Finding all solutions for a given PDE

**Combinatus** · February 9th 2011, 06:50 PM

The problem is stated as follows:

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a) Transform the differential equation

$\frac{\partial^2 f}{\partial x^2} - y^2 \frac{\partial^2 f}{\partial y^2} - y \frac{\partial f}{\partial y}=4y^4,\,$

with the substitution $u=ye^x$ and $v=ye^{-x}$ in the region y>0.

b) Determine all solutions of class $C^2$ (that is, solutions with continuous second (partial) derivatives) for the differential equation in this region. The solutions should be expressed as functions of the variables x and y.

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My question is about part b), since I don't know how I should get started on that. I'll show some of my work on part a) for some context:

$\begin{cases} u = ye^x\\ v = ye^{-x} \end{cases} \Leftrightarrow \begin{cases} x = \ln(\sqrt{\frac{u}{v}})\\ y = \sqrt{uv} \end{cases}$

(since y>0).

Also, by the looks of the PDE, we have $f(x,y) = f(x(u,v),y(u,v)) = g(u,v)$ , so we can use the chain rule:

$\begin{cases} \frac{\partial f}{\partial u} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}\\ \frac{\partial f}{\partial v} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial v} \end{cases} $

By finding the partials of x and y with respect to u and v, we get

$\begin{cases} \frac{\partial f}{\partial u} = \frac{1}{2u} \frac{\partial f}{\partial x} + \frac{\sqrt{v}}{2\sqrt{u}} \frac{\partial f}{\partial y}\\ \frac{\partial f}{\partial v} = -\frac{1}{2v} \frac{\partial f}{\partial x} + \frac{\sqrt{u}}{2\sqrt{v}} \frac{\partial f}{\partial y} \end{cases} \Leftrightarrow \begin{cases} \frac{\partial f}{\partial x} = u \frac{\partial f}{\partial u} - v \frac{\partial f}{\partial v}\\ \frac{\partial f}{\partial y} = \sqrt{\frac{u}{v}} \frac{\partial f}{\partial u} + \sqrt{\frac{v}{u}} \frac{\partial f}{\partial v} \end{cases}$ .

Differentiating again, and substituting the relevant partials back into the original PDE, we eventually, after a few pages of work, get

$\frac{\partial^2f}{\partial u\, \partial v} = -uv$

(assuming I didn't mess up somewhere).

I'm not sure what to make of that, other than how we have a function that depends on u and v, and consequently, on x and y. But, I guess that was kind of the premise for the problem. How do I approach part b)?

I suppose I could integrate with respect to one variable at a time and get $f(u,v) = -\frac{u^2 v^2}{4} + Cu +D$ or $f(u,v) = -\frac{u^2 v^2}{4} + Cv +D$ (C and D being integration constants), but I suspect that won't cover all possible cases.

**Danny** · February 10th 2011, 05:54 AM

If $f_{uv} = - uv$ (which I agree with), then integrating wrt $v$ gives

$f_u = -\dfrac{1}{2}u v^2 +A'(u)$ and then integrating wrt $u$ gives

$f = - \dfrac{1}{4} u^2 v^2 + A(u) + B(v)$

where $A$ and $B$ are arbitrary functions of their arguments. Then back substitute giving

$f = - \dfrac{1}{4} y^4 + A\left(ye^x\right) + B\left(ye^{-x}\right)$ .

**Combinatus** · February 10th 2011, 10:32 AM

Aah, of course you wouldn't get constants but rather arbitrary functions of the other variable when integrating. Thank you!

(I'm hoping the missing minus sign after the second integration was a LaTeX mishap rather than a lack of understanding on my part.)

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