist at all: $G$ is to act as an inverse of $L$ , and an inverse sends $0$ to $0$ , which forces $B_i[0] = 0$ (Hassani's Lemma 20.3.1). Inhomogeneous boundary data are handled afterwards, by superposition.

Theorem(Continuity and the jump condition)AdaptedAdapted from Arfken, Weber & Harris (2013), §10.1, Eqs. (10.8)–(10.9), p. 449; Riley, Hobson & Bence (2006), §15.2.5, Eqs. (15.63)–(15.65) and the summary (iii), pp. 512–513; Hassani (2013), §20.3.1, pp. 619–621, Eq. (20.33)

Let $G(\cdot,\xi)$ satisfy $LG = \delta(\cdot-\xi)$ as a distribution on $(a,b)$ , and suppose in addition that $G$ is twice continuously differentiable on each of $[a,\xi)$ and $(\xi,b]$ — so that, the delta being supported at $\xi$ alone, $LG = 0$ in the classical sense on each side — and that $G$ and $\partial G/\partial x$ have one-sided limits at $\xi$ . Both halves of this hypothesis are needed: without the delta equation the conclusion is false, since $G \equiv 0$ meets everything else. Then

G(\xi^+,\xi) = G(\xi^-,\xi), \qquad \left.\frac{\partial G}{\partial x}\right|_{\xi^+} - \left.\frac{\partial G}{\partial x}\right|_{\xi^-} = \frac{1}{a_2(\xi)} .

That is: $G$ is continuous across the source and its first derivative jumps by $1/a_2(\xi)$ . For the self-adjoint form $L = (p\,u')' + q\,u$ used by Arfken, $a_2 = p$ and the jump is $1/p(\xi)$ , his Eq. (10.9).

For an operator of order $n$ with leading coefficient $a_n$ , Riley, Hobson and Bence obtain the analogue: across $x=\xi$ every $x$ -derivative of $G$ of order $n-2$ or less stays continuous, and the one of order $n-1$ jumps by $1/a_n(\xi)$ .

Theorem(Existence, uniqueness and explicit construction)AdaptedAdapted from Stakgold & Holst (2011), §3.2, pp. 194–195, construction for unmixed conditions and Eqs. (3.2.9)–(3.2.10); Hassani (2013), §20.3.2, Theorem 20.3.5, p. 621

Assume the completely homogeneous problem admits no nontrivial solution, and let the boundary conditions be unmixed. Then the Green's function exists, is unique, and is given by

\boxed{\;G(x,\xi) \;=\; \frac{u_1(x_{<})\,u_2(x_{>})}{a_2(\xi)\,W(u_1,u_2;\xi)}\;}

where

$u_1$ is any nontrivial solution of $Lu = 0$ with $B_1u_1 = 0$ (the condition at $x=a$ ),
$u_2$ is any nontrivial solution of $Lu = 0$ with $B_2u_2 = 0$ (the condition at $x=b$ ),
$W(u_1,u_2;x) = u_1(x)u_2'(x) - u_1'(x)u_2(x)$ is their Wronskian, which vanishes nowhere.

This is Stakgold and Holst's (3.2.9). The formula does not depend on the normalisation of $u_1$ and $u_2$ , because rescaling either one rescales the numerator and the Wronskian by the same factor.

If $L$ is formally self-adjoint, $L = L^*$ , equivalently $a_2' = a_1$ and $L = (a_2u')' + a_0u$ , then Abel's formula gives $W(u_1,u_2;x) = C/a_2(x)$ for a constant $C$ determined by the chosen $u_1,u_2$ , so the prefactor is constant and

G(x,\xi) = \frac{1}{C}\,u_1(x_{<})\,u_2(x_{>}) ,

their (3.2.10). Uniqueness holds for general (possibly mixed) boundary conditions too, and so does existence; Hassani's Theorem 20.3.5 states existence and uniqueness for a general second-order operator under the same hypothesis that $Lu=0$ with the given homogeneous boundary conditions has no nontrivial solution.

Theorem(The Green's function solves the problem)AdaptedAdapted from Stakgold & Holst (2011), §3.2, Eq. (3.2.14) and Theorem 3.2.1 with Remark 1, pp. 196–197; Arfken, Weber & Harris (2013), §10.1, Eqs. (10.7) and (10.20)–(10.23), pp. 448, 451

Under the same hypotheses, for every piecewise continuous $f$ the problem

Lu = f \ \ (a<x<b), \qquad B_1u = 0, \quad B_2u = 0

is solvable, and its solution is unique; it is

u(x) \;=\; \int_a^b G(x,\xi)\,f(\xi)\,d\xi .

For unmixed conditions this reads, with $\sigma(\xi) = \bigl[a_2(\xi)W(u_1,u_2;\xi)\bigr]^{-1}$ ,

u(x) \;=\; u_2(x)\int_a^x \sigma(\xi)\,u_1(\xi) f(\xi)\, d\xi \;+\; u_1(x)\int_x^b \sigma(\xi)\,u_2(\xi) f(\xi)\, d\xi .

The map $f \mapsto u$ is linear, and it inverts $L$ on the set of twice continuously differentiable functions obeying the two homogeneous boundary conditions: one integral operator with kernel $G$ answers every forcing function at once. Stakgold and Holst's Theorem 3.2.1 covers inhomogeneous boundary data as well, by adding to the integral a combination of $u_1$ and $u_2$ fixed by $\gamma_1,\gamma_2$ ; the method block below records the recipe.

Theorem(Reciprocity, and symmetry in the self-adjoint case)AdaptedAdapted from Stakgold & Holst (2011), §3.2, Eqs. (3.2.23)–(3.2.26) and the reciprocity principle, pp. 199–200; Arfken, Weber & Harris (2013), §10.1, Eq. (10.15), p. 450; Hassani (2013), §20.3.1, Propositions 20.3.3 and 20.3.4, p. 619

Let $L^*$ be the formal adjoint of $L$ ,

L^*v = a_2 v'' + \bigl(2a_2' - a_1\bigr)v' + \bigl(a_2'' - a_1' + a_0\bigr)v ,

defined (for coefficients in $C^2$ ) by Green's formula

\int_a^b \bigl( v\,Lu - u\,L^*v \bigr)\, dx \;=\; \Bigl[\,J(u,v)\,\Bigr]_a^b , \qquad J(u,v) = a_2\bigl(vu' - uv'\bigr) + \bigl(a_1 - a_2'\bigr)uv .

Let $M$ be the set of functions obeying $B_1u = B_2u = 0$ , and let the adjoint boundary conditions $B_1^*v = B_2^*v = 0$ define the set $M^*$ of all $v$ with $\bigl[J(u,v)\bigr]_a^b = 0$ for every $u \in M$ . Let $G$ be the Green's function of $(L;B_1,B_2)$ and $G^*$ the adjoint Green's function, the solution of

L^*G^*(x,\xi) = \delta(x-\xi), \qquad B_1^*G^* = B_2^*G^* = 0 .

Then

G^*(x,\xi) \;=\; G(\xi,x) .

Equivalently: $G(x,\xi)$ solves the direct problem in its first argument and the adjoint problem in its second, so it obeys the adjoint boundary conditions in $\xi$ (Hassani's Proposition 20.3.3). In particular, if the problem is self-adjoint — meaning both $L^* = L$ and $M^* = M$ — then $G^* = G$ by uniqueness and

G(x,\xi) = G(\xi,x) .

A warning about the star. Throughout this unit $G^*$ means the adjoint Green's function, as it does in Stakgold and Holst. Arfken's (10.15), $G(x,t) = G(t,x)^*$ , uses the same symbol for complex conjugation: he works with a Hermitian $L$ , for which the adjoint problem is the original one, and records the symmetry of the kernel under conjugation. The two statements agree for a self-adjoint problem with real coefficients, where both reduce to $G(x,\xi) = G(\xi,x)$ , but the symbol is not doing the same work in the two books.

Stakgold and Holst call this the reciprocity principle: the response at $x$ to a unit source at $\xi$ equals the response at $\xi$ to a unit source at $x$ . A formally self-adjoint $L$ with unmixed boundary conditions is self-adjoint in this sense, and so are the periodic conditions $u(a) = u(b)$ , $a_2(a)u'(a) = a_2(b)u'(b)$ . An initial-value problem is never self-adjoint, whatever $L$ is, and its Green's function is not symmetric.

Idea of the proofOriginal

Everything follows from one observation: $\delta(x-\xi)$ is as singular as a second derivative of a kink. If $G$ itself jumped at $\xi$ , then $G'$ would contain a delta and $G''$ a derivative of a delta, and nothing else in $LG$ could cancel it. So the mildest — and, it turns out, the only — possibility is that $G$ is continuous and $G'$ has a step; then $G''$ carries the delta. Integrating $LG = \delta$ across the source once measures the size of that step, and dividing by $a_2$ first makes the measurement immediate.

The construction is then just counting. Off the source $G$ solves a second-order homogeneous equation on each of two intervals, so it has four free constants. There are four conditions: one boundary condition on each side, plus continuity and the jump. The left-hand boundary condition forces the left piece to be a multiple of the single solution $u_1$ obeying it, the right-hand one forces the right piece to be a multiple of $u_2$ , and the remaining two conditions are a $2\times 2$ linear system for the two multipliers whose determinant is the Wronskian — nonzero exactly because $u_1$ and $u_2$ are independent, which is exactly the assumption that the homogeneous problem has only the trivial solution.

That $\int Gf$ solves the problem is then a differentiation: the $x$ -dependence sits in $u_1$ and $u_2$ , which $L$ annihilates, so the only surviving term comes from differentiating the variable limits of the two integrals, and it collapses to $f$ because the prefactor was chosen as $1/(a_2W)$ .

Symmetry is Green's formula applied to $G$ and $G^*$ . Pairing the two delta equations against each other picks out the two values $G^*(\xi,\eta)$ and $G(\eta,\xi)$ , and the boundary term vanishes — because that vanishing is precisely how the adjoint boundary conditions were defined.

Proof(Continuity and the jump condition)AdaptedAdapted from Arfken, Weber & Harris (2013), §10.1, Eqs. (10.8)–(10.9), p. 449; Hassani (2013), §20.3.1, pp. 619–620

Step 1 (normalise the leading coefficient). Since $a_2$ is continuous and nowhere zero on $[a,b]$ , divide $LG = \delta(\cdot-\xi)$ by $a_2$ :

\frac{\partial^2 G}{\partial x^2} + \frac{a_1(x)}{a_2(x)}\frac{\partial G}{\partial x} + \frac{a_0(x)}{a_2(x)}\,G \;=\; \frac{\delta(x-\xi)}{a_2(x)} \;=\; \frac{\delta(x-\xi)}{a_2(\xi)} ,

the last equality by the sifting property, which lets the continuous factor $1/a_2$ be evaluated at the one point the delta sees (the Dirac delta). The coefficients $a_1/a_2$ and $a_0/a_2$ are continuous, hence bounded on $[a,b]$ ; this is where $a_2 \ne 0$ on the closed interval is used.

Step 2 (why $G$ cannot jump). Suppose $G$ had a jump of size $J$ at $\xi$ . Then, as distributions, $\partial G/\partial x$ contains the term $J\delta(x-\xi)$ and $\partial^2G/\partial x^2$ the term $J\delta'(x-\xi)$ . Collecting everything, the equation of Step 1 takes the form

J\,\delta'(x-\xi) + c\,\delta(x-\xi) + b(x) = 0

near $\xi$ , for some constant $c$ (gathered from the jump of $\partial G/\partial x$ and from $J\,a_1(\xi)/a_2(\xi)$ , the first-derivative term seeing the delta only through its value at $\xi$ ) and some function $b$ that is bounded on a neighbourhood of $\xi$ , because $G$ , $\partial G/\partial x$ , $a_0/a_2$ and $a_1/a_2$ are all bounded there.

To extract $J$ , test against the shrinking family $\varphi_\varepsilon(x) = \varepsilon\,\varphi\bigl((x-\xi)/\varepsilon\bigr)$ , where $\varphi$ is smooth, supported in $(-1,1)$ , with $\varphi(0) = 0$ and $\varphi'(0) = 1$ . Then $\varphi_\varepsilon(\xi) = 0$ and $\varphi_\varepsilon'(\xi) = \varphi'(0) = 1$ , so

\langle \delta', \varphi_\varepsilon\rangle = -\varphi_\varepsilon'(\xi) = -1 , \qquad \langle \delta, \varphi_\varepsilon\rangle = \varphi_\varepsilon(\xi) = 0 ,

while $\lvert\varphi_\varepsilon\rvert \le \varepsilon\sup\lvert\varphi\rvert$ on a support of length $2\varepsilon$ , so $\bigl|\int b\,\varphi_\varepsilon\bigr| \le 2\varepsilon^2\sup\lvert b\rvert\sup\lvert\varphi\rvert \to 0$ . The displayed equation therefore gives $-J + 0 + O(\varepsilon^2) = 0$ for every small $\varepsilon$ , whence $J = 0$ . One test function of this shape does the whole job: it is blind to $\delta$ and to bounded functions at once, which is what is needed — being "not a multiple of $\delta$ " would not suffice, since $b$ is present too.

So $G$ is continuous at $\xi$ ; this is Hassani's conclusion on p. 620, reached there by an explicit integration. Consequently $\partial G/\partial x$ has at most a finite jump.

Step 3 (integrate across the source). Integrate the equation of Step 1 over $[\xi-\varepsilon, \xi+\varepsilon]$ :

\left[\frac{\partial G}{\partial x}\right]_{\xi-\varepsilon}^{\xi+\varepsilon} + \int_{\xi-\varepsilon}^{\xi+\varepsilon}\frac{a_1}{a_2}\frac{\partial G}{\partial x}\, dx + \int_{\xi-\varepsilon}^{\xi+\varepsilon}\frac{a_0}{a_2}\,G\, dx \;=\; \frac{1}{a_2(\xi)} ,

for every $\varepsilon$ small enough that $[\xi-\varepsilon,\xi+\varepsilon] \subset [a,b]$ ; the right side is $1/a_2(\xi)$ because the delta's unit mass lies inside the interval.

Step 4 (the two integrals vanish in the limit). By Step 2, $G$ is continuous and $\partial G/\partial x$ is bounded near $\xi$ (it has one-sided limits there, by hypothesis). Each integrand is therefore bounded by some constant $K$ on the interval, so each integral is at most $2K\varepsilon$ in modulus and tends to $0$ as $\varepsilon \to 0^+$ . Here the continuity of $a_0, a_1, a_2$ is used, and the continuity of $G$ established in Step 2: had $G$ jumped, the middle integral would not have vanished.

Step 5 (conclude). Letting $\varepsilon \to 0^+$ leaves

\left.\frac{\partial G}{\partial x}\right|_{\xi^+} - \left.\frac{\partial G}{\partial x}\right|_{\xi^-} = \frac{1}{a_2(\xi)} . \qquad \square

Proof(Existence, uniqueness and the formula for

G

)AdaptedAdapted from Stakgold & Holst (2011), §3.2, pp. 194–195

Assume unmixed conditions, so that $B_1$ involves only $x=a$ and $B_2$ only $x=b$ .

Step 1 ( $u_1$ and $u_2$ exist). The row $(\alpha_{11},\alpha_{12})$ is not zero, since the two coefficient rows are independent. Let $u_1$ solve the initial-value problem $Lu = 0$ with $u_1(a) = \alpha_{12}$ , $u_1'(a) = -\alpha_{11}$ ; a solution exists and is unique because $a_2 \ne 0$ and the coefficients are continuous (existence and uniqueness). It is not identically zero, because its initial data are not both zero, and

B_1u_1 = \alpha_{11}\alpha_{12} + \alpha_{12}(-\alpha_{11}) = 0 .

The same construction at $x = b$ gives a nontrivial $u_2$ with $B_2u_2 = 0$ .

Step 2 ( $u_1$ and $u_2$ are independent, so $W \neq 0$ ). If they were dependent, then $u_2$ would be a nonzero multiple of $u_1$ , so $u_1$ would satisfy both $B_1u_1 = 0$ and $B_2u_1 = 0$ : a nontrivial solution of the completely homogeneous problem, contradicting the standing hypothesis. Two solutions of $Lu=0$ are independent if and only if their Wronskian is nonzero at one point, and then it is nonzero everywhere (linear ODE theory); so $W(u_1,u_2;x) \ne 0$ on $[a,b]$ .

Step 3 (the shape of $G$ ). On $a \le x < \xi$ , $G$ solves $Lu=0$ and satisfies $B_1G = 0$ , a condition involving only $x = a$ . Every solution of $Lu=0$ obeying $B_1u=0$ is a multiple of $u_1$ : the set of such solutions is at most one-dimensional, since two independent ones would make every solution satisfy $B_1u=0$ , in particular $u_2$ , again producing a nontrivial solution of the completely homogeneous problem. Hence $G(x,\xi) = A\,u_1(x)$ for $x < \xi$ , and likewise $G(x,\xi) = B\,u_2(x)$ for $x > \xi$ , with $A,B$ depending on $\xi$ only.

Step 4 (the two remaining conditions). Continuity at $x=\xi$ and the jump of $1/a_2(\xi)$ in $\partial G/\partial x$ give

A\,u_1(\xi) - B\,u_2(\xi) = 0 , \qquad -A\,u_1'(\xi) + B\,u_2'(\xi) = \frac{1}{a_2(\xi)} .

Step 5 (solve). The determinant of this $2\times2$ system in $(A,B)$ is $u_1(\xi)u_2'(\xi) - u_1'(\xi)u_2(\xi) = W(u_1,u_2;\xi)$ , nonzero by Step 2, so there is exactly one solution. Cramer's rule gives

A = \frac{u_2(\xi)}{a_2(\xi)W(u_1,u_2;\xi)} , \qquad B = \frac{u_1(\xi)}{a_2(\xi)W(u_1,u_2;\xi)} .

Substituting into Step 3 and noticing that for $x<\xi$ the pair $(x,\xi)$ is $(x_{<},x_{>})$ while for $x>\xi$ it is $(x_{>},x_{<})$ , both branches become the single formula

G(x,\xi) = \frac{u_1(x_{<})\,u_2(x_{>})}{a_2(\xi)\,W(u_1,u_2;\xi)} .

Step 6 (uniqueness). Two Green's functions for the same problem differ by a function that solves $Lu=0$ on each side of $\xi$ , is continuous at $\xi$ with continuous derivative there (the jumps cancel), and satisfies $B_1 = B_2 = 0$ . Such a difference solves $Lu=0$ classically on all of $(a,b)$ and meets both homogeneous boundary conditions, hence vanishes identically by hypothesis. This argument uses nothing about unmixed conditions, so uniqueness holds in general.

Step 7 (the self-adjoint form). If $a_1 = a_2'$ , then $L = (a_2u')' + a_0u$ and Abel's formula for the Wronskian of two solutions, $W(x) = C\exp\bigl(-\int a_1/a_2\bigr)$ , becomes $W(x) = C/a_2(x)$ (Stakgold and Holst's (3.1.8)). Hence $a_2(\xi)W(\xi) = C$ is independent of $\xi$ and $G(x,\xi) = u_1(x_{<})u_2(x_{>})/C$ , which is visibly symmetric in $x$ and $\xi$ . $\square$

Proof(The representation formula)AdaptedAdapted from Arfken, Weber & Harris (2013), §10.1, Eqs. (10.20)–(10.23), pp. 451; Stakgold & Holst (2011), §3.2, (3.2.13)–(3.2.14) and Theorem 3.2.1, pp. 195–196

Take unmixed conditions and write $\sigma(\xi) = \bigl[a_2(\xi)W(u_1,u_2;\xi)\bigr]^{-1}$ , a continuous function of $\xi$ since $a_2$ and $W$ are continuous and nonvanishing.

Step 1 (split the integral at the source point). For $\xi < x$ the formula for $G$ gives $G(x,\xi) = \sigma(\xi)u_1(\xi)u_2(x)$ , and for $\xi > x$ it gives $G(x,\xi) = \sigma(\xi)u_1(x)u_2(\xi)$ . Hence

u(x) = \int_a^b G(x,\xi)f(\xi)\,d\xi = u_2(x)\,I_1(x) + u_1(x)\,I_2(x) ,

I_1(x) = \int_a^x \sigma u_1 f\, d\xi , \qquad I_2(x) = \int_x^b \sigma u_2 f\, d\xi .

Step 2 (first derivative). By the fundamental theorem of calculus, $I_1' = \sigma(x)u_1(x)f(x)$ and $I_2' = -\sigma(x)u_2(x)f(x)$ at every point where $f$ is continuous. Differentiating the product,

u' = u_2'I_1 + u_1'I_2 + \underbrace{\sigma u_1 u_2 f - \sigma u_1 u_2 f}_{=\,0} = u_2'I_1 + u_1'I_2 .

The two boundary terms cancel exactly because the same function $\sigma(\xi)u_1(\xi)u_2(\xi)$ appears in both; this is the analytic content of the continuity of $G$ at $x=\xi$ .

Step 3 (second derivative). Differentiating again,

u'' = u_2''I_1 + u_1''I_2 + \sigma\bigl(u_1u_2' - u_1'u_2\bigr) f = u_2''I_1 + u_1''I_2 + \sigma W f .

The surviving term is the jump condition at work: the two boundary terms no longer cancel, and what is left is $\sigma W f$ .

Step 4 (apply $L$ ). Combining Steps 1–3,

Lu = \bigl(a_2u_2'' + a_1u_2' + a_0u_2\bigr)I_1 + \bigl(a_2u_1'' + a_1u_1' + a_0u_1\bigr)I_2 + a_2\,\sigma\,W f = I_1\,Lu_2 + I_2\,Lu_1 + a_2\sigma W f .

Here $Lu_1 = Lu_2 = 0$ , since $u_1,u_2$ solve the homogeneous equation, and $a_2(x)\sigma(x)W(x) = 1$ by the definition of $\sigma$ . Hence $Lu = f$ at every point of continuity of $f$ .

Step 5 (boundary conditions). At $x=a$ the integral $I_1$ vanishes, so $u(a) = u_1(a)I_2(a)$ and, by Step 2, $u'(a) = u_1'(a)I_2(a)$ . Therefore

B_1u = \alpha_{11}u(a) + \alpha_{12}u'(a) = I_2(a)\,\bigl[\alpha_{11}u_1(a) + \alpha_{12}u_1'(a)\bigr] = I_2(a)\,B_1u_1 = 0 .

The same computation at $x=b$ , where $I_2(b) = 0$ , gives $B_2u = I_1(b)\,B_2u_2 = 0$ . This step is where unmixedness is needed: $B_1$ had to involve only the endpoint at which one of the two integrals vanishes.

Step 6 (uniqueness). Any two solutions of the problem differ by a solution of the completely homogeneous problem, which is identically zero by hypothesis. $\square$

Proof(Reciprocity and symmetry)AdaptedAdapted from Stakgold & Holst (2011), §3.2, pp. 199–200, proof of (3.2.25)–(3.2.26)

Step 1 (pair the two problems). Fix $\xi$ and $\eta$ in $(a,b)$ . Put $u(x) = G(x,\xi)$ , which satisfies $LG(x,\xi) = \delta(x-\xi)$ and lies in $M$ , and $v(x) = G^*(x,\eta)$ , which satisfies $L^*G^*(x,\eta) = \delta(x-\eta)$ and lies in $M^*$ . Form the combination $v\,Lu - u\,L^*v$ and integrate it over $(a,b)$ :

\int_a^b \bigl(v\,Lu - u\,L^*v\bigr) dx = \int_a^b G^*(x,\eta)\,\delta(x-\xi)\,dx - \int_a^b G(x,\xi)\,\delta(x-\eta)\,dx .

Step 2 (the right-hand side). By the sifting property, treating the deltas formally as Stakgold and Holst do, the right-hand side is

G^*(\xi,\eta) - G(\eta,\xi) .

Step 3 (the left-hand side). By Green's formula, the left-hand side equals the boundary term $\bigl[J(u,v)\bigr]_a^b$ , where $J(u,v) = a_2(vu'-uv') + (a_1-a_2')uv$ .

Step 4 (the boundary term vanishes). $u \in M$ and $v \in M^*$ . By the very definition of the adjoint boundary conditions, $\bigl[J(u,v)\bigr]_a^b = 0$ for every such pair. This is the only place the boundary conditions enter, and it is the step that fails for an arbitrary pair of functions.

Step 5 (conclude). Therefore $G^*(\xi,\eta) = G(\eta,\xi)$ for all $\xi,\eta$ , which is the asserted relation after renaming the variables.

Step 6 (the self-adjoint case). If the problem is self-adjoint, $L^* = L$ and $M^* = M$ , so $G^*$ solves exactly the problem that defines $G$ . By the uniqueness proved above, $G^* = G$ , and Step 5 becomes $G(x,\xi) = G(\xi,x)$ . $\square$

Remark(Where the hypotheses are used, and what fails without them)Original

$a_2 \ne 0$ on the closed interval. Used in Step 1 of the jump proof, to divide by $a_2$ and keep the other coefficients bounded, and again in Step 5, since $1/a_2(\xi)$ must be finite. At a point where $a_2$ vanishes the equation is singular and the construction breaks down: for $L = \bigl((1-x^2)u'\bigr)'$ on $[-1,1]$ , the two solutions of $Lu=0$ are $1$ and $\tfrac12\ln\frac{1+x}{1-x}$ , and both boundary requirements " $G$ finite at $\pm1$ " select the same solution $u_1 = u_2 = 1$ . Arfken's Exercise 10.1.6 (p. 457) makes exactly this point: the candidates for $u_1$ and $u_2$ are linearly dependent, so no Green's function can be built this way.
No nontrivial solution of the completely homogeneous problem. Used in Step 2 and Step 3 of the construction (independence of $u_1,u_2$ , hence $W \ne 0$ ), in Step 6 (uniqueness) and in Step 6 of the representation proof. Without it the construction divides by zero. Stakgold and Holst's Example 2 is the model: for $u'' + a_0u = f$ on $(0,1)$ with $u(0)=u(1)=0$ and $a_0 = \pi^2$ , the homogeneous problem has the solutions $A\sin\pi x$ , the problem with $f \equiv 1$ has no solution, and the problem with $f = x - \tfrac12$ has infinitely many. What replaces the Green's function is the Fredholm alternative and the modified Green's function.
Exactly two boundary conditions. Stakgold and Holst's Remark 1 after Theorem 3.2.1: a third condition of the same type would leave the completely homogeneous problem with only the trivial solution, but would usually leave the inhomogeneous problem with no solution at all. The count four constants / four conditions in the idea block is not an accident.
Homogeneous boundary conditions. Required for $G$ to exist at all (Hassani's Lemma 20.3.1). With $B_1u = \gamma_1 \ne 0$ , the map $f \mapsto u$ could not be linear, and $f \equiv 0$ would not give $u \equiv 0$ .
Unmixedness. Used only in the construction (to make the left piece a multiple of a single $u_1$ ) and in Step 5 of the representation proof. Existence, uniqueness and the representation formula themselves hold for general boundary functionals; Stakgold and Holst build $G$ in that case from the causal solution plus a correction, and their Theorem 3.2.1 is stated for general $B_1,B_2$ .
Self-adjointness, for symmetry. Only Step 6 of the last proof uses it. Drop it and the symmetry is false: the second example below exhibits a problem with $G(x,\xi) \ne G(\xi,x)$ for which the weaker relation $G^*(x,\xi) = G(\xi,x)$ still holds. An initial-value problem is never self-adjoint — the adjoint conditions move to the other endpoint — which is why causal Green's functions are one-sided and asymmetric.

Remark(Conventions that differ between books)Original

This unit follows Stakgold and Holst, Arfken, and Riley, Hobson and Bence in writing

LG(x,\xi) = \delta(x-\xi), \qquad \text{jump in } \partial G/\partial x \ = \ \frac{1}{a_2(\xi)} .

Hassani uses a different normalisation. He works with a Sturm–Liouville weight $w(x) > 0$ , defines the Green's function by

L_xG(x,y) = \frac{\delta(x-y)}{w(x)} ,

and consequently obtains the jump $1/\bigl(p_2(y)w(y)\bigr)$ , his Eq. (20.33), with the representation $u(x) = \int_a^b w(y)\,G(x,y)f(y)\,dy$ . His differential operator is the same $p_2u'' + p_1u' + p_0u$ as ours, so the two kernels are related by the weight alone: since $\delta(x-y)/w(x) = \delta(x-y)/w(y)$ and $L$ is linear,

G_{\text{Hassani}}(x,y) = \frac{G(x,y)}{w(y)} ,

which is exactly the factor turning our jump $1/p_2(y)$ into his $1/\bigl(p_2(y)w(y)\bigr)$ . The two conventions agree when $w \equiv 1$ . They always agree on the answer: his equation carries the factor $1/w$ and his representation integral carries the compensating $w$ , so $\int w\,G_{\text{Hassani}}f\,dy = \int G f\,dy$ and the solution $u$ is the same function. What one must not do is mix them — our jump with his representation, or his kernel with our integral — which is wrong by a factor $w$ . The same caution applies to the sign: many physics texts define the Green's function of $-d^2/dx^2$ rather than $d^2/dx^2$ , which flips the sign of $G$ .

A second point of notation: the second argument of $G$ is called $\xi$ here, $t$ by Arfken, $z$ by Riley, Hobson and Bence, and $y$ by Hassani. The named and unnamed roles are the same in all four: the first argument is the field point, the second the source point.

Method(Building and using a Green's function)AdaptedAdapted from Stakgold & Holst (2011), §3.2, pp. 194–196; Riley, Hobson & Bence (2006), §15.2.5, pp. 512–513; Arfken, Weber & Harris (2013), §10.1, pp. 450–452

Given $Lu = a_2u'' + a_1u' + a_0u = f$ on $[a,b]$ with two homogeneous unmixed conditions:

Check the hypothesis. Confirm $a_2 \ne 0$ on $[a,b]$ and that $Lu = 0$ with both homogeneous boundary conditions has only $u \equiv 0$ . If not, stop: use the Fredholm alternative instead.
Find a basis $\{v_1,v_2\}$ of solutions of $Lu = 0$ .
Match each endpoint. Take $u_1$ to be the combination of $v_1,v_2$ obeying the condition at $x=a$ , and $u_2$ the one obeying the condition at $x=b$ . Each is determined up to a scale factor, which does not matter.
Compute $W(u_1,u_2;\xi) = u_1u_2' - u_1'u_2$ at $\xi$ and the prefactor $\sigma(\xi) = 1/\bigl(a_2(\xi)W(\xi)\bigr)$ . If $L$ is formally self-adjoint, $\sigma$ is a constant; a $\xi$ -dependence surviving in $a_2W$ for a self-adjoint $L$ signals an algebra error.
Assemble $G(x,\xi) = \sigma(\xi)\,u_1(x_{<})u_2(x_{>})$ , that is $\sigma(\xi)u_1(x)u_2(\xi)$ for $x<\xi$ and $\sigma(\xi)u_1(\xi)u_2(x)$ for $x>\xi$ .
Verify the four defining properties: $LG=0$ on each side, the two boundary conditions, continuity at $x=\xi$ , and the jump $1/a_2(\xi)$ . The jump check is the one that catches sign errors.
Integrate for the solution, $u(x) = u_2(x)\int_a^x \sigma u_1 f\, d\xi + u_1(x)\int_x^b \sigma u_2 f\, d\xi ,$ and note that only this step has to be repeated when $f$ changes.
Inhomogeneous boundary data. Either subtract a smooth function meeting them — for $u(a)=c_1$ , $u(b)=c_2$ Arfken uses the straight line $\bigl[c_1(b-x) + c_2(x-a)\bigr]/(b-a)$ — and solve the resulting problem with homogeneous conditions and a modified $f$ ; or add to the integral the combination $\dfrac{\gamma_2}{B_2u_1}u_1(x) + \dfrac{\gamma_1}{B_1u_2}u_2(x)$ , which is Stakgold and Holst's (3.2.16). The pairing is the crossed one, and it is worth checking: $B_1u_1 = 0$ kills the first term under $B_1$ and leaves $B_1u = \gamma_1$ , while $B_2u_2 = 0$ kills the second under $B_2$ and leaves $B_2u = \gamma_2$ .

Example(A self-adjoint problem with variable coefficients)Original

Solve $\bigl(x\,u'\bigr)' = f(x)$ on $1 \le x \le 2$ with $u(1) = u(2) = 0$ , and in particular for $f(x) = 1/x$ .

Hypotheses. Here $a_2(x) = x$ , $a_1(x) = 1 = a_2'$ , so $L$ is formally self-adjoint, and $a_2 \ne 0$ on $[1,2]$ . The homogeneous equation $(xu')' = 0$ gives $xu' = c$ , hence $u = c\ln x + d$ ; the conditions $u(1)=u(2)=0$ force $d = 0$ and $c\ln 2 = 0$ , so $c = 0$ : only the trivial solution. The construction applies.
Endpoint solutions. $u_1(x) = \ln x$ vanishes at $x=1$ ; $u_2(x) = \ln(x/2)$ vanishes at $x=2$ .
Wronskian and prefactor. $W = u_1u_2' - u_1'u_2 = \frac{\ln x}{x} - \frac{\ln (x/2)}{x} = \frac{\ln 2}{x} , \qquad a_2(\xi)W(\xi) = \xi\cdot\frac{\ln 2}{\xi} = \ln 2 ,$ a constant, as it must be for a self-adjoint $L$ . So $\sigma = 1/\ln 2$ .
The Green's function. $G(x,\xi) = \frac{\ln x_{<}\;\ln\bigl(x_{>}/2\bigr)}{\ln 2} = \begin{cases} \dfrac{\ln x\,\ln(\xi/2)}{\ln 2}, & 1 \le x \le \xi,\\[2mm] \dfrac{\ln \xi\,\ln(x/2)}{\ln 2}, & \xi \le x \le 2 . \end{cases}$ It is symmetric, as predicted. Note $G \le 0$ throughout, since $\ln x_{<} \ge 0$ and $\ln(x_{>}/2) \le 0$ .
Check the jump. For $x > \xi$ , $\partial G/\partial x = \ln\xi/(x\ln 2)$ ; for $x<\xi$ , $\partial G/\partial x = \ln(\xi/2)/(x\ln 2)$ . At $x=\xi$ the difference is $\frac{\ln\xi - \ln(\xi/2)}{\xi\ln 2} = \frac{\ln 2}{\xi \ln 2} = \frac{1}{\xi} = \frac{1}{a_2(\xi)} . \checkmark$
Solve with $f(\xi) = 1/\xi$ . $u(x) = \frac{\ln(x/2)}{\ln 2}\int_1^x \frac{\ln\xi}{\xi}\, d\xi + \frac{\ln x}{\ln 2}\int_x^2 \frac{\ln(\xi/2)}{\xi}\, d\xi .$ With $s = \ln\xi$ the first integral is $\tfrac12(\ln x)^2$ ; with $s = \ln(\xi/2)$ the second is $\bigl[\tfrac12 s^2\bigr]_{\ln(x/2)}^{0} = -\tfrac12\bigl(\ln(x/2)\bigr)^2$ . Hence $u(x) = \frac{\ln x\,\ln(x/2)}{2\ln 2}\Bigl[\ln x - \ln(x/2)\Bigr] = \frac{\ln x\,\ln(x/2)}{2\ln 2}\cdot \ln 2 = \frac{1}{2}\,\ln x\,\ln\frac{x}{2} .$
Independent check. Integrating $(xu')' = 1/x$ directly gives $xu' = \ln x + c$ , then $u = \tfrac12(\ln x)^2 + c\ln x + d$ ; $u(1)=0$ gives $d=0$ and $u(2)=0$ gives $c = -\tfrac12\ln 2$ , so $u = \tfrac12\ln x\,\ln(x/2)$ . The two routes agree.
A number. At $x = 1.5$ : $\ln 1.5 = 0.405465$ and $\ln 0.75 = -0.287682$ , so $u(1.5) = -0.0583225$ to six decimal places. The deflection is negative everywhere, as the sign of $G$ and the positive source demand.

Example(A problem that is not self-adjoint:

G

is not symmetric, but

G^*(x,\xi) = G(\xi,x)

)Original

Take $Lu = u'' + 2u'$ on $0 \le x \le 1$ with $u(0) = u(1) = 0$ . Here $a_2 = 1$ , $a_1 = 2 \ne a_2' = 0$ , so $L$ is not formally self-adjoint.

Hypotheses. $u'' + 2u' = 0$ has general solution $A + Be^{-2x}$ ; the two conditions give $A + B = 0$ and $A + Be^{-2} = 0$ , hence $A=B=0$ . Only the trivial solution, so $G$ exists.
Endpoint solutions. $u_1(x) = 1 - e^{-2x}$ vanishes at $x=0$ ; $u_2(x) = e^{-2x} - e^{-2}$ vanishes at $x=1$ .
Wronskian. $W = u_1u_2' - u_1'u_2 = \bigl(1-e^{-2x}\bigr)\bigl(-2e^{-2x}\bigr) - 2e^{-2x}\bigl(e^{-2x}-e^{-2}\bigr) = -2e^{-2x}\bigl(1 - e^{-2}\bigr) ,$ which depends on $x$ — the signature of a non-self-adjoint $L$ , since $a_2 = 1$ here.
The Green's function. With $a_2(\xi)W(\xi) = -2e^{-2\xi}(1-e^{-2})$ , $G(x,\xi) = -\,\frac{e^{2\xi}}{2\bigl(1-e^{-2}\bigr)}\;\bigl(1 - e^{-2x_{<}}\bigr)\bigl(e^{-2x_{>}} - e^{-2}\bigr) .$ Direct substitution confirms $G_{xx} + 2G_x = 0$ on each side, $G(0,\xi) = G(1,\xi) = 0$ , and a jump of exactly $1$ in $\partial G/\partial x$ at $x=\xi$ (here $1/a_2 = 1$ ).
Symmetry fails. The factor $e^{2\xi}$ depends on $\xi$ alone, while $u_1(x_{<})u_2(x_{>})$ is symmetric, so $\frac{G(x,\xi)}{G(\xi,x)} = e^{2(\xi-x)} \ne 1 \quad\text{for } x \ne \xi .$ Numerically, $G\bigl(\tfrac14,\tfrac34\bigr) = -0.0895249$ while $G\bigl(\tfrac34,\tfrac14\bigr) = -0.0329344$ , and their ratio is $e^{1} = 2.71828$ . Reciprocity genuinely fails: a unit source at $\tfrac34$ produces a response at $\tfrac14$ nearly three times the response at $\tfrac34$ to a unit source at $\tfrac14$ . The drift term $2u'$ breaks the left–right balance.
But the general relation holds. The formal adjoint is $L^*v = v'' + (2a_2' - a_1)v' + (a_2'' - a_1' + a_0)v = v'' - 2v'$ . With Dirichlet conditions, $J(u,v) = vu' - uv' + 2uv$ vanishes at both endpoints whenever $u$ and $v$ vanish there, so the adjoint boundary conditions are again $v(0)=v(1)=0$ . Solving the adjoint problem the same way, with $v_1 = e^{2x}-1$ , $v_2 = e^{2x}-e^{2}$ and $W(v_1,v_2;x) = 2e^{2x}(e^2-1)$ , $G^*(x,\xi) = \frac{\bigl(e^{2x_{<}}-1\bigr)\bigl(e^{2x_{>}}-e^{2}\bigr)}{2e^{2\xi}\bigl(e^{2}-1\bigr)} .$ Rewriting $1 - e^{-2x} = e^{-2x}(e^{2x}-1)$ and $e^{-2\xi} - e^{-2} = e^{-2\xi-2}\bigl(e^{2}-e^{2\xi}\bigr)$ in the formula of item 4 turns $G(\xi,x)$ into precisely this expression. So $G^*(x,\xi) = G(\xi,x)$ exactly, confirming the reciprocity theorem in a case where plain symmetry is false.
Why the adjoint is not the same operator. The adjoint boundary conditions coincide with the original ones here, so the failure of self-adjointness is entirely in $L$ versus $L^*$ . Multiplying $Lu$ by the integrating factor $e^{2x}$ turns it into $e^{-2x}\bigl(e^{2x}u'\bigr)'$ : the problem can be made self-adjoint, at the price of a weight $e^{2x}$ in the inner product — which is the weight $w$ of Hassani's convention.

ApplicationOriginal

The Green's function is the influence function of a linear system: the field at $x$ produced by a unit source at $\xi$ . Once it is known, the response to any distribution of sources is a single integral, and the boundary conditions have already been built in.

Statics. A string under tension $T$ with small transverse load $w(x)$ satisfies $-T\,u'' = w$ , and the Green's function of $-d^2/dx^2$ on $[0,L]$ with $u(0)=u(L)=0$ is $G(x,\xi) = x_{<}(L-x_{>})/L$ : the triangular shape a string takes under a single point load. A beam obeys the fourth-order $\bigl(EI\,u''\bigr)'' = w$ instead, so its Green's function is a different function — piecewise cubic, with three derivatives continuous and a jump in the third, by the $n$ th-order form of the conditions quoted above — but it plays the same role: it is the influence line that structural engineers tabulate so that any load can be handled by integration. The first worked example above is the radial version of this: steady one-dimensional conduction in a shell with a variable conductance $x$ .
Steady transport. For steady diffusion along a rod with source density $f$ and prescribed end temperatures, $\bigl(\kappa(x)u'\bigr)' = -f$ ; the Green's function is the temperature profile from a unit heat source at one point. The second example is the convective version: $u'' + 2u'$ is diffusion plus a drift of fixed speed, and the asymmetry of its Green's function is the physical statement that a source influences what lies downstream more than what lies upstream. Reciprocity is a property of pure diffusion, not of advection.
Reciprocity as a physical law. $G(x,\xi) = G(\xi,x)$ for self-adjoint problems is the one-dimensional case of a family of reciprocity theorems: Maxwell's and Betti's in elasticity (a unit load at $A$ deflects $B$ by the same amount as a unit load at $B$ deflects $A$ ), Rayleigh's in acoustics, and the reciprocity between transmitting and receiving antennas. When a measurement appears to violate it, the model contains a non-self-adjoint term — a flow, a magnetic field, an absorbing boundary — and locating that term is often the point of the test.
Quantum mechanics and scattering. With $L = -d^2/dx^2 + V(x) - E$ , the Green's function is the resolvent kernel $\langle x|(H-E)^{-1}|\xi\rangle$ . Writing the Schrödinger equation as $u = u_{\text{free}} + \int G\,V u$ is the starting point of the Born series (the Born approximation), and the choice of boundary condition at infinity — outgoing rather than vanishing — selects the scattering Green's function, as in Arfken's boundary-at-infinity example.
Reformulating eigenvalue problems. Since $G$ inverts $L$ , the eigenvalue problem $Lu = \lambda u$ becomes the integral equation $u(x) = \lambda\int_a^b G(x,\xi)u(\xi)\,d\xi$ . Arfken stresses the practical gain: the integral operator is bounded while the differential operator is not, which is why the sharp spectral theorems are proved on the integral side (Hilbert–Schmidt theory) and why $G$ has the bilinear eigenfunction expansion of the next unit.

Theorem

The Fredholm alternative and modified Green's functions

Adapted

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States and proves the dichotomy governing a boundary-value problem whose homogeneous problem has nontrivial solutions — either a unique solution for every forcing, or solvability exactly when the forcing is orthogonal to every solution of the homogeneous adjoint problem — gives the model matrix case with a complete proof, and builds the modified Green's function, the pseudoinverse that returns the minimum-norm solution when the problem is solvable and the minimum-norm least-squares solution when it is not.

Prerequisites: Green's function for a boundary-value problem, Adjoint and self-adjoint differential operators, Linear systems and rank, Inner products and norms, Orthogonal projection and least squares

Definition(The three problems, and solvability conditions)AdaptedAdapted from Stakgold & Holst (2011), §3.4, (3.4.1)–(3.4.3), p. 206, and p. 211

Keep the setting of the previous unit, but now for an operator of order $p$ ,

Lu = a_p(x)u^{(p)} + \dots + a_1(x)u' + a_0(x)u ,

with $a_p \ne 0$ on $[a,b]$ , together with $p$ boundary functionals $B_1,\dots,B_p$ built from the values of $u,u',\dots,u^{(p-1)}$ at $a$ and $b$ , with independent coefficient rows. Three problems are in play:

\textbf{(D)}\quad Lu = f,\ \ B_1u = \gamma_1,\dots,B_pu = \gamma_p ;

\textbf{(H)}\quad Lu = 0,\ \ B_1u = \dots = B_pu = 0 ; \qquad \textbf{(A)}\quad L^*v = 0,\ \ B_1^*v = \dots = B_p^*v = 0 .

Here $L^*$ is the formal adjoint and $B_i^*$ the adjoint boundary functionals, defined as before by Green's formula

\int_a^b\bigl(v\,Lu - u\,L^*v\bigr)\,dx = \Bigl[J(u,v)\Bigr]_a^b ,

with $B_1^*v = \dots = B_p^*v = 0$ exactly when $\bigl[J(u,v)\bigr]_a^b = 0$ for every $u$ obeying $B_1u = \dots = B_pu = 0$ . Problem (H) is the direct homogeneous problem and (A) the homogeneous adjoint problem.

When (H) has nontrivial solutions, (D) usually has none, and the conditions on the data $\{f;\gamma_1,\dots,\gamma_p\}$ that restore solvability are called solvability (or consistency) conditions. Naming them is what this unit does.

Theorem(The alternative for a linear system, the model case)AdaptedAdapted from Stakgold & Holst (2011), §3.4, Theorem (Fredholm Alternative) and the balanced case, pp. 207–209

Let $A$ be a real $m \times n$ matrix, viewed as a map $\mathbb{R}^n \to \mathbb{R}^m$ , let $A^*$ be its transpose, viewed as a map $\mathbb{R}^m \to \mathbb{R}^n$ , and use the Euclidean inner product $(w,z) = \sum_i w_iz_i$ in each space. Then

Au = f \ \text{ has a solution } \iff (f,v) = 0 \ \text{ for every } v \text{ with } A^*v = 0 .

Equivalently, $R_A$ , the range of $A$ , and $N_{A^*}$ , the kernel of $A^*$ , are orthogonal complements of one another in $\mathbb{R}^m$ . If $u_p$ is one solution, every solution is $u_p + u_H$ with $Au_H = 0$ , and conversely; so the solution is unique precisely when $Au = 0$ has only $u = 0$ .

In the balanced case $m = n$ : if $\det A \ne 0$ then also $\det A^* \ne 0$ ; neither homogeneous equation has a nonzero solution, nothing at all is imposed on $f$ , and $Au = f$ is uniquely solvable whatever $f$ may be. If $\det A = 0$ and $A$ has a $k$ -dimensional null space, then so does $A^*$ — though in general a different one — the solvability conditions are the $k$ conditions $(f,v^{(i)}) = 0$ for a basis $v^{(1)},\dots,v^{(k)}$ of the null space of $A^*$ , and when they hold the general solution is $u_p + \sum_{i=1}^k c_iu^{(i)}$ with $u^{(1)},\dots,u^{(k)}$ a basis of the null space of $A$ .

Theorem(The Fredholm alternative for a boundary-value problem)AdaptedAdapted from Stakgold & Holst (2011), §3.4, the alternative theorem for boundary value problems (a)–(b), p. 211, and the completely inhomogeneous case with (3.4.16), p. 213

Consider first homogeneous boundary data, $Lu = f$ with $B_1u = \dots = B_pu = 0$ .

(a) Should (H) admit no nontrivial solution, the same is true of (A), and the inhomogeneous problem is then uniquely solvable — the case handled by the ordinary Green's function.

(b) If (H) has $k$ independent solutions, then (A) also has $k$ independent solutions, not necessarily the same ones. Then the problem is solvable if and only if

\int_a^b f\,v^{(1)}\,dx = \dots = \int_a^b f\,v^{(k)}\,dx = 0 ,

where $v^{(1)},\dots,v^{(k)}$ are $k$ independent solutions of (A). When these hold, the general solution is

u = \bar u + \sum_{i=1}^{k}c_i\,u^{(i)} ,

with $\bar u$ one particular solution, $c_i$ arbitrary constants and $u^{(1)},\dots,u^{(k)}$ $k$ independent solutions of (H).

For inhomogeneous boundary data, $B_iu = \gamma_i$ , the same pairing gives a condition that is no longer homogeneous in the data: for $v$ a solution of (A),

\int_a^b f\,v\,dx = \Bigl[J(u,v)\Bigr]_a^b ,

and the right-hand side is completely determined by $v$ and the numbers $\gamma_1,\dots,\gamma_p$ , because once $v$ satisfies the adjoint conditions, $\bigl[J(u,v)\bigr]_a^b$ depends on $u$ only through $B_1u,\dots,B_pu$ . Stakgold and Holst write it out as $\gamma_1B_{2p}^*v + \dots + \gamma_pB_{p+1}^*v$ , where $B_{p+1}^*,\dots,B_{2p}^*$ are the complementary adjoint functionals of their §3.3, and state that this condition is sufficient as well as necessary.

If the problem is self-adjoint, (A) is (H), and the solvability condition reads: $f$ must be orthogonal to every solution of the homogeneous problem itself.

Note(What is proved here, and where the rest is proved)AdaptedAdapted from Stakgold & Holst (2011), §3.4, pp. 210–211; Kreyszig (1978), §8.7, Definition 8.7-1 and Theorems 8.7-2, 8.7-3, pp. 451–453; Evans (2010), §6.2.3, Theorem 4, pp. 302–303, and Appendix D.5, p. 682; Hassani (2013), §18.2, Theorem 18.2.2, p. 551

This unit proves the matrix alternative in full and proves the necessity of the solvability conditions for the boundary-value problem. The sufficiency, and the equality of the two dimensions $k$ , are not proved here: Stakgold and Holst defer them to their Chapter 5, where the boundary-value problem is converted into an integral equation with kernel $G$ and the alternative is proved for integral equations — part (a) and (b) of their alternative theorem for the equation $\int_a^b k(x,\xi)u(\xi)\,d\xi - u(x) = f(x)$ , stated on p. 210. That route is the subject of integral equations and Hilbert–Schmidt theory.

The same dichotomy appears in three other guises, all beyond this unit's scope:

Operators. Kreyszig defines a bounded linear $A$ on a normed space to satisfy the Fredholm alternative when exactly one of two cases holds: (I) $Ax = y$ and $A^\times f = g$ are uniquely solvable for all data, and the homogeneous equations have only the trivial solutions; or (II) the two homogeneous equations have the same finite number $n \ge 1$ of independent solutions, and the inhomogeneous equations are solvable exactly when the data annihilate those solutions. His Theorem 8.7-2: if $T$ is a compact linear operator and $\lambda \ne 0$ , then $T - \lambda I$ satisfies the Fredholm alternative.
Integral equations. Kreyszig's Theorem 8.7-3 specialises the previous bullet to Fredholm equations of the second kind on a normed space: if the kernel makes $T$ compact, then $T - \lambda I$ satisfies the alternative, so either the equation has a unique solution for every datum or the homogeneous equation has finitely many independent nontrivial solutions — with the solvability condition in the annihilation form of his case (II), since a general normed space has no inner product. Hassani's Theorem 18.2.2 is the Hilbert-space version, for a Hilbert–Schmidt $K$ : either $\lambda$ is a regular value, or the equation is solvable exactly when the data lie in the orthogonal complement of the kernel of $1 - \lambda^*K^\dagger$ . In a Hilbert space the two readings coincide, every bounded functional being an inner product.
Elliptic partial differential equations. Evans's Theorem 4 of §6.2.3 states precisely (a)/(b) for weak solutions of a second-order elliptic $Lu = f$ with $u = 0$ on the boundary, including $\dim N = \dim N^*$ and the criterion $(f,v) = 0$ for all $v \in N^*$ ; his proof reduces the problem to $u - Ku = h$ with $K$ compact and applies the compact-operator alternative of his Appendix D.5.

Definition(The modified Green's function)AdaptedAdapted from Stakgold & Holst (2011), §3.5, (3.5.1)–(3.5.7), pp. 216–217

Let $(L;B_1,B_2)$ be a self-adjoint second-order problem whose homogeneous problem has a one-dimensional space of solutions, spanned by a normalised $u_1$ :

Lu_1 = 0, \quad B_1u_1 = B_2u_1 = 0, \qquad \int_a^b u_1(x)^2\,dx = 1 .

Then $Lg = \delta(x-\xi)$ , $B_1g = B_2g = 0$ is inconsistent: the solvability condition would require $u_1(\xi) = \int_a^b\delta(x-\xi)u_1(x)\,dx = 0$ , which fails for most $\xi$ . The nearest substitute removes exactly the offending component of the source. The modified Green's function $g_M(x,\xi)$ is defined by

Lg_M = \delta(x-\xi) - u_1(x)u_1(\xi), \qquad B_1g_M = B_2g_M = 0 ,

\int_a^b g_M(x,\xi)\,u_1(x)\,dx = 0 \quad\text{for every } \xi \in (a,b) .

The first line is a consistent problem, since the modified source is orthogonal to $u_1$ by construction. Its solutions form a one-parameter family differing by multiples of $u_1(x)$ ; the second line, the normalisation, selects one of them. Stakgold and Holst note that it is not a solvability condition but a choice, that it picks the solution of minimum norm, and that this solution turns out to be symmetric, $g_M(x,\xi) = g_M(\xi,x)$ — sufficient for symmetry but not necessary, since adding $Cu_1(x)u_1(\xi)$ with $C$ constant keeps the symmetry and the equation while spoiling the normalisation.

Their motivating case is $-u'' = f$ on $0<x<1$ with $u'(0) = u'(1) = 0$ , an insulated rod: the solvability condition is $\int_0^1 f\,dx = 0$ , a concentrated unit source violates it, and compensating it by a uniform distribution of sinks of total strength $1$ gives $-g'' = \delta(x-\xi) - 1$ , with normalised solution

g_M(x,\xi) = \frac{x^2+\xi^2}{2} - x_{>} + \frac13 ,

their (3.5.5).

Theorem(Solving with the modified Green's function; the pseudoinverse)AdaptedAdapted from Stakgold & Holst (2011), §3.5, (3.5.8)–(3.5.9) and the discussion (a)–(d), pp. 217–220

With $(L;B_1,B_2)$ , $u_1$ and $g_M$ as above, consider

Lu = f \ \ (a<x<b), \qquad B_1u = B_2u = 0 .

(i) If the solvability condition $\int_a^b f\,u_1\,dx = 0$ holds, the general solution is

u(x) = C\,u_1(x) + \int_a^b g_M(x,\xi)\,f(\xi)\,d\xi , \qquad C \text{ arbitrary} ,

and the choice $C = 0$ gives the solution of minimum norm, because the two terms on the right are orthogonal.

(ii) If the solvability condition fails, the problem has no solution; among all $u$ satisfying the boundary conditions, those minimising $\lVert Lu - f\rVert$ in the mean-square norm are the least-squares solutions, and among those, $u = \int_a^b g_M(\cdot,\xi)f(\xi)\,d\xi$ is the one whose norm is smallest.

Part (i) is proved below; part (ii) is quoted, not proved here. Stakgold and Holst establish it for the single problem $u'' + \lambda_ku = f$ on $(0,1)$ with $u(0)=u(1)=0$ by expanding in the eigenfunctions $\sin n\pi x$ : the range of $L$ is orthogonal to $\sin k\pi x$ , so the element of the range nearest $f$ is $f$ with its $k$ th sine coefficient deleted, and $g_M(x,\xi) = \sum_{n \ne k} \dfrac{2\sin n\pi\xi\,\sin n\pi x}{\lambda_k - \lambda_n}$ ; they conclude from that computation that $G_M$ is the pseudoinverse of $D^2+\lambda_k$ with those boundary conditions, and refer to their Chapter 9 for the general treatment. A proof for a general self-adjoint problem would need the half of the alternative — that orthogonality to the kernel of the adjoint is sufficient for solvability, equivalently that the range of $L$ is closed — which this unit does not prove either. Taken together, (i) and (ii) say that the integral operator $G_M$ with kernel $g_M$ inverts $L$ in the sense that it returns the minimum-norm solution when there is one and the minimum-norm least-squares solution when there is not: it is the pseudoinverse of $L$ subject to the given boundary conditions.

Idea of the proofOriginal

One picture explains the whole alternative. A linear map takes its domain onto a subspace, its range. If the range is everything, every datum is attainable and the only question left is uniqueness. If the range is a proper subspace, a datum is attainable exactly when it lies in that subspace — and the clean way to say "lies in a subspace" is "is orthogonal to everything orthogonal to it". For a matrix, what is orthogonal to the range of $A$ is the null space of $A^*$ , because $(Au,v) = (u,A^*v)$ : a $v$ annihilated by $A^*$ cannot see anything $A$ produces. So solvability means that $f$ is orthogonal to the kernel of $A^*$ — and that is the alternative, in one line of inner-product algebra.

Transplanting this to a boundary-value problem needs an inner product and an adjoint. Both are supplied by Green's formula: $\int (vLu - uL^*v) = [J(u,v)]_a^b$ makes $L^*$ the adjoint of $L$ , and the adjoint boundary conditions are defined to kill the boundary term, so that $\int vLu = \int uL^*v$ for admissible $u$ and $v$ . Pairing $Lu=f$ with a solution of the homogeneous adjoint problem then gives $\int fv = \int uL^*v = 0$ at once. That is necessity, and it is all that elementary manipulation gives; sufficiency needs compactness and is quoted.

The modified Green's function is what you build when the source you want — a delta — is not in the range. Project it: subtract its component along the null direction. The remainder is in the range, so the modified problem can be solved; and the ambiguity left (any multiple of the null solution) is fixed by demanding that the answer have no null component either. What results is exactly the pseudoinverse: invert on the range, and kill the null space.

Proof(The alternative for a linear system)AdaptedAdapted from Stakgold & Holst (2011), §3.4, proof of the Fredholm Alternative, p. 208

Write $R_A = \{Au : u \in \mathbb{R}^n\} \subseteq \mathbb{R}^m$ for the range and $N_{A^*} = \{v : A^*v = 0\} \subseteq \mathbb{R}^m$ for the null space of the transpose. The identity to be used throughout is

(Au,v)_m = \sum_{i=1}^m v_i\sum_{j=1}^n a_{ij}u_j = \sum_{j=1}^n u_j\sum_{i=1}^m a_{ij}v_i = (u,A^*v)_n ,

obtained by exchanging the two finite sums; it says that $A^*$ , the transpose, is the adjoint.

Step 1 (necessity). Suppose $Au = f$ for some $u$ , and let $A^*v = 0$ . Then

(f,v)_m = (Au,v)_m = (u,A^*v)_n = (u,0)_n = 0 .

So every $f$ in the range is orthogonal to every element of $N_{A^*}$ .

Step 2 (set up the converse). Suppose $(f,v)_m = 0$ for all $v \in N_{A^*}$ ; we must show $f \in R_A$ . Since $R_A$ is a subspace of the finite-dimensional space $\mathbb{R}^m$ , every vector splits uniquely as its orthogonal projection onto $R_A$ plus a remainder orthogonal to $R_A$ (orthogonal projection). Write

f = h + h^{\perp}, \qquad h \in R_A, \quad h^{\perp} \perp R_A .

Step 3 ( $h^{\perp}$ lies in $N_{A^*}$ ). Being orthogonal to the whole range, $h^\perp$ satisfies $(Au,h^{\perp})_m = 0$ for every $u \in \mathbb{R}^n$ , so by the adjoint identity $(u,A^*h^{\perp})_n = 0$ for every $u$ . A vector orthogonal to every vector is zero, so $A^*h^{\perp} = 0$ , that is, $h^{\perp} \in N_{A^*}$ .

Step 4 (the hypothesis kills $h^\perp$ ). By hypothesis $f \perp N_{A^*}$ , and $h^{\perp} \in N_{A^*}$ by Step 3, so $(f,h^{\perp}) = 0$ . But $h \perp h^{\perp}$ by construction, so

0 = (f,h^{\perp}) = (h,h^{\perp}) + (h^{\perp},h^{\perp}) = \lVert h^{\perp}\rVert^2 ,

whence $h^{\perp} = 0$ and $f = h \in R_A$ : the system is solvable.

Step 5 (the solution set). If $Au_p = f$ , then $Au = f$ holds if and only if $A(u - u_p) = 0$ , that is, $u = u_p + u_H$ with $u_H$ in the null space of $A$ . Hence the solution is unique exactly when that null space is $\{0\}$ , and otherwise the solutions form a translate of it. $\square$

Proof(Necessity of the solvability conditions for the boundary-value problem)AdaptedAdapted from Stakgold & Holst (2011), §3.4, p. 211 and p. 213

Step 1 (pair the problem with an adjoint null solution). Let $u$ solve $Lu = f$ with $B_1u = \dots = B_pu = 0$ , and let $v$ solve the homogeneous adjoint problem (A). Form $v\,Lu - u\,L^*v$ and integrate over $(a,b)$ . On the left, $Lu = f$ and $L^*v = 0$ , so the integral is $\int_a^b f v\,dx$ .

Step 2 (Green's formula). On the right, Green's formula turns the same integral into the boundary term:

\int_a^b f\,v\, dx = \int_a^b\bigl(v\,Lu - u\,L^*v\bigr)dx = \Bigl[J(u,v)\Bigr]_a^b .

Step 3 (the boundary term vanishes). $u$ satisfies the homogeneous boundary conditions and $v$ the adjoint ones. The adjoint conditions were defined to be those making $\bigl[J(u,v)\bigr]_a^b$ vanish for every such $u$ ; so the right-hand side is zero and $\int_a^b fv\,dx = 0$ .

Step 4 (all $k$ conditions). Applying Steps 1–3 to each member of a basis $v^{(1)},\dots,v^{(k)}$ of the solution space of (A) gives the $k$ stated conditions. They are genuine restrictions on $f$ only when $k \ge 1$ ; when (A) has only the trivial solution there is nothing to check, which is case (a).

Step 5 (inhomogeneous boundary data). If instead $B_iu = \gamma_i$ , Steps 1 and 2 are unchanged but Step 3 fails: $\bigl[J(u,v)\bigr]_a^b$ need not vanish. It is, however, determined by the data. Indeed $\bigl[J(u,v)\bigr]_a^b$ is a bilinear expression in the $2p$ boundary values of $u$ and those of $v$ ; with $v$ fixed, it is a linear functional of the boundary values of $u$ which, by the definition of the adjoint conditions, vanishes on the subspace where $B_1u = \dots = B_pu = 0$ . A linear functional vanishing on the common kernel of $B_1,\dots,B_p$ is a linear combination of $B_1,\dots,B_p$ , so

\Bigl[J(u,v)\Bigr]_a^b = \sum_{i=1}^{p}\lambda_i(v)\,B_iu = \sum_{i=1}^p \lambda_i(v)\,\gamma_i ,

a quantity computable from $v$ and the given $\gamma_i$ alone. This is Stakgold and Holst's (3.4.16), with $\lambda_i(v)$ written there in terms of their complementary adjoint functionals. Hence the necessary condition $\int_a^b fv\,dx = \sum_i\lambda_i(v)\gamma_i$ . $\square$

Sufficiency is not proved here; see the note above.

Proof(The modified Green's function: consistency, and the representation)AdaptedAdapted from Stakgold & Holst (2011), §3.5, pp. 217–218

Let the problem be self-adjoint with a one-dimensional homogeneous solution space spanned by the normalised $u_1$ .

Step 1 (the modified problem is consistent). Because the problem is self-adjoint, the adjoint homogeneous problem is the homogeneous problem itself, so the solvability condition for $Lg = s$ with homogeneous boundary conditions is $\int_a^b s\,u_1\,dx = 0$ . For the modified source $s(x) = \delta(x-\xi) - u_1(x)u_1(\xi)$ ,

\int_a^b \Bigl[\delta(x-\xi) - u_1(x)u_1(\xi)\Bigr]u_1(x)\,dx = u_1(\xi) - u_1(\xi)\int_a^b u_1^2\,dx = u_1(\xi) - u_1(\xi) = 0 ,

using the sifting property and the normalisation $\int u_1^2 = 1$ . So the modified source passes the solvability test. Concluding from this that the problem defining $g_M$ really is solvable uses the sufficiency half of the alternative, which this unit quotes rather than proves (see the note above); that is the one place in this proof where something unproved is used. Granted solvability, the solutions differ by multiples of $u_1$ , and the normalisation condition fixes the multiple, since $\int (g_M + cu_1)u_1\,dx = c$ .

Step 2 (pair $g_M$ against the solution). Let $u$ solve $Lu = f$ with the homogeneous boundary conditions, which presupposes $\int fu_1\,dx = 0$ . Multiply the equation for $g_M$ by $u$ , the equation for $u$ by $g_M$ , subtract, and integrate in $x$ over $(a,b)$ :

\int_a^b\bigl(u\,Lg_M - g_M\,Lu\bigr)dx = \int_a^b u(x)\Bigl[\delta(x-\xi) - u_1(x)u_1(\xi)\Bigr]dx - \int_a^b g_M(x,\xi)f(x)\,dx .

Step 3 (the left side vanishes). Since $L^* = L$ and both $u$ and $g_M(\cdot,\xi)$ satisfy the homogeneous boundary conditions, Green's formula makes the left-hand side the boundary term $\bigl[J(u,g_M)\bigr]_a^b$ , which is zero.

Step 4 (read off $u$ ). Evaluating the first integral on the right by the sifting property and writing $C = -\int_a^b u\,u_1\,dx$ , the displayed equation becomes

0 = u(\xi) + C\,u_1(\xi) - \int_a^b g_M(x,\xi)f(x)\,dx ,

that is, $u(\xi) = -Cu_1(\xi) + \int_a^b g_M(x,\xi)f(x)\,dx$ . Relabelling $x \leftrightarrow \xi$ and using the symmetry $g_M(x,\xi) = g_M(\xi,x)$ — which is a property of the normalised choice quoted from the source, whose verification Stakgold and Holst set as their Exercise 3.5.1 and which is not proved here — gives

u(x) = C'u_1(x) + \int_a^b g_M(x,\xi)f(\xi)\,d\xi

with $C' = -C$ a constant. Conversely every such function solves the problem, since adding a multiple of $u_1$ changes neither $Lu$ nor the boundary values.

Step 5 (the choice $C' = 0$ is the minimum-norm solution). The integral term is orthogonal to $u_1$ : by the normalisation of $g_M$ and its symmetry,

\int_a^b u_1(x)\int_a^b g_M(x,\xi)f(\xi)\,d\xi\, dx = \int_a^b f(\xi)\left[\int_a^b g_M(x,\xi)u_1(x)\,dx\right]d\xi = 0 .

By the Pythagorean theorem the squared norm of $u$ is the sum of $C'^2$ (since $\lVert u_1\rVert = 1$ ) and the squared norm of the integral term, so it is least when $C' = 0$ . $\square$

Remark(Where the hypotheses are used, and what goes wrong)Original

The adjoint boundary conditions, not just the adjoint operator. Step 3 of the necessity proof uses $B_i^*v = 0$ , not merely $L^*v = 0$ . Taking the wrong homogeneous problem is the commonest way to get a wrong solvability condition. In Stakgold and Holst's Example 7, $u' = f$ on $(0,1)$ with the two conditions $u(0) = u(1) = 0$ , the boundary term $\bigl[J(u,v)\bigr]_0^1 = u(1)v(1) - u(0)v(0)$ already vanishes for every $v$ , so the adjoint problem carries no boundary conditions, its solutions are the constants, and the solvability condition is $\int_0^1 f\,dx = 0$ — which is also obvious by integrating the equation.
Self-adjointness, in the modified Green's function. Step 1 of the last proof uses $L^* = L$ and the equality of the two sets of boundary conditions, to know that the solvability condition is orthogonality to $u_1$ rather than to some other function $v_1$ . Drop it, and the condition on the modified source $s(x,\xi)$ is $\int_a^b s(x,\xi)v_1(x)\,dx = 0$ , where $v_1$ spans the solutions of the homogeneous adjoint problem. So it is the second argument that must carry $v_1$ . Taking the modified source to be $\delta(x-\xi) - \frac{v_1(x)\,v_1(\xi)}{\lVert v_1\rVert^2}$ always works, since the subtracted term integrates against $v_1$ to exactly $v_1(\xi)$ ; and the normalisation $\int_a^b g_M(x,\xi)u_1(x)\,dx = 0$ (now against $u_1$ , which spans the ambiguity) again selects one solution. The resulting $g_M$ is not symmetric. The tempting repair $u_1(x)v_1(\xi)/(u_1,v_1)$ is available only when $(u_1,v_1) \ne 0$ , which can fail: in Stakgold and Holst's Exercise 3.5.4, $L = D^2$ on $(0,1)$ with $B_1u = u(0)+u(1)$ and $B_2u = u'(0)-u'(1)$ , the direct homogeneous problem gives $u_1 = \sqrt3\,(1-2x)$ and the adjoint conditions $v(1) = v(0)$ , $v'(1)+v'(0)=0$ give $v_1 = 1$ (both normalised), and $\int_0^1 u_1v_1\,dx = 0$ . Subtracting $v_1(x)u_1(\xi)$ , as one might guess by analogy with the self-adjoint case, leaves $\int_0^1[\delta(x-\xi) - v_1(x)u_1(\xi)]v_1(x)\,dx = 1 - \sqrt3\,(1-2\xi)$ , nonzero for all but one $\xi$ : the modified problem would then be inconsistent, which is the very defect it was built to cure.
A one-dimensional null space. The definition above assumes $k = 1$ . For $k \ge 2$ the modified source is $\delta(x-\xi) - \sum_{i=1}^k u_i(x)u_i(\xi)$ with $\{u_i\}$ orthonormal, and $k$ normalisation conditions are imposed; their Exercise 3.5.2 develops the case $k = 2$ .
Normalisation of $u_1$ . If $\int u_1^2 \ne 1$ the cancellation in Step 1 fails and the modified problem is again inconsistent. The correct source is $\delta(x-\xi) - u_1(x)u_1(\xi)/\lVert u_1\rVert^2$ .
Finite dimension, in the matrix proof. Step 2 uses that every vector splits into its projection on $R_A$ and a component orthogonal to it. In infinite dimensions this needs the range to be closed, which is exactly what compactness of $K$ buys in the statements of Kreyszig and Evans, and exactly what fails for a general bounded operator. That is why the sufficiency half of the boundary-value alternative is not elementary.
Why the two dimensions agree. For an $m\times n$ matrix, rank-nullity gives $\dim N_A = n - r$ and $\dim N_{A^*} = m - r$ , which agree only when $m = n$ — the balanced case. The boundary-value problem is balanced in the same sense: $p$ conditions for a $p$ th-order equation. Stakgold and Holst's Remark 1 of §3.2, that a third boundary condition would usually destroy existence, is the unbalanced case showing itself.

Method(Using the alternative)AdaptedAdapted from Stakgold & Holst (2011), §3.4, pp. 211–213, and §3.5, pp. 216–218

Test the homogeneous problem. Solve $Lu = 0$ with the homogeneous boundary conditions. If only $u \equiv 0$ , build the ordinary Green's function and stop.
Find the adjoint problem. Compute $L^*$ and $J(u,v)$ from Green's formula, impose $B_1u = \dots = B_pu = 0$ , and read off the conditions on $v$ that make $\bigl[J(u,v)\bigr]_a^b$ vanish for all such $u$ . For a self-adjoint problem this step is a formality: the adjoint problem is the original one.
Solve the adjoint homogeneous problem for a basis $v^{(1)},\dots,v^{(k)}$ .
Write the solvability conditions $\int_a^b f v^{(i)}dx = 0$ (homogeneous boundary data), or $\int_a^b fv^{(i)}dx = \bigl[J(u,v^{(i)})\bigr]_a^b$ evaluated from the $\gamma$ 's (inhomogeneous data). Check them against the physics: they are almost always a conservation law.
If they hold, the solutions form a $k$ -parameter family. For a self-adjoint problem with $k=1$ , normalise $u_1$ , solve $Lg_M = \delta(x-\xi) - u_1(x)u_1(\xi)$ with the homogeneous boundary conditions exactly as for an ordinary Green's function — the jump condition will come out automatically satisfied, leaving one free constant — and fix that constant by $\int_a^b g_Mu_1\,dx = 0$ . Then $u = \int g_M f\,d\xi$ is the minimum-norm solution, and the general one adds $Cu_1$ .
If they fail, there is no solution. The same formula $u = \int g_M f\,d\xi$ now returns the minimum-norm least-squares solution, which is the exact solution of the problem with $f$ replaced by its projection $f - u_1\int fu_1$ .

Example(A resonant Neumann problem with variable coefficients)Original

Take $Lu = \bigl(x\,u'\bigr)'$ on $1 \le x \le e$ with the Neumann conditions $u'(1) = u'(e) = 0$ .

The homogeneous problem. $(xu')' = 0$ gives $xu' = c$ ; the condition $u'(1) = 0$ forces $c = 0$ , so $u$ is constant, and a constant satisfies both conditions. The homogeneous solution space is one-dimensional, spanned by the constants: $k = 1$ . The ordinary Green's function does not exist.
Self-adjointness and the solvability condition. $L = (a_2u')' + 0$ with $a_2 = x$ , so $L$ is formally self-adjoint, and $J(u,v) = x(vu'-uv')$ vanishes at both endpoints when $u' = v' = 0$ there: the adjoint problem is the same problem. Normalising, $\int_1^e 1^2dx = e-1$ , so $u_1 = (e-1)^{-1/2}$ and the solvability condition is $\int_1^e f(x)\,dx = 0 .$ Physically: with both ends insulated there is no steady state unless the net source strength vanishes.
The modified problem. With $u_1(x)u_1(\xi) = 1/(e-1)$ , $\bigl(x\,g_M'\bigr)' = \delta(x-\xi) - \frac{1}{e-1}, \qquad g_M'(1,\xi) = g_M'(e,\xi) = 0, \qquad \int_1^e g_M(x,\xi)\,dx = 0 .$
Integrate once. Away from $x=\xi$ $x = ξ$ , $\bigl(xg_M'\bigr)' = -1/(e-1)$ $(x g_{M}^{'})^{'} = - 1/ (e - 1)$ gives $x\,g_M' = -x/(e-1) + k$ $x g_{M}^{'} = - x / (e - 1) + k$ .
- On $(1,\xi)$ : $g_M'(1) = 0$ forces $k = 1/(e-1)$ , so $g_M' = \dfrac{1}{e-1}\Bigl(\dfrac1x - 1\Bigr)$ .
- On $(\xi,e)$ : $g_M'(e) = 0$ forces $k = e/(e-1)$ , so $g_M' = \dfrac{1}{e-1}\Bigl(\dfrac{e}{x} - 1\Bigr)$ .
The jump is automatic. The jump in $g_M'$ at $x = \xi$ is $\frac{1}{e-1}\left(\frac{e}{\xi} - \frac{1}{\xi}\right) = \frac{1}{\xi} = \frac{1}{a_2(\xi)} ,$ exactly as required — with no constant left to adjust. This is the signature of the modified problem: four conditions have become three, and one free constant survives.
Integrate again and match. Integrating, $g_M = \dfrac{\ln x - x}{e-1} + A$ on $(1,\xi)$ and $g_M = \dfrac{e\ln x - x}{e-1} + B$ on $(\xi,e)$ ; continuity at $x = \xi$ gives $A = B + \ln\xi$ . Both branches then collapse into $g_M(x,\xi) = \frac{\ln x - x}{e-1} + \ln x_{>} + B(\xi) , \qquad x_{>} = \max(x,\xi) .$
Normalise. $\int_1^e \dfrac{\ln x - x}{e-1}dx = \dfrac{3-e^2}{2(e-1)}$ , using $\int_1^e\ln x\,dx = 1$ ; and $\int_1^e \ln x_{>}\,dx = (\xi-1)\ln\xi + \bigl[x\ln x - x\bigr]_{\xi}^{e} = \xi - \ln\xi$ . Setting the total to zero gives $B(\xi) = \dfrac{e^2-3}{2(e-1)^2} + \dfrac{\ln\xi - \xi}{e-1}$ , hence $\boxed{\;g_M(x,\xi) = \frac{\ln x - x}{e-1} + \frac{\ln\xi - \xi}{e-1} + \ln\bigl(\max(x,\xi)\bigr) + \frac{e^2-3}{2(e-1)^2}\;}$ which is visibly symmetric in $x$ and $\xi$ , as the theory promises. Numerically the constant is $0.7432798$ .
Use it. Take $f(x) = 1/x - 1/(e-1)$ , which satisfies $\int_1^e f = 1 - 1 = 0$ . Then $u(x) = \int_1^e g_M(x,\xi)f(\xi)\,d\xi = \frac{(\ln x)^2}{2} - \frac{x}{e-1} + \frac{\ln x}{e-1} + c , \qquad c = \frac{3}{2(e-1)} - \frac{1}{(e-1)^2} .$
Check directly. Integrating $(xu')' = 1/x - 1/(e-1)$ gives $xu' = \ln x - x/(e-1) + k$ ; the condition $u'(1) = 0$ gives $k = 1/(e-1)$ , and then $xu'$ at $x=e$ is $1 - e/(e-1) + 1/(e-1) = 0$ , so the second condition is met automatically — the solvability condition at work. A further integration reproduces item 8 up to the additive constant, and $\int_1^e u\,dx = 0$ pins that constant to the value of $c$ above, $c = 0.5342682$ . At $x = 2$ the solution is $u(2) = 0.01393678$ .
An inconsistent source. Take instead $f(x) = 1/x$ , for which $\int_1^e f = 1 \ne 0$ : no solution exists. But $G_M$ annihilates constants, because $\int_1^e g_M(x,\xi)\,d\xi = 0$ for every $x$ by the normalisation and the symmetry. So $G_M$ applied to $1/x$ gives the same function as $G_M$ applied to $1/x - 1/(e-1)$ , namely the $u$ of item 8. That function solves the problem exactly for the projected source, and is therefore the minimum-norm least-squares solution for $f = 1/x$ .

ApplicationOriginal

The solvability condition is almost always a conservation law in disguise, and recognising it saves a great deal of fruitless computation.

Insulated systems. Steady heat conduction in a rod with both ends insulated, steady diffusion in a sealed vessel, steady flow in a closed loop: in each case the homogeneous problem has the constants as solutions, and the alternative says a steady state exists only if the total source strength is zero. Stakgold and Holst make the point dynamically: with a net source, the time-dependent temperature grows without bound rather than settling. The example above is the version with a variable conductance.
Electrostatics with Neumann data. $\nabla^2\varphi = -\rho/\varepsilon_0$ in a region with $\partial\varphi/\partial n$ prescribed on the boundary has a solution only if the prescribed normal flux integrates to the enclosed charge — Gauss's law reappearing as a solvability condition — and $\varphi$ is then fixed only to within an added constant, the null direction.
Rigid-body modes in mechanics. A free elastic body, or a truss with no supports, has a stiffness operator whose null space is the rigid translations and rotations. The alternative says an equilibrium exists only for self-equilibrated loads — zero net force and zero net torque — and the displacement is then fixed only up to a rigid motion. Finite-element codes handle this exactly as item 10 above does: they project the load onto the range and return the minimum-norm displacement, which is the pseudoinverse solution.
Resonance. For $u'' + \lambda u = f$ at an eigenvalue $\lambda = \lambda_k$ , the forcing must be orthogonal to the resonant mode or no steady response exists; a driven oscillator forced exactly at its natural frequency grows linearly in time instead. This is why the modified Green's function appears in perturbation theory: in the Rayleigh–Schrödinger expansion for a nondegenerate level, the first-order correction to the state is obtained by applying the reduced resolvent — the modified Green's function of $H_0 - E_0$ , with the unperturbed state projected out — to the perturbation, and the solvability condition is exactly the statement that $E_1 = \langle\psi_0|V|\psi_0\rangle$ .
Numerical linear algebra. The matrix case of this unit is the theory behind least squares and the Moore–Penrose pseudoinverse: $Au = f$ is solvable iff $f \perp \ker A^{\mathsf{T}}$ , and otherwise the normal equations $A^{\mathsf{T}}Au = A^{\mathsf{T}}f$ return the least-squares solution, with the singular value decomposition supplying the minimum-norm one (the singular value decomposition). Discretising a resonant boundary-value problem produces a nearly singular matrix, and the modified Green's function is what the pseudoinverse converges to.

Theorem

Eigenfunction expansion of Green's functions

Adapted

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Derives the bilinear series $G(x,\xi;\lambda) = \sum_n u_n(x)\overline{u_n(\xi)}/(\lambda_n-\lambda)$ for the resolvent of a regular Sturm–Liouville problem and the special case $G = \sum u_n(x)\overline{u_n(\xi)}/\lambda_n$ at $\lambda=0$ , shows that $G$ is meromorphic in $\lambda$ with simple poles at the eigenvalues whose residues are $-u_n(x)\overline{u_n(\xi)}$ , and uses that to read the whole spectrum off a closed-form Green's function.

Prerequisites: Green's function for a boundary-value problem, Eigenfunction expansions, Sturm-Liouville eigenvalue problems, Spectral theorem for Hermitian matrices, Classification of isolated singularities, The residue theorem

Definition(The regular problem, its resolvent, and the Green's function of

L-\lambda

)AdaptedAdapted from Stakgold & Holst (2011), §7.1, (7.1.1)–(7.1.6), (7.1.12), (7.1.15), pp. 409–413; Hassani (2013), §20.4, p. 630

Let $p \in C^1[a,b]$ with $p > 0$ , let $q$ and $s$ be continuous with $s > 0$ on $[a,b]$ , and consider the formally self-adjoint expression together with its weighted inner product

\mathcal{L}u = -\bigl(p\,u'\bigr)' + q\,u , \qquad (f,g)_s = \int_a^b s(x)\,f(x)\,\overline{g(x)}\,dx .

Impose unmixed boundary conditions written, as Stakgold and Holst do, with two angles $\alpha,\beta \in [0,\pi)$ :

B_au = u(a)\cos\alpha - u'(a)\sin\alpha = 0 , \qquad B_bu = u(b)\cos\beta + u'(b)\sin\beta = 0 .

( $\beta = 0$ is $u(b) = 0$ and $\beta = \pi/2$ is $u'(b) = 0$ ; likewise at $a$ .) The eigenvalue problem is

\mathcal{L}u = \lambda\,s\,u \ \ (a<x<b), \qquad B_au = B_bu = 0 .

The Green's function of $\mathcal{L} - \lambda s$ , also called the resolvent kernel, is the function $G(x,\xi;\lambda)$ with

-\bigl(p\,G'\bigr)' + q\,G - \lambda\,s\,G = \delta(x-\xi) \ \ (a<x,\xi<b), \qquad B_aG = B_bG = 0 ,

derivatives taken in $x$ ; this is Stakgold and Holst's (7.1.15). It is the Green's function of the previous construction applied to the operator $\mathcal{L} - \lambda s$ , whose leading coefficient is $-p$ , so the jump condition reads

\left.\frac{\partial G}{\partial x}\right|_{\xi^+} - \left.\frac{\partial G}{\partial x}\right|_{\xi^-} = -\frac{1}{p(\xi)} .

By the existence theorem it exists exactly when $\lambda$ is not an eigenvalue, and then

w(x,\lambda) = \int_a^b G(x,\xi;\lambda)\,f(\xi)\,d\xi

is the unique solution of $\mathcal{L}w - \lambda s w = f$ with the same boundary conditions. Hassani's way of putting it: $G$ is the kernel of the operator $(L - \lambda)^{-1}$ , the resolvent $R_\lambda(L)$ , and $G$ at $\lambda = 0$ is the kernel of $L^{-1}$ . Write $G(x,\xi) = G(x,\xi;0)$ for that case.

Note(Facts about the spectrum used but not proved here)AdaptedAdapted from Stakgold & Holst (2011), §7.1, item 3 and (7.1.14), p. 412

For the regular problem above the following hold, and are quoted rather than proved:

the eigenvalues are real, simple, and can be listed as $\lambda_1 < \lambda_2 < \dots \to +\infty$ , with only finitely many negative;
the corresponding normalised eigenfunctions $\{u_n\}$ , with $(u_n,u_m)_s = \delta_{nm}$ , form an orthonormal basis of the space with inner product $(\cdot,\cdot)_s$ ;
hence the completeness relation, valid as an identity between distributions, $\frac{\delta(x-\xi)}{s(x)} = \sum_n u_n(x)\,\overline{u_n(\xi)} ,$ Stakgold and Holst's (7.1.20).

Their proof converts the eigenvalue problem into the integral equation $v = \lambda Kv$ with the symmetric Hilbert–Schmidt kernel $k(x,\xi) = \sqrt{s(x)}\,g_0(x,\xi)\sqrt{s(\xi)}$ , where $g_0$ is the Green's function at $\lambda=0$ , and applies the spectral theory of self-adjoint Hilbert–Schmidt operators; the simplicity of the eigenvalues uses the unmixedness of the boundary conditions. That route is Sturm–Liouville theory and Hilbert–Schmidt theory. Note that the Green's function is used to prove the completeness that is used below to expand it; the two statements are not circular, because the expansion below assumes completeness as given.

Theorem(The bilinear expansion)AdaptedAdapted from Stakgold & Holst (2011), §7.1, (7.1.16) and (7.1.18), p. 413; Hassani (2013), §20.4, (20.42) and (20.44), pp. 631–632; Arfken, Weber & Harris (2013), §10.1, (10.14)–(10.15), p. 450; Riley, Hobson & Bence (2006), §17.5, (17.50)–(17.52), p. 570

Let $\lambda$ be different from every eigenvalue. Then

\boxed{\;G(x,\xi;\lambda) \;=\; \sum_{n} \frac{u_n(x)\,\overline{u_n(\xi)}}{\lambda_n - \lambda}\;}

the bilinear series for the resolvent kernel, and the solution of $\mathcal{L}w - \lambda sw = f$ with the given boundary conditions is

w(x,\lambda) = \sum_n \frac{\langle f,u_n\rangle}{\lambda_n - \lambda}\,u_n(x) , \qquad \langle f,u_n\rangle = \int_a^b f(\xi)\,\overline{u_n(\xi)}\,d\xi .

In particular, provided $0$ is not an eigenvalue,

G(x,\xi) = \sum_n \frac{u_n(x)\,\overline{u_n(\xi)}}{\lambda_n} ,

which is Arfken's (10.14), Hassani's (20.44) and Riley, Hobson and Bence's (17.51). Two consequences are immediate from the form of the series:

Symmetry. $G(x,\xi;\lambda) = \overline{G(\xi,x;\overline\lambda)}$ ; for real $\lambda$ and real eigenfunctions, $G(x,\xi;\lambda) = G(\xi,x;\lambda)$ . Arfken's (10.15) and Riley, Hobson and Bence's (17.52) record this, and it agrees with the reciprocity proved directly in the construction unit.
Where the solution is large. The coefficient of $u_n$ in $w$ carries the factor $(\lambda_n-\lambda)^{-1}$ , so driving the system near an eigenvalue amplifies the component of the response along that eigenfunction without limit — resonance, read off the formula.

If an eigenvalue $\lambda_n$ is degenerate, the numerator becomes $\sum_k u_n^{(k)}(x)\overline{u_n^{(k)}(\xi)}$ , summed over an orthonormal basis of its eigenspace, as in Hassani's (20.42). The regular problem of the definition has simple eigenvalues, so a single term suffices there.

Theorem(Poles, residues, and recovery of the spectrum)AdaptedAdapted from Stakgold & Holst (2011), §7.1, pp. 413–415, the residue statement, (7.1.19) and (7.1.22)–(7.1.25); Hassani (2013), §20.4, the residue and (20.43), p. 631

Regard $G(x,\xi;\lambda)$ as a function of the complex variable $\lambda$ with $x,\xi$ fixed.

(i) Meromorphy. $G$ is meromorphic in $\lambda$ , with simple poles exactly at the eigenvalues $\lambda_n$ , and

\operatorname*{Res}_{\lambda = \lambda_n} G(x,\xi;\lambda) = -\,u_n(x)\,\overline{u_n(\xi)} .

(ii) Closed form. Let $v(\cdot,\lambda)$ solve $\mathcal{L}v = \lambda s v$ with $v(a,\lambda) = \sin\alpha$ , $v'(a,\lambda) = \cos\alpha$ , and let $z(\cdot,\lambda)$ solve the same equation with $z(b,\lambda) = \sin\beta$ , $z'(b,\lambda) = -\cos\beta$ . Then $B_av = 0$ , $B_bz = 0$ , both are entire functions of $\lambda$ , and

G(x,\xi;\lambda) = -\,\frac{v(x_{<},\lambda)\,z(x_{>},\lambda)}{C(\lambda)} , \qquad p(x)\,W\bigl[v,z;x\bigr] = p\,(vz' - zv') = C(\lambda) ,

where $C$ , the characteristic function, is entire and independent of $x$ . The zeros of $C$ are exactly the eigenvalues, and they are simple.

(iii) Eigenfunctions from residues. Writing $v_n = v(\cdot,\lambda_n)$ , $z_n = z(\cdot,\lambda_n)$ , these are proportional, $v_n = k_nz_n$ with $k_n$ a nonzero real constant, and the normalised eigenfunction is

u_n(x) = \pm\,\frac{v_n(x)}{\bigl\{k_n\,C'(\lambda_n)\bigr\}^{1/2}} = \pm\left\{\frac{k_n}{C'(\lambda_n)}\right\}^{1/2}z_n(x) ,

Stakgold and Holst's (7.1.25). So a closed-form expression for $G$ , obtained without any knowledge of the spectrum, determines the eigenvalues as the poles and the normalised eigenfunctions from the residues — the normalisation constant included, which is the part usually computed by a separate integral.

(iv) Completeness as a contour integral. Integrating $G$ counterclockwise around a circle $C_\infty$ in the $\lambda$ plane enclosing all the poles and summing the residues gives

\frac{1}{2\pi i}\oint_{C_\infty} G(x,\xi;\lambda)\,d\lambda = -\sum_n u_n(x)\overline{u_n(\xi)} = -\frac{\delta(x-\xi)}{s(x)} ,

Stakgold and Holst's (7.1.19) with (7.1.20), and Hassani's (20.43). Both authors note that the series does not converge in the ordinary sense; the identity holds in the sense of distributions. Stakgold and Holst add that for a problem with continuous spectrum the same contour formula survives with the sum replaced by an integral over a continuous parameter, because $G$ then has a branch cut instead of poles.

Idea of the proofOriginal

A Green's function inverts a differential operator; an orthonormal basis of eigenfunctions diagonalises it. Put the two together and the inverse must be diagonal in the same basis, with the reciprocals of the eigenvalues on the diagonal — exactly as a symmetric matrix $A = \sum\lambda_n e_ne_n^{\mathsf T}$ has $A^{-1} = \sum\lambda_n^{-1}e_ne_n^{\mathsf T}$ . The bilinear series is that identity written as a kernel, and $(\mathcal{L}-\lambda s)^{-1}$ replaces $\lambda_n$ by $\lambda_n - \lambda$ .

Two short routes prove it. Arfken's: expand both the unknown kernel and the delta in the basis, apply $\mathcal{L}$ , and compare coefficients. Stakgold's and Hassani's: pair the defining equation for $G$ with $u_n$ and use the symmetry of $\mathcal{L}$ to move the operator onto $u_n$ , which turns $(\mathcal{L}-\lambda s)G$ into $(\lambda_n-\lambda)$ times the coefficient; the delta on the other side supplies $\overline{u_n(\xi)}$ . The second route is used below, because it needs completeness only at the last step.

The pole structure is then forced: a sum of simple poles with residues $-u_n(x)\overline{u_n(\xi)}$ . The surprise is that the same information can be extracted from the constructed Green's function, which knows nothing about eigenfunctions. Build $G$ out of two solutions normalised by initial data independent of $\lambda$ ; those depend analytically on $\lambda$ , so all the $\lambda$ -dependence of the poles sits in the one scalar $C(\lambda) = pW[v,z]$ in the denominator. $C$ vanishes exactly when $v$ and $z$ are proportional — that is, when a single function satisfies the equation and both boundary conditions, which is what an eigenfunction is. Taking the residue, the two pieces $v(x_<)z(x_>)$ fuse into the symmetric product $z_n(x)z_n(\xi)$ : the kink at $x=\xi$ leaves no trace, because at an eigenvalue the two branches are multiples of the same function.

Proof(The bilinear expansion)AdaptedAdapted from Hassani (2013), §20.4, the computation of $a_n^{(k)}(y)$, pp. 630–631; Stakgold & Holst (2011), §7.1, derivation of (7.1.16), p. 413

Fix $\xi \in (a,b)$ and let $\lambda$ be no eigenvalue, so that $G(\cdot,\xi;\lambda)$ exists.

Step 1 (expand in the basis). Since $\{u_n\}$ is an orthonormal basis for $(\cdot,\cdot)_s$ and $G(\cdot,\xi;\lambda)$ satisfies the boundary conditions,

G(x,\xi;\lambda) = \sum_n g_n(\xi,\lambda)\,u_n(x), \qquad g_n = \bigl(G(\cdot,\xi;\lambda),\,u_n\bigr)_s = \int_a^b s(x)\,G(x,\xi;\lambda)\,\overline{u_n(x)}\,dx .

Step 2 (pair the defining equation with $u_n$ ). Multiply $\mathcal{L}G - \lambda sG = \delta(\cdot-\xi)$ by $\overline{u_n(x)}$ and integrate over $(a,b)$ . The right-hand side gives $\overline{u_n(\xi)}$ by the sifting property. The left-hand side is

\int_a^b \overline{u_n}\,\mathcal{L}G\,dx \;-\; \lambda\int_a^b s\,\overline{u_n}\,G\,dx = \bigl(\mathcal{L}G,u_n\bigr)_{\!*} - \lambda\,g_n ,

where $\bigl(\mathcal{L}G,u_n\bigr)_{\!*}$ denotes the unweighted integral $\int \overline{u_n}\,\mathcal{L}G$ .

Step 3 (move $\mathcal{L}$ onto the eigenfunction). For the formally self-adjoint $\mathcal{L}$ , Lagrange's identity gives $\overline{u_n}\,\mathcal{L}G - G\,\overline{\mathcal{L}u_n} = \bigl[p\,W(G,\overline{u_n})\bigr]'$ with $W(G,\overline{u_n}) = G\,\overline{u_n}' - G'\,\overline{u_n}$ — directly: the two terms in $q$ cancel, and $-\overline{u_n}(pG')' + G(p\overline{u_n}')'$ is the derivative of $p(G\overline{u_n}' - G'\overline{u_n})$ , the cross terms $p\,G'\overline{u_n}'$ cancelling. (This is Stakgold and Holst's (7.1.8), which carries their $1/s$ because their $L$ does; multiplying by $s$ gives the form above.) Hence

\int_a^b\bigl(\overline{u_n}\,\mathcal{L}G - G\,\overline{\mathcal{L}u_n}\bigr)dx = \Bigl[p\,\bigl(G\,\overline{u_n}' - G'\,\overline{u_n}\bigr)\Bigr]_a^b = 0 ,

because $G(\cdot,\xi;\lambda)$ and $u_n$ both satisfy the same unmixed homogeneous conditions, which make the Wronskian vanish at each endpoint separately. Here the self-adjointness of the problem — both $\mathcal{L}^* = \mathcal{L}$ and the matching boundary conditions — is used. Since $\mathcal{L}u_n = \lambda_n s u_n$ and $\lambda_n$ is real,

\int_a^b \overline{u_n}\,\mathcal{L}G\,dx = \int_a^b G\,\overline{\mathcal{L}u_n}\,dx = \lambda_n\int_a^b s\,G\,\overline{u_n}\,dx = \lambda_n\,g_n .

Step 4 (solve for the coefficient). Steps 2 and 3 combine to $\bigl(\lambda_n - \lambda\bigr)g_n = \overline{u_n(\xi)}$ . As $\lambda$ is no eigenvalue, $\lambda_n - \lambda \ne 0$ and

g_n(\xi,\lambda) = \frac{\overline{u_n(\xi)}}{\lambda_n - \lambda} .

Substituting into Step 1 gives the bilinear series.

Step 5 (the solution series). For $w = \int_a^b G(\cdot,\xi;\lambda)f(\xi)\,d\xi$ , insert the series and integrate term by term:

w(x,\lambda) = \sum_n \frac{u_n(x)}{\lambda_n-\lambda}\int_a^b \overline{u_n(\xi)}\,f(\xi)\,d\xi = \sum_n\frac{\langle f,u_n\rangle}{\lambda_n-\lambda}\,u_n(x) .

Step 6 (symmetry). The series displays $G(x,\xi;\lambda)$ as $\sum u_n(x)\overline{u_n(\xi)}/(\lambda_n-\lambda)$ ; interchanging $x$ and $\xi$ and conjugating sends each term to itself with $\lambda$ replaced by $\overline\lambda$ , so $G(x,\xi;\lambda) = \overline{G(\xi,x;\overline\lambda)}$ . $\square$

Proof(Poles, residues and the characteristic function)AdaptedAdapted from Stakgold & Holst (2011), §7.1, pp. 413–415

Step 1 (the two entire solutions). Let $v(\cdot,\lambda)$ and $z(\cdot,\lambda)$ be as in the statement. Each is the solution of an initial-value problem whose data do not involve $\lambda$ and whose equation depends on $\lambda$ analytically. That $v$ , $z$ and their $x$ -derivatives are then entire functions of $\lambda$ is quoted from Stakgold and Holst, not proved here: the argument converts the initial-value problem into a Volterra integral equation and sums its Neumann series, which converges for every complex $\lambda$ , and neither the Volterra form nor the convergence of that series is among this unit's prerequisites. Everything below uses only the conclusion. The choice of initial data makes $B_av = 0$ and $B_bz = 0$ : for instance $B_av = \sin\alpha\cos\alpha - \cos\alpha\sin\alpha = 0$ .

Step 2 (the shape of $G$ ). For $x < \xi$ , $G$ solves the homogeneous equation and satisfies the condition at $a$ , so it is a multiple of $v$ ; for $x>\xi$ it is a multiple of $z$ . Continuity at $x=\xi$ forces the common form $G = A\,v(x_{<},\lambda)z(x_{>},\lambda)$ with $A$ depending on $\xi$ and $\lambda$ but not on $x$ .

Step 3 (the jump fixes $A$ ). The jump condition is $\bigl[\partial G/\partial x\bigr]_{\xi^-}^{\xi^+} = -1/p(\xi)$ , since the leading coefficient of $\mathcal{L}-\lambda s$ is $-p$ . Computing the two one-sided derivatives,

A\bigl[v(\xi,\lambda)z'(\xi,\lambda) - v'(\xi,\lambda)z(\xi,\lambda)\bigr] = -\frac{1}{p(\xi)} ,

and the bracket is the Wronskian $W[v,z;\xi]$ . By Abel's formula for a formally self-adjoint operator, $p(x)W[v,z;x]$ is independent of $x$ ; call it $C(\lambda)$ . Then $A\,C(\lambda)/p(\xi) = -1/p(\xi)$ , so $A = -1/C(\lambda)$ and

G(x,\xi;\lambda) = -\frac{v(x_{<},\lambda)\,z(x_{>},\lambda)}{C(\lambda)} .

Being a Wronskian of entire functions times $p$ , $C$ is entire.

Step 4 (zeros of $C$ are the eigenvalues). If $C(\mu) = 0$ then $W[v,z;x]$ vanishes identically, so $v(\cdot,\mu)$ and $z(\cdot,\mu)$ are linearly dependent. Neither vanishes identically, because their initial data $(\sin\alpha,\cos\alpha)$ and $(\sin\beta,-\cos\beta)$ are not both zero for $\alpha,\beta \in [0,\pi)$ . Hence $v(\cdot,\mu)$ is a nonzero multiple of $z(\cdot,\mu)$ and therefore satisfies both boundary conditions as well as the equation: $\mu$ is an eigenvalue. Conversely, the bilinear series shows $G$ is singular at each $\lambda_n$ , and the only possible source of a singularity in Step 3 is a zero of $C$ ; so $C(\lambda_n) = 0$ . The bilinear series also shows the poles are simple, so the zeros of $C$ are simple.

Step 5 (the residue). Fix $n$ and abbreviate $v_n = v(\cdot,\lambda_n)$ , $z_n = z(\cdot,\lambda_n)$ , with $v_n = k_nz_n$ from Step 4. Since $C(\lambda_n) = 0$ and $C'(\lambda_n) \ne 0$ ,

\operatorname*{Res}_{\lambda=\lambda_n} G = -\frac{v_n(x_{<})\,z_n(x_{>})}{C'(\lambda_n)} = -\frac{k_n\,z_n(x_{<})\,z_n(x_{>})}{C'(\lambda_n)} = -\frac{k_n\,z_n(x)\,z_n(\xi)}{C'(\lambda_n)} ,

the last step because $z_n(x_<)z_n(x_>) = z_n(x)z_n(\xi)$ whichever of $x,\xi$ is the larger. The discontinuity of $\partial G/\partial x$ at $x=\xi$ has left no trace: at an eigenvalue the two branches are multiples of one and the same function.

Step 6 (identify the eigenfunction). Comparing with the residue $-u_n(x)\overline{u_n(\xi)}$ read off the bilinear series, and recalling that for a real problem the eigenfunctions may be taken real,

u_n(x)u_n(\xi) = \frac{k_n\,z_n(x)z_n(\xi)}{C'(\lambda_n)} \quad\text{for all } x,\xi ,

so $u_n = \pm\{k_n/C'(\lambda_n)\}^{1/2}z_n$ , equivalently $u_n = \pm v_n/\{k_nC'(\lambda_n)\}^{1/2}$ . The sign is the usual arbitrary phase of an eigenfunction.

Step 7 (the contour integral). Let $C_\infty$ be a large circle in the $\lambda$ plane enclosing all the poles. Summing residues,

\frac{1}{2\pi i}\oint_{C_\infty} G(x,\xi;\lambda)\,d\lambda = \sum_n\operatorname*{Res}_{\lambda_n}G = -\sum_n u_n(x)\overline{u_n(\xi)} = -\frac{\delta(x-\xi)}{s(x)} ,

by the completeness relation quoted in the note. Stakgold and Holst and Hassani both record that the last series diverges pointwise and the identity is to be read distributionally. $\square$

Remark(Where the hypotheses are used, and what the series does and does not do)Original

$\lambda$ not an eigenvalue. Used in Step 4 of the first proof to divide by $\lambda_n-\lambda$ , and it is also exactly the condition for $G$ to exist (Hassani states the equivalence explicitly). At $\lambda = \lambda_n$ the resolvent does not exist and its substitute is the modified Green's function, whose bilinear series is the same one with the offending term deleted.
Zero as an eigenvalue. The $\lambda=0$ series $\sum u_nu_n/\lambda_n$ is the special case $\lambda=0$ and requires $0 \notin\{\lambda_n\}$ . Arfken flags this explicitly and declines to pursue it; the insulated-rod problem of the previous unit is the standard instance.
Self-adjointness. Step 3 of the first proof uses both $\mathcal{L}^*=\mathcal{L}$ and the fact that $G$ and $u_n$ satisfy the same boundary conditions. For a non-self-adjoint problem the expansion needs the eigenfunctions of $L$ in one argument and those of $L^*$ in the other, and the eigenfunctions need not be complete at all.
Convergence. The series converges in the mean-square sense for each fixed $\xi$ , since it is the expansion of a function in an orthonormal basis. Pointwise convergence is better than that because the coefficients decay and the eigenfunctions stay bounded: for a regular problem $\lambda_n \sim c\,n^2$ and the normalised $u_n$ are bounded uniformly in $n$ and in $x$ — both facts belong to Sturm–Liouville theory and are used here, not proved — so the terms are $O(n^{-2})$ uniformly and the series converges absolutely and uniformly, with $G$ continuous. It cannot, however, be differentiated twice term by term: the result would be the completeness relation, a distribution, not a function — which is precisely the kink at $x=\xi$ reappearing.
Practical weight. Arfken is blunt that the bilinear series is of "dubious value as a computational tool": computing a few eigenfunctions and summing a slowly converging series is usually worse than solving the two-point problem in closed form. Its value is structural — symmetry, resonance, the pole picture — and in the reverse direction of part (iii), where a closed-form $G$ hands over the whole spectrum.
Where simplicity of the eigenvalues matters. Steps 5 and 6 of the second proof assume each eigenvalue has a one-dimensional eigenspace, which holds for the unmixed conditions above. With periodic conditions eigenvalues can be double, and the residue is then the sum over an orthonormal basis of the eigenspace, as in Hassani's (20.42).

Remark(Conventions)Original

Four books, four normalisations of the same object. This unit uses Stakgold and Holst's: the equation carries no weight on the delta, the weight sits in the inner product, and the solution is $\int Gf\,d\xi$ .

Source	Eigenvalue problem	Equation for $G$	Solution
here, Stakgold–Holst	$\mathcal{L}u = \lambda s u$	$(\mathcal{L}-\lambda s)G = \delta$	$\int G f\,d\xi$
Arfken	$L\varphi_n = \lambda_n\varphi_n$	$LG = \delta$	$\int Gf\,dt$
Riley–Hobson–Bence	$Ly_n = \lambda_n\rho\,y_n$	$LG = \delta$	$\int Gf\,dz$
Hassani	$L_xu = \lambda u$	$(L_x-\lambda)G_\lambda = \delta/w$	$\int w\,Gf\,dy$

All four produce the same bilinear series $\sum u_n(x)\overline{u_n(\xi)}/(\lambda_n-\lambda)$ , each in its own symbols, with eigenfunctions normalised in the inner product that goes with the weight.

Two cautions, about the weight and about the sign.

The weight. Hassani's $1/w$ on the delta does not make his kernel ours divided by $w$ : his (20.24) defines $L_x = \frac1w\frac{d}{dx}\bigl(p\frac{d}{dx}\bigr) + q$ , with the $1/w$ already inside the operator, so multiplying $(L_x-\lambda)G_\lambda = \delta/w$ through by $w$ returns the same unweighted delta the other three rows carry. (Take $p=1$ , $q=0$ , $w=s=2$ on $(0,1)$ with Dirichlet conditions at $\lambda=0$ : ours solves $-G''=\delta$ , giving $G = x_<(1-x_>)$ ; his solves $\tfrac12\,\partial_x^2G_\lambda = \delta/2$ , giving $G_\lambda = -x_<(1-x_>)$ . The two differ by a sign — the sign discussed next — and by nothing else; there is no factor $2$ anywhere.) Where the weight really sits is on the source: his equation multiplied through by $w$ reads $\bigl[(pu')' + qwu - \lambda wu\bigr] = wf$ when the forcing is $f$ , so it is $wf$ , not $f$ , that his representation $u(x) = \int w\,G_\lambda f\,dy$ is matched to. Dropping his kernel into our integral $\int G f\,d\xi$ therefore loses that $w$ : the mistake is a factor $w$ in the forcing, not in the kernel.

The sign. Arfken writes $L$ for the self-adjoint $(pu')' + qu$ , whose eigenvalues decrease to $-\infty$ , whereas $\mathcal{L} = -(pu')'+qu$ used here has eigenvalues increasing to $+\infty$ : the two spectra run in opposite directions. They are related by a plain change of sign, $\lambda_n^{\text{Arfken}} = -\lambda_n$ , only when $q \equiv 0$ , for only then is $(pu')'+qu$ the negative of $-(pu')'+qu$ ; with $q \ne 0$ the two operators are genuinely different and the eigenvalues are not sign-flips of one another. Hassani's $L_x$ of (20.24) carries Arfken's sign too, so the same caveat applies to his row.

Method(Reading the spectrum off a Green's function)AdaptedAdapted from Stakgold & Holst (2011), §7.1, pp. 413–415

Build $G(x,\xi;\lambda)$ in closed form, keeping $\lambda$ as a parameter: solve $\mathcal{L}u = \lambda su$ twice, once with the initial data at $a$ that enforce $B_au=0$ and once with the data at $b$ that enforce $B_bu = 0$ , giving $v(\cdot,\lambda)$ and $z(\cdot,\lambda)$ . Use the data of part (ii) literally, $\bigl(\sin\alpha,\cos\alpha\bigr)$ at $a$ and $\bigl(\sin\beta,-\cos\beta\bigr)$ at $b$ : they must not involve $\lambda$ .
Compute the characteristic function $C(\lambda) = p\,(vz'-zv')$ , which must come out independent of $x$ ; if it does not, the operator was not in formally self-adjoint form.
Assemble $G = -v(x_{<},\lambda)z(x_{>},\lambda)/C(\lambda)$ .
Find the eigenvalues as the zeros of $C$ . They are the poles of $G$ , and for a regular problem they are real and simple. This reading depends on step 1 having been followed: if $v$ or $z$ was rescaled by a factor depending on $\lambda$ — easily done when a convenient closed form is picked up instead — then $C$ acquires extra zeros that are not eigenvalues, as item 7 of the example shows. The check is to confirm that $G$ really has a pole at each candidate.
Normalise the eigenfunctions by $u_n = \pm v_n/\{k_nC'(\lambda_n)\}^{1/2}$ , where $v_n = k_nz_n$ ; no separate normalisation integral is needed.
Or go the other way: given the spectrum, write $G(x,\xi;\lambda) = \sum u_n(x)\overline{u_n(\xi)}/(\lambda_n-\lambda)$ and $w = \sum\langle f,u_n\rangle u_n/(\lambda_n-\lambda)$ . Use this to locate resonances and to see which modes a given forcing excites, rather than to compute $G$ numerically.

Example(An Euler operator on

[1,e]

: closed form, series, poles and residues)Original

Take

\mathcal{L}u = -\bigl(x\,u'\bigr)' \ \ \text{on } 1 \le x \le e , \qquad u(1) = u(e) = 0 , \qquad s(x) = \frac1x .

So $p(x) = x$ , $q = 0$ , and $\alpha = \beta = 0$ in the angle notation.

The substitution that trivialises it. Put $t = \ln x$ , so $x\,d/dx = d/dt$ and $t$ runs over $[0,1]$ . Then $\bigl(xu'\bigr)' = \frac1x u_{tt}$ and $\delta(x-\xi) = \delta(t-\tau)/x$ with $\tau = \ln\xi$ , so the eigenvalue problem $-\bigl(xu'\bigr)' = \lambda u/x$ becomes $-u_{tt} = \lambda u$ with $u = 0$ at $t = 0,1$ .
Spectrum. Hence $\lambda_n = n^2\pi^2$ for $n = 1,2,\dots$ , with eigenfunctions $\sin(n\pi t)$ . Normalising in $(\cdot,\cdot)_s$ : $\int_1^e u^2\,\frac{dx}{x} = \int_0^1u^2\,dt$ , so $u_n(x) = \sqrt2\,\sin\bigl(n\pi\ln x\bigr) , \qquad \lambda_n = n^2\pi^2 .$
The Green's function at $\lambda = 0$ , built without eigenfunctions. The homogeneous equation $(xu')'=0$ gives $u = c\ln x + d$ ; take $u_1 = \ln x$ (zero at $x=1$ ) and $u_2 = \ln x - 1$ (zero at $x=e$ ). Their Wronskian is $W = u_1u_2' - u_1'u_2 = \frac{\ln x}{x} - \frac{\ln x - 1}{x} = \frac1x$ , and the leading coefficient of $\mathcal{L}$ is $a_2 = -x$ , so $a_2(\xi)W(\xi) = -1$ and $G(x,\xi) = \frac{u_1(x_{<})u_2(x_{>})}{-1} = \ln x_{<}\,\bigl(1 - \ln x_{>}\bigr) .$ Checks: $-\bigl(xG'\bigr)' = 0$ on each side; $G$ vanishes at $x=1$ and $x=e$ ; $G$ is continuous at $x=\xi$ ; and the jump in $\partial G/\partial x$ there is $-\frac{1}{\xi}\ln\xi - \frac{1-\ln\xi}{\xi} = -\frac1\xi = -\frac{1}{p(\xi)}$ , as required.
The bilinear series. The theorem predicts $\ln x_{<}\,\bigl(1-\ln x_{>}\bigr) = \frac{2}{\pi^2}\sum_{n=1}^{\infty}\frac{\sin(n\pi\ln x)\,\sin(n\pi\ln\xi)}{n^2} .$ At $x = 1.5$ , $\xi = 2$ : the left side is $\ln 1.5\,(1-\ln 2) = 0.4054651\times0.3068528 = 0.12441811$ ; summing $4000$ terms of the right side gives $0.12441811$ . (At $x = \xi$ the identity is the classical $\sum n^{-2}\sin^2 n\pi t = \tfrac{\pi^2}{2}t(1-t)$ .)
Solving with it. For $f(x) = 1/x$ , substituting $\tau = \ln\xi$ turns $u(x) = \int_1^e G(x,\xi)f(\xi)\,d\xi$ into $\int_0^1 t_{<}(1-t_{>})\,d\tau$ , which is $\tfrac12 t(1-t)$ ; that is $u(x) = \tfrac12\,\ln x\,\bigl(1 - \ln x\bigr) , \qquad u(2) = 0.10634708 .$ Direct check: $-\bigl(xu'\bigr)' = -\tfrac1x u_{tt} = \tfrac1x = f(x)$ , and $u(1)=u(e)=0$ .
The same answer as a series. Here $\langle f,u_n\rangle = \int_1^e \frac1\xi\sqrt2\sin(n\pi\ln\xi)\,d\xi = \sqrt2\int_0^1\sin n\pi\tau\,d\tau = \dfrac{\sqrt2\,\bigl(1-(-1)^n\bigr)}{n\pi}$ , so $u(x) = \sum_n\frac{\langle f,u_n\rangle}{\lambda_n}u_n(x) = \frac{4}{\pi^3}\sum_{n\ \text{odd}}\frac{\sin(n\pi\ln x)}{n^3} ,$ and the classical sum $\sum_{n\ \text{odd}}n^{-3}\sin n\pi t = \tfrac{\pi^3}{8}t(1-t)$ reproduces item 5 exactly.
The resolvent and its poles. Keeping $\lambda$ , the same substitution gives $-g_{tt} - \lambda g = \delta(t-\tau)$ , so with $k = \sqrt\lambda$ , $G(x,\xi;\lambda) = \frac{\sin\bigl(k\,t_{<}\bigr)\,\sin\bigl(k(1-t_{>})\bigr)}{k\,\sin k} , \qquad t = \ln x,\ \tau = \ln\xi .$ In the language of part (ii), $v = \sin kt$ satisfies the condition at $x=1$ and $z = \sin k(1-t)$ the one at $x=e$ . These are not the initial-value normalisations part (ii) prescribes. With $\alpha=\beta=0$ those ask for $v(1,\lambda)=0$ , $v'(1,\lambda)=1$ and $z(e,\lambda)=0$ , $z'(e,\lambda)=-1$ (primes in $x$ , and $d/dx = \frac1x\,d/dt$ ), which give $v_{\mathrm{ii}} = \frac{\sin kt}{k} , \qquad z_{\mathrm{ii}} = \frac{e}{k}\,\sin\bigl(k(1-t)\bigr) .$ The factors $k$ and $k/e$ relating them to $v,z$ depend on $\lambda$ , not merely on scale. For $G$ itself and for the residues this is harmless — rescaling $v$ and $z$ rescales $C = p\,W[v,z]$ by the same product, and the residue formula of part (iii) is invariant under it, as item 8 will use. It is not harmless for part (i)'s reading of $C$ . Because $p\,W_x[v,z] = W_t[v,z]$ under $t=\ln x$ , the convenient pair gives $C(\lambda) = -k\sin k ,$ whose zeros include $k = n\pi$ , that is $\lambda = n^2\pi^2$ — the eigenvalues, recovered without ever solving the eigenvalue problem — but also $\lambda = 0$ , since $-k\sin k = -\lambda + \lambda^2/6 - \dots$ has a simple zero there. That zero is spurious: $\lambda = 0$ is not an eigenvalue, and $G$ has no pole there but tends to the $\ln x_<(1-\ln x_>)$ of item 3. The normalisation of part (ii) is what removes the trap, since $C_{\mathrm{ii}}(\lambda) = \frac{e}{k^2}\bigl(-k\sin k\bigr) = -\frac{e\sin k}{k} \longrightarrow -e \ne 0 \quad\text{as } \lambda \to 0 .$ The moral for step 4 of the method: read eigenvalues off the zeros of $C$ only when $v$ and $z$ carry part (ii)'s $\lambda$ -independent initial data; a rescaling by a function of $\lambda$ can manufacture zeros that are not eigenvalues.
Residues return the normalised eigenfunctions. Here the convenient pair $v,z$ of item 7 may be used together with its own $C = -k\sin k$ : replacing $(v,z)$ by $(av,bz)$ sends $C \mapsto abC$ , $k_n = v_n/z_n \mapsto (a/b)k_n$ and $z_n \mapsto bz_n$ , so the combination $k_nz_n(x)z_n(\xi)/C'(\lambda_n)$ is unchanged. Now $C'(\lambda) = -\frac{d}{d\lambda}\bigl(k\sin k\bigr) = -\frac{\sin k + k\cos k}{2k}$ , which at $k = n\pi$ is $-\tfrac12(-1)^n$ . Also $z_n(t) = \sin n\pi(1-t) = -(-1)^n\sin n\pi t$ , so $k_n = v_n/z_n = -(-1)^n$ and $\operatorname*{Res}_{\lambda=\lambda_n}G = -\frac{k_nz_n(t)z_n(\tau)}{C'(\lambda_n)} = -\frac{(-1)^{2n}\sin n\pi t\,\sin n\pi\tau}{\tfrac12} = -2\sin(n\pi\ln x)\sin(n\pi\ln\xi) ,$ which is $-u_n(x)u_n(\xi)$ , normalisation included. Numerically at $x=1.5$ , $\xi=2$ , $n=2$ : multiplying $G$ by $\lambda - \lambda_2$ at $\lambda = 4\pi^2 + 10^{-6}$ gives $1.0486688$ , against the predicted $-u_2(1.5)u_2(2) = 1.0486688$ .
Resonance. The solution of $\mathcal{L}w - \lambda sw = f$ with $f = 1/x$ is $\sum_n\langle f,u_n\rangle u_n/(\lambda_n-\lambda)$ . Only odd $n$ contribute, so pushing $\lambda$ towards $4\pi^2$ — an even eigenvalue — leaves $w$ bounded, while pushing it towards $\pi^2$ makes $w$ blow up along $u_1$ . Whether a near-resonance is felt depends on whether the forcing overlaps the mode, which is exactly what the numerator $\langle f,u_n\rangle$ records.

ApplicationOriginal

Reading a spectrum from a response function. Part (iii) is the mathematics behind spectroscopy in its broadest sense. A measured frequency response is a resolvent; its poles in the complex frequency plane are the natural frequencies of the system and the residues give the mode shapes together with their normalisation. Experimental modal analysis of a structure, impedance spectroscopy of a circuit, and the extraction of resonance parameters from a scattering amplitude are all this procedure. The pole--residue form $\sum u_n(x)u_n(\xi)/(\lambda_n-\lambda)$ is what a fitting routine fits.
Resonance and mode selection. The factor $\langle f,u_n\rangle/(\lambda_n-\lambda)$ says that a drive excites a mode in proportion to its overlap with that mode and inversely with the detuning. An organ pipe blown at a node of a given harmonic does not sound it; a load applied at a nodal line of a plate mode does not excite it; a laser cavity tuned between two longitudinal modes couples weakly to both. Engineers use the same formula backwards, placing actuators where $\langle f,u_n\rangle$ is large for the mode they want and small for the one they fear.
Quantum mechanics. With $\mathcal{L} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V$ , the resolvent $(H-E)^{-1}$ is the energy Green's function, the bilinear series is the spectral decomposition $\sum_n\psi_n(x)\overline{\psi_n(\xi)}/(E_n-E)$ , and the poles on the real axis are the bound-state energies with residues giving the wavefunctions. Continuing to complex $E$ turns a continuous spectrum into a branch cut, and resonances appear as poles on the second sheet — the continuous-spectrum version Stakgold and Holst point to. The contour formula of part (iv) is the completeness relation $\sum_n|\psi_n\rangle\langle\psi_n| = 1$ in disguise, and the same contour integral with an extra factor $e^{-iEt/\hbar}$ yields the propagator.
Heat and diffusion. Expanding the initial datum in eigenfunctions and attaching $e^{-\lambda_nt}$ to each solves the heat equation, with the slowest mode $\lambda_1$ controlling the long-time behaviour; a negative $\lambda_1$ signals instability. The Green's function of the time-dependent problem is the Laplace transform of the resolvent, which is why the poles of $G(\cdot,\cdot;\lambda)$ are the decay rates (Green's functions for the wave and heat equations).
Converting to an integral equation. Because $G$ inverts $\mathcal{L}$ , the eigenvalue problem becomes $u(x) = \lambda\int_a^b G(x,\xi)s(\xi)u(\xi)\,d\xi$ . Arfken's reason for preferring this form is that the integral operator is bounded while the differential operator is not, so stronger theorems apply; it is also the form a numerical scheme can discretise stably, since discretising $G$ gives a well-conditioned matrix where discretising $d^2/dx^2$ gives one whose condition number grows like the square of the mesh refinement.

Method

Causal Green's functions for initial-value problems

Adapted

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Builds the retarded Green's function $G(t,\tau) = H(t-\tau)\,u_\tau(t)$ of a linear initial-value problem, proves that $u(t) = \int_a^t G(t,\tau)f(\tau)\,d\tau$ solves it with zero initial data, specialises to the impulse response of a constant-coefficient system so that the solution becomes a convolution with transfer function $\hat G(\omega)$ , works the damped oscillator in full, and shows that vanishing before the impulse makes $\hat G$ analytic in the upper half-plane.

Prerequisites: Green's function for a boundary-value problem, The convolution theorem, Existence and uniqueness of solutions, The Dirac delta in calculations, The damped and driven harmonic oscillator, The Fourier transform, Differentiation under the integral sign, Cauchy's integral formula, Classification of isolated singularities

Definition(Causal and advanced Green's functions)AdaptedAdapted from Stakgold & Holst (2011), §2.5, Example 10, (2.5.47)–(2.5.49), pp. 176–177, and §3.2, (3.2.11), p. 195; Arfken, Weber & Harris (2013), §10.1, pp. 452–453

Let

Lu = a_p(t)\,u^{(p)} + \dots + a_1(t)\,u' + a_0(t)\,u ,

with continuous coefficients and $a_p \ne 0$ , and consider the initial-value problem

Lu = f \ \ (t > a), \qquad u(a) = u'(a) = \dots = u^{(p-1)}(a) = 0 .

All $p$ accessory conditions sit at one endpoint. The causal (or retarded) Green's function, which Stakgold and Holst call the causal fundamental solution, is the function $G(t,\tau)$ with

LG(t,\tau) = \delta(t-\tau) , \qquad G(t,\tau) = 0 \ \text{ for } t < \tau ,

derivatives taken in $t$ . Unpacked as in the boundary-value construction, this says: $G$ solves $LG = 0$ on each side of $\tau$ ; $G, \partial_tG,\dots,\partial_t^{\,p-2}G$ are continuous at $t=\tau$ ; $\partial_t^{\,p-1}G$ jumps there by $1/a_p(\tau)$ ; and $G$ vanishes to the left. Because $G$ is zero for $t<\tau$ and continuous up to order $p-2$ , all those derivatives vanish at $\tau^+$ as well.

The advanced Green's function is the other one-sided choice,

LG_{\mathrm{adv}}(t,\tau) = \delta(t-\tau), \qquad G_{\mathrm{adv}}(t,\tau) = 0 \ \text{ for } t > \tau .

When $t$ is time, $G$ is the response at time $t$ to a unit impulse delivered at the earlier time $\tau$ : an effect that does not precede its cause. In engineering it is the impulse response, and Stakgold and Holst use that name too. For constant coefficients $G$ depends only on the elapsed time, $G(t,\tau) = G_0(t-\tau)$ .

Method(Construction and the solution formula)AdaptedAdapted from Stakgold & Holst (2011), §2.5, (2.5.48)–(2.5.49), pp. 176–177; Stakgold & Holst (2011), §3.1, (3.1.9) and (3.1.13), pp. 188–190; Stakgold & Holst (2011), §3.2, (3.2.11), p. 195

Let $u_\tau$ be the solution of the homogeneous initial-value problem

Lu_\tau = 0 \ \ \text{on the whole interval where the coefficients are defined},

u_\tau(\tau) = u_\tau'(\tau) = \dots = u_\tau^{(p-2)}(\tau) = 0, \qquad u_\tau^{(p-1)}(\tau) = \frac{1}{a_p(\tau)} ,

which exists and is unique because an initial-value problem for a linear equation with continuous coefficients and nonvanishing $a_p$ has exactly one solution. Following Stakgold and Holst's (3.1.10), $u_\tau$ is posed on both sides of $\tau$ , not only on $t > \tau$ ; the advanced Green's function below needs its values for $t < \tau$ . Then:

(i) The causal Green's function exists, is unique, and is

\boxed{\;G(t,\tau) = H(t-\tau)\,u_\tau(t)\;}

with $H$ the Heaviside step. For $p=2$ this is $G(t,\tau) = H(t-\tau)u_\tau(t)$ with $u_\tau(\tau) = 0$ , $u_\tau'(\tau) = 1/a_2(\tau)$ . The advanced Green's function is $G_{\mathrm{adv}}(t,\tau) = -H(\tau-t)\,u_\tau(t)$ , and the two differ by the homogeneous solution $u_\tau$ .

(ii) For $f$ piecewise continuous, the initial-value problem with zero initial data has the unique solution

u(t) = \int_a^{b} G(t,\tau)\,f(\tau)\,d\tau = \int_a^{t} u_\tau(t)\,f(\tau)\,d\tau ,

Stakgold and Holst's (3.1.9), where $b$ is any point with $b > t$ (the problem is posed on $t > a$ , with no right-hand endpoint; $b$ is introduced only to make the comparison). The upper limit collapses from $b$ to $t$ because $G$ vanishes for $\tau > t$ : the value of $u$ at time $t$ uses only the forcing up to time $t$ . Arfken's Example 10.1.2 is the instance $u'' + u = f$ , $u(0) = u'(0) = 0$ , with $G(t,\tau) = \sin(t-\tau)$ for $t > \tau$ and $0$ otherwise, and he notes that this is exactly causality.

(iii) If the coefficients are constant, then $u_\tau(t) = v(t-\tau)$ with $v$ the solution for $\tau = 0$ , so

u(t) = \int_a^{t} v(t-\tau)\,f(\tau)\,d\tau ,

their (3.1.13): the response is the convolution of the forcing with the impulse response.

(iv) Step response. Let $F(t,\tau)$ solve $LF = H(t-\tau)$ with $F = 0$ for $t<\tau$ . Then $-\partial F/\partial\tau = G$ , and the solution can also be written

u(t) = F(t,a)\,f(a) + \int_a^{t} F(t,\tau)\,f'(\tau)\,d\tau ,

Stakgold and Holst's (3.1.15), convenient when $f$ is given as a sequence of steps.

(v) No symmetry. An initial-value problem is never self-adjoint: whatever $L$ is, the adjoint boundary conditions move to the other endpoint, so they never coincide with the original ones (Stakgold and Holst's Example 6). Correspondingly $G(t,\tau) \ne G(\tau,t)$ — one of them is zero whenever $t \ne \tau$ — and the reciprocity of the self-adjoint boundary-value problem fails completely. Arfken makes the same observation.

Method(Transfer function, and what causality does to it)AdaptedAdapted from Arfken, Weber & Harris (2013), §20.5, Eqs. (20.104)–(20.106) and (20.110)–(20.113), pp. 999–1001; Arfken, Weber & Harris (2013), §12.8, Eqs. (12.116)–(12.118), pp. 591–593

Let $L$ have constant coefficients and let $G_0(t)$ be the impulse response, so $LG_0 = \delta$ and $G_0(t) = 0$ for $t<0$ . Define the transfer function (or susceptibility, or frequency response) by the Fourier transform

\hat G(\omega) = \int_{-\infty}^{\infty} G_0(t)\,e^{i\omega t}\,dt = \int_0^{\infty} G_0(t)\,e^{i\omega t}\,dt .

Then, with $P(\,\cdot\,)$ the characteristic polynomial of $L$ , that is $L = P(d/dt)$ ,

\hat G(\omega) = \frac{1}{P(-i\omega)} ,

and the solution of $Lu = f$ with zero data in the far past is the convolution $u = G_0 * f$ , equivalently $\hat u(\omega) = \hat G(\omega)\,\hat f(\omega)$ (the convolution theorem). Three consequences:

Causality in the time domain. $G_0(t-\tau) = 0$ for $\tau > t$ , so the output at time $t$ is built only from inputs at earlier times, $u(t) = \int_{-\infty}^t G_0(t-\tau)f(\tau)\,d\tau$ . These are Arfken's Eqs. (20.105)–(20.106); he derives them by imposing that a cause precede its effect.
Causality in the frequency domain. Because the integral runs only over $t \ge 0$ , $e^{i\omega t}$ is bounded there when $\operatorname{Im}\omega \ge 0$ , so $\hat G$ extends to a function analytic in the upper half-plane $\operatorname{Im}\omega > 0$ (assuming $G_0$ is bounded and integrable, as it is for a stable system). All the poles of $\hat G$ , which are the zeros of $P(-i\omega)$ , lie in the closed lower half-plane. Neither cited passage of Arfken states this step: his §12.8 assumes analyticity in the upper half-plane as a hypothesis, and his §20.5 uses the opposite transform sign, which would put the analyticity in the lower half-plane. The bridge from one-sidedness to analyticity is supplied by Step 4 of the second proof below, and the sign clash is spelt out in the remark on conventions.
Real and imaginary parts are not independent. Writing $\hat G = u + iv$ on the real axis and assuming $\hat G \to 0$ fast enough at infinity, Cauchy's formula on a contour closed in the upper half-plane gives the dispersion relations, the Hilbert-transform pair $u(\omega_0) = \frac{1}{\pi}\,\mathrm{P}\!\!\int_{-\infty}^{\infty}\frac{v(\omega)}{\omega-\omega_0}\,d\omega , \qquad v(\omega_0) = -\frac{1}{\pi}\,\mathrm{P}\!\!\int_{-\infty}^{\infty}\frac{u(\omega)}{\omega-\omega_0}\,d\omega ,$ Arfken's (12.118), where $\mathrm{P}$ denotes the Cauchy principal value. In optics these are the Kramers–Kronig relations: refraction and absorption determine one another. Arfken attributes this pair to causality alone, which is why it constrains a response function whatever the underlying interaction happens to be; he derives the corresponding interdependence directly from the transfer function in his Eqs. (20.111)–(20.113). The derivation of the pair itself from analyticity is dispersion relations.

Idea of the proofOriginal

Think of the forcing as a line of hammer blows. A blow of strength $f(\tau)\,d\tau$ delivered at time $\tau$ leaves the system, immediately afterwards, with every state variable unchanged except the highest one, which receives a kick of size $f(\tau)d\tau/a_p(\tau)$ — that is the jump condition. After that the system runs freely. So the response to one blow is $u_\tau(t)$ scaled by $f(\tau)d\tau$ , and the response to all of them is the sum, which is the integral. The only reason the upper limit is $t$ and not $b$ is that blows yet to come have not been struck.

That is why the boundary-value apparatus simplifies so much here. In a boundary-value problem the two conditions sit at opposite ends and have to be met by two different branches of $G$ ; here both conditions sit at the left end, they force the left branch to vanish identically, and the right branch is then fixed by continuity and the jump alone — with nothing left over. Stakgold and Holst put it that causality removes all arbitrariness from the fundamental solution.

For constant coefficients the system cannot tell what time it is, only how long ago the blow landed, so $G$ depends on $t-\tau$ and the superposition becomes a convolution. A convolution is a product after Fourier transform, and the multiplier is the transfer function. The last step is pure complex analysis: a function that is the transform of something supported on $t\ge0$ cannot be an arbitrary function of $\omega$ — the integral still converges when $\omega$ is pushed upward into the complex plane, so the transfer function is analytic in the whole upper half-plane, and analyticity ties its real and imaginary parts together.

Proof(The causal Green's function and the solution formula)AdaptedAdapted from Stakgold & Holst (2011), §2.5, Example 10, pp. 176–177, and §3.1, pp. 188–190

Take $p = 2$ for concreteness and indicate the general case at the end; write $L u = a_2u'' + a_1u' + a_0u$ .

Step 1 (the one-sided condition and the initial conditions are the same condition). The definition imposes $G = 0$ for $t < \tau$ ; what the problem imposes is instead the $p$ homogeneous initial conditions $G(a,\tau) = \partial_tG(a,\tau) = 0$ at $t=a$ . On $[a,\tau)$ these are equivalent. One way: $G \equiv 0$ there certainly satisfies them. The other way: on $[a,\tau)$ , $G$ solves $LG = 0$ with vanishing data at $t=a$ , and by uniqueness for the initial-value problem (existence and uniqueness) the only such solution is $G \equiv 0$ . So nothing is lost by taking either as the definition, and we may use both. This equivalence has no analogue in a boundary-value problem, and it is exactly where the two conditions being at the same end is used.

Step 2 (matching at $\tau$ ). Continuity of $G$ at $t=\tau$ forces $G(\tau^+,\tau) = 0$ . The jump condition forces $\partial_tG(\tau^+,\tau) - \partial_tG(\tau^-,\tau) = 1/a_2(\tau)$ , and the second term is $0$ by Step 1, so $\partial_tG(\tau^+,\tau) = 1/a_2(\tau)$ .

Step 3 (the right branch is $u_\tau$ ). For $t>\tau$ , $G$ solves $LG = 0$ with the initial data just computed at $t=\tau$ . That is exactly the problem defining $u_\tau$ , which has one and only one solution. Hence $G(t,\tau) = u_\tau(t)$ for $t>\tau$ and $0$ for $t<\tau$ , which is $H(t-\tau)u_\tau(t)$ . Existence and uniqueness of $G$ follow from existence and uniqueness for that initial-value problem.

Step 4 (the advanced solution). Put $G_{\mathrm{adv}} = -H(\tau-t)u_\tau(t)$ . It solves $LG=0$ on each side of $\tau$ , vanishes for $t>\tau$ , is continuous at $\tau$ because $u_\tau(\tau)=0$ , and its derivative jumps there by $0 - \bigl(-u_\tau'(\tau)\bigr) = 1/a_2(\tau)$ . So it is the advanced Green's function, and $G - G_{\mathrm{adv}} = u_\tau(t)\bigl[H(t-\tau)+H(\tau-t)\bigr] = u_\tau(t)$ , a solution of the homogeneous equation, as it must be.

Step 5 (the integral solves the problem). Set $u(t) = \int_a^t u_\tau(t)f(\tau)\,d\tau$ and differentiate, using the rule for an integral with a variable limit and a parameter in the integrand (differentiation under the integral sign); the integrand and its $t$ -derivatives are continuous in $(t,\tau)$ , which is what that rule needs. The boundary term at $\tau = t$ involves $u_t(t)$ , which is $0$ :

u'(t) = \underbrace{u_t(t)}_{=\,0}f(t) + \int_a^t \partial_tu_\tau(t)\,f(\tau)\,d\tau = \int_a^t \partial_tu_\tau(t)\,f(\tau)\,d\tau .

Step 6 (the second derivative is where the source appears). Differentiating again, the boundary term now involves $\partial_tu_\tau(t)$ at $\tau = t$ , which is $1/a_2(t)$ :

u''(t) = \frac{f(t)}{a_2(t)} + \int_a^t \partial_t^2u_\tau(t)\,f(\tau)\,d\tau .

Step 7 (apply $L$ ). Combining Steps 5 and 6,

Lu = a_2(t)\frac{f(t)}{a_2(t)} + \int_a^t \Bigl[a_2\partial_t^2u_\tau + a_1\partial_tu_\tau + a_0u_\tau\Bigr]f(\tau)\,d\tau = f(t) + \int_a^t \bigl(Lu_\tau\bigr)(t)\,f(\tau)\,d\tau = f(t) ,

since $Lu_\tau = 0$ . The initial conditions hold because both integrals in Steps 5 and 6 are empty at $t = a$ . Uniqueness is uniqueness for the initial-value problem.

Step 8 (general order). For order $p$ the same computation runs with $p$ differentiations: the first $p-1$ boundary terms vanish because $u_\tau^{(j)}(\tau) = 0$ for $j \le p-2$ , and the $p$ th produces $u_\tau^{(p-1)}(t)f(t) = f(t)/a_p(t)$ , which $a_p$ turns into $f$ . Stakgold and Holst verify the same formula distributionally, pairing $E = H(t-\tau)u_\tau(t)$ against a test function and using Green's formula; every term of the resulting boundary expression vanishes except the one carrying $u_\tau^{(p-1)}(\tau)a_p(\tau) = 1$ . $\square$

Proof(Transfer function and analyticity in the upper half-plane)AdaptedAdapted from Arfken, Weber & Harris (2013), §20.5, (20.104)–(20.110), pp. 999–1000

Let the coefficients be constant, $L = P(d/dt)$ with $P(\zeta) = a_p\zeta^p + \dots + a_0$ , and let $G_0$ be the impulse response.

Step 1 (transform the defining equation). The equation $LG_0 = \delta$ holds in the sense of distributions, and it has to: $G_0$ is not $p$ times classically differentiable at $t = 0$ — the jump in $G_0^{(p-1)}$ there is precisely what produces the delta. So the rule to use is the distributional one. For a tempered distribution $g$ and the convention $\hat g(\omega) = \int g(t)e^{i\omega t}dt$ , the transform of the distributional derivative is $\widehat{g'} = -i\omega\,\hat g$ , hence $\widehat{g^{(k)}} = (-i\omega)^k\hat g$ . (No decay hypothesis enters; decay is what Step 4 will need, for a different purpose.) Transforming $LG_0 = \delta$ and using $\hat\delta = 1$ ,

P(-i\omega)\,\hat G(\omega) = 1 , \qquad \hat G(\omega) = \frac{1}{P(-i\omega)} .

Step 2 (the solution is a convolution). Transforming $Lu = f$ the same way gives $P(-i\omega)\hat u = \hat f$ , so $\hat u = \hat G\hat f$ , and by the convolution theorem $u = G_0 * f$ , that is $u(t) = \int_{-\infty}^{\infty}G_0(t-\tau)f(\tau)\,d\tau$ .

Step 3 (causality truncates the convolution). $G_0(t-\tau) = 0$ unless $\tau \le t$ , so the integral runs only over $\tau \le t$ . Arfken runs this step in reverse, taking the vanishing of the kernel for $\tau > t$ as the definition of a causal system.

Step 4 (analytic continuation upward). Assume the system is stable, so that $G_0$ is bounded and absolutely integrable on $[0,\infty)$ . For complex $\omega = \sigma + i\eta$ with $\eta > 0$ ,

\bigl|G_0(t)e^{i\omega t}\bigr| = \bigl|G_0(t)\bigr|\,e^{-\eta t} \le \bigl|G_0(t)\bigr| \qquad (t \ge 0) ,

so the integral $\int_0^\infty G_0(t)e^{i\omega t}dt$ converges absolutely and uniformly on $\{\operatorname{Im}\omega \ge 0\}$ , and converges even better as $\eta$ grows. Differentiating under the integral sign (legitimate by the same uniform bound, with the extra factor $t$ absorbed by $e^{-\eta t}$ for $\eta>0$ ) shows $\hat G$ is holomorphic there. Here the one-sidedness is everything: had $G_0$ been nonzero for $t<0$ , the factor $e^{-\eta t}$ would have grown on that part and nothing would converge.

Step 5 (poles). By Step 1 the singularities of $\hat G$ are the zeros of $P(-i\omega)$ , that is $\omega = i\zeta_j$ where $\zeta_j$ runs over the roots of $P$ . Stability means every $\zeta_j$ has $\operatorname{Re}\zeta_j < 0$ , hence every pole has $\operatorname{Im}\omega = \operatorname{Re}\zeta_j < 0$ : all poles lie strictly below the real axis, consistent with Step 4. $\square$

The passage from analyticity in the upper half-plane to the Hilbert-transform pair is Arfken's §12.8 and is not repeated here; it is dispersion relations.

Remark(Where the hypotheses are used, and what differs from the boundary-value case)Original

All conditions at one endpoint. Used in Step 1 of the first proof. Move one condition to the other end and the left branch no longer vanishes; one gets the two-sided Green's function of the previous units instead, and the solution at $t$ then depends on the forcing at later times — perfectly sensible for a steady-state problem in space, nonsense for a problem in time.
Zero initial data. The formula $u = \int_a^tGf$ solves the problem with $u(a) = \dots = u^{(p-1)}(a) = 0$ . Nonzero initial data are handled by superposition: add the solution of the homogeneous equation with those data. Stakgold and Holst's Exercise 1.4.5 (pp. 89–90) writes out both pieces for $m\ddot u + ku = f$ with $u(0)=\alpha$ , $\dot u(0)=\beta$ .
$a_p \ne 0$ . The jump is $1/a_p(\tau)$ , so a zero of the leading coefficient at $\tau$ destroys the construction, as it does for boundary-value problems.
Stability, for the frequency-domain statements. Step 4 of the second proof needs $G_0$ integrable on $[0,\infty)$ . For an undamped oscillator, $G_0(t) = \sin\omega_0t/m\omega_0$ does not decay, $\hat G$ has poles on the real axis at $\pm\omega_0$ , and the transfer function exists only as a distribution. The standard repair is to give the poles a small imaginary part — add an infinitesimal damping — which is the same device as the $i\epsilon$ prescription that selects the retarded propagator in field theory.
Retarded versus advanced is a choice, not a theorem. Both solve $LG=\delta$ ; the differential equation does not know the arrow of time. What picks the retarded one is the physical requirement that nothing respond before it is struck. For an equation in a space variable the opposite choice may be the right one, and in scattering theory the advanced Green's function describes the time-reversed process.
No symmetry, no eigenfunction expansion. Because the problem is not self-adjoint (Stakgold and Holst's Example 6), $G(t,\tau) \ne G(\tau,t)$ and the bilinear series of the previous unit does not apply. The frequency-domain representation $\hat G(\omega) = 1/P(-i\omega)$ is its replacement: the poles are still the natural frequencies, but they are complex, and their imaginary parts are decay rates rather than a discrete real spectrum.

Remark(Sign conventions in the transform)Original

This unit uses $\hat g(\omega) = \int g(t)e^{i\omega t}\,dt$ , the convention under which a causal response function is analytic in the upper half-plane and its poles lie below the real axis. It is the convention of the Kramers–Kronig literature and of Arfken's §12.8, where the dispersion relations are derived for a function analytic in the upper half-plane.

Arfken's §20.5 uses the opposite sign, $\varphi(\omega) = \int_0^\infty\Phi(t)e^{-i\omega t}\,dt$ (his Eq. (20.110)), under which a causal $\Phi$ makes $\varphi$ analytic in the lower half-plane and the poles sit above the real axis; he notes only that the resulting constraints on the real and imaginary parts are similar to those of §12.8. Both are correct; mixing them flips the half-plane and the sign of every dispersion integral. Whenever a text asserts "causality implies analyticity in the upper half-plane", check which transform it means. The engineering alternative is the Laplace transform $\bar g(s) = \int_0^\infty g(t)e^{-st}dt$ , for which causality is automatic, stability is "all poles in $\operatorname{Re}s<0$ ", and $\bar G(s) = 1/P(s)$ ; the substitution $s = -i\omega$ converts to the convention used here.

Method(Solving an initial-value problem with the causal Green's function)AdaptedAdapted from Stakgold & Holst (2011), §3.1, pp. 188–191; Arfken, Weber & Harris (2013), §10.1, pp. 452–454, and §20.5, pp. 999–1001

Given $Lu = f$ for $t > a$ with $p$ initial conditions at $t=a$ :

Split off the initial data. Write $u = u_{\mathrm{h}} + u_{\mathrm{p}}$ , where $u_{\mathrm{h}}$ solves $Lu_{\mathrm{h}}=0$ with the given initial data and $u_{\mathrm{p}}$ solves $Lu_{\mathrm{p}}=f$ with zero data. Only the second needs a Green's function.
Find the impulse response. Solve $Lu_\tau = 0$ with $u_\tau^{(j)}(\tau) = 0$ for $j \le p-2$ and $u_\tau^{(p-1)}(\tau) = 1/a_p(\tau)$ . For constant coefficients, solve once at $\tau = 0$ and translate: $u_\tau(t) = v(t-\tau)$ .
Check the normalisation. $u_\tau(\tau)=0$ and $u_\tau'(\tau)=1/a_2(\tau)$ for a second-order problem. Forgetting the $1/a_2$ is the commonest error; it rescales the whole answer.
Integrate. $u_{\mathrm{p}}(t) = \displaystyle\int_a^t u_\tau(t)f(\tau)\,d\tau$ , or $\int_a^t v(t-\tau)f(\tau)\,d\tau$ in the constant-coefficient case. Substituting $s = t-\tau$ usually makes the integral easier, since the exponentials and trigonometric functions then depend on $s$ alone.
Piecewise forcing. If $f$ is switched on and off, split the integral at the switching times; the Green's function itself does not change. Equivalently use the step response $F$ and Stakgold and Holst's form $u = F(t,a)f(a) + \int_a^tF(t,\tau)f'(\tau)\,d\tau$ , which turns a staircase forcing into a finite sum.
For constant coefficients, consider the frequency domain instead. Read off $\hat G(\omega) = 1/P(-i\omega)$ with no work at all, multiply by $\hat f$ , and invert — or just use $\hat G$ directly to answer questions about resonance, bandwidth and phase lag, which is what it is really for.

Example(The damped oscillator under a pulse)Original

Take a mass on a spring with linear damping,

m\ddot u + c\,\dot u + k\,u = f(t), \qquad u(0) = \dot u(0) = 0 ,

and the particular values $m = 1$ , $c = 2$ , $k = 5$ (SI units throughout).

Impulse response. Here $a_2 = m$ , so $u_\tau$ solves $m\ddot u + c\dot u + ku = 0$ with $u_\tau(\tau)=0$ , $\dot u_\tau(\tau) = 1/m$ . With $\gamma = c/2m$ and $\omega_d = \sqrt{k/m - \gamma^2}$ — Arfken's $\omega_1$ in his Example 20.8.5 — the underdamped solution is $G(t,\tau) = H(t-\tau)\,\frac{1}{m\,\omega_d}\,e^{-\gamma(t-\tau)}\,\sin\bigl[\omega_d(t-\tau)\bigr] .$ Check of the normalisation: at $t = \tau$ the value is $0$ , and the $t$ -derivative there is $\frac{1}{m\omega_d}\cdot\omega_d = \frac1m$ , as required.
Numbers. $\gamma = 1$ , $k/m = 5$ , $\omega_d = \sqrt{5-1} = 2$ , so $G(t,\tau) = \tfrac12\,e^{-(t-\tau)}\sin\bigl[2(t-\tau)\bigr] \quad (t>\tau) .$
A rectangular pulse. Let $f(t) = 1$ for $0 \le t \le 1$ and $f(t) = 0$ afterwards. Substituting $s = t-\tau$ and using $\int e^{-s}\sin 2s\,ds = -\tfrac15e^{-s}\bigl(\sin 2s + 2\cos 2s\bigr)$ , which is checked by differentiating, $u(t) = \tfrac12\int_0^{t}e^{-s}\sin 2s\,ds = \frac15 - \frac{e^{-t}\bigl(\sin 2t + 2\cos 2t\bigr)}{10} \qquad (0 \le t \le 1) ,$ $u(t) = \tfrac12\int_{t-1}^{t}e^{-s}\sin 2s\,ds = \frac{e^{-(t-1)}\bigl(\sin 2(t-1) + 2\cos 2(t-1)\bigr) - e^{-t}\bigl(\sin 2t + 2\cos 2t\bigr)}{10} \qquad (t \ge 1) .$
Checkable numbers. $u(1) = \tfrac15 - e^{-1}(\sin 2 + 2\cos 2)/10 = 0.19716719$ , and $u(3) = -0.03610410$ . Integrating the differential equation numerically from $t=0$ to $t=3$ with the same pulse reproduces both to eight decimal places. The sign of $u(3)$ is the system ringing back through zero after the push stops.
Sanity at the two ends. For small $t$ , expanding the first formula gives $u(t) = \tfrac12t^2 - \tfrac13t^3+O(t^4)$ : initial acceleration $f/m = 1$ , exactly as Newton requires, with the damping entering only at third order. If instead the pulse never ended, $u \to 1/5 = f/k$ , the static deflection.
Transfer function. With $\hat G(\omega) = 1/P(-i\omega)$ and $P(\zeta) = m\zeta^2 + c\zeta + k$ , $\hat G(\omega) = \frac{1}{k - m\omega^2 - ic\omega} = \frac{1}{5-\omega^2-2i\omega} .$ Its poles are the roots of $\omega^2 + 2i\omega - 5 = 0$ , that is $\omega = \pm 2 - i$ : both strictly in the lower half-plane, so $\hat G$ is analytic in the upper half-plane, as causality demands. Their real parts $\pm\omega_d = \pm2$ are the ringing frequency, their common imaginary part $-\gamma = -1$ the decay rate.
Resonance from the formula. $|\hat G(\omega)|^2 = \bigl[(k-m\omega^2)^2 + c^2\omega^2\bigr]^{-1}$ is largest at $\omega^2 = k/m - 2\gamma^2 = 3$ . There the bracket is $c^2\omega_d^2 = 16$ , so $|\hat G|^2 = 1/(c\,\omega_d)^2 = 1/16 = 0.0625$ and the amplitude itself is $|\hat G| = 1/(c\,\omega_d) = 1/4$ . (A numerical scan of $|5-\omega^2-2i\omega|^{-2}$ in steps of $10^{-3}$ peaks at $\omega = 1.732$ with value $0.0625$ .) So the peak response is at $\omega = \sqrt3 = 1.7320508$ , slightly below the ringing frequency $\omega_d = 2$ and further below the undamped $\omega_0 = \sqrt5 = 2.2360680$ — three distinct "resonant frequencies" that coincide only when the damping vanishes. At zero frequency $\hat G(0) = 1/k = 0.2$ , the static compliance of item 5, and this also equals $\int_0^\infty G_0(t)\,dt$ .
Advanced solution, for contrast. $G_{\mathrm{adv}}(t,\tau) = -\tfrac12e^{-(t-\tau)}\sin[2(t-\tau)]$ for $t<\tau$ and $0$ for $t>\tau$ also satisfies $LG = \delta(t-\tau)$ . Used in place of $G$ it would predict a mass that begins oscillating — with exponentially growing amplitude as one runs backwards — before the hammer strikes. The mathematics permits it; physics does not.

ApplicationOriginal

The causal Green's function is the impulse response, and essentially every linear time-invariant system is specified by it.

Mechanical and electrical systems. The damped oscillator above is simultaneously a mass–spring–dashpot, a series $RLC$ circuit with $L,R,1/C$ in place of $m,c,k$ — Arfken's " $RLC$ Analog" of p. 1022 — a galvanometer, and an accelerometer. Its transfer function is the complex impedance or compliance, and the pole locations are what a control engineer moves to tune a response. Riley, Hobson and Bence obtain the impulse response of the first-order system $T\dot x + x = \delta(t)$ both as a limit of rectangular pulses of shrinking width and directly by Laplace transform, which is the simplest instance of the whole scheme.
Signal processing. A filter is the convolution operator $u = G_0 * f$ ; its frequency response is $\hat G$ . Arfken's high-pass filter with $\varphi(\omega) = i\omega RC/(1+i\omega RC)$ is one line of circuit analysis away from the general theory. Causality is the reason an ideal brick-wall filter is unrealisable: a frequency response that is exactly zero on an interval cannot come from a function analytic in a half-plane, so a real filter must trade sharpness against phase distortion and pre-ringing.
Response theory in physics. In linear response, the perturbation $f$ is an applied field and $G_0$ is the retarded correlation function; $\hat G$ is the generalised susceptibility — electric polarisability, magnetic susceptibility, dielectric function, conductivity. Causality forces it to be analytic in the upper half-plane, hence the Kramers–Kronig relations: measure absorption across the whole spectrum and the refractive index follows, and vice versa. This is the standard way optical constants are extracted from reflectivity data, and the same logic constrains scattering amplitudes in particle physics.
Field theory and electromagnetism. The retarded Green's function of the wave operator gives the Liénard–Wiechert potentials: the field at $(\mathbf{r},t)$ depends on the source at the retarded time $t - |\mathbf{r}-\mathbf{r}'|/c$ . The choice of retarded over advanced is the same choice made in item 8 above, and the $i\epsilon$ prescription that implements it is the shift of poles described in the remarks (Green's functions for the wave and heat equations).
Numerics and identification. Duhamel's formula of part (iv) is the basis of step-response methods in structural dynamics and in control: excite with a step, record $F$ , and every other input is handled by one integral against $f'$ . Conversely, measuring the response to an approximate impulse — a hammer tap, a chirp, a pseudorandom sequence — and deconvolving gives $G_0$ experimentally, which is system identification.