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Mathematics LibreTexts

4.2.2: Case n=2

  • Page ID
    2178
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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.

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