- #1

pondzo

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## Homework Statement

Find a solution of $$\frac{1}{x^2}\frac{\partial u(x,y)}{\partial x}+\frac{1}{y^3}\frac{\partial u(x,y)}{\partial y}=0$$

Which satisfies the condition ##\frac{\partial u(x,y)}{\partial x}\big |_{y=0}=x^3## for all ##x##.

## The Attempt at a Solution

I get the following general solution for the PDE:

$$u(x,y)=f(\frac{x^3}{3}-\frac{y^4}{4}):=f(z)$$

For some function f which satisfies the BC. I'm not too sure how to deal with this type of boundary condition. But here is an attempt:

##\partial_x u=\frac{\text{d}f}{\text{d}z}\frac{\partial z}{\partial x}=\frac{\text{d}f}{\text{d}z} x^2##. Thus ##\frac{\text{d}f}{\text{d}z} x^2=x^3\implies \frac{\text{d}f}{\text{d}z} =x\implies f(z)=zx##

At ##y=0,~ z=\frac{x^3}{3}\implies x=(3z)^{\frac{1}{3}}##. So ##f(z)=(3z)^{\frac{1}{3}}z##, and changing variables back gives

##u(x,y)=(3(\frac{x^3}{3}-\frac{y^4}{4}))^{\frac{1}{3}}(\frac{x^3}{3}-\frac{y^4}{4})=3^{\frac{1}{3}}(\frac{x^3}{3}-\frac{y^4}{4})^{\frac{4}{3}}##

This is a solution to the PDE but doesn't satisfy the given boundary condition, it is out by a factor 3/4 and i'm not sure why.

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