# Linear 1st order PDE (boundary conditions)

1. Feb 24, 2010

### twizzy

1. The problem statement, all variables and given/known data
Solve the equation $$u_{x}+2xy^{2}u_{y}=0$$ with $$u(x,0)=\phi(x)$$

2. Relevant equations
Implicit function theorem
$$\frac{dy}{dx}=-\frac{\partial u/\partial x}{\partial u/\partial y}$$

3. The attempt at a solution
$$-\frac{u_x}{u_y}=\frac{dy}{dx}=2xy^2$$
Separating variables
$$\frac{dy}{y^2}=2xdx$$
$$\frac{-1}{y}=x^2+c$$
$$C=x^2+\frac{1}{y}$$
So $$u(x,y)=f(x^2+\frac{1}{y})$$
The boundary condition is given as evaluating at $$y=0$$ which doesn't seem to make sense. Any thoughts? Thanks!

Last edited: Feb 24, 2010
2. Feb 24, 2010

### Dick

Wouldn't be boundary condition mean you need to pick an f such that f(u)->0 as u->infinity?

3. Feb 24, 2010

### twizzy

Since $$u(x,y)=f(x^2+\frac{1}{y})$$ and $$u(x,0)=\phi(x)$$,
we have $$u(x,0)=f(x^2+\frac{1}{\epsilon})=\phi(x)$$ where $$\epsilon\rightarrow 0$$. This would imply that $$\lim_{u\rightarrow \infty}f(u)=\phi(x)$$. Then again, that can't be right because the limit as $$u\rightarrow \infty$$ should not depend on x. I'm not sure if I see why you have "f(u)->0".

4. Feb 24, 2010

### Dick

I have "f(u)->0" because I was reading the boundary condition as u(x,0)=0, not phi(x). My mistake.

5. Feb 24, 2010

### twizzy

So then if $$\phi(x)$$ were not a constant, would the question even make sense?

6. Feb 24, 2010

### Dick

I'm having trouble seeing how. But then maybe I'm missing something...