Quasi-linear pde $u_t + x u u_x = 0$, find shock time












2












$begingroup$



Using the characteristics method, show that the Cauchy Problem for the
quasi-linear equation $$u_t + x u u_x = 0 qquad u(0, x) = phi(x) =
frac pi 2 - arctan(x)$$ has two shock times, $t^*_pm $, one in the
future and one in the past. Determine $t^*_pm $.




The characteristic curves are the solutions to the ODE system
$$ dot t = 1 qquad t(0) = 0$$
$$ dot x = ux qquad x(0) = x_0$$
$$ dot u = 0 qquad u(0) = phi (x_0)$$
This yields the solutions $$x(t) = exp(tphi(x_0))+x_0+1$$
If this expression can be inverted into a function $x_0(t, x)$ then there is no shock, but if that is the case then $x(t)$ must be monotone as function of $x_0$. Thus, define $F_t(x) = exp(tphi(x))+x+1$ : I need to find the values of $t$ for which $F' >0$ for all $x$. This reduces to:
$$ frac s {1+x^2} expleft (sleft (frac pi 2-arctan(x)right )right ) < 1$$
At which point I am stuck. I cannot solve this inequality, but also I have the nagging feeling that I made a mistake somewhere earlier. Did I? And what are the shock times?






pde characteristics







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    2












    $begingroup$



    Using the characteristics method, show that the Cauchy Problem for the
    quasi-linear equation $$u_t + x u u_x = 0 qquad u(0, x) = phi(x) =
    frac pi 2 - arctan(x)$$ has two shock times, $t^*_pm $, one in the
    future and one in the past. Determine $t^*_pm $.




    The characteristic curves are the solutions to the ODE system
    $$ dot t = 1 qquad t(0) = 0$$
    $$ dot x = ux qquad x(0) = x_0$$
    $$ dot u = 0 qquad u(0) = phi (x_0)$$
    This yields the solutions $$x(t) = exp(tphi(x_0))+x_0+1$$
    If this expression can be inverted into a function $x_0(t, x)$ then there is no shock, but if that is the case then $x(t)$ must be monotone as function of $x_0$. Thus, define $F_t(x) = exp(tphi(x))+x+1$ : I need to find the values of $t$ for which $F' >0$ for all $x$. This reduces to:
    $$ frac s {1+x^2} expleft (sleft (frac pi 2-arctan(x)right )right ) < 1$$
    At which point I am stuck. I cannot solve this inequality, but also I have the nagging feeling that I made a mistake somewhere earlier. Did I? And what are the shock times?






    pde characteristics







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      2



      $begingroup$



      Using the characteristics method, show that the Cauchy Problem for the
      quasi-linear equation $$u_t + x u u_x = 0 qquad u(0, x) = phi(x) =
      frac pi 2 - arctan(x)$$ has two shock times, $t^*_pm $, one in the
      future and one in the past. Determine $t^*_pm $.




      The characteristic curves are the solutions to the ODE system
      $$ dot t = 1 qquad t(0) = 0$$
      $$ dot x = ux qquad x(0) = x_0$$
      $$ dot u = 0 qquad u(0) = phi (x_0)$$
      This yields the solutions $$x(t) = exp(tphi(x_0))+x_0+1$$
      If this expression can be inverted into a function $x_0(t, x)$ then there is no shock, but if that is the case then $x(t)$ must be monotone as function of $x_0$. Thus, define $F_t(x) = exp(tphi(x))+x+1$ : I need to find the values of $t$ for which $F' >0$ for all $x$. This reduces to:
      $$ frac s {1+x^2} expleft (sleft (frac pi 2-arctan(x)right )right ) < 1$$
      At which point I am stuck. I cannot solve this inequality, but also I have the nagging feeling that I made a mistake somewhere earlier. Did I? And what are the shock times?






      pde characteristics







      share|cite|improve this question











      $endgroup$





      Using the characteristics method, show that the Cauchy Problem for the
      quasi-linear equation $$u_t + x u u_x = 0 qquad u(0, x) = phi(x) =
      frac pi 2 - arctan(x)$$ has two shock times, $t^*_pm $, one in the
      future and one in the past. Determine $t^*_pm $.




      The characteristic curves are the solutions to the ODE system
      $$ dot t = 1 qquad t(0) = 0$$
      $$ dot x = ux qquad x(0) = x_0$$
      $$ dot u = 0 qquad u(0) = phi (x_0)$$
      This yields the solutions $$x(t) = exp(tphi(x_0))+x_0+1$$
      If this expression can be inverted into a function $x_0(t, x)$ then there is no shock, but if that is the case then $x(t)$ must be monotone as function of $x_0$. Thus, define $F_t(x) = exp(tphi(x))+x+1$ : I need to find the values of $t$ for which $F' >0$ for all $x$. This reduces to:
      $$ frac s {1+x^2} expleft (sleft (frac pi 2-arctan(x)right )right ) < 1$$
      At which point I am stuck. I cannot solve this inequality, but also I have the nagging feeling that I made a mistake somewhere earlier. Did I? And what are the shock times?






      pde characteristics




      pde characteristics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 13:52









      Harry49

      6,17331132




      6,17331132










      asked Sep 14 '18 at 15:19









      Riccardo OrlandoRiccardo Orlando

      1,654517




      1,654517






















          2 Answers
          2






          active

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          1












          $begingroup$

          As outlined by OP's self-answer,
          $$
          x(t) = x_0exp(tphi(x_0)) , .
          $$
          Several methods can be used to find the shock times. Here we examine the non-ambiguous dependence to the initial data. One observes that $text d x/text d x_0$ vanishes at $t=t_S$ such that
          $$
          t_S = frac{-1}{x_0phi'(x_0)} = x_0 + frac{1}{x_0} , .
          $$
          Thus, for positive times, the classical solution breaks at the breaking time
          $$t^*_+ = inf_{x_0>0} t_S = 2, .$$
          Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
            $endgroup$
            – Riccardo Orlando
            Sep 15 '18 at 11:10



















          1












          $begingroup$

          There is no error in the problem statement. Instead, I made a silly mistake in solving the ODE system for th echaracteristics, wich turn out to be
          $$ x = x_0 exp(tphi) $$
          From here on, the method I was trying to use, and that Harry49 explained in his answer.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            As outlined by OP's self-answer,
            $$
            x(t) = x_0exp(tphi(x_0)) , .
            $$
            Several methods can be used to find the shock times. Here we examine the non-ambiguous dependence to the initial data. One observes that $text d x/text d x_0$ vanishes at $t=t_S$ such that
            $$
            t_S = frac{-1}{x_0phi'(x_0)} = x_0 + frac{1}{x_0} , .
            $$
            Thus, for positive times, the classical solution breaks at the breaking time
            $$t^*_+ = inf_{x_0>0} t_S = 2, .$$
            Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
              $endgroup$
              – Riccardo Orlando
              Sep 15 '18 at 11:10
















            1












            $begingroup$

            As outlined by OP's self-answer,
            $$
            x(t) = x_0exp(tphi(x_0)) , .
            $$
            Several methods can be used to find the shock times. Here we examine the non-ambiguous dependence to the initial data. One observes that $text d x/text d x_0$ vanishes at $t=t_S$ such that
            $$
            t_S = frac{-1}{x_0phi'(x_0)} = x_0 + frac{1}{x_0} , .
            $$
            Thus, for positive times, the classical solution breaks at the breaking time
            $$t^*_+ = inf_{x_0>0} t_S = 2, .$$
            Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
              $endgroup$
              – Riccardo Orlando
              Sep 15 '18 at 11:10














            1












            1








            1





            $begingroup$

            As outlined by OP's self-answer,
            $$
            x(t) = x_0exp(tphi(x_0)) , .
            $$
            Several methods can be used to find the shock times. Here we examine the non-ambiguous dependence to the initial data. One observes that $text d x/text d x_0$ vanishes at $t=t_S$ such that
            $$
            t_S = frac{-1}{x_0phi'(x_0)} = x_0 + frac{1}{x_0} , .
            $$
            Thus, for positive times, the classical solution breaks at the breaking time
            $$t^*_+ = inf_{x_0>0} t_S = 2, .$$
            Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.






            share|cite|improve this answer











            $endgroup$



            As outlined by OP's self-answer,
            $$
            x(t) = x_0exp(tphi(x_0)) , .
            $$
            Several methods can be used to find the shock times. Here we examine the non-ambiguous dependence to the initial data. One observes that $text d x/text d x_0$ vanishes at $t=t_S$ such that
            $$
            t_S = frac{-1}{x_0phi'(x_0)} = x_0 + frac{1}{x_0} , .
            $$
            Thus, for positive times, the classical solution breaks at the breaking time
            $$t^*_+ = inf_{x_0>0} t_S = 2, .$$
            Similarly one gets $t^*_- = -2$ for negative times. This may be confirmed by a plot of the characteristics for several $x_0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 18 '18 at 18:33

























            answered Sep 15 '18 at 10:47









            Harry49Harry49

            6,17331132




            6,17331132












            • $begingroup$
              Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
              $endgroup$
              – Riccardo Orlando
              Sep 15 '18 at 11:10


















            • $begingroup$
              Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
              $endgroup$
              – Riccardo Orlando
              Sep 15 '18 at 11:10
















            $begingroup$
            Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
            $endgroup$
            – Riccardo Orlando
            Sep 15 '18 at 11:10




            $begingroup$
            Thank you for the answer. The exercise is supposed to be doable in 30 mins and with no notes or calculators. There might be an error in the statement.
            $endgroup$
            – Riccardo Orlando
            Sep 15 '18 at 11:10











            1












            $begingroup$

            There is no error in the problem statement. Instead, I made a silly mistake in solving the ODE system for th echaracteristics, wich turn out to be
            $$ x = x_0 exp(tphi) $$
            From here on, the method I was trying to use, and that Harry49 explained in his answer.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              There is no error in the problem statement. Instead, I made a silly mistake in solving the ODE system for th echaracteristics, wich turn out to be
              $$ x = x_0 exp(tphi) $$
              From here on, the method I was trying to use, and that Harry49 explained in his answer.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                There is no error in the problem statement. Instead, I made a silly mistake in solving the ODE system for th echaracteristics, wich turn out to be
                $$ x = x_0 exp(tphi) $$
                From here on, the method I was trying to use, and that Harry49 explained in his answer.






                share|cite|improve this answer









                $endgroup$



                There is no error in the problem statement. Instead, I made a silly mistake in solving the ODE system for th echaracteristics, wich turn out to be
                $$ x = x_0 exp(tphi) $$
                From here on, the method I was trying to use, and that Harry49 explained in his answer.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 18 '18 at 14:05









                Riccardo OrlandoRiccardo Orlando

                1,654517




                1,654517






























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