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First order Partial differential equation

  1. May 21, 2016 #1
    1. The problem statement, all variables and given/known data

    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##.

    3. 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: May 21, 2016
  2. jcsd
  3. May 21, 2016 #2

    stevendaryl

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    You got to the point:

    [itex]\frac{df}{dz} = x[/itex]

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

    [itex]z = \frac{x^3}{3}[/itex], which means [itex]x = (3z)^\frac{1}{3}[/itex]

    So your equation becomes:

    [itex]\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}}[/itex]
     
  4. May 21, 2016 #3

    Ray Vickson

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    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##.
     
  5. May 22, 2016 #4
    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.
     
  6. May 22, 2016 #5

    Ray Vickson

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    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
    [tex] x^2 g\left( \frac{x^3}{3} \right) = x^3 [/tex]
    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.
     
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