Homogeneous Quasi-linear PDE, Method of Characteristics.












4












$begingroup$


I want to solve the PDE:



$$frac{partial u}{partial t}+(1-2u)frac{partial u}{partial x}=0
$$



for $x leq0 $ and $t geq 0$, with initial condition $u(0,x)=frac{1}{2}$, and boundary condition;



$u(t,0)=left{begin{matrix}
frac{t}{4}+frac{1}{2}, 0leq tleq 2 & \
1, t geq 2&
end{matrix}right.$



by using the method of characteristics.



The method of characteristics gives us that $u$ is constant along characteristics which have equation defined by $frac{dx}{dt}=1-2u$.



For a characteristic originating from the x axis at a point $(x,t)=(x_0,0)$, this tells us that on characteristics defined by $x(t)=x_0$, we have $u(x(t),t)=u(x(0),0)=frac{1}{2}$. i.e. the characteristics originating on the x axis are vertical straight lines on which $u=frac{1}{2}$.



For a characteristic originating from the t axis at a point $(x,t)=(0,tau)$, we have 2 cases, depending on the value of $tau$.



In both cases $frac{du}{dt}=0$, on $frac{dx}{dt}=1-2u$, so u is constant on the characteristics, so $u(x(tau),tau)=u(0,tau)$ on characteristics defined by $x(t)=(1-2u(0,tau))t+C$, where $C$ is constant. Applying our boundary condition this gives us:



$u=frac{tau}{4}+frac{1}{2}$, on $x(t)=-frac{tau}{2}(t-tau)$, for $0leqtauleq2$, and:



$u=1$ on $x(t)=-(t-tau)$, for $taugeq2$.



Up to here I'm fine, but when I draw up a plot of the characteristics in the t-x plane I get very confused. The x-axis characteristics are all vertical lines. The t-axis characteristics for $0leqtauleq2$ are straight line originating from $(x,t)=(0,tau)$ with slope that varies. As $tau$, decreases from 2 to 0, the slope of these characteristics (in the t-x plane), goes from -1 and approaces $-infty$, as $tau$, approaches zero, meaning that each of these characteristics intersect.



As $u$ is constant on a characteristic, this is a contradiction, and so we expect a shock wave (implying a discontinuous solution).



Also for $taugeq2$, the characteristics are all parallel lines of slope -1, simply translated in the positive t direction (upwards in the t-x plane).



What is really confusing me is that the three types of characteristics are all intersecting, and that the $0leqtauleq2$ characteristics are intersecting themselves infinitely often, and this is maling it hard for me to apply the shock criterion. At the origin, the characteristics firast intersect, so we expect a shock will immediately form. It will propagate with shock velocity given by:
$frac{dx_s}{dt}=frac{(u(1-u))|_+-(u(1-u))|_-}{(u)|_+-(u)|_-}$



Where the subscripts denote the value of u just to the right or jsut to the left of the shock.



I really don't know how to continue from here, especially because of the three types of intersecting characteristics and the $0leqtauleq2$ characteristics intersecting with themselves. I would really be grateful for any hints or ideas or help. I've tried a lot of things to resolve this issue, but none have gotten anywhere and it would take a long time to write them out. I think if I can just undbrstnad the shock condition/formation here, I can solve this problem.










share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    I want to solve the PDE:



    $$frac{partial u}{partial t}+(1-2u)frac{partial u}{partial x}=0
    $$



    for $x leq0 $ and $t geq 0$, with initial condition $u(0,x)=frac{1}{2}$, and boundary condition;



    $u(t,0)=left{begin{matrix}
    frac{t}{4}+frac{1}{2}, 0leq tleq 2 & \
    1, t geq 2&
    end{matrix}right.$



    by using the method of characteristics.



    The method of characteristics gives us that $u$ is constant along characteristics which have equation defined by $frac{dx}{dt}=1-2u$.



    For a characteristic originating from the x axis at a point $(x,t)=(x_0,0)$, this tells us that on characteristics defined by $x(t)=x_0$, we have $u(x(t),t)=u(x(0),0)=frac{1}{2}$. i.e. the characteristics originating on the x axis are vertical straight lines on which $u=frac{1}{2}$.



    For a characteristic originating from the t axis at a point $(x,t)=(0,tau)$, we have 2 cases, depending on the value of $tau$.



    In both cases $frac{du}{dt}=0$, on $frac{dx}{dt}=1-2u$, so u is constant on the characteristics, so $u(x(tau),tau)=u(0,tau)$ on characteristics defined by $x(t)=(1-2u(0,tau))t+C$, where $C$ is constant. Applying our boundary condition this gives us:



    $u=frac{tau}{4}+frac{1}{2}$, on $x(t)=-frac{tau}{2}(t-tau)$, for $0leqtauleq2$, and:



    $u=1$ on $x(t)=-(t-tau)$, for $taugeq2$.



    Up to here I'm fine, but when I draw up a plot of the characteristics in the t-x plane I get very confused. The x-axis characteristics are all vertical lines. The t-axis characteristics for $0leqtauleq2$ are straight line originating from $(x,t)=(0,tau)$ with slope that varies. As $tau$, decreases from 2 to 0, the slope of these characteristics (in the t-x plane), goes from -1 and approaces $-infty$, as $tau$, approaches zero, meaning that each of these characteristics intersect.



    As $u$ is constant on a characteristic, this is a contradiction, and so we expect a shock wave (implying a discontinuous solution).



    Also for $taugeq2$, the characteristics are all parallel lines of slope -1, simply translated in the positive t direction (upwards in the t-x plane).



    What is really confusing me is that the three types of characteristics are all intersecting, and that the $0leqtauleq2$ characteristics are intersecting themselves infinitely often, and this is maling it hard for me to apply the shock criterion. At the origin, the characteristics firast intersect, so we expect a shock will immediately form. It will propagate with shock velocity given by:
    $frac{dx_s}{dt}=frac{(u(1-u))|_+-(u(1-u))|_-}{(u)|_+-(u)|_-}$



    Where the subscripts denote the value of u just to the right or jsut to the left of the shock.



    I really don't know how to continue from here, especially because of the three types of intersecting characteristics and the $0leqtauleq2$ characteristics intersecting with themselves. I would really be grateful for any hints or ideas or help. I've tried a lot of things to resolve this issue, but none have gotten anywhere and it would take a long time to write them out. I think if I can just undbrstnad the shock condition/formation here, I can solve this problem.










    share|cite|improve this question











    $endgroup$















      4












      4








      4


      4



      $begingroup$


      I want to solve the PDE:



      $$frac{partial u}{partial t}+(1-2u)frac{partial u}{partial x}=0
      $$



      for $x leq0 $ and $t geq 0$, with initial condition $u(0,x)=frac{1}{2}$, and boundary condition;



      $u(t,0)=left{begin{matrix}
      frac{t}{4}+frac{1}{2}, 0leq tleq 2 & \
      1, t geq 2&
      end{matrix}right.$



      by using the method of characteristics.



      The method of characteristics gives us that $u$ is constant along characteristics which have equation defined by $frac{dx}{dt}=1-2u$.



      For a characteristic originating from the x axis at a point $(x,t)=(x_0,0)$, this tells us that on characteristics defined by $x(t)=x_0$, we have $u(x(t),t)=u(x(0),0)=frac{1}{2}$. i.e. the characteristics originating on the x axis are vertical straight lines on which $u=frac{1}{2}$.



      For a characteristic originating from the t axis at a point $(x,t)=(0,tau)$, we have 2 cases, depending on the value of $tau$.



      In both cases $frac{du}{dt}=0$, on $frac{dx}{dt}=1-2u$, so u is constant on the characteristics, so $u(x(tau),tau)=u(0,tau)$ on characteristics defined by $x(t)=(1-2u(0,tau))t+C$, where $C$ is constant. Applying our boundary condition this gives us:



      $u=frac{tau}{4}+frac{1}{2}$, on $x(t)=-frac{tau}{2}(t-tau)$, for $0leqtauleq2$, and:



      $u=1$ on $x(t)=-(t-tau)$, for $taugeq2$.



      Up to here I'm fine, but when I draw up a plot of the characteristics in the t-x plane I get very confused. The x-axis characteristics are all vertical lines. The t-axis characteristics for $0leqtauleq2$ are straight line originating from $(x,t)=(0,tau)$ with slope that varies. As $tau$, decreases from 2 to 0, the slope of these characteristics (in the t-x plane), goes from -1 and approaces $-infty$, as $tau$, approaches zero, meaning that each of these characteristics intersect.



      As $u$ is constant on a characteristic, this is a contradiction, and so we expect a shock wave (implying a discontinuous solution).



      Also for $taugeq2$, the characteristics are all parallel lines of slope -1, simply translated in the positive t direction (upwards in the t-x plane).



      What is really confusing me is that the three types of characteristics are all intersecting, and that the $0leqtauleq2$ characteristics are intersecting themselves infinitely often, and this is maling it hard for me to apply the shock criterion. At the origin, the characteristics firast intersect, so we expect a shock will immediately form. It will propagate with shock velocity given by:
      $frac{dx_s}{dt}=frac{(u(1-u))|_+-(u(1-u))|_-}{(u)|_+-(u)|_-}$



      Where the subscripts denote the value of u just to the right or jsut to the left of the shock.



      I really don't know how to continue from here, especially because of the three types of intersecting characteristics and the $0leqtauleq2$ characteristics intersecting with themselves. I would really be grateful for any hints or ideas or help. I've tried a lot of things to resolve this issue, but none have gotten anywhere and it would take a long time to write them out. I think if I can just undbrstnad the shock condition/formation here, I can solve this problem.










      share|cite|improve this question











      $endgroup$




      I want to solve the PDE:



      $$frac{partial u}{partial t}+(1-2u)frac{partial u}{partial x}=0
      $$



      for $x leq0 $ and $t geq 0$, with initial condition $u(0,x)=frac{1}{2}$, and boundary condition;



      $u(t,0)=left{begin{matrix}
      frac{t}{4}+frac{1}{2}, 0leq tleq 2 & \
      1, t geq 2&
      end{matrix}right.$



      by using the method of characteristics.



      The method of characteristics gives us that $u$ is constant along characteristics which have equation defined by $frac{dx}{dt}=1-2u$.



      For a characteristic originating from the x axis at a point $(x,t)=(x_0,0)$, this tells us that on characteristics defined by $x(t)=x_0$, we have $u(x(t),t)=u(x(0),0)=frac{1}{2}$. i.e. the characteristics originating on the x axis are vertical straight lines on which $u=frac{1}{2}$.



      For a characteristic originating from the t axis at a point $(x,t)=(0,tau)$, we have 2 cases, depending on the value of $tau$.



      In both cases $frac{du}{dt}=0$, on $frac{dx}{dt}=1-2u$, so u is constant on the characteristics, so $u(x(tau),tau)=u(0,tau)$ on characteristics defined by $x(t)=(1-2u(0,tau))t+C$, where $C$ is constant. Applying our boundary condition this gives us:



      $u=frac{tau}{4}+frac{1}{2}$, on $x(t)=-frac{tau}{2}(t-tau)$, for $0leqtauleq2$, and:



      $u=1$ on $x(t)=-(t-tau)$, for $taugeq2$.



      Up to here I'm fine, but when I draw up a plot of the characteristics in the t-x plane I get very confused. The x-axis characteristics are all vertical lines. The t-axis characteristics for $0leqtauleq2$ are straight line originating from $(x,t)=(0,tau)$ with slope that varies. As $tau$, decreases from 2 to 0, the slope of these characteristics (in the t-x plane), goes from -1 and approaces $-infty$, as $tau$, approaches zero, meaning that each of these characteristics intersect.



      As $u$ is constant on a characteristic, this is a contradiction, and so we expect a shock wave (implying a discontinuous solution).



      Also for $taugeq2$, the characteristics are all parallel lines of slope -1, simply translated in the positive t direction (upwards in the t-x plane).



      What is really confusing me is that the three types of characteristics are all intersecting, and that the $0leqtauleq2$ characteristics are intersecting themselves infinitely often, and this is maling it hard for me to apply the shock criterion. At the origin, the characteristics firast intersect, so we expect a shock will immediately form. It will propagate with shock velocity given by:
      $frac{dx_s}{dt}=frac{(u(1-u))|_+-(u(1-u))|_-}{(u)|_+-(u)|_-}$



      Where the subscripts denote the value of u just to the right or jsut to the left of the shock.



      I really don't know how to continue from here, especially because of the three types of intersecting characteristics and the $0leqtauleq2$ characteristics intersecting with themselves. I would really be grateful for any hints or ideas or help. I've tried a lot of things to resolve this issue, but none have gotten anywhere and it would take a long time to write them out. I think if I can just undbrstnad the shock condition/formation here, I can solve this problem.







      pde transport-equation






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      edited Dec 20 '18 at 14:01









      Harry49

      6,98631238




      6,98631238










      asked Apr 28 '17 at 6:33









      J.DoughJ.Dough

      232




      232






















          1 Answer
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          $begingroup$

          An alternative way for solving thanks to the method of characteristics. So, you can compare with your result.



          enter image description hereenter image description here



          The next figure shows $u$ as a function of $x$ at various time.



          enter image description here






          share|cite|improve this answer











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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            An alternative way for solving thanks to the method of characteristics. So, you can compare with your result.



            enter image description hereenter image description here



            The next figure shows $u$ as a function of $x$ at various time.



            enter image description here






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              An alternative way for solving thanks to the method of characteristics. So, you can compare with your result.



              enter image description hereenter image description here



              The next figure shows $u$ as a function of $x$ at various time.



              enter image description here






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                An alternative way for solving thanks to the method of characteristics. So, you can compare with your result.



                enter image description hereenter image description here



                The next figure shows $u$ as a function of $x$ at various time.



                enter image description here






                share|cite|improve this answer











                $endgroup$



                An alternative way for solving thanks to the method of characteristics. So, you can compare with your result.



                enter image description hereenter image description here



                The next figure shows $u$ as a function of $x$ at various time.



                enter image description here







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Apr 30 '17 at 9:21

























                answered Apr 29 '17 at 14:16









                JJacquelinJJacquelin

                44k21853




                44k21853






























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