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# 7.5: Inhomogeneous Equation

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Here we consider solutions $$u\in C^2(\Omega)\cap C(\overline{\Omega})$$ of
\begin{eqnarray}
\label{DI1}
-\triangle u&=&f(x)\ \ \mbox{in}\ \Omega\\
\label{DI2}
u&=&0\ \ \ \mbox{on}\ \partial\Omega,
\end{eqnarray}
where $$f$$ is given.

We need the following lemma concerning volume potentials. We assume that $$\Omega$$ is bounded and sufficiently regular such that all the following integrals exist. See  for generalizations concerning these assumptions.

Let for $$x\in\mathbb{R}^n$$, $$n\ge3$$,
$$V(x)=\int_\Omega\ f(y)\frac{1}{|x-y|^{n-2}}\ dy$$
and set in the two-dimensional case
$$V(x)=\int_\Omega\ f(y)\ln\left(\frac{1}{|x-y|}\right)\ dy.$$
We recall that $$\omega_n=|\partial B_1(0)|$$.

Lemma.
(i) Assume $$f\in C(\Omega)$$. Then $$V\in C^1(\mathbb{R}^n)$$ and
\begin{eqnarray*}
V_{x_i}(x)&=& \int_\Omega\ f(y)\frac{\partial}{\partial x_i}\left(\frac{1}{|x-y|^{n-2}}\right)\ dy,\ \ \mbox{if}\ n\ge3,\\
V_{x_i}(x)&=&\int_\Omega\ f(y)\frac{\partial}{\partial x_i} \left(\ln\left(\frac{1}{|x-y|}\right)\right)\ dy\ \ \mbox{if}\ n=2.
\end{eqnarray*}

(ii) If $$f\in C^1(\Omega)$$, then $$V\in C^2(\Omega)$$ and
\begin{eqnarray*}
\triangle V&=&-(n-2)\omega_n f(x),\ x\in\Omega,\ n\ge 3\\
\triangle V&=&-2\pi f(x),\ x\in\Omega,\ n=2.
\end{eqnarray*}

Proof. To simplify the presentation, we consider the case $$n=3$$.
(i) The first assertion follows since we can change differentiation with integration since the differentiate integrand is weakly singular, see an exercise.

(ii) We will differentiate at $$x\in\Omega$$. Let $$B_\rho$$ be a fixed ball such that $$x\in B_\rho$$, $$\rho$$ sufficiently small such that $$B_\rho\subset\Omega$$. Then, according to (i)
and since we have the identity
$$\frac{\partial}{\partial x_i}\left(\frac{1}{|x-y|}\right)=-\frac{\partial}{\partial y_i}\left(\frac{1}{|x-y|}\right)$$
which implies that
$$f(y)\frac{\partial}{\partial x_i}\left(\frac{1}{|x-y|}\right)=-\frac{\partial}{\partial y_i}\left(f(y)\frac{1}{|x-y|}\right)+f_{y_i}(y)\frac{1}{|x-y|},$$
we obtain
\begin{eqnarray*}
V_{x_i}(x)&=&\int_\Omega\ f(y)\frac{\partial}{\partial x_i}\left(\frac{1}{|x-y|}\right)\ dy\\
&=&\int_{\Omega\setminus B_\rho}\ f(y)\frac{\partial}{\partial x_i}\left(\frac{1}{|x-y|}\right)\ dy+\int_{B_\rho}\ f(y)\frac{\partial}{\partial x_i}\left(\frac{1}{|x-y|}\right)\ dy\\
&=&\int_{\Omega\setminus B_\rho}\ f(y)\frac{\partial}{\partial x_i}\left(\frac{1}{|x-y|}\right)\ dy\\
&&+\int_{B_\rho}\ \left(-\frac{\partial}{\partial y_i}\left(f(y)\frac{1}{|x-y|}\right)+f_{y_i}(y)\frac{1}{|x-y|}\right)\ dy\\
&=&\int_{\Omega\setminus B_\rho}\ f(y)\frac{\partial}{\partial x_i}\left(\frac{1}{|x-y|}\right)\ dy\\
&&+\int_{B_\rho}\ f_{y_i}(y)\frac{1}{|x-y|}\ dy-\int_{\partial B_\rho}\ f(y)\frac{1}{|x-y|}n_i\ dS_y,
\end{eqnarray*}
where $$n$$ is the exterior unit normal at $$\partial B_\rho$$. It follows that the first and second integral is in $$C^1(\Omega)$$. The second integral is also in $$C^1(\Omega)$$ according to (i) and since $$f\in C^1(\Omega)$$ by assumption.

Because of $$\triangle_x(|x-y|^{-1})=0,\ x\not=y$$, it follows
\begin{eqnarray*}
\triangle V&=&\int_{B_\rho}\ \sum_{i=1}^n f_{y_i}(y)\frac{\partial}{\partial x_i}\left(\frac{1}{|x-y|}\right)\ dy\\
&&\ -\int_{\partial B_\rho}\ f(y)\sum_{i=1}^n \frac{\partial}{\partial x_i}\left(\frac{1}{|x-y|}\right)n_i\ dS_y.
\end{eqnarray*}
Now we choose for $$B_\rho$$ a ball with the center at $$x$$, then
$$\triangle V=I_1+I_2,$$
where
\begin{eqnarray*}
I_1&=&\int_{B_\rho(x)}\ \sum_{i=1}^n f_{y_i}(y)\frac{y_i-x_i}{|x-y|^3}\ dy\\
I_2&=&-\int_{\partial B_\rho(x)}\ f(y)\frac{1}{\rho^2} \ dS_y.
\end{eqnarray*}
We recall that $$n\cdot(y-x)=\rho$$ if $$y\in\partial B_\rho(x)$$. It is $$I_1=O(\rho)$$ as $$\rho\to 0$$ and for $$I_2$$ we obtain from the mean value theorem of the integral calculus that for a $$\overline{y}\in\partial B_\rho(x)$$
\begin{eqnarray*}
I_2&=&-\frac{1}{\rho^2}f(\overline{y})\int_{\partial B_\rho(x)}\ dS_y\\
&=&-\omega_nf(\overline{y}),
\end{eqnarray*}
which implies that $$\lim_{\rho\to0} I_2=-\omega_nf(x)$$.

$$\Box$$

In the following we assume that Green's function exists for the domain $$\Omega$$, which is the case if $$\Omega$$ is a ball.

Theorem 7.3. Assume $$f\in C^1(\Omega)\cap C(\overline{\Omega})$$. Then
$$u(x)=\int_\Omega\ G(x,y)f(y)\ dy$$
is the solution of the inhomogeneous problem
(\ref{DI1}), (\ref{DI2}).

Proof. For simplicity of the presentation let $$n=3$$. We will show that
$$u(x):=\int_\Omega\ G(x,y) f(y)\ dy$$ is a solution of (7.3.1.1), (7.3.1.2). Since
$$G(x,y)=\frac{1}{4\pi|x-y|}+\phi(x,y),$$
where $$\phi$$ is a potential function with respect to $$x$$ or $$y$$, we obtain from the above lemma that
\begin{eqnarray*}
\triangle u&=&\frac{1}{4\pi}\triangle\int_\Omega\ f(y)\frac{1}{|x-y|}\ dy+\int_\Omega\triangle_x\phi (x,y)f(y)\ dy\\
&=&-f(x),
\end{eqnarray*}
where $$x\in\Omega$$. It remains to show that $$u$$ achieves its boundary values. That is, for fixed $$x_0\in\partial\Omega$$ we will prove that
$$\lim_{x\to x_0,\ x\in\Omega} u(x)=0.$$
Set
$$u(x)=I_1+I_2,$$
where
\begin{eqnarray*}
I_1(x)&=&\int_{\Omega\setminus B_\rho(x_0)}\ G(x,y)f(y)\ dy,\\
I_2(x)&=&\int_{\Omega\cap B_\rho(x_0)}\ G(x,y)f(y)\ dy.
\end{eqnarray*}
Let $$M=\max_{\overline{\Omega}}|f(x)|$$. Since
$$G(x,y)= \frac{1}{4\pi}\frac{1}{|x-y|}+\phi(x,y),$$
we obtain, if $$x\in B_\rho(x_0)\cap\Omega$$,
\begin{eqnarray*}
|I_2|&\le&\frac{M}{4\pi}\int_{\Omega\cap B_\rho(x_0)}\ \frac{dy}{|x-y|}+O(\rho^2)\\
&\le&\frac{M}{4\pi}\int_{B_{2\rho(x)}}\ \frac{dy}{|x-y|}+O(\rho^2)\\
&=&O(\rho^2)
\end{eqnarray*}
as $$\rho\to0$$. Consequently for given $$\epsilon$$ there is a $$\rho_0=\rho_0(\epsilon)>0$$ such that
$$|I_2|<\frac{\epsilon}{2}\ \ \mbox{for all}\ \ 0<\rho\le\rho_0.$$
For each fixed $$\rho$$, $$0<\rho\le\rho_0$$, we have
$$\lim_{x\to x_0,\ x\in\Omega} I_1(x)=0$$
since $$G(x_0,y)=0$$ if $$y\in\Omega\setminus B_\rho(x_0)$$ and $$G(x,y)$$ is uniformly continuous in $$x\in B_{\rho/2}(x_0)\cap\Omega$$ and $$y\in\Omega\setminus B_\rho(x_0)$$, see Figure 7.5.1. Figure 7.5.1: Proof of Theorem 7.3

$$\Box$$

Remark. For the proof of (ii) in the above lemma it is sufficient to assume that $$f$$ is H\"older continuous. More precisely, let
$$f\in C^{\lambda}(\Omega)$$, $$0<\lambda<1$$, then $$V\in C^{2,\lambda}(\Omega)$$, see for instance .

## Contributors

• Integrated by Justin Marshall.