# First order Partial differential equation

## 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.

Last edited:

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stevendaryl
Staff Emeritus
You got to the point:

$\frac{df}{dz} = x$

But to integrate that, you have to have the right side as a function of $z$. On the border $y=0$, you have:

$z = \frac{x^3}{3}$, which means $x = (3z)^\frac{1}{3}$

$\frac{df}{dz} = (3)^{\frac{1}{3}} z^{\frac{1}{3}} \Rightarrow f(z) = 3^{\frac{1}{3}} \frac{z^\frac{4}{3}}{\frac{4}{3}}$

Ray Vickson
Homework Helper
Dearly Missed

## 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.
You are being led astray by poor notation! It is better in problems such as this to avoid the $df/dx$ notation for derivatives, and use $f'$ instead. The point is that $f'(u) = df(u)/du$ is just a function of $u$---never mind for the moment that it is a derivative. Your equation $x^2 f'(x^3/3) = x^3$ becomes $f'(x^3/3) = x$, or $f'(u) = (3u)^{1/3}$. Now do $\int f'(u) \, du$ to find $f(u)$. Then, and only then, put back $u$ in terms of $x$ and $y$.

Thank you guys, I knew it must have been something silly like that.

You are being led astray by poor notation!
Hi Ray, I really don't like the notation I use to solve these types of questions because as you say it can become confusing at times. What would your solution to a question like this look like? with particular emphasis on the notation used.

Ray Vickson
Homework Helper
Dearly Missed
Thank you guys, I knew it must have been something silly like that.

Hi Ray, I really don't like the notation I use to solve these types of questions because as you say it can become confusing at times. What would your solution to a question like this look like? with particular emphasis on the notation used.
Sometimes it helps to define another function temporarily, so you could say something like "let $g(u) = df(u)/du$... ". Then your boundary equation would be
$$x^2 g\left( \frac{x^3}{3} \right) = x^3$$
Of course, using a new notation $g( \cdot)$ is not really necessary, since using $f'$ instead works perfectly well, if you realize that $f'$ is a whole new function on its own. The fact that it happens also to be a derivative if $f$ is not important in some parts of the working.