Lax entropy condition for convex flux and sign of shock speed












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Assume that $u$ is an entropy solution to $u_t+partial_x(A(u))=0$, where $A$ is a non-decreasing function on $mathbb{R}$ such that $A(0)=0$ and $$A'(u_ell)>gamma'(t)=frac{A(u_ell)-A(u_r)}{u_ell-u_r}>A'(u_r)$$ (namely the Lax' entropy condition, where $u_ell, u_r$ are of the usual meanings). Prove that the shocks travel from left to right if $A$ is convex.




I got frustrated after spending a long time on it (not sure if one is supposed to consider the shock speed $gamma'(t)$, which is always non-negative since $A$ is non-decreasing). I believe that I missed something here and that there is some very simple way to solve this problem. Can someone help me? Thanks.










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    3












    $begingroup$



    Assume that $u$ is an entropy solution to $u_t+partial_x(A(u))=0$, where $A$ is a non-decreasing function on $mathbb{R}$ such that $A(0)=0$ and $$A'(u_ell)>gamma'(t)=frac{A(u_ell)-A(u_r)}{u_ell-u_r}>A'(u_r)$$ (namely the Lax' entropy condition, where $u_ell, u_r$ are of the usual meanings). Prove that the shocks travel from left to right if $A$ is convex.




    I got frustrated after spending a long time on it (not sure if one is supposed to consider the shock speed $gamma'(t)$, which is always non-negative since $A$ is non-decreasing). I believe that I missed something here and that there is some very simple way to solve this problem. Can someone help me? Thanks.










    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$



      Assume that $u$ is an entropy solution to $u_t+partial_x(A(u))=0$, where $A$ is a non-decreasing function on $mathbb{R}$ such that $A(0)=0$ and $$A'(u_ell)>gamma'(t)=frac{A(u_ell)-A(u_r)}{u_ell-u_r}>A'(u_r)$$ (namely the Lax' entropy condition, where $u_ell, u_r$ are of the usual meanings). Prove that the shocks travel from left to right if $A$ is convex.




      I got frustrated after spending a long time on it (not sure if one is supposed to consider the shock speed $gamma'(t)$, which is always non-negative since $A$ is non-decreasing). I believe that I missed something here and that there is some very simple way to solve this problem. Can someone help me? Thanks.










      share|cite|improve this question











      $endgroup$





      Assume that $u$ is an entropy solution to $u_t+partial_x(A(u))=0$, where $A$ is a non-decreasing function on $mathbb{R}$ such that $A(0)=0$ and $$A'(u_ell)>gamma'(t)=frac{A(u_ell)-A(u_r)}{u_ell-u_r}>A'(u_r)$$ (namely the Lax' entropy condition, where $u_ell, u_r$ are of the usual meanings). Prove that the shocks travel from left to right if $A$ is convex.




      I got frustrated after spending a long time on it (not sure if one is supposed to consider the shock speed $gamma'(t)$, which is always non-negative since $A$ is non-decreasing). I believe that I missed something here and that there is some very simple way to solve this problem. Can someone help me? Thanks.







      pde hyperbolic-equations






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      edited Dec 20 '18 at 9:54









      Harry49

      8,55431344




      8,55431344










      asked Oct 3 '16 at 8:28









      Liebster JugendtraumLiebster Jugendtraum

      1649




      1649






















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          $begingroup$

          The problem statement is slightly misleading. Indeed, it is assumed that the flux $A$ is convex and increasing. The speed of a shock wave is given by the Rankine-Hugoniot condition
          $$
          gamma'(t) = frac{A(u_ell) - A(u_r)}{u_ell - u_r}
          $$

          so as to ensure that the shock is a weak solution. Since $A$ is convex, the shock must also satisfy the Lax entropy condition
          $$
          A'(u_ell) > gamma'(t) > A'(u_ell)
          $$

          to be an entropy solution. To prove that $gamma'(t) > 0$, we use the fact that $A$ is non-decreasing.
          Therefore, admissible shocks propagate to the right if $A$ is convex and non-decreasing (note that $A(0)=0$ is not required). In the case of the inviscid Burgers' equation $A(u) = frac{1}{2}u^2$, this is only true for $u geq 0$.






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            $begingroup$

            The problem statement is slightly misleading. Indeed, it is assumed that the flux $A$ is convex and increasing. The speed of a shock wave is given by the Rankine-Hugoniot condition
            $$
            gamma'(t) = frac{A(u_ell) - A(u_r)}{u_ell - u_r}
            $$

            so as to ensure that the shock is a weak solution. Since $A$ is convex, the shock must also satisfy the Lax entropy condition
            $$
            A'(u_ell) > gamma'(t) > A'(u_ell)
            $$

            to be an entropy solution. To prove that $gamma'(t) > 0$, we use the fact that $A$ is non-decreasing.
            Therefore, admissible shocks propagate to the right if $A$ is convex and non-decreasing (note that $A(0)=0$ is not required). In the case of the inviscid Burgers' equation $A(u) = frac{1}{2}u^2$, this is only true for $u geq 0$.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              The problem statement is slightly misleading. Indeed, it is assumed that the flux $A$ is convex and increasing. The speed of a shock wave is given by the Rankine-Hugoniot condition
              $$
              gamma'(t) = frac{A(u_ell) - A(u_r)}{u_ell - u_r}
              $$

              so as to ensure that the shock is a weak solution. Since $A$ is convex, the shock must also satisfy the Lax entropy condition
              $$
              A'(u_ell) > gamma'(t) > A'(u_ell)
              $$

              to be an entropy solution. To prove that $gamma'(t) > 0$, we use the fact that $A$ is non-decreasing.
              Therefore, admissible shocks propagate to the right if $A$ is convex and non-decreasing (note that $A(0)=0$ is not required). In the case of the inviscid Burgers' equation $A(u) = frac{1}{2}u^2$, this is only true for $u geq 0$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                The problem statement is slightly misleading. Indeed, it is assumed that the flux $A$ is convex and increasing. The speed of a shock wave is given by the Rankine-Hugoniot condition
                $$
                gamma'(t) = frac{A(u_ell) - A(u_r)}{u_ell - u_r}
                $$

                so as to ensure that the shock is a weak solution. Since $A$ is convex, the shock must also satisfy the Lax entropy condition
                $$
                A'(u_ell) > gamma'(t) > A'(u_ell)
                $$

                to be an entropy solution. To prove that $gamma'(t) > 0$, we use the fact that $A$ is non-decreasing.
                Therefore, admissible shocks propagate to the right if $A$ is convex and non-decreasing (note that $A(0)=0$ is not required). In the case of the inviscid Burgers' equation $A(u) = frac{1}{2}u^2$, this is only true for $u geq 0$.






                share|cite|improve this answer











                $endgroup$



                The problem statement is slightly misleading. Indeed, it is assumed that the flux $A$ is convex and increasing. The speed of a shock wave is given by the Rankine-Hugoniot condition
                $$
                gamma'(t) = frac{A(u_ell) - A(u_r)}{u_ell - u_r}
                $$

                so as to ensure that the shock is a weak solution. Since $A$ is convex, the shock must also satisfy the Lax entropy condition
                $$
                A'(u_ell) > gamma'(t) > A'(u_ell)
                $$

                to be an entropy solution. To prove that $gamma'(t) > 0$, we use the fact that $A$ is non-decreasing.
                Therefore, admissible shocks propagate to the right if $A$ is convex and non-decreasing (note that $A(0)=0$ is not required). In the case of the inviscid Burgers' equation $A(u) = frac{1}{2}u^2$, this is only true for $u geq 0$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 20 '18 at 13:30

























                answered Dec 20 '18 at 10:22









                Harry49Harry49

                8,55431344




                8,55431344






























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