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Math Help - Rankine Hugoniot Satisfy Weak Solution

  1. #1
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    Rankine Hugoniot Satisfy Weak Solution

    Consider the Riemann Problem
     u_t + (f(u))_x = 0\qquad -\infty<x<\infty,\;0<t<\infty
     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 ( 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
    u(x,t) = \begin{cases}u_{\ell}, & x < s\cdot t \\ u_r & x > s\cdot t\end{cases}
    where
    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 \phi with compact support and multiply it by the PDE and then integrate over \mathbb{R}, i.e.
    \int_{\mathbb{R}}\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx
    = \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
    = \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
    = \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.
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  2. #2
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    Quote Originally Posted by lvleph View Post
    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 \phi with compact support and multiply it by the PDE and then integrate over \mathbb{R}, i.e.
    \int_{\mathbb{R}}\! \left(u_t + f'(u)\cdot u_x\right)\cdot \phi\, dx
    = \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
    = \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
    = \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 (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 u_r=u_{\ell}).

    Edit: Another thing, when you split the integral shouldn't you have st intead of 0 in the limits of integration?
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  3. #3
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    Yes, you are correct that it should be split at s\cdot t. However, u_{\ell},\; u_r are constants. But, I am unsure if I should be placing them in the integral at that point.
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  4. #4
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    Quote Originally Posted by lvleph View Post
    Yes, you are correct that it should be split at s\cdot t. However, 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 H^1 or H_0^1.

    Another thing that's bugging me is: if 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 W^{1,p} )
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  5. #5
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    There is no problem with 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 s\cdot t.
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  6. #6
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    Quote Originally Posted by lvleph View Post
    There is no problem with 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 s\cdot t.
    How do you define an admissible solution?

    Also, shouldn't your integration be over the whole semi-plane t>0?
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  7. #7
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    Well, now I just have a bunch of mistakes don't I.

    I defined the admissible solution for shock curves in the original post.
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  8. #8
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    Weak Solution: A weak solution of

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

    \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 \vec{\phi}(x,t) \in C^1_c(\mathbb{R}\times \mathbb{R}^+), where

    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\}
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  9. #9
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    Quote Originally Posted by lvleph View Post
    Weak Solution: A weak solution of

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

    \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 \vec{\phi}(x,t) \in C^1_c(\mathbb{R}\times \mathbb{R}^+), where

    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 \phi (x,0)=0 for all \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)
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