Prove Lax entropy condition for conservation law with convex flux












1












$begingroup$


A conservation law $u_t + phi(u)_x = 0$ is considered.




For a flux $phi(u)$ satisfying $phi'' (u) > 0$, show that the entropy
condition in the form: $u(x + a, t) − u(x, t) leq frac{aE}{t}$, for
some $E > 0$ and all $x, t, a > 0$, implies the inequality,
$phi'(u^- ) > gamma'(t) > phi'(u^+ )$, where $u^-$ and $u^+$ are the values of $u$ behind and in front of the shock respectively, and $gamma'(t)$ is the shock speed.




So we have $u(x+a,t)-u(x,t)=(phi')^{-1}(frac{x+a}{t})-(phi')^{-1}(frac{x}{t})$. I don't know how to proceed. Any help would be appreciated.






pde hyperbolic-equations







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    A conservation law $u_t + phi(u)_x = 0$ is considered.




    For a flux $phi(u)$ satisfying $phi'' (u) > 0$, show that the entropy
    condition in the form: $u(x + a, t) − u(x, t) leq frac{aE}{t}$, for
    some $E > 0$ and all $x, t, a > 0$, implies the inequality,
    $phi'(u^- ) > gamma'(t) > phi'(u^+ )$, where $u^-$ and $u^+$ are the values of $u$ behind and in front of the shock respectively, and $gamma'(t)$ is the shock speed.




    So we have $u(x+a,t)-u(x,t)=(phi')^{-1}(frac{x+a}{t})-(phi')^{-1}(frac{x}{t})$. I don't know how to proceed. Any help would be appreciated.






    pde hyperbolic-equations







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      A conservation law $u_t + phi(u)_x = 0$ is considered.




      For a flux $phi(u)$ satisfying $phi'' (u) > 0$, show that the entropy
      condition in the form: $u(x + a, t) − u(x, t) leq frac{aE}{t}$, for
      some $E > 0$ and all $x, t, a > 0$, implies the inequality,
      $phi'(u^- ) > gamma'(t) > phi'(u^+ )$, where $u^-$ and $u^+$ are the values of $u$ behind and in front of the shock respectively, and $gamma'(t)$ is the shock speed.




      So we have $u(x+a,t)-u(x,t)=(phi')^{-1}(frac{x+a}{t})-(phi')^{-1}(frac{x}{t})$. I don't know how to proceed. Any help would be appreciated.






      pde hyperbolic-equations







      share|cite|improve this question











      $endgroup$




      A conservation law $u_t + phi(u)_x = 0$ is considered.




      For a flux $phi(u)$ satisfying $phi'' (u) > 0$, show that the entropy
      condition in the form: $u(x + a, t) − u(x, t) leq frac{aE}{t}$, for
      some $E > 0$ and all $x, t, a > 0$, implies the inequality,
      $phi'(u^- ) > gamma'(t) > phi'(u^+ )$, where $u^-$ and $u^+$ are the values of $u$ behind and in front of the shock respectively, and $gamma'(t)$ is the shock speed.




      So we have $u(x+a,t)-u(x,t)=(phi')^{-1}(frac{x+a}{t})-(phi')^{-1}(frac{x}{t})$. I don't know how to proceed. Any help would be appreciated.






      pde hyperbolic-equations




      pde hyperbolic-equations






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 20 '18 at 9:56









      Harry49

      6,21331132




      6,21331132










      asked Dec 8 '18 at 15:16









      dxdydzdxdydz

      33110




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          $begingroup$

          Let us consider that $u$ has a single discontinuity with jump $u^+-u^-$ at $x=gamma(t)$. Since $u$ is nonsmooth, we go back to the integral definition of the conservation law
          $$
          frac{text d}{text d t} int_{x_1}^{x_2} u,text d x = phi (u|_{x=x_1})- phi (u|_{x=x_2}) .
          $$
          If $x_1<gamma(t)<x_2$, the identity
          $$
          frac{text d}{text d t} int_{x_1}^{x_2} u,text d x = int_{x_1}^{gamma(t)} u_t,text d x + int_{gamma(t)}^{x_2} u_t,text d x + gamma'(t)left(u^--u^+right)
          $$
          and the conservation law over $[x_1, gamma (t)]$, $[gamma (t), x_2]$ yield the Rankine-Hugoniot condition.
          Now, using the convexity of the flux $phi$ over $[minlbrace u^-, u^+rbrace, maxlbrace u^-, u^+rbrace]$, we have the inequalities
          $$
          phi' (minlbrace u^-, u^+rbrace) < overbrace{frac{phi (u^+) - phi (u^-)}{u^+-u^-}}^{gamma'(t)} < phi' (maxlbrace u^-, u^+rbrace) .
          $$
          Taking the limit $epsilonto 0^+$ in the entropy condition
          $$
          u|_{x=gamma (t)+epsilon} - u|_{x=gamma (t)-epsilon} leq frac {Eepsilon}{t}
          $$
          imposes $u^+ < u^-$. Hence, the Lax entropy condition is obtained:
          $$
          phi' (u^-) > {gamma'(t)} > phi' (u^+) .
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what is $min u^±$? I suppose $min(u^-,u^+)$?
            $endgroup$
            – dxdydz
            Dec 9 '18 at 3:02












          • $begingroup$
            @dxdydz yes, exactly. Answer edited
            $endgroup$
            – Harry49
            Dec 9 '18 at 9:51











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          1 Answer
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          0












          $begingroup$

          Let us consider that $u$ has a single discontinuity with jump $u^+-u^-$ at $x=gamma(t)$. Since $u$ is nonsmooth, we go back to the integral definition of the conservation law
          $$
          frac{text d}{text d t} int_{x_1}^{x_2} u,text d x = phi (u|_{x=x_1})- phi (u|_{x=x_2}) .
          $$
          If $x_1<gamma(t)<x_2$, the identity
          $$
          frac{text d}{text d t} int_{x_1}^{x_2} u,text d x = int_{x_1}^{gamma(t)} u_t,text d x + int_{gamma(t)}^{x_2} u_t,text d x + gamma'(t)left(u^--u^+right)
          $$
          and the conservation law over $[x_1, gamma (t)]$, $[gamma (t), x_2]$ yield the Rankine-Hugoniot condition.
          Now, using the convexity of the flux $phi$ over $[minlbrace u^-, u^+rbrace, maxlbrace u^-, u^+rbrace]$, we have the inequalities
          $$
          phi' (minlbrace u^-, u^+rbrace) < overbrace{frac{phi (u^+) - phi (u^-)}{u^+-u^-}}^{gamma'(t)} < phi' (maxlbrace u^-, u^+rbrace) .
          $$
          Taking the limit $epsilonto 0^+$ in the entropy condition
          $$
          u|_{x=gamma (t)+epsilon} - u|_{x=gamma (t)-epsilon} leq frac {Eepsilon}{t}
          $$
          imposes $u^+ < u^-$. Hence, the Lax entropy condition is obtained:
          $$
          phi' (u^-) > {gamma'(t)} > phi' (u^+) .
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what is $min u^±$? I suppose $min(u^-,u^+)$?
            $endgroup$
            – dxdydz
            Dec 9 '18 at 3:02












          • $begingroup$
            @dxdydz yes, exactly. Answer edited
            $endgroup$
            – Harry49
            Dec 9 '18 at 9:51
















          0












          $begingroup$

          Let us consider that $u$ has a single discontinuity with jump $u^+-u^-$ at $x=gamma(t)$. Since $u$ is nonsmooth, we go back to the integral definition of the conservation law
          $$
          frac{text d}{text d t} int_{x_1}^{x_2} u,text d x = phi (u|_{x=x_1})- phi (u|_{x=x_2}) .
          $$
          If $x_1<gamma(t)<x_2$, the identity
          $$
          frac{text d}{text d t} int_{x_1}^{x_2} u,text d x = int_{x_1}^{gamma(t)} u_t,text d x + int_{gamma(t)}^{x_2} u_t,text d x + gamma'(t)left(u^--u^+right)
          $$
          and the conservation law over $[x_1, gamma (t)]$, $[gamma (t), x_2]$ yield the Rankine-Hugoniot condition.
          Now, using the convexity of the flux $phi$ over $[minlbrace u^-, u^+rbrace, maxlbrace u^-, u^+rbrace]$, we have the inequalities
          $$
          phi' (minlbrace u^-, u^+rbrace) < overbrace{frac{phi (u^+) - phi (u^-)}{u^+-u^-}}^{gamma'(t)} < phi' (maxlbrace u^-, u^+rbrace) .
          $$
          Taking the limit $epsilonto 0^+$ in the entropy condition
          $$
          u|_{x=gamma (t)+epsilon} - u|_{x=gamma (t)-epsilon} leq frac {Eepsilon}{t}
          $$
          imposes $u^+ < u^-$. Hence, the Lax entropy condition is obtained:
          $$
          phi' (u^-) > {gamma'(t)} > phi' (u^+) .
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what is $min u^±$? I suppose $min(u^-,u^+)$?
            $endgroup$
            – dxdydz
            Dec 9 '18 at 3:02












          • $begingroup$
            @dxdydz yes, exactly. Answer edited
            $endgroup$
            – Harry49
            Dec 9 '18 at 9:51














          0












          0








          0





          $begingroup$

          Let us consider that $u$ has a single discontinuity with jump $u^+-u^-$ at $x=gamma(t)$. Since $u$ is nonsmooth, we go back to the integral definition of the conservation law
          $$
          frac{text d}{text d t} int_{x_1}^{x_2} u,text d x = phi (u|_{x=x_1})- phi (u|_{x=x_2}) .
          $$
          If $x_1<gamma(t)<x_2$, the identity
          $$
          frac{text d}{text d t} int_{x_1}^{x_2} u,text d x = int_{x_1}^{gamma(t)} u_t,text d x + int_{gamma(t)}^{x_2} u_t,text d x + gamma'(t)left(u^--u^+right)
          $$
          and the conservation law over $[x_1, gamma (t)]$, $[gamma (t), x_2]$ yield the Rankine-Hugoniot condition.
          Now, using the convexity of the flux $phi$ over $[minlbrace u^-, u^+rbrace, maxlbrace u^-, u^+rbrace]$, we have the inequalities
          $$
          phi' (minlbrace u^-, u^+rbrace) < overbrace{frac{phi (u^+) - phi (u^-)}{u^+-u^-}}^{gamma'(t)} < phi' (maxlbrace u^-, u^+rbrace) .
          $$
          Taking the limit $epsilonto 0^+$ in the entropy condition
          $$
          u|_{x=gamma (t)+epsilon} - u|_{x=gamma (t)-epsilon} leq frac {Eepsilon}{t}
          $$
          imposes $u^+ < u^-$. Hence, the Lax entropy condition is obtained:
          $$
          phi' (u^-) > {gamma'(t)} > phi' (u^+) .
          $$






          share|cite|improve this answer











          $endgroup$



          Let us consider that $u$ has a single discontinuity with jump $u^+-u^-$ at $x=gamma(t)$. Since $u$ is nonsmooth, we go back to the integral definition of the conservation law
          $$
          frac{text d}{text d t} int_{x_1}^{x_2} u,text d x = phi (u|_{x=x_1})- phi (u|_{x=x_2}) .
          $$
          If $x_1<gamma(t)<x_2$, the identity
          $$
          frac{text d}{text d t} int_{x_1}^{x_2} u,text d x = int_{x_1}^{gamma(t)} u_t,text d x + int_{gamma(t)}^{x_2} u_t,text d x + gamma'(t)left(u^--u^+right)
          $$
          and the conservation law over $[x_1, gamma (t)]$, $[gamma (t), x_2]$ yield the Rankine-Hugoniot condition.
          Now, using the convexity of the flux $phi$ over $[minlbrace u^-, u^+rbrace, maxlbrace u^-, u^+rbrace]$, we have the inequalities
          $$
          phi' (minlbrace u^-, u^+rbrace) < overbrace{frac{phi (u^+) - phi (u^-)}{u^+-u^-}}^{gamma'(t)} < phi' (maxlbrace u^-, u^+rbrace) .
          $$
          Taking the limit $epsilonto 0^+$ in the entropy condition
          $$
          u|_{x=gamma (t)+epsilon} - u|_{x=gamma (t)-epsilon} leq frac {Eepsilon}{t}
          $$
          imposes $u^+ < u^-$. Hence, the Lax entropy condition is obtained:
          $$
          phi' (u^-) > {gamma'(t)} > phi' (u^+) .
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 10:00

























          answered Dec 8 '18 at 19:40









          Harry49Harry49

          6,21331132




          6,21331132












          • $begingroup$
            what is $min u^±$? I suppose $min(u^-,u^+)$?
            $endgroup$
            – dxdydz
            Dec 9 '18 at 3:02












          • $begingroup$
            @dxdydz yes, exactly. Answer edited
            $endgroup$
            – Harry49
            Dec 9 '18 at 9:51


















          • $begingroup$
            what is $min u^±$? I suppose $min(u^-,u^+)$?
            $endgroup$
            – dxdydz
            Dec 9 '18 at 3:02












          • $begingroup$
            @dxdydz yes, exactly. Answer edited
            $endgroup$
            – Harry49
            Dec 9 '18 at 9:51
















          $begingroup$
          what is $min u^±$? I suppose $min(u^-,u^+)$?
          $endgroup$
          – dxdydz
          Dec 9 '18 at 3:02






          $begingroup$
          what is $min u^±$? I suppose $min(u^-,u^+)$?
          $endgroup$
          – dxdydz
          Dec 9 '18 at 3:02














          $begingroup$
          @dxdydz yes, exactly. Answer edited
          $endgroup$
          – Harry49
          Dec 9 '18 at 9:51




          $begingroup$
          @dxdydz yes, exactly. Answer edited
          $endgroup$
          – Harry49
          Dec 9 '18 at 9:51


















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