# Thread: PDE; change of variables

1. ## PDE; change of variables

I have the PDE a(t)*(df/dx) + df/dt + b(t)f=0.
I have to make a change of variables now where g=fc(t), y=xd(t) and s=e(t). I have no idea how this works with coefficients depending on t.

2. Maybe I wasn't clear in my first post, f is a function of x and t and with df/dx I mean the partial derivative of f to x. The question is to rewrite the PDE into a PDE in g(y,s), y and s.

3. It wasn't clear to me. In general, for the PDE:

$a(x,y)\frac{\partial u}{\partial x}+b(x,y)\frac{\partial u}{\partial y}+c(x,y)u(x,y)=f(x,y)$, we solve the characteristic DE:

$\frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}$

Assume the characteristic solution can be put into the form $h(x,y)=k$. Then the change of variable $w=h(x,y),\; z=y$ will convert the PDE to an ODE in $z$ for fixed $w$. Probably not what you want. That's why I didn't say anything. Also while I'm sayin', this is right out of "Basic Partial Differential Equations" by Bleecker and Csordas. It's a nice PDE book that's easy to read. They go over first order PDEs like this pretty good.

4. I just want to rewrite my original PDE in a new PDE of the form h(s)*(dg/dy) + k(s)*(dg/ds) + l(s)*g=0.
Again with 'dg/dy' I mean the partial derivative.

5. This is how I'd do it; maybe someone can suggest a better approach to both of us. I first write it in standard form:

$\frac{a(t)}{b(t)}\frac{\partial f}{\partial x}+\frac{1}{b(t)}\frac{\partial f}{\partial t}+f=0$

The characteristic equation becomes then:

$\frac{dt}{dx}=\frac{1}{a(t)}\Rightarrow x-\int a(t)dt=K$

So I make the change of variables:

$y=x-\int a(t)dt \quad\quad s=t$

Then define: $g(y,s)\equiv f(x,t)$

and under these transformations it can be shown:

$\frac{a}{b}\frac{\partial f}{\partial x}+\frac{1}{b}\frac{\partial f}{\partial t}=\frac{1}{b(s)}\frac{\partial g}{\partial y}$

Thus I have:

$\frac{1}{b(s)}\frac{\partial g}{\partial y}+g=0$