Recently I am thinking the solution of the following 3-dimensional wave equation with Cauchy data:
\begin{align*}&\frac{\partial^2 u}{\partial t^2}=4 \Big(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}\Big),(x,y,z)\in\mathbf{R}^3, t>0,\\
&u(x,y,z,0)=\phi(x,y,z):=(3x-y+z)e^{3x-y+z}, u_t(x,y,z,0)=0, (x,y,z)\in\mathbb{R}^3.
\end{align*}
Someone has found the solution of this Cauchy problem, that is:
\begin{gather}
u(x,y,z,t)=\frac{\left( 3\,x-y+z+2\,\sqrt {11}t \right) {{\rm e}^{3\,x-y+z+2\,
\sqrt {11}t}}+\left( 3\,x-y+z-2\,\sqrt {11}t \right) {{\rm e}^{3
\,x-y+z-2\,\sqrt {11}t}}}{2}
\end{gather}
But I do not know how did he(she) find it. Then I tried my best to calculate by myself. Actually I have tried many times to figure out all the calculations of derivation of the above solution, but the result is of no use. I have tried of using Kirchhoff's formula:
\begin{gather*}
u(x,y,z,t)=\frac{\partial}{\partial t}\bigg(\frac{1}{8\pi}\int_{\partial B((x,y,z),2t)}\frac{\phi(\xi,\eta,\zeta)}{2t} dS(\xi,\eta,\zeta)\bigg),
\end{gather*}
but the surface integral is too hard to estimate for me. I have also tried to use Gauss formula by using the fact $\Delta_3 e^{3x-y+z}=11e^{3x-y+z}$, but the result is very hopeless.
Can anyone help me to the derivation of the above solution formula? Thanks a lot.
Best Answer
You are given a very special initial data. Let $f(x,y,z) = 3x - y + z$, note that this is a linear function on $\mathbb{R}^3$. Your initial data is
$$ \phi = f e^f $$
Why is linearity important? that means we can change coordinates with isometries.
Let $v_1 = \frac{1}{\sqrt{11}}\begin{pmatrix} 3 \\ -1 \\ 1\end{pmatrix}$ and complete this to an orthonormal basis $\{v_1, v_2, v_3\}$. Let $O$ be the orthogonal matrix whose column vectors are $v_1, v_2, v_3$. Now define the coordinates $x',y',z'$ by
$$ \begin{pmatrix} x' \\ y' \\ z' \end{pmatrix} = O \begin{pmatrix} x \\ y \\ z\end{pmatrix} $$
we have that in this coordinate, the function $f$ can be written as $f(x',y',z') = \sqrt{11} x'$.
Since this change of coordinates is orthogonal in $\mathbb{R}^3$, it preserves the Laplace operator. So in this new coordinate system your function $u$ still solves
$$ \frac{\partial^2}{\partial t^2} u = 4 \left( \frac{\partial^2}{\partial x'^2} u + \frac{\partial^2}{\partial y'^2} u + \frac{\partial^2}{\partial z'^2} u \right) $$
But now you are given initial data $\phi$ which is independent of $y'$ and $z'$. This implies that the solution will also be independent of $y'$ and $z'$. This means that $u$ in fact solves
$$ \partial^2_t u = 4 \partial^2_{x'} u $$
a one dimensional wave equation with initial data $\phi(x',y',z') = \phi(x') = \sqrt{11}x' e^{\sqrt{11} x'} $. Now you can use the one-dimensional representation formula for the solution of the wave equation to get precisely that
$$ u(t,x') = \frac12\phi( x' + 2 t) + \frac12\phi(x' - 2t) $$
which after you change the variables back to $x,y,z$ is precisely what you are looking for.