# Thread: Rankine Hugoniot Satisfy Weak Solution

1. ## Rankine Hugoniot Satisfy Weak Solution

Consider the Riemann Problem
$\displaystyle u_t + (f(u))_x = 0\qquad -\infty<x<\infty,\;0<t<\infty$
$\displaystyle u(x,0) = \begin{cases}u_{\ell}, & x<0\\ u_r, & x>0.\end{cases}$

I am only going to deal with the case that there is a shock curve ($\displaystyle f'(u_{\ell})>f'(u_r)$), since I can figure out the rarefaction wave once I understand the shock. The solution for this PDE using the Rankine Hugoniot Condition is
$\displaystyle u(x,t) = \begin{cases}u_{\ell}, & x < s\cdot t \\ u_r & x > s\cdot t\end{cases}$
where
$\displaystyle s = \frac{f(u_{\ell}) - f(u_r)}{u_{\ell} - u_r}$
Now, I want to determine if this solution satisfies the definition of the weak solution. So I assume that I would want to take a test function $\displaystyle \phi$ with compact support and multiply it by the PDE and then integrate over $\displaystyle \mathbb{R}$, i.e.
$\displaystyle \int_{\mathbb{R}}\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx$
$\displaystyle = \int_{-\infty}^0\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx + \int_0^{\infty}\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx$
$\displaystyle = \int_{-\infty}^0\! \left((u_{\ell})_t + f'(u_{\ell})\cdot (u_{\ell})_x\right)\cdot \phi\, dx + \int_0^{\infty}\! \left((u_r)_t + f'(u_r)\cdot (u_r)_x\right)\cdot \phi\, dx$
$\displaystyle = \int_{-\infty}^0\! \left((0 + f'(u_{\ell})\cdot 0\right)\cdot \phi\, dx + \int_0^{\infty}\! \left(0 + f'(u_r)\cdot 0\right)\cdot \phi\, dx = 0$

For some reason this just doesn't seem right.

2. Originally Posted by lvleph
Now, I want to determine if this solution satisfies the definition of the weak solution. So I assume that I would want to take a test function $\displaystyle \phi$ with compact support and multiply it by the PDE and then integrate over $\displaystyle \mathbb{R}$, i.e.
$\displaystyle \int_{\mathbb{R}}\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx$
$\displaystyle = \int_{-\infty}^0\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx + \int_0^{\infty}\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx$
$\displaystyle = \int_{-\infty}^0\! \left((u_{\ell})_t + f'(u_{\ell})\cdot (u_{\ell})_x\right)\cdot \phi\, dx + \int_0^{\infty}\! \left((u_r)_t + f'(u_r)\cdot (u_r)_x\right)\cdot \phi\, dx$
$\displaystyle = \int_{-\infty}^0\! \left((0 + f'(u_{\ell})\cdot 0\right)\cdot \phi\, dx + \int_0^{\infty}\! \left(0 + f'(u_r)\cdot 0\right)\cdot \phi\, dx = 0$

For some reason this just doesn't seem right.
Well, how is it that $\displaystyle (u_r)_x=0=(u_r)_t$? Surely these are not constant, since then it would be difficult to apply your boundary conditions in the weak solution (unless $\displaystyle u_r=u_{\ell}$).

Edit: Another thing, when you split the integral shouldn't you have $\displaystyle st$ intead of 0 in the limits of integration?

3. Yes, you are correct that it should be split at $\displaystyle s\cdot t$. However, $\displaystyle u_{\ell},\; u_r$ are constants. But, I am unsure if I should be placing them in the integral at that point.

4. Originally Posted by lvleph
Yes, you are correct that it should be split at $\displaystyle s\cdot t$. However, $\displaystyle u_{\ell},\; u_r$ are constants. But, I am unsure if I should be placing them in the integral at that point.
Hm, could you please give your definition of weak solution. Usually you multiply and do all you did before giving a solution, therefore arriving at a necessary condition which involves a larger class of functions, for example $\displaystyle H^1$ or $\displaystyle H_0^1$.

Another thing that's bugging me is: if $\displaystyle u_r \neq u_{\ell }$ then your solution looks a lot like Heaviside's function (which at a first glance would have me believe it will not be a weak solution in the sense above, or at least not in $\displaystyle W^{1,p}$ )

5. There is no problem with $\displaystyle u_r \ne u_{\ell}$ this will result in what is known as an admissible solution. Typically a shock-curve or rarefaction wave happens. In the case that a shock curve happens, the shock follows the line $\displaystyle s\cdot t$.

6. Originally Posted by lvleph
There is no problem with $\displaystyle u_r \ne u_{\ell}$ this will result in what is known as an admissible solution. Typically a shock-curve or rarefaction wave happens. In the case that a shock curve happens, the shock follows the line $\displaystyle s\cdot t$.
How do you define an admissible solution?

Also, shouldn't your integration be over the whole semi-plane t>0?

7. Well, now I just have a bunch of mistakes don't I.

I defined the admissible solution for shock curves in the original post.

8. Weak Solution: A weak solution of

$\displaystyle \vec{u}_t + (\vec{f}(\vec{u}))_x = 0, \quad \vec{u}|_{t=0} = \vec{u}_0(x)$
is a function $\displaystyle \vec{u}(x,t):\mathbb{R}\times\mathbb{R}^+ \rightarrow \mathbb{R}^n$ such that

$\displaystyle \int_0^{\infty}\!\!\!\int_{-\infty}^{\infty}\!\left[\vec{u}(x,t)\cdot \vec{\phi}_t(x,t) + \vec{f}(\vec{u}(x,t))\cdot \vec{\phi}_x(x,t)\right]\dx\,dt + \int_{-\infty}^{\infty}\!\vec{u}_0(x)\vec{\phi}(x,0)\, dx = 0$
for all $\displaystyle \vec{\phi}(x,t) \in C^1_c(\mathbb{R}\times \mathbb{R}^+)$, where

$\displaystyle C^1_c(\mathbb{R}\times \mathbb{R}^+) = \left\{\vec{\phi} \in C^1_c(\mathbb{R}\times \mathbb{R}^+) | \phi \equiv 0 \text{ for } (x,t)\not \in B_r(0,0) \bigcap (\mathbb{R}\times \mathbb{R}^+) \text{ for some } r>0\right\}$

9. Originally Posted by lvleph
Weak Solution: A weak solution of

$\displaystyle \vec{u}_t + (\vec{f}(\vec{u}))_x = 0, \quad \vec{u}|_{t=0} = \vec{u}_0(x)$
is a function $\displaystyle \vec{u}(x,t):\mathbb{R}\times\mathbb{R}^+ \rightarrow \mathbb{R}^n$ such that

$\displaystyle \int_0^{\infty}\!\!\!\int_{-\infty}^{\infty}\!\left[\vec{u}(x,t)\cdot \vec{\phi}_t(x,t) + \vec{f}(\vec{u}(x,t))\cdot \vec{\phi}_x(x,t)\right]\dx\,dt + \int_{-\infty}^{\infty}\!\vec{u}_0(x)\vec{\phi}(x,0)\, dx = 0$
for all $\displaystyle \vec{\phi}(x,t) \in C^1_c(\mathbb{R}\times \mathbb{R}^+)$, where

$\displaystyle C^1_c(\mathbb{R}\times \mathbb{R}^+) = \left\{\vec{\phi} \in C^1_c(\mathbb{R}\times \mathbb{R}^+) | \phi \equiv 0 \text{ for } (x,t)\not \in B_r(0,0) \bigcap (\mathbb{R}\times \mathbb{R}^+) \text{ for some } r>0\right\}$
Well, it's easy to see that given your space for "test" functions, we must have $\displaystyle \phi (x,0)=0$ for all $\displaystyle \phi \in C_c^1 (\mathbb{R} \times \mathbb{R} ^+)$, so the last term doesn't count. So your calculations are correct (taking into account that the integration is over the semi-plane)