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Thread: How to solve this pde?

  1. #1
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    How to solve this pde?

    du/dt-t(du/dx)=-u^2

    d is meant to be partial derivatives.
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  2. #2
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    Quote Originally Posted by theoretician View Post
    du/dt-t(du/dx)=-u^2

    d is meant to be partial derivatives.
    I am going to try. I cannot gaurentee my solution is correct.

    $\displaystyle \frac{\partial u}{\partial t} - t\cdot \frac{\partial u}{\partial x} = - u^2$

    Say we want to solve this subject to some curve $\displaystyle x=x(s) \mbox{ and }t=t(s)$ with $\displaystyle x(0)=x_0 \mbox{ and }t(0)=0$.

    Then $\displaystyle u(x,t) = u(x(s),t(s))$
    Use the chain rule,
    $\displaystyle \frac{du}{ds} = \frac{\partial u}{\partial x} \cdot \frac{dx}{ds} + \frac{\partial u}{\partial t} \cdot \frac{dt}{ds}$
    By the method of charachteristics we set,
    1)$\displaystyle \frac{dt}{ds} = 1 \mbox{ with }t(0)=0$
    2)$\displaystyle \frac{dx}{ds} = -t \mbox{ with }x(0) = x_0$
    3)$\displaystyle \frac{du}{ds} = -u^2$

    The solution to the IVP problems are:
    1)$\displaystyle t = s$ (**)
    2)$\displaystyle x = x_0 - ts$ (*)

    The solution to #3 is simple:
    $\displaystyle \frac{du}{dx} = -u^2 \Rightarrow -\frac{1}{u^2}u' = 1$
    This is a Seperable equation,
    $\displaystyle \frac{1}{u} = s + F(x_0) \Rightarrow u =\frac{1}{s+F(x_0)}$
    Where $\displaystyle F(x_0)$ is an arbitrary starting point on the solution curve.
    By (*) we see that $\displaystyle x_0 = x+ts$.
    Thus by (**),
    $\displaystyle u(x,t) = \frac{1}{t+F(x+ts)}$
    Where $\displaystyle F$ is any arbitrary differencial function (and not equal to $\displaystyle -s$).
    ----
    Which seems to work !!! Because if $\displaystyle F = 0$ everywhere then $\displaystyle u(x,t) = \frac{1}{t+0} = \frac{1}{t}$. If you substitute that into the PDE you get:
    $\displaystyle \frac{\partial u}{\partial t} = - \frac{1}{t^2} = - u^2$
    Which works.
    Last edited by ThePerfectHacker; Jun 11th 2007 at 12:14 PM.
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