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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.

doraemonpaul
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nuage
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1 Answers1

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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.

Willie Wong
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  • Thank you, Willie Wong, your answer is very interesting. And it is very helpful. Can it be possible to use Kirchhoff's formula directly, without using your interesting method of change of coordinates? – nuage May 03 '13 at 10:24
  • Well, yes. But you still have to use the fact that $u$ will be constant on the planes for which $3x - y + z$ is constant. Physically speaking: you are given a plane-wave initial data, so you will get a plane-wave final solution. Once you identify the direction of the plane-wave, the Kirchhoff integral in those directions simplify. – Willie Wong May 03 '13 at 11:16