Linear 1st order PDE (boundary conditions)

[twizzy]

Homework Statement

Solve the equation u_{x}+2xy^{2}u_{y}=0 with u(x,0)=\phi(x)

Homework Equations

Implicit function theorem
\frac{dy}{dx}=-\frac{\partial u/\partial x}{\partial u/\partial y}

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!

 

[Dick]

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

 

[twizzy]

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

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".

 

[Dick]

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".

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

 

[twizzy]

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

 

[Dick]

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

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