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Mathematics LibreTexts

7.5: Inhomogeneous Equation

  • Page ID
    2166
  • [ "article:topic" ]

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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 [6] 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 [9].

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