Green’s Functions, Separation of Variables and Transforms

The theorems of Book 6A say that solutions exist, are unique and are smooth. Engineers, physicists and quantitative analysts usually need more: a formula, or at least a recipe that a computer can follow. This chapter collects the classical recipes, each of which turns a PDE into something simpler.

Green's functions turn a boundary-value problem into an integral: the solution is the data integrated against a kernel that depends only on the domain. Separation of variables turns a PDE into a family of ODEs, one for each eigenfunction, which is the spectral theorem of 4A.8 Compact Operators and Spectra made concrete. Transforms (Fourier in space, Laplace in time) turn derivatives into multiplications. And similarity and travelling-wave solutions use the symmetries of an equation to reduce the number of variables. None of these works for every equation, and each has a precise domain of validity; together they produce almost every explicit solution in the literature, and the Gaussian, the Poisson kernel and the Bessel function that appear throughout the guide all come from here.

By the end of this chapter you will be able to:

• prove Green's representation formula, define the Green's function of a domain, and prove that it is symmetric;
• construct Green's functions for the half-space and the ball by reflection, and prove that the Poisson integral solves the Dirichlet problem on a half-space;
• solve the Laplace, heat and wave equations on rectangles and discs by separation of variables, and recognise Bessel's equation;
• use the Laplace transform to solve the heat equation with time-dependent boundary data, and recognise Duhamel's principle in the result;
• find travelling-wave and soliton solutions, and compute the damping depth of temperature waves in the ground.

4.1Temperature waves in the ground

4.2Green's representation formula

Let $U \subset \mathbb{R}^n$ be bounded with $C^1$ boundary and outward normal $\nu$, and let $\Phi$ be the fundamental solution of the Laplacian (6A.2 Harmonic Functions), so that $-\Delta\Phi = \delta_0$ and $\nabla\Phi(z) = -\frac{z}{n\alpha(n)|z|^n}$.

The formula expresses $u$ inside $U$ through three things: $\Delta u$ inside, and both $u$ and $\partial_\nu u$ on the boundary. But for the Dirichlet problem only $u$ is given on the boundary, and $\partial_\nu u$ is unknown; prescribing both would overdetermine the problem (Hadamard's example, 6A.1 What a PDE Is). The way out is to subtract a correction that removes the $\partial_\nu u$ term.

4.3Green's function

For each $x \in U$, let $\phi^x$ be the corrector: the harmonic function on $U$ with boundary values $\phi^x(y) = \Phi(y - x)$ for $y \in \partial U$. (It exists for reasonable $U$, by 6B.6 Perron’s Method and Barriers or by the weak theory of 6A.6 Weak Solutions and Elliptic Regularity applied to $\phi^x - \Phi(\cdot - x)\zeta$ for a cut-off $\zeta$; for the half-space and the ball it is written down below.)

So $-\Delta_yG(x, \cdot) = \delta_x$ in $U$ and $G(x, \cdot) = 0$ on $\partial U$: $G(x, y)$ is the potential at $y$ of a unit charge at $x$ inside a grounded conductor $\partial U$, or the steady temperature at $y$ from a unit heat source at $x$ in a body whose surface is held at $0$.

The kernel $K(x, y) = -\frac{\partial G}{\partial\nu_y}(x, y)$ on $\partial U$ is the Poisson kernel of $U$: it solves the problem with $f = 0$, and it is the density of harmonic measure, the distribution of the point where a Brownian motion from $x$ first leaves $U$ (6A.3 The Heat Equation on ℝⁿ, 6D.1 Brownian Motion). The kernel $G$ solves the problem with $g = 0$, and as an integral operator it is the inverse of $-\Delta$ with Dirichlet conditions, the compact self-adjoint operator of 4A.8 Compact Operators and Spectra. Self-adjointness corresponds to a symmetry of the kernel.

Physically: the potential at $y$ due to a unit charge at $x$ equals the potential at $x$ due to a unit charge at $y$, a reciprocity principle (Green, 1828; in acoustics and elasticity it goes by the names of Helmholtz and Maxwell–Betti).

4.4The half-space and the ball: images

For a few domains the corrector can be guessed by reflection, the method of images: put a fictitious opposite charge outside $U$, placed so that the two potentials cancel on $\partial U$.

The half-space. Let $\mathbb{R}^n_+ = \{x_n > 0\}$ and, for $x \in \mathbb{R}^n_+$, let $\tilde x = (x_1, \ldots, x_{n-1}, -x_n)$ be its mirror image. For $y$ on the boundary $\{y_n = 0\}$, $|y - x| = |y - \tilde x|$, so $\Phi(y - \tilde x) = \Phi(y - x)$ there, and $y \mapsto \Phi(y - \tilde x)$ is harmonic in $\mathbb{R}^n_+$ (its singularity is outside). So $\phi^x(y) = \Phi(y - \tilde x)$, and

Differentiating, with $\partial_{y_n}\Phi(y - x) = -\frac{y_n - x_n}{n\alpha(n)|y - x|^n}$, at $y_n = 0$

since the outward normal of the half-space is $-e_n$. This suggests the Poisson formula for the half-space

The half-space is unbounded, so Theorem 4.3 does not apply as stated; instead we verify directly.

The structure is that of an approximate identity (3A.8 Convolution and Mollifiers): positive kernels of mass one that concentrate at $x^0$ as $x \to x^0$. The heat kernel as $t \to 0$ (6A.3 The Heat Equation on ℝⁿ) and the Poisson kernel as $x_n \to 0$ are the two most important examples, and the proofs are the same.

The ball. For the unit ball $B$, reflection in the sphere is inversion, $x \mapsto \tilde x = \frac{x}{|x|^2}$. For $|y| = 1$ one checks $|x|\,|y - \tilde x| = |y - x|$ (Exercise 4.10), so the function $y \mapsto \Phi\big(|x|(y - \tilde x)\big)$, which is harmonic in $B$ (its singularity $\tilde x$ is outside), agrees with $\Phi(y - x)$ on the sphere. Hence

and a computation like the one for the half-space gives the Poisson kernel and Poisson's formula for the ball

which 6A.2 Harmonic Functions quoted, and which solves the Dirichlet problem on the ball by the same approximate-identity argument (Evans, section 2.2.4, Theorem 15). In the plane, writing $x = re^{i\theta}$ and $y = e^{i\varphi}$, it is

which separation of variables recovers below.

4.5Separation of variables

Separation of variables looks for solutions that are products of functions of one variable, and builds general solutions as sums of them. It works when the domain is a product in some coordinate system (a rectangle, a disc in polar coordinates, a cylinder) and the operator respects that product. Its justification is the spectral theorem: the products are eigenfunctions, and they form a basis (4A.8 Compact Operators and Spectra).

Laplace's equation on a rectangle. On $R = (0, a)\times(0, b)$, solve $\Delta u = 0$ with $u(x, b) = f(x)$ and $u = 0$ on the other three sides. Try $u = X(x)Y(y)$: then $X''Y + XY'' = 0$, so $\frac{X''}{X} = -\frac{Y''}{Y}$, a function of $x$ equal to a function of $y$, hence a constant $-\mu$. The boundary conditions $X(0) = X(a) = 0$ force $X = \sin\frac{k\pi x}{a}$ with $\mu = (\frac{k\pi}{a})^2$, and then $Y'' = \mu Y$ with $Y(0) = 0$ gives $Y = \sinh\frac{k\pi y}{a}$. So

At $y = b$ this is the sine series of $f$ (2B.7 Fourier Series and the First Heat Equation). Below the top, the $k$-th term is damped by roughly $e^{-k\pi(b - y)/a}$: fine detail in the boundary data dies off within a distance comparable to its wavelength, the elliptic counterpart of the heat equation's smoothing.

Laplace's equation on the disc. In polar coordinates $\Delta u = u_{rr} + \frac1ru_r + \frac{1}{r^2}u_{\theta\theta}$ (6A.2 Harmonic Functions). Separating, $u = R(r)\Theta(\theta)$ with $\Theta'' = -m^2\Theta$ (periodicity forces integer $m$), so $\Theta = e^{im\theta}$, and $r^2R'' + rR' - m^2R = 0$, an Euler equation with solutions $r^{\pm m}$ (and $1$, $\log r$ for $m = 0$). Boundedness at the centre keeps $r^{|m|}$. So

Substituting the formula for $\hat g_m$ and summing the two geometric series $\sum_{m\geq0}(re^{i(\theta-\varphi)})^m$ and $\sum_{m\geq1}(re^{-i(\theta-\varphi)})^m$ gives exactly the planar Poisson kernel above (Exercise 4.11). Two methods, one answer.

The heat equation on an interval. On $(0, \pi)$ with $u = 0$ at both ends, separation gives $u = \sum_kb_ke^{-k^2t}\sin kx$ (2B.7 Fourier Series and the First Heat Equation, 6A.7 Parabolic Regularity). The same problem can be solved by images: extend the data to be odd and $2\pi$-periodic and convolve with the heat kernel on the line (6A.3 The Heat Equation on ℝⁿ, Exercise 4.13). So the Green's function (heat kernel) of the interval has two expressions,

the first converging fast for large $t$ and the second for small $t$. That they are equal is the Poisson summation formula of 4A.6 The Fourier Transform: the eigenfunction expansion and the image sum are Fourier duals.

Bessel functions. Separating the heat or wave equation on a disc of radius $a$ in polar coordinates, $u = e^{-\lambda t}R(r)e^{im\theta}$ (or with $\cos(\sqrt\lambda t)$), leads to

Bessel's equation after the substitution $s = \sqrt\lambda r$. Its solution bounded at $r = 0$ is the Bessel function $J_m(\sqrt\lambda\,r)$, where

(a power series solution, as in 6B.5 Power Series and the Cauchy–Kovalevskaya Theorem). The boundary condition $R(a) = 0$ forces $\sqrt\lambda\,a$ to be a zero $j_{m,k}$ of $J_m$, so the Dirichlet eigenvalues of the disc are $j_{m,k}^2/a^2$ (6B.3 The Wave Equation's drum, 4A.8 Compact Operators and Spectra). Bessel functions behave like damped cosines for large $s$, $J_m(s) \approx \sqrt{\frac{2}{\pi s}}\cos\big(s - \frac{m\pi}{2} - \frac\pi4\big)$, which is why the zeros become nearly equally spaced (Taylor I, §3.6).

4.6Transforms

A transform replaces a derivative by multiplication. The Fourier transform in space did this for the heat equation on $\mathbb{R}^n$ (4A.6 The Fourier Transform, 6A.3 The Heat Equation on ℝⁿ). In time, on a half-line $t > 0$, the natural tool is the Laplace transform

defined when $u$ grows at most exponentially. Integrating by parts, $\overline{u'}(s) = s\bar u(s) - u(0)$: a derivative becomes multiplication by $s$, and the initial value appears automatically. The transform of a convolution in time, $(f \star g)(t) = \int_0^tf(t - \tau)g(\tau)\,d\tau$, is the product $\bar f\bar g$ (Evans, section 4.3.2).

The temperature waves in the ground are the same computation with $h(t) = A\cos\omega t$, run for a long time: replacing $s$ by $i\omega$, the factor $e^{-x\sqrt{i\omega}} = e^{-x\sqrt{\omega/2}(1 + i)}$ is exactly the damped, delayed wave $e^{-x/d}e^{-ix/d}$ with $d = \sqrt{2/\omega}$.

4.7Similarity and travelling waves

6A.1 What a PDE Is found the heat kernel and the error function by looking for solutions unchanged by the scaling symmetry of the heat equation. Two other symmetries give two other families.

Travelling waves, $u(x, t) = v(x - ct)$, use the symmetry under translation in space and time together. They reduce a PDE to an ODE for $v$. 6A.5 Maximum Principles used them for the Fisher–KPP front and 6B.2 Conservation Laws and Shocks for viscous shocks.

Solitons. The Korteweg–de Vries equation

models long waves in shallow water: the nonlinear term steepens waves (as in Burgers' equation) and the third-derivative term disperses them (different wavelengths travel at different speeds). A travelling wave $u = v(x - ct)$ must satisfy $-cv' + 6vv' + v''' = 0$; integrating twice with decay at infinity gives

(Exercise 4.15). Steepening and dispersion balance exactly, and the hump travels without changing shape, faster the taller it is. John Scott Russell watched such a "wave of translation" travel along the Union Canal near Edinburgh in 1834, followed it on horseback, and reported it in 1844; Korteweg and de Vries derived the equation in 1895. In 1965 Norman Zabusky and Martin Kruskal found numerically that two such waves pass through each other and emerge with their shapes intact, and named them solitons; the explanation, the inverse scattering transform (Gardner, Greene, Kruskal and Miura, 1967), turned KdV into a linear problem, a transform method of a far more sophisticated kind.

Converting nonlinear to linear. The Hopf–Cole transformation of 6B.2 Conservation Laws and Shocks turned viscous Burgers into the heat equation. It is the prototype of a family of tricks (Evans, section 4.4): potential functions, the hodograph and Legendre transforms, all of which exchange dependent and independent variables or take logarithms so that a nonlinear equation becomes linear. They are rare, but when they apply they give complete information.

4.8History

George Green's Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism (1828) introduced the representation formula, the Green's function and the reciprocity principle, and was largely ignored until William Thomson (Lord Kelvin) found it in 1845 and arranged its republication in Crelle's journal from 1850; Thomson also introduced the method of images, in the 1840s. Siméon Denis Poisson's formula for the ball dates from 1820. Separation of variables is Fourier's method (1807, 1822). Friedrich Bessel studied the functions named after him in 1824, in work on planetary perturbations, though Daniel Bernoulli and Euler had met them earlier in problems about hanging chains and membranes. Pierre-Simon Laplace introduced his transform in work on probability around 1780–1812, and its use for differential equations was made systematic by Oliver Heaviside's operational calculus in the 1890s. Russell's wave of translation, Korteweg–de Vries and Zabusky–Kruskal date from 1834, 1895 and 1965.

4.9Exercises

Solution

(a) $i\omega Z = \kappa Z''$. (b) $Z = e^{\pm\sqrt{i\omega/\kappa}\,z}$ with $\sqrt{i\omega/\kappa} = \frac{1 + i}{\sqrt2}\sqrt{\omega/\kappa} = \frac{1 + i}{d}$; boundedness picks the minus sign. $\operatorname{Re}\big(Ae^{i\omega t}e^{-z/d}e^{-iz/d}\big) = Ae^{-z/d}\cos(\omega t - z/d)$. (c) Daily: $\omega = 2\pi/86400 \approx 7.27 \times 10^{-5}$ s⁻¹, $d = \sqrt{10^{-6}/7.27 \times 10^{-5}} \approx 0.117$ m. Annual: $\omega \approx 1.99 \times 10^{-7}$ s⁻¹, $d \approx 2.24$ m. At $2$ m: amplitude $e^{-0.89} \approx 0.41$, lag $0.89$ rad, which is $0.89/2\pi$ of a year, about $52$ days. Half a year out of phase when $z/d = \pi$, at $z \approx 7.0$ m, where the amplitude is $e^{-\pi} \approx 4\%$.

Solution

After $y' = x_nz$: $\frac{2}{n\alpha(n)}\int_{\mathbb{R}^{n-1}}\frac{dz}{(1 + |z|^2)^{n/2}}$. In polar coordinates this is $\frac{2}{n\alpha(n)}\cdot(n - 1)\alpha(n - 1)\int_0^\infty\frac{r^{n-2}}{(1 + r^2)^{n/2}}dr$, and with $r = \tan\theta$ the integral is $\int_0^{\pi/2}\sin^{n-2}\theta\,d\theta$. The identity $(n - 1)\alpha(n - 1)\int_0^{\pi/2}\sin^{n-2}\theta\,d\theta = \frac n2\alpha(n)$ is the formula for the volume of the unit ball by slicing (3A.5 Product Measures and Change of Variables), which makes the total $1$. For $n = 2$: $\frac1\pi[\arctan(y/x_2)]_{-\infty}^\infty = 1$.

Solution

$\Delta x_n = 0$, $\nabla x_n = e_n$, $\nabla|x|^{-n} = -n|x|^{-n-2}x$, and $\Delta|x|^{-n} = v'' + \frac{n - 1}{r}v'$ with $v = r^{-n}$: $n(n + 1)r^{-n-2} - n(n - 1)r^{-n-2} = 2nr^{-n-2}$. So $\Delta K = 2e_n\cdot(-n|x|^{-n-2}x) + x_n\cdot2n|x|^{-n-2} = 0$.

Solution

$|x|^2|y - \tilde x|^2 = |x|^2\big(|y|^2 - 2\frac{x\cdot y}{|x|^2} + \frac{1}{|x|^2}\big) = |x|^2 - 2x\cdot y + 1$ when $|y| = 1$, and $|y - x|^2 = 1 - 2x\cdot y + |x|^2$.

Solution

$\sum_{m\geq0}(re^{i\psi})^m + \sum_{m\geq1}(re^{-i\psi})^m = \frac{1}{1 - re^{i\psi}} + \frac{re^{-i\psi}}{1 - re^{-i\psi}}$. Over the common denominator $|1 - re^{i\psi}|^2 = 1 - 2r\cos\psi + r^2$, the numerator is $(1 - re^{-i\psi}) + re^{-i\psi}(1 - re^{i\psi}) = 1 - r^2$. Then $u = \sum\hat g_mr^{|m|}e^{im\theta} = \frac{1}{2\pi}\int g(\varphi)\sum r^{|m|}e^{im(\theta - \varphi)}d\varphi$ (interchange justified by uniform convergence for fixed $r < 1$).

Solution

$G(x, y) = y(1 - x)$ for $y \leq x$ and $x(1 - y)$ for $y \geq x$: in $y$ it is linear on each side of $x$, vanishes at $0$ and $1$, is continuous, and its derivative jumps by $-1$ at $y = x$, which is $-\partial_y^2G = \delta_x$. It is symmetric by inspection. Then $u(x) = (1 - x)\int_0^xyf(y)\,dy + x\int_x^1(1 - y)f(y)\,dy$, and differentiating twice gives $u'' = -f$. A string under unit tension, pinned at both ends, pushed at one point, takes this tent shape.

Solution

The convolution of the even kernel with the odd function $\tilde g$ is odd about $0$; since $\tilde g$ is also odd about $\pi$ (being odd and $2\pi$-periodic), $u$ is odd about $\pi$ as well, so $u(0, t) = u(\pi, t) = 0$. Writing $\int_{\mathbb{R}}\Phi(x - y)\tilde g(y)\,dy$ as a sum over the intervals $[2\pi m, 2\pi m + \pi]$ (where $\tilde g(y) = g(y - 2\pi m)$) and $[2\pi m - \pi, 2\pi m]$ (where $\tilde g(y) = -g(2\pi m - y)$), and substituting back into $[0, \pi]$, gives the kernel $\sum_m\big(\Phi(x - y - 2\pi m, t) - \Phi(x + y - 2\pi m, t)\big)$, which is the formula in the text after replacing $m$ by $-m$.

Solution

(a) With $s = \frac{x}{2\sqrt\tau}$, $ds = -\frac{x}{4}\tau^{-3/2}d\tau$, so $k\,d\tau = -\frac{2}{\sqrt\pi}e^{-s^2}ds$, and as $\tau$ runs from $0$ to $t$, $s$ runs from $\infty$ to $\frac{x}{2\sqrt t}$: the integral is $\frac{2}{\sqrt\pi}\int_{x/2\sqrt t}^\infty e^{-s^2}ds = \operatorname{erfc}\big(\frac{x}{2\sqrt t}\big)$. (b) The function in (a) is the integral of $k$ from $0$, whose transform is $\frac1s\bar k$ (since the transform of $\int_0^tk$ is $\bar k/s$). So $\bar k = e^{-x\sqrt s}$.

Solution

With $v = a\operatorname{sech}^2(b\xi)$: $v' = -2ab\operatorname{sech}^2(b\xi)\tanh(b\xi)$, so $v'^2 = 4a^2b^2\operatorname{sech}^4\tanh^2 = 4a^2b^2\operatorname{sech}^4(1 - \operatorname{sech}^2)$. And $cv^2 - 2v^3 = ca^2\operatorname{sech}^4 - 2a^3\operatorname{sech}^6$. Matching: $4a^2b^2 = ca^2$ and $4a^2b^2 = 2a^3$, so $b^2 = \frac c4$ and $a = 2b^2 = \frac c2$.

Solution

(a) $-\Delta e^{2\pi ik\cdot x} = 4\pi^2|k|^2e^{2\pi ik\cdot x}$, so the Fourier coefficients satisfy $4\pi^2|k|^2\hat u_k = \hat f_k$; for $k = 0$ the equation forces $\hat f_0 = 0$, and $\hat u_0$ is free (set it to $0$). (b) Convolution multiplies Fourier coefficients (2B.7 Fourier Series and the First Heat Equation). The coefficients of $-\Delta G$ are $1$ for $k \neq 0$ and $0$ for $k = 0$, which are those of $\delta_0 - 1$.