Linear 1st order PDE (boundary conditions)

[twizzy]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

[tex]-\frac{u_x}{u_y}=\frac{dy}{dx}=2xy^2[/tex]
Separating variables
[tex]\frac{dy}{y^2}=2xdx[/tex]
[tex]\frac{-1}{y}=x^2+c[/tex]
[tex]C=x^2+\frac{1}{y}[/tex]
So [tex]u(x,y)=f(x^2+\frac{1}{y})[/tex]
The boundary condition is given as evaluating at [tex]y=0[/tex] 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 [tex]u(x,y)=f(x^2+\frac{1}{y})[/tex] and [tex]u(x,0)=\phi(x) [/tex],
we have [tex]u(x,0)=f(x^2+\frac{1}{\epsilon})=\phi(x)[/tex] where [tex]\epsilon\rightarrow 0[/tex]. This would imply that [tex]\lim_{u\rightarrow \infty}f(u)=\phi(x)[/tex]. Then again, that can't be right because the limit as [tex]u\rightarrow \infty[/tex] should not depend on x. I'm not sure if I see why you have "f(u)->0".

 

[Dick]

Since [tex]u(x,y)=f(x^2+\frac{1}{y})[/tex] and [tex]u(x,0)=\phi(x) [/tex],
we have [tex]u(x,0)=f(x^2+\frac{1}{\epsilon})=\phi(x)[/tex] where [tex]\epsilon\rightarrow 0[/tex]. This would imply that [tex]\lim_{u\rightarrow \infty}f(u)=\phi(x)[/tex]. Then again, that can't be right because the limit as [tex]u\rightarrow \infty[/tex] 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 [tex]\phi(x)[/tex] were not a constant, would the question even make sense?

 

[Dick]

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

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