An initial- and boundary-value problem for Burgers' equation with no solution











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Prove that there is no solution to the following Cauchy problem:
$$begin{align}
u_t+uu_x& =0&quad x&in(-1,1), tgt0 label{1}tag{1}\
u(x,0)&=x&quad x&in[-1,1] label{2}tag{2}\
u(-1,t)&=-1& quad t&geq0 label{3}tag{3}\
u(1,t)&=1&quad t&geq0 label{4}tag{4}
end{align}$$



My attempt: Using the method of characteristics, I found the classical solution
$$
u=frac{x}{1+t} quad tgt{-1}
$$
which satisfies conditions eqref{1} and eqref{2}, but not eqref{3} and eqref{4}.




But how can I show that there is also no weak solution to this problem?




Thanks in advance.






pde hyperbolic-equations







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    down vote

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    Prove that there is no solution to the following Cauchy problem:
    $$begin{align}
    u_t+uu_x& =0&quad x&in(-1,1), tgt0 label{1}tag{1}\
    u(x,0)&=x&quad x&in[-1,1] label{2}tag{2}\
    u(-1,t)&=-1& quad t&geq0 label{3}tag{3}\
    u(1,t)&=1&quad t&geq0 label{4}tag{4}
    end{align}$$



    My attempt: Using the method of characteristics, I found the classical solution
    $$
    u=frac{x}{1+t} quad tgt{-1}
    $$
    which satisfies conditions eqref{1} and eqref{2}, but not eqref{3} and eqref{4}.




    But how can I show that there is also no weak solution to this problem?




    Thanks in advance.






    pde hyperbolic-equations







    share|cite|improve this question


























      up vote
      3
      down vote

      favorite
      1









      up vote
      3
      down vote

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      1






      1





      Prove that there is no solution to the following Cauchy problem:
      $$begin{align}
      u_t+uu_x& =0&quad x&in(-1,1), tgt0 label{1}tag{1}\
      u(x,0)&=x&quad x&in[-1,1] label{2}tag{2}\
      u(-1,t)&=-1& quad t&geq0 label{3}tag{3}\
      u(1,t)&=1&quad t&geq0 label{4}tag{4}
      end{align}$$



      My attempt: Using the method of characteristics, I found the classical solution
      $$
      u=frac{x}{1+t} quad tgt{-1}
      $$
      which satisfies conditions eqref{1} and eqref{2}, but not eqref{3} and eqref{4}.




      But how can I show that there is also no weak solution to this problem?




      Thanks in advance.






      pde hyperbolic-equations







      share|cite|improve this question















      Prove that there is no solution to the following Cauchy problem:
      $$begin{align}
      u_t+uu_x& =0&quad x&in(-1,1), tgt0 label{1}tag{1}\
      u(x,0)&=x&quad x&in[-1,1] label{2}tag{2}\
      u(-1,t)&=-1& quad t&geq0 label{3}tag{3}\
      u(1,t)&=1&quad t&geq0 label{4}tag{4}
      end{align}$$



      My attempt: Using the method of characteristics, I found the classical solution
      $$
      u=frac{x}{1+t} quad tgt{-1}
      $$
      which satisfies conditions eqref{1} and eqref{2}, but not eqref{3} and eqref{4}.




      But how can I show that there is also no weak solution to this problem?




      Thanks in advance.






      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 5 at 13:17









      Harry49

      5,87421030




      5,87421030










      asked Dec 4 at 19:53









      dmtri

      1,3001521




      1,3001521






















          1 Answer
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          accepted










          In facts, it is impossible for a classical solution to solve the initial- and boundary-value problem. A plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data is given below*:



          characteristics



          The possibility of admissible discontinuities near the boundaries (weak solutions) should be examined. A look at the characteristics in the $x$-$t$ plane shows that such solutions are not admissible in the sense of Lax** (characteristics do not intersect appropriately). For example, at $xsimeq 1$, we have $u_R = 1$ and $u_L = 1/(1+t)<1$, so that $u_L < u_R$. No shock wave is admissible. Alternatively, we could have investigated the boundary $xsimeq -1$, where $u_L = -1$ and $u_R = -1/(1+t) > -1$. Here too, no shock wave is admissible, since $u_L < u_R$.




          *The lines drawn outside $]-1,1[$ are the characteristics starting at the boundaries $x=pm 1$, which have been represented outside the interior domain to keep the figure readable.



          **The Lax entropy condition reads $u_L>s>u_R$ where $u_L$ is the value on the left of the discontinuity, $u_R$ is the value on the right of the discontinuity, and $s$ is the speed of shock given by Rankine-Hugoniot.






          share|cite|improve this answer























          • Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
            – dmtri
            Dec 5 at 17:27










          • one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
            – dmtri
            Dec 5 at 18:06












          • I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
            – dmtri
            2 days ago








          • 1




            @dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
            – Harry49
            2 days ago











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          In facts, it is impossible for a classical solution to solve the initial- and boundary-value problem. A plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data is given below*:



          characteristics



          The possibility of admissible discontinuities near the boundaries (weak solutions) should be examined. A look at the characteristics in the $x$-$t$ plane shows that such solutions are not admissible in the sense of Lax** (characteristics do not intersect appropriately). For example, at $xsimeq 1$, we have $u_R = 1$ and $u_L = 1/(1+t)<1$, so that $u_L < u_R$. No shock wave is admissible. Alternatively, we could have investigated the boundary $xsimeq -1$, where $u_L = -1$ and $u_R = -1/(1+t) > -1$. Here too, no shock wave is admissible, since $u_L < u_R$.




          *The lines drawn outside $]-1,1[$ are the characteristics starting at the boundaries $x=pm 1$, which have been represented outside the interior domain to keep the figure readable.



          **The Lax entropy condition reads $u_L>s>u_R$ where $u_L$ is the value on the left of the discontinuity, $u_R$ is the value on the right of the discontinuity, and $s$ is the speed of shock given by Rankine-Hugoniot.






          share|cite|improve this answer























          • Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
            – dmtri
            Dec 5 at 17:27










          • one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
            – dmtri
            Dec 5 at 18:06












          • I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
            – dmtri
            2 days ago








          • 1




            @dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
            – Harry49
            2 days ago















          up vote
          1
          down vote



          accepted










          In facts, it is impossible for a classical solution to solve the initial- and boundary-value problem. A plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data is given below*:



          characteristics



          The possibility of admissible discontinuities near the boundaries (weak solutions) should be examined. A look at the characteristics in the $x$-$t$ plane shows that such solutions are not admissible in the sense of Lax** (characteristics do not intersect appropriately). For example, at $xsimeq 1$, we have $u_R = 1$ and $u_L = 1/(1+t)<1$, so that $u_L < u_R$. No shock wave is admissible. Alternatively, we could have investigated the boundary $xsimeq -1$, where $u_L = -1$ and $u_R = -1/(1+t) > -1$. Here too, no shock wave is admissible, since $u_L < u_R$.




          *The lines drawn outside $]-1,1[$ are the characteristics starting at the boundaries $x=pm 1$, which have been represented outside the interior domain to keep the figure readable.



          **The Lax entropy condition reads $u_L>s>u_R$ where $u_L$ is the value on the left of the discontinuity, $u_R$ is the value on the right of the discontinuity, and $s$ is the speed of shock given by Rankine-Hugoniot.






          share|cite|improve this answer























          • Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
            – dmtri
            Dec 5 at 17:27










          • one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
            – dmtri
            Dec 5 at 18:06












          • I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
            – dmtri
            2 days ago








          • 1




            @dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
            – Harry49
            2 days ago













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          In facts, it is impossible for a classical solution to solve the initial- and boundary-value problem. A plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data is given below*:



          characteristics



          The possibility of admissible discontinuities near the boundaries (weak solutions) should be examined. A look at the characteristics in the $x$-$t$ plane shows that such solutions are not admissible in the sense of Lax** (characteristics do not intersect appropriately). For example, at $xsimeq 1$, we have $u_R = 1$ and $u_L = 1/(1+t)<1$, so that $u_L < u_R$. No shock wave is admissible. Alternatively, we could have investigated the boundary $xsimeq -1$, where $u_L = -1$ and $u_R = -1/(1+t) > -1$. Here too, no shock wave is admissible, since $u_L < u_R$.




          *The lines drawn outside $]-1,1[$ are the characteristics starting at the boundaries $x=pm 1$, which have been represented outside the interior domain to keep the figure readable.



          **The Lax entropy condition reads $u_L>s>u_R$ where $u_L$ is the value on the left of the discontinuity, $u_R$ is the value on the right of the discontinuity, and $s$ is the speed of shock given by Rankine-Hugoniot.






          share|cite|improve this answer














          In facts, it is impossible for a classical solution to solve the initial- and boundary-value problem. A plot of the characteristic curves in the $x$-$t$ plane deduced from the initial data is given below*:



          characteristics



          The possibility of admissible discontinuities near the boundaries (weak solutions) should be examined. A look at the characteristics in the $x$-$t$ plane shows that such solutions are not admissible in the sense of Lax** (characteristics do not intersect appropriately). For example, at $xsimeq 1$, we have $u_R = 1$ and $u_L = 1/(1+t)<1$, so that $u_L < u_R$. No shock wave is admissible. Alternatively, we could have investigated the boundary $xsimeq -1$, where $u_L = -1$ and $u_R = -1/(1+t) > -1$. Here too, no shock wave is admissible, since $u_L < u_R$.




          *The lines drawn outside $]-1,1[$ are the characteristics starting at the boundaries $x=pm 1$, which have been represented outside the interior domain to keep the figure readable.



          **The Lax entropy condition reads $u_L>s>u_R$ where $u_L$ is the value on the left of the discontinuity, $u_R$ is the value on the right of the discontinuity, and $s$ is the speed of shock given by Rankine-Hugoniot.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered Dec 5 at 13:16









          Harry49

          5,87421030




          5,87421030












          • Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
            – dmtri
            Dec 5 at 17:27










          • one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
            – dmtri
            Dec 5 at 18:06












          • I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
            – dmtri
            2 days ago








          • 1




            @dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
            – Harry49
            2 days ago


















          • Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
            – dmtri
            Dec 5 at 17:27










          • one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
            – dmtri
            Dec 5 at 18:06












          • I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
            – dmtri
            2 days ago








          • 1




            @dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
            – Harry49
            2 days ago
















          Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
          – dmtri
          Dec 5 at 17:27




          Thanks for your answer, but the only criterion I know about weak solutions is the Runkine - Hugionot theorem, which I cannot apply here as I can not figure one possible curve...should it be characteristic?
          – dmtri
          Dec 5 at 17:27












          one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
          – dmtri
          Dec 5 at 18:06






          one more question please, what are the lines outside of the area $xge1$ and $-1ge{x}$, all the characteristics should pass by the point $(0,-1)$.
          – dmtri
          Dec 5 at 18:06














          I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
          – dmtri
          2 days ago






          I would like to ask something more, as I think I do not get clearly what is really asked in an initial and boundary problem (IBP). In the IBP, I posted, are we looking for a classical solution of equation $(1)$ valid in the interior (open set) of the area given and this solution should take at boundary of this set the values described by equations $(2), (3), (4)$ ? If this is the case, we can easily conclude that $u=frac{x}{t+1}$ is the only solution satisfying $(1) , (2)$ but not $(3)$. Why do we need $(4)$ then to show that there is no solution?
          – dmtri
          2 days ago






          1




          1




          @dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
          – Harry49
          2 days ago




          @dmtri One boundary (e.g., $(3)$) is enough to show that there is no admissible solution compatible with all conditions.
          – Harry49
          2 days ago


















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