Homework 07

1 Green’s function for Poisson’s equation

Use Fourier analysis to compute a Green’s function for Poisson’s equation on $\mathbb{R}^3$, satisfying

\[ \left(\partial_x^2+\partial_y^2+\partial_z^2\right) G(x, y, z)=\delta^{(3)}(x, y, z) \]

Prove there is a unique such Green’s function which goes to zero at infinity (hint: use Liouville’s theorem for harmonic functions).

Find the Green’s function but now for functions on the unit ball around the origin, with Dirichlet boundary conditions $\phi(x, y, z)=0$ for $x^2+y^2+z^2=1$. Hint: Schwartz reflection principle/method of images.

1.1 Fourier Analysis

Let $\hat{G}(k_x, k_y, k_z)$ be the Fourier transform of $G(x, y, z)$. After Fourier transform, the equation becomes:

\[ -(k_x^2 + k_y^2 + k_z^2) \hat{G}(k_x, k_y, k_z) = 1 \]

since the Fourier transform of $\delta^{(3)}(x, y, z)$ is 1.

Thus, the solution in Fourier Space:

\[ \hat{G}(k_x, k_y, k_z) = -\frac{1}{k_x^2 + k_y^2 + k_z^2} \]

To find $G(x, y, z)$, perform the inverse Fourier transform:

\[ G(x, y, z) = \int_{\mathbb{R}^3} -\frac{e^{i(k_x x + k_y y + k_z z)}}{k_x^2 + k_y^2 + k_z^2} \, dk_x \, dk_y \, dk_z \]

This integral yields $G(x, y, z) = \frac{-1}{4\pi\sqrt{x^2 + y^2 + z^2}}$.

1.2 Uniqueness of the Green’s Function

Liouville’s theorem states that a bounded harmonic function on $\mathbb{R}^n$ is constant. Given any two Green’s functions $G_1$ and $G_2$ that vanish at infinity, their difference $G_1 - G_2$ is harmonic (satisfies Laplace’s equation) and bounded. By Liouville’s theorem, $G_1 - G_2$ is constant, and since both go to zero at infinity, this constant must be zero. Thus, $G_1 = G_2$, proving uniqueness.

1.3 Green’s Function on the Unit Ball with Dirichlet Boundary Conditions

To find a function $G(x, y, z; x', y', z')$ that satisfies:

$(\partial_x^2 + \partial_y^2 + \partial_z^2) G(x, y, z; x', y', z') = -\delta(x - x', y - y', z - z')$ within the unit ball $x^2 + y^2 + z^2 < 1$.
$G(x, y, z; x', y', z') = 0$ for $x^2 + y^2 + z^2 = 1$, enforcing the Dirichlet boundary conditions.

We place an “image” point source at $(x'', y'', z'')$ where $(x'', y'', z'') = \frac{(x', y', z')}{|x'|^2 + |y'|^2 + |z'|^2}$ outside the unit ball in such a way that the combined effect of the real source and the image source satisfies the boundary conditions.

The Green’s function then is a combination of the solution from both the real and the image source

\[ G(x, y, z; x', y', z') = \frac{1}{4\pi\sqrt{(x - x')^2 + (y - y')^2 + (z - z')^2}} - \frac{1}{4\pi\sqrt{(x - x'')^2 + (y - y'')^2 + (z - z'')^2}} \]

More generally, we define the point $x^* = \frac{x}{|x|^2}$ dual to $x$. Therefore, a Green’s function for $B^n(0, 1)$ is given by $G(x, y) = Φ(y − x) − Φ(|x|(y − x^*))$.

2 Green’s function for the heat equation

Find the Green’s function for the heat equation

\[ \partial_t u=\partial_x^2 u \]

by Fourier analysis.

We want to find a Green’s function $G(x, t; x', t')$ that satisfies the following properties:

$\partial_t G = \partial_x^2 G$
$G(x, t; x', t')$ behaves like a delta function as $t \to t'^+$, i.e., $G(x, t; x', t') \to \delta(x - x')$ as $t \to t'^+$.

By taking the Fourier transform of $G$ with respect to $x$, the heat equation in Fourier space becomes

\[ \partial_t \hat{G} = -k^2 \hat{G} \]

This is a first-order linear differential equation in $t$, the solution is

\[ \hat{G}(k, t; x', t') = A(k, x', t') e^{-k^2 (t - t')} \]

where $A(k, x', t')$ is to be determined.

Now we apply the initial condition. The initial condition is that $G$ approaches $\delta(x - x')$ as $t \to t'^+$. In Fourier space, this translates to $\hat{G}(k, t; x', t') \to e^{-ik(x - x')}$ as $t \to t'^+$. Thus

\[ A(k, x', t') = e^{-ikx'} \]

Then take the inverse Fourier transform of $\hat{G}(k, t; x', t')$ to get back to the spatial domain:

\[ G(x, t; x', t') = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ikx'} e^{-ikx} e^{-k^2 (t - t')} dk \]

This integral yields a function of the form:

\[ G(x, t; x', t') = \frac{1}{\sqrt{4\pi (t - t')}} e^{-\frac{(x - x')^2}{4(t - t')}} \]

for $t > t'$.

So the Green’s function for the heat equation is:

\[ G(x, t; x', t') = \begin{cases} \frac{1}{\sqrt{4\pi (t - t')}} e^{-\frac{(x - x')^2}{4(t - t')}}, & t > t' \\ 0, & \text{otherwise} \end{cases} \]

3 Resolvent $R(z, A)=(z \mathbb{1}-A)^{-1}$

Given a finite dimensional complex matrix $A$, we can define the resolvent

\[ R(z, A)=(z \mathbb{1}-A)^{-1}=\frac{1}{z-A}, \]

where $\mathbb{1}$ is the identity matrix. This equation makes sense whenever $z$ is not an eigenvalue of $A$, and so we can consider $R(z, A)$ as a meromorphic, matrix-valued function of $z$, with poles at the eigenvalues of $A$. For these problems, assume $A$ is Hermitian:

Show that

\[ R(z, A)=\frac{P_\lambda}{z-\lambda}, \]

where $P_\lambda$ is the projector onto the eigenspace of $\lambda$ eigenvectors.

Suppose $w$ is orthogonal to the kernel of $A$. Prove that

\[ v=\int_C \frac{d z}{2 \pi i} \frac{R(z, A)}{z} w \]

solves the equation

\[ A v=w, \]

where $C$ is a contour winding once around $z=0$ and not enclosing any other eigenvalues.

Show the same as in (2) for a contour enclosing all eigenvalues except for 0.

3.1 Expression for $R(z, A)$

Given a Hermitian matrix $A$, its eigenvalues are real, and it can be diagonalized. Let $\lambda$ be an eigenvalue of $A$ and $P_\lambda$ the projector onto the eigenspace of $\lambda$. We have spectral decomposition

\[ A = \sum_{\lambda} \lambda P_\lambda \]

the expression for $R(z, A)$ can be simplified:

\[ R(z, A) = (z\mathbb{1} - A)^{-1} = \left( z \sum_\lambda P_\lambda - \sum_\lambda \lambda P_\lambda \right)^{-1} \]

Because the projectors $P_\lambda$ are orthogonal and sum to the identity: $R(z, A) = \sum_\lambda (zP_\lambda - \lambda P_\lambda)^{-1}$.

Thus, $R(z, A) = \sum_\lambda \frac{P_\lambda}{z - \lambda}$.

3.2 Solving $A v = w$

Substitute the expression for $R(z, A)$ into the integral:

\[ v = \int_C \frac{d z}{2 \pi i} \frac{\sum_\lambda \frac{P_\lambda}{z - \lambda}}{z} w \]

Evaluate it using the residue theorem. Since $C$ winds around $z = 0$ and does not enclose any other eigenvalues, the only contribution comes from the pole at $z = 0$.

The residue at $z = 0$ gives $P_0 w$. So $A v = A P_0 w = w$

3.3 Contour Enclosing All Eigenvalues Except 0

Use the same integral expression for $v$. But now the integral will pick up residues from all poles $\lambda \neq 0$ within the contour $C$.

For each $\lambda \neq 0$, the residue at $z = \lambda$ is $\frac{P_\lambda w}{\lambda}$.

Summing up the residues, $v = \sum_{\lambda \neq 0} \frac{P_\lambda w}{\lambda}$.

Multiplying both sides by $A$ (and using $AP_\lambda = \lambda P_\lambda$), we get $A v = \sum_{\lambda \neq 0} P_\lambda w = w - P_0 w$.

Since $w$ is orthogonal to the kernel of $A$, $P_0 w = 0$. Therefore, $A v = w$.

--- title: Homework 07 format: html: default typst: output-file: phys231_hw07 number-sections: true --- ## Green's function for Poisson's equation > * Use Fourier analysis to compute a Green's function for Poisson's equation on $\mathbb{R}^3$, satisfying > > $$ > \left(\partial_x^2+\partial_y^2+\partial_z^2\right) G(x, y, z)=\delta^{(3)}(x, y, z) > $$ > * Prove there is a unique such Green's function which goes to zero at infinity (hint: use Liouville's theorem for harmonic functions). > * Find the Green's function but now for functions on the unit ball around the origin, with Dirichlet boundary conditions $\phi(x, y, z)=0$ for $x^2+y^2+z^2=1$. Hint: Schwartz reflection principle/method of images. ### Fourier Analysis Let $\hat{G}(k_x, k_y, k_z)$ be the Fourier transform of $G(x, y, z)$. After Fourier transform, the equation becomes: $$ -(k_x^2 + k_y^2 + k_z^2) \hat{G}(k_x, k_y, k_z) = 1 $$ since the Fourier transform of $\delta^{(3)}(x, y, z)$ is 1. Thus, the solution in Fourier Space: $$ \hat{G}(k_x, k_y, k_z) = -\frac{1}{k_x^2 + k_y^2 + k_z^2} $$ To find $G(x, y, z)$, perform the inverse Fourier transform: $$ G(x, y, z) = \int_{\mathbb{R}^3} -\frac{e^{i(k_x x + k_y y + k_z z)}}{k_x^2 + k_y^2 + k_z^2} \, dk_x \, dk_y \, dk_z $$ This integral yields $G(x, y, z) = \frac{-1}{4\pi\sqrt{x^2 + y^2 + z^2}}$. ### Uniqueness of the Green's Function Liouville's theorem states that a bounded harmonic function on $\mathbb{R}^n$ is constant. Given any two Green's functions $G_1$ and $G_2$ that vanish at infinity, their difference $G_1 - G_2$ is harmonic (satisfies Laplace's equation) and bounded. By Liouville's theorem, $G_1 - G_2$ is constant, and since both go to zero at infinity, this constant must be zero. Thus, $G_1 = G_2$, proving uniqueness. ### Green's Function on the Unit Ball with Dirichlet Boundary Conditions To find a function $G(x, y, z; x', y', z')$ that satisfies: 1. $(\partial_x^2 + \partial_y^2 + \partial_z^2) G(x, y, z; x', y', z') = -\delta(x - x', y - y', z - z')$ within the unit ball $x^2 + y^2 + z^2 < 1$. 2. $G(x, y, z; x', y', z') = 0$ for $x^2 + y^2 + z^2 = 1$, enforcing the Dirichlet boundary conditions. We place an "image" point source at $(x'', y'', z'')$ where $(x'', y'', z'') = \frac{(x', y', z')}{|x'|^2 + |y'|^2 + |z'|^2}$ outside the unit ball in such a way that the combined effect of the real source and the image source satisfies the boundary conditions. The Green's function then is a combination of the solution from both the real and the image source $$ G(x, y, z; x', y', z') = \frac{1}{4\pi\sqrt{(x - x')^2 + (y - y')^2 + (z - z')^2}} - \frac{1}{4\pi\sqrt{(x - x'')^2 + (y - y'')^2 + (z - z'')^2}} $$ More generally, we define the point $x^* = \frac{x}{|x|^2}$ dual to $x$. Therefore, a Green’s function for $B^n(0, 1)$ is given by $G(x, y) = Φ(y − x) − Φ(|x|(y − x^*))$. ## Green's function for the heat equation > Find the Green's function for the heat equation > > $$ > \partial_t u=\partial_x^2 u > $$ > > by Fourier analysis. We want to find a Green's function $G(x, t; x', t')$ that satisfies the following properties: 1. $\partial_t G = \partial_x^2 G$ 2. $G(x, t; x', t')$ behaves like a delta function as $t \to t'^+$, i.e., $G(x, t; x', t') \to \delta(x - x')$ as $t \to t'^+$. By taking the Fourier transform of $G$ with respect to $x$, the heat equation in Fourier space becomes $$ \partial_t \hat{G} = -k^2 \hat{G} $$ This is a first-order linear differential equation in $t$, the solution is $$ \hat{G}(k, t; x', t') = A(k, x', t') e^{-k^2 (t - t')} $$ where $A(k, x', t')$ is to be determined. Now we apply the initial condition. The initial condition is that $G$ approaches $\delta(x - x')$ as $t \to t'^+$. In Fourier space, this translates to $\hat{G}(k, t; x', t') \to e^{-ik(x - x')}$ as $t \to t'^+$. Thus $$ A(k, x', t') = e^{-ikx'} $$ Then take the inverse Fourier transform of $\hat{G}(k, t; x', t')$ to get back to the spatial domain: $$ G(x, t; x', t') = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ikx'} e^{-ikx} e^{-k^2 (t - t')} dk $$ This integral yields a function of the form: $$ G(x, t; x', t') = \frac{1}{\sqrt{4\pi (t - t')}} e^{-\frac{(x - x')^2}{4(t - t')}} $$ for $t > t'$. So the Green's function for the heat equation is: $$ G(x, t; x', t') = \begin{cases} \frac{1}{\sqrt{4\pi (t - t')}} e^{-\frac{(x - x')^2}{4(t - t')}}, & t > t' \\ 0, & \text{otherwise} \end{cases} $$ ## Resolvent $R(z, A)=(z \mathbb{1}-A)^{-1}$ > Given a finite dimensional complex matrix $A$, we can define the resolvent > > $$ > R(z, A)=(z \mathbb{1}-A)^{-1}=\frac{1}{z-A}, > $$ > > where $\mathbb{1}$ is the identity matrix. This equation makes sense whenever $z$ is not an eigenvalue of $A$, and so we can consider $R(z, A)$ as a meromorphic, matrix-valued function of $z$, with poles at the eigenvalues of $A$. For these problems, assume $A$ is Hermitian: > > 1. Show that > > $$ > R(z, A)=\frac{P_\lambda}{z-\lambda}, > $$ > > where $P_\lambda$ is the projector onto the eigenspace of $\lambda$ eigenvectors. > 2. Suppose $w$ is orthogonal to the kernel of $A$. Prove that > > $$ > v=\int_C \frac{d z}{2 \pi i} \frac{R(z, A)}{z} w > $$ > > solves the equation > > $$ > A v=w, > $$ > > where $C$ is a contour winding once around $z=0$ and not enclosing any other eigenvalues. > 3. Show the same as in (2) for a contour enclosing all eigenvalues except for 0. ### Expression for $R(z, A)$ Given a Hermitian matrix $A$, its eigenvalues are real, and it can be diagonalized. Let $\lambda$ be an eigenvalue of $A$ and $P_\lambda$ the projector onto the eigenspace of $\lambda$. We have spectral decomposition $$ A = \sum_{\lambda} \lambda P_\lambda $$ the expression for $R(z, A)$ can be simplified: $$ R(z, A) = (z\mathbb{1} - A)^{-1} = \left( z \sum_\lambda P_\lambda - \sum_\lambda \lambda P_\lambda \right)^{-1} $$ Because the projectors $P_\lambda$ are orthogonal and sum to the identity: $R(z, A) = \sum_\lambda (zP_\lambda - \lambda P_\lambda)^{-1}$. * Thus, $R(z, A) = \sum_\lambda \frac{P_\lambda}{z - \lambda}$. ### Solving $A v = w$ Substitute the expression for $R(z, A)$ into the integral: $$ v = \int_C \frac{d z}{2 \pi i} \frac{\sum_\lambda \frac{P_\lambda}{z - \lambda}}{z} w $$ Evaluate it using the residue theorem. Since $C$ winds around $z = 0$ and does not enclose any other eigenvalues, the only contribution comes from the pole at $z = 0$. The residue at $z = 0$ gives $P_0 w$. So $A v = A P_0 w = w$ ### Contour Enclosing All Eigenvalues Except 0 Use the same integral expression for $v$. But now the integral will pick up residues from all poles $\lambda \neq 0$ within the contour $C$. For each $\lambda \neq 0$, the residue at $z = \lambda$ is $\frac{P_\lambda w}{\lambda}$. Summing up the residues, $v = \sum_{\lambda \neq 0} \frac{P_\lambda w}{\lambda}$. Multiplying both sides by $A$ (and using $AP_\lambda = \lambda P_\lambda$), we get $A v = \sum_{\lambda \neq 0} P_\lambda w = w - P_0 w$. Since $w$ is orthogonal to the kernel of $A$, $P_0 w = 0$. Therefore, $A v = w$.