jump discontinuity proof for PDE

• Sep 30th 2011, 04:25 PM
hatsoff
jump discontinuity proof for PDE
So here is the problem statement:

Quote:

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:

Quote:

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 !
• Oct 2nd 2011, 09:00 AM
hatsoff
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) $\displaystyle \displaystyle 0=\frac{1}{2}\left[u(b,y)^2-u(a,y)^2\right]+\frac{d\xi}{dy}[u(a,y)-u(b,y)]$

$\displaystyle \displaystyle \frac{1}{2}\left[u(b,y)+u(a,y)\right]=\frac{d\xi}{dy}$
and taking limits as $\displaystyle a,b\to\xi(y)$ will give us the desired result $\displaystyle u=\frac{d\xi}{dy}$.