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Thread: jump discontinuity proof for PDE

  1. #1
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    jump discontinuity proof for PDE

    So here is the problem statement:

    Consider the equation

    (6.12) $\displaystyle u_y+uu_x=0$.

    Let $\displaystyle u$ be a $\displaystyle C^1$ solution of (6.12) in each of two regions separated by a curve $\displaystyle x=\xi(y)$. Let $\displaystyle u$ be continuous, but $\displaystyle u_x$ have a jump discontinuity along the curve. Prove that

    $\displaystyle \frac{d\xi}{dy}=u$

    and hence that the curve is characteristic.
    In addition, the following "hint" is given:

    Hint: By (6.12)

    $\displaystyle (u_y^+-u_y^-)+u(u_x^+-u_x^-)=0$.

    Moreover, $\displaystyle u(\xi(y),y)$ and $\displaystyle (d/dy)u(\xi(y),y)$ are continuous on the curve.
    I believe $\displaystyle u_y^+,u_y^-$ denote the limits of $\displaystyle u_y$ from the right and left, respectively, and similarly for $\displaystyle u_x^+,u_x^-$.

    This is all from Fritz John's Partial Differential Equations, exercise 1.6.3, p19.

    I'm pretty lost on this one. Any help would be much appreciated. Thanks !
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  2. #2
    Senior Member
    Joined
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    Re: jump discontinuity proof for PDE

    I've been working on this for a bit, and I think I have an idea. However, I still need some help making the idea work.

    According to the textbook (Fritz John, p17), if we choose $\displaystyle R(u),S(u)$ satisfying $\displaystyle S'(u)=uR'(u)$, then we have the following "conservation law":

    $\displaystyle \displaystyle 0=\frac{d}{dy}\int_a^b R(u(x,y))\;dx+S(u(b,y))-S(u(a,y))$

    and therefore if we choose $\displaystyle a<\xi(y)<b$ then

    $\displaystyle \displaystyle 0=S(u(b,y))-S(u(a,y))+\frac{d}{dy}\left(\int_a^{\xi(y)} R(u)\;dx+\int_{\xi(y)}^b R(u)\;dx\right)$

    Now, if we choose $\displaystyle R(u)=u$ and $\displaystyle S(u)=u^2/2$ then this will satisfy the conservation law, giving us

    (1) $\displaystyle \displaystyle 0=\frac{1}{2}\left[u(b,y)^2-u(a,y)^2\right]+\frac{d}{dy}\left(\int_a^{\xi(y)} u\;dx+\int_{\xi(y)}^b u\;dx\right)$

    If I can show that this implies

    (2) $\displaystyle \displaystyle 0=\frac{1}{2}\left[u(b,y)^2-u(a,y)^2\right]+\frac{d\xi}{dy}[u(a,y)-u(b,y)]$

    then it will follow that

    $\displaystyle \displaystyle \frac{1}{2}\left[u(b,y)+u(a,y)\right]=\frac{d\xi}{dy}$

    and taking limits as $\displaystyle a,b\to\xi(y)$ will give us the desired result $\displaystyle u=\frac{d\xi}{dy}$.

    But the problem is, I don't know how to show that (1) implies (2). Any suggestions ?

    Thanks again guys !
    Last edited by hatsoff; Oct 2nd 2011 at 09:41 AM.
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