# Finding all solutions for a given PDE

• Feb 9th 2011, 06:50 PM
Combinatus
Finding all solutions for a given PDE
The problem is stated as follows:

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a) Transform the differential equation

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

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

b) Determine all solutions of class $\displaystyle C^2$ (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:

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

(since y>0).

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

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

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

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

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

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

(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 $\displaystyle f(u,v) = -\frac{u^2 v^2}{4} + Cu +D$ or $\displaystyle f(u,v) = -\frac{u^2 v^2}{4} + Cv +D$ (C and D being integration constants), but I suspect that won't cover all possible cases.
• Feb 10th 2011, 05:54 AM
Jester
If $\displaystyle f_{uv} = - uv$ (which I agree with), then integrating wrt $\displaystyle v$ gives

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

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

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

$\displaystyle f = - \dfrac{1}{4} y^4 + A\left(ye^x\right) + B\left(ye^{-x}\right)$.
• Feb 10th 2011, 10:32 AM
Combinatus
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.)