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Thread: First Order, Linear, PDE

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    31

    First Order, Linear, PDE

    $\displaystyle u_x + u_y + u = 1 $
    $\displaystyle u = sin(x)$ on $\displaystyle y = x + x^2, x>0$
    Let $\displaystyle x = s $ be the parameter along the initial line $\displaystyle y = x + x^2$

    Solution:

    Characteristic Equation:
    $\displaystyle \frac{dx}{dt} = 1, x = t + s$
    $\displaystyle \frac{dy}{dt} = 1, y = t + s +s^2 $
    $\displaystyle \frac{du}{dt} = 1 - u, u(0) = sin(s) $

    Hence:

    $\displaystyle y = s + x^2$
    $\displaystyle u = 1 + (sin(s) - 1)e^{-t}$

    and the solution continues on...
    ------------------------------------------------------

    What I don't understand is how to get $\displaystyle u = 1 + (sin(s) - 1)e^{-t}$.

    I sort of understand the $\displaystyle \frac{dx}{dt} = 1, x = t + s$ part, but can someone give me an explanation as well?

    Thanks =]
    Last edited by Creebe; Oct 9th 2010 at 09:43 AM.
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  2. #2
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    $\displaystyle
    \displaystyle{\frac{du}{1-u}=dt}
    $

    $\displaystyle
    \displaystyle{\frac{d(1-u)}{1-u}=-dt}
    $

    $\displaystyle
    ln(1-u)=-t+C
    $

    $\displaystyle
    1-u=Ce^{-t}
    $

    $\displaystyle
    **(1)** \; \: \: u=1-Ce^{-t}
    $

    $\displaystyle
    u(0)=sin(s)
    $

    $\displaystyle
    u(0)=sin(s)=1-C
    $

    $\displaystyle
    C=1-sin(s)
    $

    $\displaystyle
    Inserting \; to \; (1)
    $

    $\displaystyle
    u=1-(1-sin(s))e^{-t}
    $
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