Show that a solution of $u_t+(|u|^alpha)_x=0$ violates entropy condition












1












$begingroup$



Consider $$u_t+(|u|^alpha)_x=0, quadalpha>1$$ Given the initial condition
$$u(x,0)=begin{cases} 0, x<0\1,x>0end{cases}$$



a) Find a solution for $u(x, t)$ that is continuous for all $t > 0$ and satisfies the initial condition.



b) Find a solution for $u(x, t)$ that includes a shock obeying the
appropriate jump condition.



(c) Show that one of these solutions does not satisfy the entropy
condition.




My attempt:



We can rewrite the equation $$u_t+(|u|^alpha)_x=u_t+alpha u |u|^{alpha-2}u_x$$ then we parametrize begin{align} x(0,r) &= r \ t(0,r) &= 1 \ u(0,r) &= u_0 end{align}



The characteristics satisfying IVP are



begin{align}
x_s &= alpha u |u|^{alpha-2} \
implies x &= alpha u |u|^{alpha-2}t+r \
t_s &= 1 \
implies t &= s \
u_s &=0 \
implies u &= u_0 end{align}



The projection on $(x,t)$-plane is given by $$t = frac{x-r}{alpha u |u|^{alpha-2}}$$ So if $x<0$, the denominator is $0$. How do I find the solution?






pde hyperbolic-equations







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Consider $$u_t+(|u|^alpha)_x=0, quadalpha>1$$ Given the initial condition
    $$u(x,0)=begin{cases} 0, x<0\1,x>0end{cases}$$



    a) Find a solution for $u(x, t)$ that is continuous for all $t > 0$ and satisfies the initial condition.



    b) Find a solution for $u(x, t)$ that includes a shock obeying the
    appropriate jump condition.



    (c) Show that one of these solutions does not satisfy the entropy
    condition.




    My attempt:



    We can rewrite the equation $$u_t+(|u|^alpha)_x=u_t+alpha u |u|^{alpha-2}u_x$$ then we parametrize begin{align} x(0,r) &= r \ t(0,r) &= 1 \ u(0,r) &= u_0 end{align}



    The characteristics satisfying IVP are



    begin{align}
    x_s &= alpha u |u|^{alpha-2} \
    implies x &= alpha u |u|^{alpha-2}t+r \
    t_s &= 1 \
    implies t &= s \
    u_s &=0 \
    implies u &= u_0 end{align}



    The projection on $(x,t)$-plane is given by $$t = frac{x-r}{alpha u |u|^{alpha-2}}$$ So if $x<0$, the denominator is $0$. How do I find the solution?






    pde hyperbolic-equations







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Consider $$u_t+(|u|^alpha)_x=0, quadalpha>1$$ Given the initial condition
      $$u(x,0)=begin{cases} 0, x<0\1,x>0end{cases}$$



      a) Find a solution for $u(x, t)$ that is continuous for all $t > 0$ and satisfies the initial condition.



      b) Find a solution for $u(x, t)$ that includes a shock obeying the
      appropriate jump condition.



      (c) Show that one of these solutions does not satisfy the entropy
      condition.




      My attempt:



      We can rewrite the equation $$u_t+(|u|^alpha)_x=u_t+alpha u |u|^{alpha-2}u_x$$ then we parametrize begin{align} x(0,r) &= r \ t(0,r) &= 1 \ u(0,r) &= u_0 end{align}



      The characteristics satisfying IVP are



      begin{align}
      x_s &= alpha u |u|^{alpha-2} \
      implies x &= alpha u |u|^{alpha-2}t+r \
      t_s &= 1 \
      implies t &= s \
      u_s &=0 \
      implies u &= u_0 end{align}



      The projection on $(x,t)$-plane is given by $$t = frac{x-r}{alpha u |u|^{alpha-2}}$$ So if $x<0$, the denominator is $0$. How do I find the solution?






      pde hyperbolic-equations







      share|cite|improve this question











      $endgroup$





      Consider $$u_t+(|u|^alpha)_x=0, quadalpha>1$$ Given the initial condition
      $$u(x,0)=begin{cases} 0, x<0\1,x>0end{cases}$$



      a) Find a solution for $u(x, t)$ that is continuous for all $t > 0$ and satisfies the initial condition.



      b) Find a solution for $u(x, t)$ that includes a shock obeying the
      appropriate jump condition.



      (c) Show that one of these solutions does not satisfy the entropy
      condition.




      My attempt:



      We can rewrite the equation $$u_t+(|u|^alpha)_x=u_t+alpha u |u|^{alpha-2}u_x$$ then we parametrize begin{align} x(0,r) &= r \ t(0,r) &= 1 \ u(0,r) &= u_0 end{align}



      The characteristics satisfying IVP are



      begin{align}
      x_s &= alpha u |u|^{alpha-2} \
      implies x &= alpha u |u|^{alpha-2}t+r \
      t_s &= 1 \
      implies t &= s \
      u_s &=0 \
      implies u &= u_0 end{align}



      The projection on $(x,t)$-plane is given by $$t = frac{x-r}{alpha u |u|^{alpha-2}}$$ So if $x<0$, the denominator is $0$. How do I find the solution?






      pde hyperbolic-equations




      pde hyperbolic-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 10:09









      Harry49

      6,18331132




      6,18331132










      asked Dec 9 '18 at 5:06









      dxdydzdxdydz

      3009




      3009






















          1 Answer
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          0












          $begingroup$

          The characteristics are the curves $x=x_0+alpha u|u|^{alpha-2}t$ along which $u=u(x_0,0)=F(x_0)$ is constant.
          With the given initial data, we have
          $$
          u(x,t) = leftlbrace
          begin{aligned}
          &0 &&text{if}quad x <0\
          &U (x/t) &&text{if}quad 0 leq x leq alpha t\
          &1 &&text{if}quad alpha t <x
          end{aligned}
          right.
          $$
          where $U (x/t)$ is a rarefaction wave solution. The latter is obtained from assuming a smooth solution of the form $u (x,t) = U (xi)$ with $xi=x/t$. Using this self-similarity Ansatz and the PDE, one obtains
          $$
          U (x/t) = left.left (xitoalpha xi|xi|^{alpha-2}right)^{-1}right|_{xi=x/t} = left(frac{x}{alpha t}right)^frac {1}{alpha-1} .
          $$
          Concerning the shock wave solution
          $$
          u(x,t) = leftlbrace
          begin{aligned}
          &0 &&text{if}quad x <st\
          &1 &&text{if}quad st <x
          end{aligned}
          right.
          $$
          with
          $$
          s = frac{|1|^alpha - |0|^alpha}{1 - 0} = 1
          $$
          following from the Rankine-Hugoniot condition, one shows that this solution does not satisfy the Lax entropy condition since
          $$
          alpha, 0, |0|^{alpha-2} < s < alpha, 1, |1|^{alpha-2} .
          $$
          Indeed, the Lax entropy condition requires opposite inequalities.
          Finally, the entropy solution is the rarefaction wave.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How can we find s, the jump condition gives us $s=frac{|u^-|^alpha-|u^+|^alpha}{u^- - u^+}$, and how can we show that the shockwave solution is a weak solution? I know the formula I do not see how that condition is easy to check
            $endgroup$
            – dxdydz
            Dec 9 '18 at 16:26












          • $begingroup$
            thnx, got it, how can we show that the shock wave solution is a weak solution?
            $endgroup$
            – dxdydz
            Dec 9 '18 at 18:14










          • $begingroup$
            @dxdydz to prove that the shock is a weak solution, we can follow the steps in this post
            $endgroup$
            – Harry49
            Dec 9 '18 at 19:42











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          0












          $begingroup$

          The characteristics are the curves $x=x_0+alpha u|u|^{alpha-2}t$ along which $u=u(x_0,0)=F(x_0)$ is constant.
          With the given initial data, we have
          $$
          u(x,t) = leftlbrace
          begin{aligned}
          &0 &&text{if}quad x <0\
          &U (x/t) &&text{if}quad 0 leq x leq alpha t\
          &1 &&text{if}quad alpha t <x
          end{aligned}
          right.
          $$
          where $U (x/t)$ is a rarefaction wave solution. The latter is obtained from assuming a smooth solution of the form $u (x,t) = U (xi)$ with $xi=x/t$. Using this self-similarity Ansatz and the PDE, one obtains
          $$
          U (x/t) = left.left (xitoalpha xi|xi|^{alpha-2}right)^{-1}right|_{xi=x/t} = left(frac{x}{alpha t}right)^frac {1}{alpha-1} .
          $$
          Concerning the shock wave solution
          $$
          u(x,t) = leftlbrace
          begin{aligned}
          &0 &&text{if}quad x <st\
          &1 &&text{if}quad st <x
          end{aligned}
          right.
          $$
          with
          $$
          s = frac{|1|^alpha - |0|^alpha}{1 - 0} = 1
          $$
          following from the Rankine-Hugoniot condition, one shows that this solution does not satisfy the Lax entropy condition since
          $$
          alpha, 0, |0|^{alpha-2} < s < alpha, 1, |1|^{alpha-2} .
          $$
          Indeed, the Lax entropy condition requires opposite inequalities.
          Finally, the entropy solution is the rarefaction wave.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How can we find s, the jump condition gives us $s=frac{|u^-|^alpha-|u^+|^alpha}{u^- - u^+}$, and how can we show that the shockwave solution is a weak solution? I know the formula I do not see how that condition is easy to check
            $endgroup$
            – dxdydz
            Dec 9 '18 at 16:26












          • $begingroup$
            thnx, got it, how can we show that the shock wave solution is a weak solution?
            $endgroup$
            – dxdydz
            Dec 9 '18 at 18:14










          • $begingroup$
            @dxdydz to prove that the shock is a weak solution, we can follow the steps in this post
            $endgroup$
            – Harry49
            Dec 9 '18 at 19:42
















          0












          $begingroup$

          The characteristics are the curves $x=x_0+alpha u|u|^{alpha-2}t$ along which $u=u(x_0,0)=F(x_0)$ is constant.
          With the given initial data, we have
          $$
          u(x,t) = leftlbrace
          begin{aligned}
          &0 &&text{if}quad x <0\
          &U (x/t) &&text{if}quad 0 leq x leq alpha t\
          &1 &&text{if}quad alpha t <x
          end{aligned}
          right.
          $$
          where $U (x/t)$ is a rarefaction wave solution. The latter is obtained from assuming a smooth solution of the form $u (x,t) = U (xi)$ with $xi=x/t$. Using this self-similarity Ansatz and the PDE, one obtains
          $$
          U (x/t) = left.left (xitoalpha xi|xi|^{alpha-2}right)^{-1}right|_{xi=x/t} = left(frac{x}{alpha t}right)^frac {1}{alpha-1} .
          $$
          Concerning the shock wave solution
          $$
          u(x,t) = leftlbrace
          begin{aligned}
          &0 &&text{if}quad x <st\
          &1 &&text{if}quad st <x
          end{aligned}
          right.
          $$
          with
          $$
          s = frac{|1|^alpha - |0|^alpha}{1 - 0} = 1
          $$
          following from the Rankine-Hugoniot condition, one shows that this solution does not satisfy the Lax entropy condition since
          $$
          alpha, 0, |0|^{alpha-2} < s < alpha, 1, |1|^{alpha-2} .
          $$
          Indeed, the Lax entropy condition requires opposite inequalities.
          Finally, the entropy solution is the rarefaction wave.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How can we find s, the jump condition gives us $s=frac{|u^-|^alpha-|u^+|^alpha}{u^- - u^+}$, and how can we show that the shockwave solution is a weak solution? I know the formula I do not see how that condition is easy to check
            $endgroup$
            – dxdydz
            Dec 9 '18 at 16:26












          • $begingroup$
            thnx, got it, how can we show that the shock wave solution is a weak solution?
            $endgroup$
            – dxdydz
            Dec 9 '18 at 18:14










          • $begingroup$
            @dxdydz to prove that the shock is a weak solution, we can follow the steps in this post
            $endgroup$
            – Harry49
            Dec 9 '18 at 19:42














          0












          0








          0





          $begingroup$

          The characteristics are the curves $x=x_0+alpha u|u|^{alpha-2}t$ along which $u=u(x_0,0)=F(x_0)$ is constant.
          With the given initial data, we have
          $$
          u(x,t) = leftlbrace
          begin{aligned}
          &0 &&text{if}quad x <0\
          &U (x/t) &&text{if}quad 0 leq x leq alpha t\
          &1 &&text{if}quad alpha t <x
          end{aligned}
          right.
          $$
          where $U (x/t)$ is a rarefaction wave solution. The latter is obtained from assuming a smooth solution of the form $u (x,t) = U (xi)$ with $xi=x/t$. Using this self-similarity Ansatz and the PDE, one obtains
          $$
          U (x/t) = left.left (xitoalpha xi|xi|^{alpha-2}right)^{-1}right|_{xi=x/t} = left(frac{x}{alpha t}right)^frac {1}{alpha-1} .
          $$
          Concerning the shock wave solution
          $$
          u(x,t) = leftlbrace
          begin{aligned}
          &0 &&text{if}quad x <st\
          &1 &&text{if}quad st <x
          end{aligned}
          right.
          $$
          with
          $$
          s = frac{|1|^alpha - |0|^alpha}{1 - 0} = 1
          $$
          following from the Rankine-Hugoniot condition, one shows that this solution does not satisfy the Lax entropy condition since
          $$
          alpha, 0, |0|^{alpha-2} < s < alpha, 1, |1|^{alpha-2} .
          $$
          Indeed, the Lax entropy condition requires opposite inequalities.
          Finally, the entropy solution is the rarefaction wave.






          share|cite|improve this answer











          $endgroup$



          The characteristics are the curves $x=x_0+alpha u|u|^{alpha-2}t$ along which $u=u(x_0,0)=F(x_0)$ is constant.
          With the given initial data, we have
          $$
          u(x,t) = leftlbrace
          begin{aligned}
          &0 &&text{if}quad x <0\
          &U (x/t) &&text{if}quad 0 leq x leq alpha t\
          &1 &&text{if}quad alpha t <x
          end{aligned}
          right.
          $$
          where $U (x/t)$ is a rarefaction wave solution. The latter is obtained from assuming a smooth solution of the form $u (x,t) = U (xi)$ with $xi=x/t$. Using this self-similarity Ansatz and the PDE, one obtains
          $$
          U (x/t) = left.left (xitoalpha xi|xi|^{alpha-2}right)^{-1}right|_{xi=x/t} = left(frac{x}{alpha t}right)^frac {1}{alpha-1} .
          $$
          Concerning the shock wave solution
          $$
          u(x,t) = leftlbrace
          begin{aligned}
          &0 &&text{if}quad x <st\
          &1 &&text{if}quad st <x
          end{aligned}
          right.
          $$
          with
          $$
          s = frac{|1|^alpha - |0|^alpha}{1 - 0} = 1
          $$
          following from the Rankine-Hugoniot condition, one shows that this solution does not satisfy the Lax entropy condition since
          $$
          alpha, 0, |0|^{alpha-2} < s < alpha, 1, |1|^{alpha-2} .
          $$
          Indeed, the Lax entropy condition requires opposite inequalities.
          Finally, the entropy solution is the rarefaction wave.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 17:43

























          answered Dec 9 '18 at 10:54









          Harry49Harry49

          6,18331132




          6,18331132












          • $begingroup$
            How can we find s, the jump condition gives us $s=frac{|u^-|^alpha-|u^+|^alpha}{u^- - u^+}$, and how can we show that the shockwave solution is a weak solution? I know the formula I do not see how that condition is easy to check
            $endgroup$
            – dxdydz
            Dec 9 '18 at 16:26












          • $begingroup$
            thnx, got it, how can we show that the shock wave solution is a weak solution?
            $endgroup$
            – dxdydz
            Dec 9 '18 at 18:14










          • $begingroup$
            @dxdydz to prove that the shock is a weak solution, we can follow the steps in this post
            $endgroup$
            – Harry49
            Dec 9 '18 at 19:42


















          • $begingroup$
            How can we find s, the jump condition gives us $s=frac{|u^-|^alpha-|u^+|^alpha}{u^- - u^+}$, and how can we show that the shockwave solution is a weak solution? I know the formula I do not see how that condition is easy to check
            $endgroup$
            – dxdydz
            Dec 9 '18 at 16:26












          • $begingroup$
            thnx, got it, how can we show that the shock wave solution is a weak solution?
            $endgroup$
            – dxdydz
            Dec 9 '18 at 18:14










          • $begingroup$
            @dxdydz to prove that the shock is a weak solution, we can follow the steps in this post
            $endgroup$
            – Harry49
            Dec 9 '18 at 19:42
















          $begingroup$
          How can we find s, the jump condition gives us $s=frac{|u^-|^alpha-|u^+|^alpha}{u^- - u^+}$, and how can we show that the shockwave solution is a weak solution? I know the formula I do not see how that condition is easy to check
          $endgroup$
          – dxdydz
          Dec 9 '18 at 16:26






          $begingroup$
          How can we find s, the jump condition gives us $s=frac{|u^-|^alpha-|u^+|^alpha}{u^- - u^+}$, and how can we show that the shockwave solution is a weak solution? I know the formula I do not see how that condition is easy to check
          $endgroup$
          – dxdydz
          Dec 9 '18 at 16:26














          $begingroup$
          thnx, got it, how can we show that the shock wave solution is a weak solution?
          $endgroup$
          – dxdydz
          Dec 9 '18 at 18:14




          $begingroup$
          thnx, got it, how can we show that the shock wave solution is a weak solution?
          $endgroup$
          – dxdydz
          Dec 9 '18 at 18:14












          $begingroup$
          @dxdydz to prove that the shock is a weak solution, we can follow the steps in this post
          $endgroup$
          – Harry49
          Dec 9 '18 at 19:42




          $begingroup$
          @dxdydz to prove that the shock is a weak solution, we can follow the steps in this post
          $endgroup$
          – Harry49
          Dec 9 '18 at 19:42


















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