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1. The problem 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 $x$.

2.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 $f(z)=(3z)^{\frac{1}{3}}z$.

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 off by a factor 3/4 and i'm not sure why.

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  • $\begingroup$ When you integrated $f'$, you forgot to divide by $4/3$. $\endgroup$ – Rodrigo de Azevedo May 22 '16 at 0:44
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If $u (x,y) = f \left(\frac{x^3}{3} - \frac{y^4}{4}\right)$, then the boundary condition gives us $f ' \left(\frac{x^3}{3}\right) = x$. Hence, we have

$$f ' (z) = \sqrt[3]{3z}$$

Integrating, we obtain

$$f (z) = \frac{3\sqrt[3]{3}}{4} \, z^{\frac{4}{3}}$$

and, thus,

$$u (x,y) = \frac{3\sqrt[3]{3}}{4} \left(\frac{x^3}{3} - \frac{y^4}{4}\right)^{\frac{4}{3}}$$

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