Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

4.3: Inhomogeneous Equations

Here we consider the initial value problem

\begin{eqnarray}
\label{inh1}
\Box u&=&w(x,t)\ \ \mbox{on}\ x\in\mathbb{R}^n,\ t\in\mathbb{R}^1\\
\label{inh2}
u(x,0)&=&f(x)\\
\label{inh3}
u_t(x,0)&=&g(x),
\end{eqnarray}

where \(\Box u:=u_{tt}-c^2\triangle u\). We assume \(f\in C^3\), \(g\in C^2\) and \(w\in C^1\), which are given.

Set \(u=u_1+u_2\), where \(u_1\) is a solution of problem (\ref{inh1})-(\ref{inh3}) with \(w:=0\) and \(u_2\) is the solution where \(f=0\) and \(g=0\) in (\ref{inh1})-(\ref{inh3}). Since we have explicit solutions \(u_1\) in the cases \(n=1\), \(n=2\) and \(n=3\), it remains to solve

\begin{eqnarray}
\label{duhu2gl}
\Box u&=&w(x,t)\ \ \mbox{on}\ x\in\mathbb{R}^n,\ t\in\mathbb{R}^1\\
\label{duhu2in1}
u(x,0)&=&0\\
\label{duhu2in2}
u_t(x,0)&=&0.
\end{eqnarray}

The following method is called Duhamel's principle which can be considered as a generalization of the method of variations of constants in the theory of ordinary differential equations.

To solve this problem, we make the ansatz

\begin{equation}
\label{duh1}
u(x,t)=\int_0^t\ v(x,t,s)\ ds,
\end{equation}
where \(v\) is a function satisfying
\begin{equation}
\label{duh2}
\Box v=0\ \ \mbox{for all}\ s
\end{equation}

and

\begin{equation}
\label{duh3}
v(x,s,s)=0.
\end{equation}

From ansatz (\ref{duh1}) and assumption (\ref{duh3}) we get

\begin{eqnarray}
u_t&=&v(x,t,t)+\int_0^t\ v_t(x,t,s)\ ds,\nonumber\\
\label{duh4}
&=&\int_0^t\ v_t(x,t,s).
\end{eqnarray}

It follows \(u_t(x,0)=0\). The initial condition \(u(x,t)=0\) is satisfied because of the ansatz (\ref{duh1}). From (\ref{duh4}) and ansatz (\ref{duh1}) we see that

\begin{eqnarray*}
u_{tt}&=&v_t(x,t,t)+\int_0^t\ v_{tt}(x,t,s)\ ds,\\
\triangle_x u&=&\int_0^t\ \triangle_x v(x,t,s)\ ds.
\end{eqnarray*}

Therefore, since \(u\) is an ansatz for (\ref{duhu2gl})-(\ref{duhu2in2}),

\begin{eqnarray*}
u_{tt}-c^2\triangle_x u&=&v_t(x,t,t)+\int_0^t(\Box v)(x,t,s)\ ds\\
&=&w(x,t).
\end{eqnarray*}

Thus necessarily \(v_t(x,t,t)=w(x,t)\), see (\ref{duh2}). We have seen that the ansatz provides a solution of (\ref{duhu2gl})-(\ref{duhu2in2}) if for all \(s\)

\begin{equation}
\label{duh5}
\Box v=0,\ \ v(x,s,s)=0,\ \ v_t(x,s,s)=w(x,s).
\end{equation}
Let \(v^*(x,t,s)\) be a solution of
\begin{equation}
\label{duh6}
\Box v=0,\ \ v(x,0,s)=0,\ \ v_t(x,0,s)=w(x,s),
\end{equation}
then

$$v(x,t,s):=v^*(x,t-s,s)$$

is a solution of (\ref{duh5}).
In the case \(n=3\), where \(v^*\) is given by, see Theorem 4.2,

$$v^*(x,t,s)=\frac{1}{4\pi c^2 t}\int_{\partial B_{ct}(x)}\ w(\xi,s)\ dS_\xi.$$

Then

\begin{eqnarray*}
v(x,t,s)&=&v^*(x,t-s,s)\\
  &=&\frac{1}{4\pi c^2 (t-s)}\int_{\partial B_{c(t-s)}(x)}\ w(\xi,s)\ dS_\xi.
\end{eqnarray*}

from ansatz (\ref{duh1}) it follows

\begin{eqnarray*}
u(x,t)&=&\int_0^t\ v(x,t,s)\ ds\\
&=&\frac{1}{4\pi c^2}\int_0^t\ \int_{\partial B_{c(t-s)}(x)}\ \frac{w(\xi,s)}{t-s}\ dS_\xi ds.
\end{eqnarray*}

Changing variables by \(\tau=c(t-s)\) yields

\begin{eqnarray*}
u(x,t)&=&\frac{1}{4\pi c^2}\int_0^{ct}\ \int_{\partial B_{\tau}(x)}\ \frac{w(\xi,t-\tau/c)}{\tau}\ dS_\xi d\tau\\
&=&\frac{1}{4\pi c^2} \int_{ B_{ct}(x)}\ \frac{w(\xi,t-r/c)}{r}\ d\xi,
\end{eqnarray*}

where \(r=|x-\xi|\).

Formulas for the cases \(n=1\) and \(n=2\) follow from formulas for the associated homogeneous equation with inhomogeneous initial values for these cases.

Theorem 4.4. The solution of

$$\Box u=w(x,t),\ \ u(x,0)=0,\ \ u_t(x,0)=0,$$

where \(w\in C^1\), is given by:

Case \(n=3\):

$$u(x,t)=\frac{1}{4\pi c^2} \int_{ B_{ct}(x)}\ \frac{w(\xi,t-r/c)}{r}\ d\xi,$$

where \(r=|x-\xi|\), \(x=(x_1,x_2,x_3)\), \(\xi=(\xi_1,\xi_2,\xi_3)\).

Case \(n=2\):

$$u(x,t)=\frac{1}{4\pi c}\int_0^t\ \left( \int_{ B_{c(t-\tau)}(x)}\ \frac{w(\xi,\tau)}{\sqrt{c^2(t-\tau)^2-r^2}}\ d\xi\right)\ d\tau,$$

\(x=(x_1,x_2)\), \(\xi=(\xi_1,\xi_2)\).

Case \(n=1\):

$$u(x,t)=\frac{1}{2c}\int_0^t\ \left(\int_{x-c(t-\tau)}^{x+c(t-\tau)}\ w(\xi,\tau)\ d\xi\right)\ d\tau.$$

Remark. The integrand on the right in formula for \(n=3\) is called retarded potential. The integrand is taken not at \(t\), it is taken at an earlier time \(t-r/c\).

Contributors