# Solve Linear PDE via a change of variables

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• Feb 12th 2011, 04:50 AM
bugatti79
I was hoping that by solving the pde the normal way which is part of the question that it might give a clue as to how the change of variables method should should turn out.

Solving the original PDE $\displaystyle y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1$ I get

$\displaystyle \frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}=\frac{-x}{y} \implies \frac{y^2}{2}=\frac{-x^2}{2}+k$. Taking the positive solution of y

$\displaystyle \frac{du}{dx}=\frac{h(x,y)}{a(x,y)}=\frac{1}{y} \implies u=\int \frac{1dx}{\sqrt{2k-x^2}}$

which gives

$\displaystyle u=Sin^{-1} [\frac{x}{\sqrt {2k}}]+f(\frac{y^2}{2} +\frac{x^2}{2})$

The IC's are u(x,0)=0, therefore

$\displaystyle 0=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2})}}]+f(\frac{x^2}{2})$. Let $\displaystyle t=\frac{x^2}{2}$ therefore

$\displaystyle f(t)=\sqrt {2t}=-\frac {\pi}{2}$

$\displaystyle u(x,y)=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2}+\frac{y^2}{2})}}}]-\frac {\pi}{2}$

I am not sure of this is right either...
• Feb 12th 2011, 05:20 AM
Ackbeet
I'm afraid you take me beyond my province there. You see how little I know about pde's!
• Feb 15th 2011, 04:35 AM
bugatti79
Quote:

Originally Posted by bugatti79
I was hoping that by solving the pde the normal way which is part of the question that it might give a clue as to how the change of variables method should should turn out.

Solving the original PDE $\displaystyle y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1$ I get

$\displaystyle \frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}=\frac{-x}{y} \implies \frac{y^2}{2}=\frac{-x^2}{2}+k$. Taking the positive solution of y

$\displaystyle \frac{du}{dx}=\frac{h(x,y)}{a(x,y)}=\frac{1}{y} \implies u=\int \frac{1dx}{\sqrt{2k-x^2}}$

which gives

$\displaystyle u=Sin^{-1} [\frac{x}{\sqrt {2k}}]+f(\frac{y^2}{2} +\frac{x^2}{2})$

The IC's are u(x,0)=0, therefore

$\displaystyle 0=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2})}}]+f(\frac{x^2}{2})$. Let $\displaystyle t=\frac{x^2}{2}$ therefore

$\displaystyle f(t)=\sqrt {2t}=-\frac {\pi}{2}$

$\displaystyle u(x,y)=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2}+\frac{y^2}{2})}}}]-\frac {\pi}{2}$

I am not sure of this is right either...

Referring to post #15

$\displaystyle \omega_\theta=-\theta+f(r)$

IC given is $\omega(rCos \theta,0)=0$

Using polar coordinates, when y is 0 (hence along the x line) theta is 0 therefore r Cos 0 =1 implies the f(r) =0. therefore

$\omega(r,\theta)=-\theta$
for $-\pi <\theta<\pi$

Not sure how this domain of influence came about though.

regarding post #16, Iv no idea how to solve it. I will try dx/dy instead of dy/dx and see where that gets me.
• Feb 15th 2011, 04:43 AM
Ackbeet
Quote:

Using polar coordinates, when y is 0 (hence along the x line) theta is 0 therefore r Cos 0 =1 implies the f(r) =0.
Why wouldn't it be $r\cos(\theta)=r?$ Why does $r=1?$ And why does $\theta=0?$ Couldn't it also equal $\pi?$ In that case, wouldn't

$r\cos(\theta)=\pm r?$
• Feb 15th 2011, 04:58 AM
bugatti79
Quote:

Originally Posted by Ackbeet
Why wouldn't it be $r\cos(\theta)=r?$ Why does $r=1?$ And why does $\theta=0?$ Couldn't it also equal $\pi?$ In that case, wouldn't

$r\cos(\theta)=\pm r?$

That is a typo, your right it should be r Cos 0 =r

Good point...im not sure...this is where I get bogged down defining domain of influences.
• Feb 15th 2011, 06:24 AM
Ackbeet
Yeah, I think it's time for Danny, if he's willing.
• Feb 15th 2011, 03:50 PM
Jester
Quote:

Originally Posted by bugatti79
I was hoping that by solving the pde the normal way which is part of the question that it might give a clue as to how the change of variables method should should turn out.

Solving the original PDE $\displaystyle y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1$ I get

$\displaystyle \frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}=\frac{-x}{y} \implies \frac{y^2}{2}=\frac{-x^2}{2}+k$. Taking the positive solution of y

$\displaystyle \frac{du}{dx}=\frac{h(x,y)}{a(x,y)}=\frac{1}{y} \implies u=\int \frac{1dx}{\sqrt{2k-x^2}}$

which gives

$\displaystyle u=Sin^{-1} [\frac{x}{\sqrt {2k}}]+f(\frac{y^2}{2} +\frac{x^2}{2})$

The IC's are u(x,0)=0, therefore

$\displaystyle 0=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2})}}]+f(\frac{x^2}{2})$. Let $\displaystyle t=\frac{x^2}{2}$ therefore

$\displaystyle f(t)=\sqrt {2t}=-\frac {\pi}{2}$

$\displaystyle u(x,y)=Sin^{-1} [\frac{x}{\sqrt {2(\frac{x^2}{2}+\frac{y^2}{2})}}}]-\frac {\pi}{2}$

I am not sure of this is right either...

What you have here is fine except replace the $\sqrt{2k}$ with $\sqrt{x^2+y^2}$ and simplify your final answer.

You have a solution that satisifes the PDE and BC.

Note: You can transform to polar coordinates if you like - the PDE becomes simple but you must interpret the BC, i.e.

when $y = 0$, $u = 0$ for all $x$.

Thus, $r \sin \theta = 0$ for all $r \cos \theta$ (This means that $r \ne 0$)
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