basics

Apr 30, 2025

1 min read

12. Green’s function method

12.1. Poisson equation

General Poisson’s problem

\[\begin{split}\begin{cases} - \nabla^2 \mathbf{u}(\mathbf{r}, t) = \mathbf{f}(\mathbf{r},t) \\ \text{+ b.c.} \end{cases}\end{split}\]

with common boundary conditions

\[\begin{split}\begin{cases} \mathbf{u} = \mathbf{g} & \quad \text{on $S_D$} \\ \hat{\mathbf{n}} \cdot \nabla \mathbf{u} = \mathbf{h} & \quad \text{on $S_N$} \end{cases}\end{split}\]

over Dirichlet and Neumann regions of the boundary.

Poisson’s problem for Green’s function, in infinite domain

\[\begin{split} - \nabla_{\mathbf{r}}^2 G(\mathbf{r}; \mathbf{r}_0) = \delta(\mathbf{r} - \mathbf{r}_0) \\ \end{split}\]

Green’s function method

\[\begin{split}\begin{aligned} E(\mathbf{r}_0, t) u_i(\mathbf{r}_0, t) & = \int_{\mathbf{r} \in \Omega} u_i(\mathbf{r},t) \delta(\mathbf{r}-\mathbf{r}_0) = \\ & = - \int_{\mathbf{r} \in \Omega} u_i(\mathbf{r},t) \nabla_{\mathbf{r}}^2 G(\mathbf{r}-\mathbf{r}_0) = \\ & = - \int_{\mathbf{r} \in \Omega} \nabla_{\mathbf{r}} \cdot ( u_i \nabla_{\mathbf{r}} G - G \nabla_{\mathbf{r}} u_i) - \int_{\mathbf{r} \in \Omega} G \nabla^2 u_i= \\ & = - \oint_{\mathbf{r} \in \partial \Omega} \hat{\mathbf{n}} \cdot ( u_i \nabla_{\mathbf{r}} G - G \nabla_{\mathbf{r}} u_i) + \int_{\mathbf{r} \in \Omega} G(\mathbf{r}-\mathbf{r}_0) f_i(\mathbf{r}, t) . \\ \end{aligned}\end{split}\]

An integro-differential boundary problem can be written using boundary conditions. As an example, using Dirichlet and Neumann boundary conditions, the integro-differential problem reads

\[\begin{split}\begin{aligned} & E(\mathbf{r}_0, t) \mathbf{u}(\mathbf{r}_0, t) + \int_{\mathbf{r} \in S_N} \mathbf{u}(\mathbf{r},t) \, \hat{\mathbf{n}} \cdot \nabla_{\mathbf{r}} G(\mathbf{r}-\mathbf{r}_0) - \int_{\mathbf{r} \in S_D} G(\mathbf{r}-\mathbf{r}_0) \, \hat{\mathbf{n}} \cdot \nabla_{\mathbf{r}} \mathbf{u}(\mathbf{r},t) = \\ & = - \int_{\mathbf{r} \in S_D} \mathbf{g}(\mathbf{r},t) \, \hat{\mathbf{n}} \cdot \nabla_{\mathbf{r}} G(\mathbf{r}-\mathbf{r}_0) + \int_{\mathbf{r} \in S_N} G(\mathbf{r}-\mathbf{r}_0) \, \mathbf{h}(\mathbf{r},t) + \int_{\mathbf{r} \in \Omega} G(\mathbf{r}-\mathbf{r}_0) \, \mathbf{f}(\mathbf{r}, t) . \\ \end{aligned}\end{split}\]

Green’s function of the Poisson-Laplace equation reads

\[G(\mathbf{r};\mathbf{r}_0) = \frac{1}{4 \pi} \frac{1}{\left| \mathbf{r}-\mathbf{r}_0 \right|} \ .\]

Green’s function of the Laplace equation

\[-\nabla^2 G = 0 \qquad \text{for $\mathbf{r} \ne \mathbf{r}_0$}\]

Solutions with spherical symmetry,

\[0 = \nabla^2 G = \frac{1}{r^2} \left( r^2 G' \right)' \quad \rightarrow \quad G'(r) = \frac{A}{r^2} \quad \rightarrow \quad G(r) = - \frac{A}{r} + B \]

Choosing \(B = 0\) s.t. \(G(r) \rightarrow 0\) as \(r \rightarrow \infty\), and integrating over a sphere centered in \(r=0\) to get \(A = -\frac{1}{4 \pi}\),

\[1 = \int_{V} \delta(r) = - \int_{V} \nabla^2 G = - \oint_{\partial V} \hat{\mathbf{n}} \cdot \nabla G = -\oint_{\partial V} \hat{\mathbf{r}} \cdot \hat{\mathbf{r}} \frac{A}{r^2} = - 4 \pi \, A \]

12.2. Helmholtz equation

todo from Fourier to Laplace trasnform in the first lines of this section

A Helmholtz’s equation can be thouoght as the time Fourier transform of a wave equation,

\[\begin{split}\begin{cases} \dfrac{1}{c^2} \partial_{tt} \mathbf{u}(\mathbf{r},t) - \nabla^2 \mathbf{u}(\mathbf{r},t) = \mathbf{f}(\mathbf{r},t) \\ \text{+ b.c.} \\ \text{+ i.c.} \ , \end{cases}\end{split}\]

Fourier transform in time of field \(\mathbf{u}(\mathbf{r},t)\) reads

\[\tilde{\mathbf{u}}(\mathbf{r}, \omega) = \mathscr{F}\{ \mathbf{u}(\mathbf{r},t) \} = \int_{t=-\infty}^{+\infty} \mathbf{u}(\mathbf{r},t) e^{-i \omega t} \, d \omega\]

and, if \(\mathbf{u}(\mathbf{r},t)\) is compact in time, Fourier transform of its time partial derivatives read

\[\begin{split}\begin{aligned} \mathscr{F}\{ \dot{\mathbf{u}}(\mathbf{r},t) \} & = \int_{t=-\infty}^{+\infty} \dot{\mathbf{u}}(\mathbf{r},t) e^{-i \omega t} \, d \omega = \\ & = \mathbf{u}(\mathbf{r},t) e^{-i \omega t} \big|_{t=-\infty}^{+\infty} + i \omega \int_{t=-\infty}^{+\infty} \mathbf{u}(\mathbf{r},t) e^{-i \omega t} \, d \omega = \\ & = i \omega \mathscr{F}\{ \mathbf{u}(\mathbf{r},t) \} \end{aligned}\end{split}\]

\[\mathscr{F}\{ \partial_t^n \mathbf{u}(\mathbf{r},t) \} = (i \omega)^n \tilde{\mathbf{u}} \ .\]

The differential problem in the transformed domain thus reads

\[- \frac{\omega^2}{c^2} \tilde{\mathbf{u}} - \nabla^2 \tilde{\mathbf{u}} = \tilde{\mathbf{f}}\]

Green’s function of Helmholtz’e equation reads

\[G(\mathbf{r}, s) = \alpha^+ \frac{e^{ \frac{s|\mathbf{r} - \mathbf{r}_0|}{c}}}{|\mathbf{r} - \mathbf{r}_0|} + \alpha^- \frac{e^{-\frac{s|\mathbf{r} - \mathbf{r}_0|}{c}}}{|\mathbf{r} - \mathbf{r}_0|} \]

with \(\alpha^+ + \alpha^- = \frac{1}{4 \pi}\).

Being the Laplace transform,

\[\mathscr{L}\{ f(t) \} = \int_{t=0^-}^{+\infty} f(t) e^{-st} dt \ ,\]

the Laplace transform of a causal function with time delay \(\tau \ge 0\) reads

\[\mathscr{L}\{ f(t-\tau) \} = \int_{t=0^-}^{+\infty} f(t-\tau) e^{-st} dt = \int_{z = - \tau}^{+\infty} f(z) e^{-s(z+\tau)} \, dz = e^{-s\tau} \, \int_{z = 0}^{+\infty} f(z) e^{-s z} \, dz = e^{-s \tau} \, \mathscr{L}\{ f(t) \}\]

having used causality \(f(t) = 0\) for \(t < 0\). Laplace transform of Dirac’s delta \(\delta(t)\) reads

\[\mathscr{L}\{ \delta(t) \} = \int_{t=0^-}^{+\infty} \delta(t) \, dt = 1 \ ,\]

so that \(e^{-s \tau} = e^{- s \tau} \, 1 = \mathscr{L}\{ \delta(t-\tau) \}\).

Thus, Green’s function for the wave equation reads

\[G(\mathbf{r},t; \mathbf{r}_0, t_0) = \alpha^+ \frac{ \delta \left( t - t_0 + \frac{|\mathbf{r}-\mathbf{r}_0|}{c} \right)}{|\mathbf{r} - \mathbf{r}_0|} + \alpha^- \frac{ \delta \left( t - t_0 - \frac{|\mathbf{r}-\mathbf{r}_0|}{c} \right)}{|\mathbf{r} - \mathbf{r}_0|} \]

If \(t \ge t_0\), and \(G(\mathbf{r}, t; \mathbf{r}_0, t_0)\) connects the past \(t_0\) with the future \(t\), the first term is not causal, and thus \(\alpha^+ = 0\) and

\[G(\mathbf{r},t; \mathbf{r}_0, t_0) = \frac{1}{4 \pi} \frac{ \delta \left( t - t_0 - \frac{|\mathbf{r}-\mathbf{r}_0|}{c} \right)}{|\mathbf{r} - \mathbf{r}_0|} \ .\]

Green’s function of Helmholtz’s equation

\[\frac{s^2}{c^2} G - \nabla^2 G = \delta(r)\]

\[G(r) = \frac{\alpha e^{k r} + \beta e^{-kr}}{r}\]

Proof:

• Gradient

  \[\nabla G(r) = \hat{\mathbf{r}} \partial_r G = \hat{\mathbf{r}} \frac{\alpha (k r - 1) e^{kr} + \beta(-k r - 1)e^{-kr}}{r^2}\]

• Laplacian

  \[\begin{split}\begin{aligned} \nabla^2 G(r) & = \frac{1}{r^2} \left( r^2 G'(r) \right)' = \\ & = \frac{1}{r^2} \left( \alpha (k r - 1) e^{kr} + \beta(-k r - 1)e^{-kr}\right)' = \\ & = \frac{1}{r^2} \left( \alpha k e^{kr} + \alpha k^2 r e^{kr} - \alpha k e^{kr} - \beta k e^{-kr} + \beta k^2 r e^{-kr} + \beta k e^{-kr} \right) = \\ & = \frac{1}{r} \left( \alpha e^{kr} + \beta e^{-kr} \right) k^2 = k^2 G(r) \ . \end{aligned}\end{split}\]

  and thus \(k^2 G(r) - \nabla^2 G = 0\), for \(r \ne 0\);

• Unity

  \[1 = \int_{V} \delta(r) = \int_V \left( k^2 G - \nabla^2 G \right) = \int_V k^2 G - \oint_{\partial V} \hat{\mathbf{n}} \cdot \nabla G \]

  the second term is the sum of two contributions of the form

  \[\oint_{\partial V} \hat{\mathbf{n}} \cdot \nabla G^{\pm} = \oint_{\partial V} \frac{\alpha^{\pm}(\pm k r - 1) e^{\pm k r}}{r^2} = 4 \pi \alpha^{\pm} (\pm k r - 1) e^{\pm k r}\]

  the first term is the sum of two contributions of the form

  \[\begin{split}\begin{aligned} k^2 \int_{V} G(r) & = k^2 \int_{V} \frac{\alpha^{\pm} e^{\pm k r}}{r} = \\ & = k^2 \alpha^{\pm} \int_{R = 0}^{r} \int_{\phi=0}^{\pi} \int_{\theta=0}^{2 \pi} \frac{e^{\pm k R}}{R} R^2 \sin \phi \, dR \, d \phi \, d \theta = \\ & = k^2 \alpha^{\pm} \, 4 \pi \int_{R = 0}^{r} R \, e^{\pm k R} \, dR \ . \end{aligned}\end{split}\]

  the last integral can be evaluated with integration by parts

  \[\begin{split}\begin{aligned} \int_{R = 0}^{r} R \, e^{\pm k R} \, dR & = \left.\left[ \frac{1}{\pm k} e^{\pm k R } R \right]\right|_{R=0}^{r} \mp \frac{1}{k} \int_{R=0}^{r} e^{\pm k R} \, dR = \\ & = \frac{1}{\pm k} e^{\pm k r } r - \frac{1}{k^2} e^{\pm k R} + \frac{1}{k^2} = \\ \end{aligned}\end{split}\]

  Thus summing everything together,

  \[\begin{split}\begin{aligned} 1 & = \alpha^+ \left[ 4 \pi k^2 \left( \frac{r}{k} e^{k r} - \frac{1}{k^2} e^{kr} + \frac{1}{k^2} \right) - 4 \pi \left( k r - 1 \right) e^{kr} \right] + \alpha^- \left[ \dots \right] = \\ & = 4 \pi \left( \alpha^+ + \alpha^- \right) \ . \end{aligned}\end{split}\]

12.3. Wave equation

Wave equation general problem

\[\begin{split}\begin{cases} \dfrac{1}{c^2} \partial_{tt} \mathbf{u}(\mathbf{r},t) - \nabla^2 \mathbf{u}(\mathbf{r},t) = \mathbf{f}(\mathbf{r},t) \\ \text{+ b.c.} \\ \text{+ i.c.} \\ \end{cases}\end{split}\]

Green’s problem of the wave equation

\[\frac{1}{c^2} \partial_{tt} G(\mathbf{r},t;\mathbf{r}_0,t_0) - \nabla_{\mathbf{r}}^2 G(\mathbf{r},t;\mathbf{r}_0,t_0) = \delta(\mathbf{r}-\mathbf{r}_0) \delta(t-t_0)\]

Integration by parts

\[\begin{split}\begin{aligned} E(\mathbf{r}_{\alpha}, t_{\alpha}) \mathbf{u}(\mathbf{r}_{\alpha},t_{\alpha}) & = \int_{t \in T} \int_{\mathbf{r} \in V} \delta(t-t_{\alpha}) \delta(\mathbf{r}-\mathbf{r}_{\alpha}) \mathbf{u}(\mathbf{r},t) = \\ & = \int_{t \in T} \int_{\mathbf{r} \in V} \left\{ \frac{1}{c^2} \partial_{tt} G - \nabla^2_{\mathbf{r}} G \right\} \mathbf{u} = \\ & = \int_{t \in T} \int_{\mathbf{r} \in V} \left\{ \frac{1}{c^2} \left[ \partial_t \left( \mathbf{u} \partial_t G - G \partial_t \mathbf{u} \right) + G \partial_{tt} \mathbf{u} \right] - \nabla_{\mathbf{r}} \cdot \left( \nabla_{\mathbf{r}} G \, \mathbf{u} - G \nabla_{\mathbf{r}} \mathbf{u} \right) - G \, \nabla^2_{\mathbf{r}} \mathbf{u} \right\} = \\ & = \int_{\mathbf{r} \in V} \frac{1}{c^2} \left[ \mathbf{u}(\mathbf{r},t) \partial_t G(\mathbf{r},t; \mathbf{r}_{\alpha},t_{\alpha}) - G(\mathbf{r},t; \mathbf{r}_{\alpha},t_{\alpha}) \partial_t \mathbf{u}(\mathbf{r},t) \right] \bigg|_{t_0}^{t_1} + \\ & \quad + \int_{t \in T} \oint_{\mathbf{r} \in \partial V} \left\{ - \hat{\mathbf{n}}(\mathbf{r},t) \cdot \nabla_{\mathbf{r}} G(\mathbf{r},t; \mathbf{r}_{\alpha},t_{\alpha}) \, \mathbf{u}(\mathbf{r},t) + G(\mathbf{r},t; \mathbf{r}_{\alpha},t_{\alpha}) \, \hat{\mathbf{n}}(\mathbf{r},t) \cdot \nabla_{\mathbf{r}} \mathbf{u}(\mathbf{r},t) \right\} + \\ & \quad + \int_{t \in T} \int_{\mathbf{r} \in V} G(\mathbf{r},t; \mathbf{r}_{\alpha},t_{\alpha}) \underbrace{ \left\{ \frac{1}{c^2} \partial_{tt} \mathbf{u}(\mathbf{r},t) - \nabla^2_{\mathbf{r}} \mathbf{u}(\mathbf{r},t) \right\}}_{= \mathbf{f}(\mathbf{r},t)} \\ \end{aligned}\end{split}\]

\[\begin{aligned} \int_{t \in T} \int_{\mathbf{r} \in V} \frac{1}{4 \pi}\frac{\delta\left( t-t_{\alpha} + \frac{|\mathbf{r} - \mathbf{r}_{\alpha}|}{c} \right)}{|\mathbf{r} - \mathbf{r}_{\alpha}|} \, \mathbf{f}(\mathbf{r},t) & = \int_{\mathbf{r} \in V \cap B_{|\mathbf{r} - \mathbf{r}_{\alpha}| \le c (t_\alpha - t)}} \frac{1}{4 \pi |\mathbf{r} - \mathbf{r}_{\alpha}|} \mathbf{f}\left(\mathbf{r}, t_{\alpha} - \frac{|\mathbf{r}-\mathbf{r}_{\alpha}|}{c}\right) \end{aligned}\]