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# 4.2.2: Case n=2

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Consider the initial value problem

\begin{eqnarray}
\label{n2dgl}\tag{4.2.2.1}
v_{xx}+v_{yy}&=&c^{-2}v_{tt}\\
\label{initial1} \tag{4.2.2.2}
v(x,y,0)&=&f(x,y)\\
\label{initial2} \tag{4.2.2.3}
v_t(x,y,0)&=&g(x,y),
\end{eqnarray}

where $$f\in C^3,\ g\in C^2$$.

Using the formula for the solution of the three-dimensional initial value problem we will derive a formula for the two-dimensional case. The following consideration is called Hadamard's method of decent.

Let $$v(x,y,t)$$ be a solution of (\ref{n2dgl})-(\ref{initial2}), then

$$u(x,y,z,t):=v(x,y,t)$$

is a solution of the three-dimensional initial value problem with initial data $$f(x,y)$$, $$g(x,y)$$, independent of $$z$$, since $$u$$ satisfies (\ref{n2dgl})-(\ref{initial2}). Hence, since $$u(x,y,z,t)=u(x,y,0,t)+u_z(x,y,\delta z,t)z$$, $$0<\delta<1$$, and $$u_z=0$$, we have

$$v(x,y,t)=u(x,y,0,t).$$

Poisson's formula in the three-dimensional case implies

\begin{eqnarray}
v(x,y,t)&=&\frac{1}{4\pi c^2}\frac{\partial}{\partial t}\left(\frac{1}{t}\int_{\partial B_{ct}(x,y,0)}\ f(\xi,\eta)\ dS\right)\nonumber\\
\label{poissonhilf1} \tag{4.2.2.4}
&&+\frac{1}{4\pi c^2 t} \int_{\partial B_{ct}(x,y,0)}\ g(\xi,\eta)\ dS.
\end{eqnarray} Figure 4.2.2.1: Domains of integration

The integrands are independent on $$\zeta$$. The surface $$S$$ is defined by $$\chi(\xi,\eta,\zeta):=(\xi-x)^2+(\eta-y)^2+\zeta^2-c^2 t^2=0$$. Then the exterior normal $$n$$ at $$S$$ is $$n=\nabla\chi/|\nabla\chi|$$ and the surface element is given by $$dS=(1/|n_3|)d\xi d\eta$$, where the third coordinate of $$n$$ is

$$n_3=\pm\frac{\sqrt{c^2 t^2-(\xi-x)^2-(\eta-y)^2}}{ct}.$$

The positive sign applies on $$S^+$$, where $$\zeta>0$$ and the sign is negative on $$S^-$$ where $$\zeta<0$$, see Figure 4.2.2.1. We have $$S=S^+\cup\overline{S^-}$$.

Set $$\rho=\sqrt{(\xi-x)^2+(\eta-y)^2}$$. Then it follows from (\ref{poissonhilf1})

Theorem 4.3. The solution of the Cauchy initial value problem (\ref{n2dgl})-(\ref{initial2}) is given by

\begin{eqnarray*}
v(x,y,t)&=&\frac{1}{2\pi c}\frac{\partial}{\partial t}\int_{B_{ct}(x,y)}\ \frac{f(\xi,\eta)}{\sqrt{c^2 t^2-\rho^2}}\ d\xi d\eta\\
&&+\frac{1}{2\pi c}\int_{B_{ct}(x,y)}\ \frac{g(\xi,\eta)}{\sqrt{c^2 t^2-\rho^2}}\ d\xi d\eta. \end{eqnarray*} Figure 4.2.2.2: Interval of dependence, case $$n=2$$

Corollary. In contrast to the three dimensional case, the domain of dependence is here the disk $$B_{ct_o}(x_0,y_0)$$ and not the boundary only. Therefore, see formula of Theorem 4.3, if $$f,\ g$$ have supports in a compact domain $$D\subset\mathbb{R}^2$$, then these functions have influence on the value $$v(x,y,t)$$ for all time $$t>T$$, $$T$$ sufficiently large.

## Contributors

• Integrated by Justin Marshall.