Green's Functions and the Fredholm Alternative, differential equations lesson

A taut string of unit tension is fixed at both ends of $[0, 1]$ and pushed sideways by a load $f(x)$ , so its displacement solves $-y'' = f$ with $y(0) = y(1) = 0$ . Here $x$ is position along the string and a prime means $d/dx$ . If you knew the shape the string takes under a unit point load at each location $s$ , could you add those shapes to get the response to any load? When the unloaded problem has only the zero solution the answer is yes, and that shape is the Green’s function, an integral kernel that builds the equation and both boundary conditions into one solution formula. This lesson constructs it, checks the jump in slope at the source, and uses the Fredholm alternative to settle what happens when the homogeneous problem has a nonzero solution and no such kernel exists.

Key terms

Green’s function

Definition Let $p$ be continuously differentiable with $p > 0$ on $[a, b]$ , let $q$ be continuous and real there, and consider the boundary-value problem

Ly = -(p\,y')' + q\,y = f

on $[a, b]$ with the real separated boundary conditions

\begin{gathered} \alpha_0\,y(a) + \beta_0\,y'(a) = 0, \\[0.4em] \alpha_1\,y(b) + \beta_1\,y'(b) = 0, \end{gathered}

where $(\alpha_0, \beta_0) \ne (0, 0)$ and $(\alpha_1, \beta_1) \ne (0, 0)$ . Assume that the homogeneous problem, with $f = 0$ , has only the solution $y = 0$ . The Green’s function of the problem is the function $G(x, s)$ on the square $[a, b] \times [a, b]$ such that for each source point $s$ in $(a, b)$ the function $x \mapsto G(x, s)$

solves $LG = 0$ on $[a, s)$ and on $(s, b]$ ,
satisfies both boundary conditions,
is continuous at $x = s$ ,
has the slope jump $G_x(s^+, s) - G_x(s^-, s) = -1/p(s)$ .

Here $G_x$ is the partial derivative in $x$ , and $G_x(s^\pm, s)$ are its limits as $x \to s$ from the right and from the left. These conditions determine $G$ uniquely, and $G$ is symmetric, $G(x, s) = G(s, x)$ . To build it, take a nonzero solution $y_1$ of $Ly = 0$ that satisfies the condition at $a$ and a nonzero solution $y_2$ that satisfies the condition at $b$ . By Abel’s identity the number $c = p\,(y_1'y_2 - y_1y_2')$ is a constant. It is not zero, because otherwise $y_1$ and $y_2$ would be multiples of each other and $y_1$ would be a nonzero solution of the homogeneous problem. For $x \le s$ the Green’s function is

G(x, s) = \frac{y_1(x)\,y_2(s)}{c},

and for $x \ge s$ it is $G(x, s) = y_1(s)\,y_2(x)/c$ .

In plain words The Green’s function is the response of the boundary problem to a unit point source at $s$ . Away from the source it solves the unforced equation, at the ends it obeys the boundary conditions, and at the source its graph has a corner but no break.

Example For $-y'' = f$ on $[0, 1]$ with $y(0) = y(1) = 0$ , the coefficients are $p = 1$ and $q = 0$ . Away from $s$ the kernel has zero second derivative, so each branch is a straight line. The left branch must vanish at $0$ and the right branch at $1$ , so they are $Ax$ and $B(1 - x)$ . Continuity at $s$ gives $As = B(1 - s)$ , and the jump condition gives $-B - A = -1$ . Solving, $A = 1 - s$ and $B = s$ , so

\begin{gathered} G(x, s) = x(1 - s), \quad x \le s, \\[0.4em] G(x, s) = s(1 - x), \quad x \ge s. \end{gathered}

The formula gives the same kernel with $y_1 = x$ , $y_2 = 1 - x$ and $c = 1 \cdot (1 - x) - x \cdot (-1) = 1$ . The graph is a tent with its peak $s(1 - s)$ at the source, and exchanging $x$ and $s$ exchanges the two branches, which is the symmetry.

Integral kernel

Definition An integral operator with kernel $K(x, s)$ sends a continuous function $f$ on $[a, b]$ to the function

x \mapsto \int_a^b K(x, s)\,f(s)\,ds.

Under the hypotheses of the previous card, for every continuous $f$ on $[a, b]$ the boundary-value problem $Ly = f$ has exactly one solution, and it is

y(x) = \int_a^b G(x, s)\,f(s)\,ds.

So the Green’s function is the kernel of the inverse of $L$ together with its boundary conditions. It belongs to one operator, one interval and one complete set of boundary conditions, and changing any of them changes the kernel. A boundary kernel collects sources from both sides of $x$ . The kernel of an initial-value problem, as in convolution, is causal. It uses only earlier times and gives the solution with zero initial data, which in general does not satisfy a condition at the far end.

In plain words Split the load into many small pieces. The piece near $s$ has strength $f(s)\,ds$ and produces the response $G(x, s)\,f(s)\,ds$ at $x$ , and linearity lets you add all these responses. Once the kernel is known, a new load needs only a new integral, not a new boundary calculation.

Example Solve $-y'' = 2$ on $[0, 1]$ with $y(0) = y(1) = 0$ using the tent kernel. The kernel changes formula at $s = x$ , so split the integral there. Sources with $s \le x$ use $s(1 - x)$ , and sources with $s \ge x$ use $x(1 - s)$ , so

\begin{aligned} y &= 2(1 - x)\int_0^x s\,ds \\[0.4em] &\quad + 2x\int_x^1 (1 - s)\,ds \\[0.4em] &= x^2(1 - x) + x(1 - x)^2. \end{aligned}

Factoring out $x(1 - x)$ leaves $x + (1 - x) = 1$ , so $y = x(1 - x)$ . As a check, $y'' = -2$ , so $-y'' = 2$ , and $y(0) = y(1) = 0$ .

Jump condition at the source

Definition For $Ly = -(p\,y')' + q\,y$ , the Green’s function satisfies, at each source point $s$ in $(a, b)$ ,

\begin{aligned} &G(s^+, s) = G(s^-, s), \\[0.4em] &G_x(s^+, s) - G_x(s^-, s) \\[0.2em] &\qquad = -\frac{1}{p(s)}. \end{aligned}

Formally $G$ solves $LG = \delta(x - s)$ , where the delta is a distribution, a unit point source, and not an ordinary function with one exceptional value. The four conditions of the first card are the precise meaning of this equation. Integrating it over $[s - \varepsilon, s + \varepsilon]$ gives

\begin{aligned} &-\big[p\,G_x\big]_{x = s - \varepsilon}^{x = s + \varepsilon} \\[0.4em] &\quad + \int_{s - \varepsilon}^{s + \varepsilon} q\,G\,dx = 1. \end{aligned}

As $\varepsilon \to 0$ the integral of the bounded function $qG$ tends to $0$ , and what remains is $-p(s)$ times the jump in $G_x$ , which must equal $1$ . The sign comes from the operator. For the opposite operator $(p\,y')' - q\,y$ the jump is $+1/p(s)$ and the Green’s function is $-G$ .

In plain words A point source cannot break the graph, but it bends it sharply. The size of the kink is fixed by the strength of the source and by $p$ at the source, and whether the slope drops or rises depends on the sign in front of the second derivative.

Example For the tent kernel, both branches equal $s(1 - s)$ at $x = s$ . The slope is $1 - s$ on the left and $-s$ on the right, so the right slope minus the left slope is $-s - (1 - s) = -1$ , which is $-1/p$ with $p = 1$ . A slope that drops by $1$ is exactly what $-d^2/dx^2$ turns into a positive unit source. For $y'' = f$ with the same boundary conditions the kernel is $-G$ , whose slope rises by $1$ at the source, and for $f = 2$ it gives $y = -x(1 - x) = x^2 - x$ , which indeed has $y'' = 2$ .

Eigenvalue obstruction

Definition Suppose the homogeneous problem $L\varphi = 0$ with the same boundary conditions has a nonzero solution $\varphi$ . If $y$ solves $Ly = f$ with the boundary conditions, then so does $y + C\varphi$ for every constant $C$ , since adding $C\varphi$ changes neither the equation nor the boundary values. Solutions, when they exist, are therefore never unique, so the problem has no inverse and no Green’s function. The construction of the first card breaks down as well, because $y_1$ and $y_2$ are then both multiples of $\varphi$ and $c = 0$ . For an eigenvalue problem $-(p\,y')' + q\,y = \lambda w\,y$ with a continuous weight $w > 0$ and a real $\lambda$ , the problem $-(p\,y')' + q\,y - \lambda w\,y = f$ has a Green’s function exactly when $\lambda$ is not an eigenvalue. When $\lambda$ is an eigenvalue, its eigenfunctions are the nonzero solutions $\varphi$ .

In plain words A Green’s function is an inverse, and an inverse cannot exist when some nonzero shape is invisible to the problem. That shape, a solution of the homogeneous problem, can be added to any solution without changing the forcing or the boundary values.

Example For $-y'' - k^2y = f$ on $[0, 1]$ with $y(0) = y(1) = 0$ and $k > 0$ , take $y_1 = \sin kx$ and $y_2 = \sin k(1 - x)$ . Then $c = k\sin k$ , and

G(x, s) = \frac{\sin kx\,\sin k(1 - s)}{k\sin k}

for $x \le s$ , with $x$ and $s$ exchanged for $x \ge s$ . This kernel exists whenever $\sin k \ne 0$ , and letting $k \to 0$ recovers the tent kernel $x(1 - s)$ . As $k$ approaches $\pi$ the denominator tends to $0$ , and the kernel blows up at every pair $x, s$ in $(0, 1)$ . At $k = \pi$ the function $\varphi = \sin \pi x$ solves the homogeneous problem, so $\lambda = \pi^2$ is an eigenvalue of $-y'' = \lambda y$ with these conditions, and no Green’s function exists. The same happens at every $k = n\pi$ .

Fredholm alternative

Definition Let $L$ and the separated boundary conditions be as in the first card, and let $f$ be continuous on $[a, b]$ . Exactly one of two cases holds.

In the first case the homogeneous problem has only the zero solution. Then for every continuous $f$ the problem $Ly = f$ has exactly one solution, given by the Green’s function.

In the second case the homogeneous problem has a nonzero solution $\varphi$ . The condition at $a$ alone leaves only multiples of one solution of $Ly = 0$ , so every solution of the homogeneous problem is a multiple of $\varphi$ . The problem $Ly = f$ has a solution if and only if

\int_a^b f(x)\,\varphi(x)\,dx = 0,

and then the solutions are exactly $y_p + C\varphi$ , for one particular solution $y_p$ and every constant $C$ . When the integral is not zero there is no solution. The same statement holds for $-(p\,y')' + q\,y - \lambda w\,y = f$ with a real $\lambda$ , since that is the same kind of problem with $q$ replaced by $q - \lambda w$ . The test function is $\varphi$ itself because real separated conditions make the problem self-adjoint. For a problem that is not self-adjoint the test functions come from an adjoint problem instead.

In plain words A linear boundary problem of this kind has only two possible behaviors. Either every forcing gives exactly one solution, or the homogeneous problem has a free shape, and then each forcing gives either no solution or a whole line of solutions. A problem that is solvable for every forcing but without uniqueness, or unique but not solvable for some forcing, does not occur here.

Example Take $-y'' - \pi^2y = f$ on $[0, 1]$ with $y(0) = y(1) = 0$ , so $\varphi = \sin \pi x$ . The forcing $\cos \pi x$ passes the test, since $\int_0^1 \cos \pi x\,\sin \pi x\,dx = 0$ . A particular solution is

y_p = -\frac{x\sin \pi x}{2\pi}.

Its second derivative is $y_p'' = -\cos \pi x + \tfrac{\pi}{2}x\sin \pi x$ , so $-y_p'' - \pi^2y_p = \cos \pi x$ , and $y_p(0) = y_p(1) = 0$ . Every function $y_p + C\sin \pi x$ is a solution as well. The forcing $\sin \pi x$ fails the test, since $\int_0^1 \sin^2 \pi x\,dx = \tfrac12$ , so that problem has no solution at all. An extra requirement such as $\int_0^1 y\,\varphi\,dx = 0$ singles out the member with $C = 1/(4\pi)$ , which is $(1 - 2x)\sin(\pi x)/(4\pi)$ . That is a normalization added to the problem, not uniqueness of the original problem.

Fredholm compatibility condition

Definition In the second case of the Fredholm alternative, the requirement $\int_a^b f\varphi\,dx = 0$ for a nonzero solution $\varphi$ of the homogeneous problem is the Fredholm compatibility condition. It is then true for every such solution, since they are all multiples of $\varphi$ . Its necessity comes from Lagrange’s identity, obtained by integrating by parts twice,

\begin{aligned} &\int_a^b (\varphi\,Ly - y\,L\varphi)\,dx \\[0.4em] &\quad = \Big[p\,(y\varphi' - y'\varphi)\Big]_a^b. \end{aligned}

If $y$ and $\varphi$ satisfy the same separated condition $\alpha\,y + \beta\,y' = 0$ at an endpoint, the vectors $(y, y')$ and $(\varphi, \varphi')$ there are both orthogonal to the nonzero vector $(\alpha, \beta)$ . So they are parallel, and the bracket vanishes at that endpoint. With $Ly = f$ and $L\varphi = 0$ the identity leaves $\int_a^b f\varphi\,dx = 0$ . For the eigenvalue form

-(p\,y')' + q\,y - \lambda w\,y = f,

with $\lambda$ an eigenvalue and $\varphi$ an eigenfunction, the condition is still the unweighted integral $\int_a^b f\varphi\,dx = 0$ . If the forcing is written as $f = w\,g$ , the same condition reads

\langle g, \varphi \rangle_w = \int_a^b w\,g\,\varphi\,dx = 0.

In plain words The forcing must have no component along the free shape. Which integral expresses this depends only on what you call the forcing. The test has no weight when the right side is called $f$ and carries the weight once when it is written as $w\,g$ , and mixing the two conventions gives a wrong test.

Example Take $p = w = e^x$ and $q = 0$ on $[0, \pi]$ with $y(0) = y(\pi) = 0$ . As in regular Sturm-Liouville problems, the first eigenvalue is $\lambda_1 = 5/4$ with eigenfunction $\varphi_1 = e^{-x/2}\sin x$ . Consider

\begin{gathered} -(e^x y')' - \tfrac54 e^x y = e^x g, \\[0.4em] g = e^{-x/2}\sin 2x. \end{gathered}

The weighted test gives $\int_0^\pi e^x g\,\varphi_1\,dx = \int_0^\pi \sin 2x\,\sin x\,dx = 0$ , so solutions exist. Substituting $y = e^{-x/2}v$ turns the equation into $-v'' - v = \sin 2x$ with $v(0) = v(\pi) = 0$ , and the solutions are

y = e^{-x/2}\left(\tfrac13\sin 2x + C\sin x\right)

for every constant $C$ . The unweighted integral is not zero,

\begin{aligned} \int_0^\pi g\,\varphi_1\,dx &= \tfrac15\left(1 + e^{-\pi}\right) \\[0.4em] &\approx 0.209, \end{aligned}

so testing $g$ without the weight would wrongly declare the problem unsolvable.

Check yourself

Use the tent kernel to solve $-y'' = x$ on $[0, 1]$ with $y(0) = y(1) = 0$ , and check the answer.

Show the answer

Split the integral at $s = x$ , using $s(1 - x)$ for $s \le x$ and $x(1 - s)$ for $s \ge x$ . Then

\begin{aligned} y(x) &= (1 - x)\int_0^x s^2\,ds \\[0.4em] &\quad + x\int_x^1 (1 - s)\,s\,ds \\[0.4em] &= \frac{(1 - x)\,x^3}{3} \\[0.4em] &\quad + x\left(\frac16 - \frac{x^2}{2} + \frac{x^3}{3}\right) \\[0.4em] &= \frac{x - x^3}{6}. \end{aligned}

As a check, $y'' = -x$ , so $-y'' = x$ , and $y(0) = 0$ and $y(1) = (1 - 1)/6 = 0$ .

For which constant $c$ does $-y'' = x - c$ on $[0, 1]$ with $y'(0) = y'(1) = 0$ have a solution? Find all solutions for that $c$ , and explain why this problem has no Green’s function.

Show the answer

Every constant solves $-y'' = 0$ with $y'(0) = y'(1) = 0$ , so $\varphi = 1$ is a nonzero solution of the homogeneous problem. That is the eigenvalue obstruction, so there is no Green’s function. The compatibility condition is

\int_0^1 (x - c)\,dx = \frac12 - c = 0,

so $c = \tfrac12$ . The same follows directly. Integrating the equation over $[0, 1]$ gives $y'(0) - y'(1) = \tfrac12 - c$ , and the boundary conditions make the left side zero. For $c = \tfrac12$ , the equation $y'' = \tfrac12 - x$ gives $y' = \tfrac12x - \tfrac12x^2 + C_1$ , and $y'(0) = 0$ forces $C_1 = 0$ . Then $y'(1) = 0$ holds automatically, and the solutions are

y = \frac{x^2}{4} - \frac{x^3}{6} + C

for every constant $C$ , a whole line of solutions as the Fredholm alternative predicts.

Construct the Green’s function of $-y'' = f$ on $[0, 1]$ with $y(0) = 0$ and $y'(1) = 0$ , and use it to solve the problem with $f = 1$ .

Show the answer

The homogeneous problem has only the zero solution, since $y = A + Bx$ with $y(0) = 0$ and $y'(1) = 0$ forces $A = B = 0$ . So the Green’s function exists. The left branch is a line through the origin, $Ax$ , and the right branch is a line with zero slope at $1$ , a constant $B$ . Continuity gives $As = B$ , and the jump condition gives $0 - A = -1$ . So $A = 1$ and $B = s$ , which means

G(x, s) = \begin{cases} x, & x \le s, \\ s, & x \ge s, \end{cases}

that is, $G(x, s) = \min(x, s)$ . For $f = 1$ ,

\begin{aligned} y(x) &= \int_0^x s\,ds + \int_x^1 x\,ds \\[0.4em] &= \frac{x^2}{2} + x(1 - x) \\[0.4em] &= x - \frac{x^2}{2}. \end{aligned}

As a check, $-y'' = 1$ , $y(0) = 0$ and $y'(1) = 1 - 1 = 0$ .

Green's Functions and the Fredholm Alternative

Key terms

Green’s function

Integral kernel

Jump condition at the source

Eigenvalue obstruction

Fredholm alternative

Fredholm compatibility condition

Check yourself

How this lesson connects

Builds on