2nd Order PDE hyperbolic

**harkapobi** · November 17th 2009, 08:13 AM

Hi, I have to find the general solution to this second order pde.

$x^2Z_{xx} - y^2Z_{yy} = 0$ Where x and y not equal to 0.

I have found characteristics to be n=xy and v=y/x and I have used these to change the variables to get

$Z_{vn} =$ 1/(2n) $Z_v$

And now i'm stuck, for some reason I just dont know how to solve this with the first order partial derivative on the RHS.

Please help.
Thanks Katy

**shawsend** · November 17th 2009, 10:11 AM

I'm a little unsure. Here goes though. You have:

$\frac{\partial^2 z}{\partial v \partial n}=\frac{1}{2n}\frac{\partial z}{\partial v}$

and if the function is well-behaved, then:

$\frac{\partial^2 z}{\partial n \partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}=\frac{\partial}{\partial n}\frac{\partial z}{\partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}$

Let $\frac{\partial z}{\partial v}=g(n,v)$ then we have $\frac{\partial g}{\partial n}=\frac{1}{2n}g$ which is an ordinary DE in the variable n with v kept constant. Then:

$\frac{dg}{dn}-\frac{1}{2n}g=0$

Solving via integrating factor and noting that the constant of integration is now an arbitrary function of v, I obtain:

$d\left[n^{1/2}g\right]=0$

and so:

$g(n,v)=n^{-1/2} H(v)$ with $H(v)$ arbitrary.

Now:

$\frac{\partial z}{\partial v}=n^{-1/2} H(v)$

so that:

$z(n,v)=n^{-1/2}\int H(v) dv+G(n)$

**harkapobi** · November 17th 2009, 10:20 AM

Thats fantastic, thats the answer I'm supposed to get.

Thank you very much

**Danny** · November 17th 2009, 10:29 AM

Originally Posted by shawsend

I'm a little unsure. Here goes though. You have:

$\frac{\partial^2 z}{\partial v \partial n}=\frac{1}{2n}\frac{\partial z}{\partial v}$

and if the function is well-behaved, then:

$\frac{\partial^2 z}{\partial n \partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}=\frac{\partial}{\partial n}\frac{\partial z}{\partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}$

Let $\frac{\partial z}{\partial v}=g(n,v)$ then we have $\frac{\partial g}{\partial n}=\frac{1}{2n}g$ which is an ordinary DE in the variable n with v kept constant. Then:

$\frac{dg}{dn}-\frac{1}{2n}g=0$ (**)

Solving via integrating factor and noting that the constant of integration is now an arbitrary function of v, I obtain:

$d\left[n^{1/2}g\right]=0$

and so:

$g(n,v)=n^{-1/2} H(v)$ with $H(v)$ arbitrary.

Now:

$\frac{\partial z}{\partial v}=n^{-1/2} H(v)$

so that:

$z(n,v)=n^{-1/2}\int H(v) dv+G(n)$

I agree in the most part except for a negative (**) separates. Also since $H(v)$ is arbitrary we can let $\tilde{H} = \int H(v)\,dv$

@ Katy. You could also introduce the coords

$n = \ln x + \ln y,\;\; v = \ln x - \ln y$

to obtain the new PDE

$u_{nv} = \frac{1}{2} u_v$ - maybe a little easier.

**shawsend** · November 17th 2009, 11:32 AM

Hi. I made a mistake with the integrating factor. It should be:

$z(n,v)=n^{1/2}\left(\int H(v) dv+G(n)\right)$

or As Danny mentioned:

$z(n,v)=n^{1/2}\left(h(v)+g(n)\right)$

and I checked it with:

$H(v)=v^2+\sin(v)$

$G(n)=3n-4 \cos(n)$

which equates to:

$z(x,y)=(xy)^{1/2}\left[\frac{1}{3}\left(\frac{y}{x}\right)^3-\cos(y/x)+3 x y-4 \cos(xy)\right]$

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