2
$\begingroup$

Please help me in solving: $$a^2z+\frac{\partial^2z}{\partial x^2}-\frac{\partial^2 z}{\partial y^2}=0$$ ($a$ is a constant)

I plugged this in Wolfram Alpha and it outputs that this is a second order linear partial differential equation but unfortunately, I have never solved these type of equations before. I came across this equation while solving some other problem. I tried to look up the methods to solve such equations but they seem too advanced to me. :( Can somebody please please solve this? I am not even sure if it is possible to solve this equation.

Any help is appreciated. Thanks!

$\endgroup$
2
  • $\begingroup$ $\partial^2$ or $\partial$ in the second derivative? $\endgroup$ Commented Jun 26, 2014 at 14:29
  • $\begingroup$ @Riccardo.Alestra: Sorry about that. I have fixed it. :) $\endgroup$ Commented Jun 26, 2014 at 14:29

2 Answers 2

Reset to default
2
$\begingroup$

Suggestion (for a domain bounded in $x$):

Try separation of variables, $z(x,y) = X(x)Y(y)$, leading to two ODE eigenvalue problems

$$X''(x) = cX(x) \\\ Y''(y) =(c+\alpha^2)Y(y)$$

For a semi-infinite domain ($0 \leq x<\infty,0 \leq y<\infty$) we have a wave equation

$$z_{yy}-z_{xx}= a^2z.$$

If initial/boundary conditions are correctly specified -- making the problem well-posed --then a method for solution is to take the Laplace transform with respect to $x$ and solve the resulting second-order ODE for the tranform.

$\endgroup$
7
  • $\begingroup$ that's the canonical way, but I doubt the OP knows about it. Besides the fact that, without initial conditions, separation of variables will not take you too far (in the sense that you will not be able to write "a" solution). $\endgroup$ Commented Jun 26, 2014 at 14:39
  • $\begingroup$ Yes - it depend on boundary conditions -- also is c positive or negative $\endgroup$ Commented Jun 26, 2014 at 14:41
  • $\begingroup$ @RRL: I am glad to know that you tried to help but I unfortunately cannot solve such equations. I have some more data. I thought that the data was not necessary to solve the equation. It is as follows: $z(0,y)=0$, $z(x,\infty)=0$. The partial derivative of $z$ wrt $x$ is zero at $x=0$ and also when $y$ tends to infinity. Similar condition is with the partial derivative of $z$ wrt $y$. Moreover, these conditions follow for second derivatives too. Do you need more information? Please let me know. Thank you once again. :) $\endgroup$ Commented Jun 26, 2014 at 14:46
  • $\begingroup$ You have a wave equation $z_{yy}-z_{xx}=a^2z$ on a semi-infinite domain. With well-posed conditions you can solve by taking the Laplace transform with respect to $x$ and get an ODE for the transform. I did not know originally that the domain was unbounded. $\endgroup$ Commented Jun 26, 2014 at 15:51
  • 2
    $\begingroup$ You're welcome. With PDEs the conditions are critical -- otherwise we can say $z=0$ is a solution among many. $\endgroup$ Commented Jun 26, 2014 at 16:03
0
$\begingroup$

Similar to Solve initial value problem for $u_{tt} - u_{xx} - u = 0$ using characteristics,

Let $\begin{cases}p=\dfrac{x+y}{2}\\q=\dfrac{x-y}{2}\end{cases}$ ,

Then $\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial z}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)$

$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial z}{\partial q}\dfrac{\partial q}{\partial y}=\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)$

$\dfrac{\partial^2z}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)\right)=\dfrac{\partial}{\partial p}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}+\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial q}{\partial x}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}$

$\dfrac{\partial^2z}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)\right)=\dfrac{\partial}{\partial p}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial p}{\partial y}+\dfrac{\partial}{\partial q}\left(\dfrac{1}{2}\left(\dfrac{\partial z}{\partial p}-\dfrac{\partial z}{\partial q}\right)\right)\dfrac{\partial q}{\partial y}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}=\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}-\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}$

$\therefore a^2z+\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}+\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial p^2}+\dfrac{1}{2}\dfrac{\partial^2z}{\partial p\partial q}-\dfrac{1}{4}\dfrac{\partial^2z}{\partial q^2}=0$

$a^2z+\dfrac{\partial^2z}{\partial p\partial q}=0$

$\dfrac{\partial^2z}{\partial p\partial q}=-a^2z$

Similar to Second order hyperbolic PDE,

The general solution is $z(p,q)=\int_0^pf(s)J_0\left(\dfrac{2\sqrt{q(p-s)}}{a}\right)~ds+\int_0^qg(s)J_0\left(\dfrac{2\sqrt{p(q-s)}}{a}\right)~ds$

$z(x,y)=\int_0^\frac{x+y}{2}f(s)J_0\left(\dfrac{2}{a}\sqrt{\dfrac{x-y}{2}\left(\dfrac{x+y}{2}-s\right)}\right)~ds+\int_0^\frac{x-y}{2}g(s)J_0\left(\dfrac{2}{a}\sqrt{\dfrac{x+y}{2}\left(\dfrac{x-y}{2}-s\right)}\right)~ds$

$z(x,y)=\int_0^{x+y}F(s)J_0\left(\dfrac{\sqrt{(x-y)(x+y-s)}}{a}\right)~ds+\int_0^{x-y}G(s)J_0\left(\dfrac{\sqrt{(x+y)(x-y-s)}}{a}\right)~ds$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.