Transform of a wave equation to a hyperbolic system












2














We consider the wave equation $$y_{tt}=y_{xx}+a(t,x)y, text{ x$in$(0,1)}, tin (0,infty).$$
with Dirichlet boundary conditions.
I want to transform this equation to a hyperbolic system of the form
$$z_t=Az_x+Bz.$$
So, I introduced the following substitutions: $z^1=y_t$, $z^2=y_x$, I obtained
$$z^1_t=z^2_{x}+az^1+a_ty$$
$$z^2_t=z^1_{x}$$



The question here:

How I get rid of the $y$ in the $z^1$ formula? Is there more adequate substitution then this ? thank you.






pde systems-of-equations hyperbolic-equations







share|cite|improve this question





























    2














    We consider the wave equation $$y_{tt}=y_{xx}+a(t,x)y, text{ x$in$(0,1)}, tin (0,infty).$$
    with Dirichlet boundary conditions.
    I want to transform this equation to a hyperbolic system of the form
    $$z_t=Az_x+Bz.$$
    So, I introduced the following substitutions: $z^1=y_t$, $z^2=y_x$, I obtained
    $$z^1_t=z^2_{x}+az^1+a_ty$$
    $$z^2_t=z^1_{x}$$



    The question here:

    How I get rid of the $y$ in the $z^1$ formula? Is there more adequate substitution then this ? thank you.






    pde systems-of-equations hyperbolic-equations







    share|cite|improve this question



























      2












      2








      2







      We consider the wave equation $$y_{tt}=y_{xx}+a(t,x)y, text{ x$in$(0,1)}, tin (0,infty).$$
      with Dirichlet boundary conditions.
      I want to transform this equation to a hyperbolic system of the form
      $$z_t=Az_x+Bz.$$
      So, I introduced the following substitutions: $z^1=y_t$, $z^2=y_x$, I obtained
      $$z^1_t=z^2_{x}+az^1+a_ty$$
      $$z^2_t=z^1_{x}$$



      The question here:

      How I get rid of the $y$ in the $z^1$ formula? Is there more adequate substitution then this ? thank you.






      pde systems-of-equations hyperbolic-equations







      share|cite|improve this question















      We consider the wave equation $$y_{tt}=y_{xx}+a(t,x)y, text{ x$in$(0,1)}, tin (0,infty).$$
      with Dirichlet boundary conditions.
      I want to transform this equation to a hyperbolic system of the form
      $$z_t=Az_x+Bz.$$
      So, I introduced the following substitutions: $z^1=y_t$, $z^2=y_x$, I obtained
      $$z^1_t=z^2_{x}+az^1+a_ty$$
      $$z^2_t=z^1_{x}$$



      The question here:

      How I get rid of the $y$ in the $z^1$ formula? Is there more adequate substitution then this ? thank you.






      pde systems-of-equations hyperbolic-equations




      pde systems-of-equations hyperbolic-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 15 hours ago









      Harry49

      5,99121031




      5,99121031










      asked 2 days ago









      Gustave

      720211




      720211






















          1 Answer
          1






          active

          oldest

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          1














          You need to include $y$ in your $z$ vector. Set $z^1 =y, z^2 = y_t, z^3 = y_x$. Then
          $$
          partial_t
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}
          =
          begin{pmatrix}
          y_t \ y_{xx} + a y \ y_{xt}
          end{pmatrix}
          =
          begin{pmatrix}
          0 & 0& 0\
          0 & 0& 1\
          0 & 1& 0\
          end{pmatrix}
          partial_x
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}
          +
          begin{pmatrix}
          0 & 1& 0\
          a & 0& 0\
          0 & 0& 0\
          end{pmatrix}
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}.
          $$
          This is the form you want.






          share|cite|improve this answer





















          • Thank you @Glitch for replaying. Is there any alternative substitutions to write the above system under a $2 times 2$ hyperbolic system with $A$ is diagonal matrix? I h'ave tried but I have not secceeded.
            – Gustave
            yesterday








          • 1




            @Gustave You're welcome. I don't think it's possible to reduce to a $2 times 2$ first-order system when you have the lower order term $a y$ in your original equation. If you want to capture the PDE then $z$ must contain both $y_t$ and $y_x$, but $y$ itself cannot be obtained from these without integration. If you wanted to integrate, then you could obtain $y$, but you would break the structure of the first-order PDE system. Is there a particular reason you don't like the $3 times 3$ system? The tricks to solve a $2times 2$ should work here just as well...
            – Glitch
            yesterday










          • Thank you sir..
            – Gustave
            yesterday











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          You need to include $y$ in your $z$ vector. Set $z^1 =y, z^2 = y_t, z^3 = y_x$. Then
          $$
          partial_t
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}
          =
          begin{pmatrix}
          y_t \ y_{xx} + a y \ y_{xt}
          end{pmatrix}
          =
          begin{pmatrix}
          0 & 0& 0\
          0 & 0& 1\
          0 & 1& 0\
          end{pmatrix}
          partial_x
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}
          +
          begin{pmatrix}
          0 & 1& 0\
          a & 0& 0\
          0 & 0& 0\
          end{pmatrix}
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}.
          $$
          This is the form you want.






          share|cite|improve this answer





















          • Thank you @Glitch for replaying. Is there any alternative substitutions to write the above system under a $2 times 2$ hyperbolic system with $A$ is diagonal matrix? I h'ave tried but I have not secceeded.
            – Gustave
            yesterday








          • 1




            @Gustave You're welcome. I don't think it's possible to reduce to a $2 times 2$ first-order system when you have the lower order term $a y$ in your original equation. If you want to capture the PDE then $z$ must contain both $y_t$ and $y_x$, but $y$ itself cannot be obtained from these without integration. If you wanted to integrate, then you could obtain $y$, but you would break the structure of the first-order PDE system. Is there a particular reason you don't like the $3 times 3$ system? The tricks to solve a $2times 2$ should work here just as well...
            – Glitch
            yesterday










          • Thank you sir..
            – Gustave
            yesterday
















          1














          You need to include $y$ in your $z$ vector. Set $z^1 =y, z^2 = y_t, z^3 = y_x$. Then
          $$
          partial_t
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}
          =
          begin{pmatrix}
          y_t \ y_{xx} + a y \ y_{xt}
          end{pmatrix}
          =
          begin{pmatrix}
          0 & 0& 0\
          0 & 0& 1\
          0 & 1& 0\
          end{pmatrix}
          partial_x
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}
          +
          begin{pmatrix}
          0 & 1& 0\
          a & 0& 0\
          0 & 0& 0\
          end{pmatrix}
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}.
          $$
          This is the form you want.






          share|cite|improve this answer





















          • Thank you @Glitch for replaying. Is there any alternative substitutions to write the above system under a $2 times 2$ hyperbolic system with $A$ is diagonal matrix? I h'ave tried but I have not secceeded.
            – Gustave
            yesterday








          • 1




            @Gustave You're welcome. I don't think it's possible to reduce to a $2 times 2$ first-order system when you have the lower order term $a y$ in your original equation. If you want to capture the PDE then $z$ must contain both $y_t$ and $y_x$, but $y$ itself cannot be obtained from these without integration. If you wanted to integrate, then you could obtain $y$, but you would break the structure of the first-order PDE system. Is there a particular reason you don't like the $3 times 3$ system? The tricks to solve a $2times 2$ should work here just as well...
            – Glitch
            yesterday










          • Thank you sir..
            – Gustave
            yesterday














          1












          1








          1






          You need to include $y$ in your $z$ vector. Set $z^1 =y, z^2 = y_t, z^3 = y_x$. Then
          $$
          partial_t
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}
          =
          begin{pmatrix}
          y_t \ y_{xx} + a y \ y_{xt}
          end{pmatrix}
          =
          begin{pmatrix}
          0 & 0& 0\
          0 & 0& 1\
          0 & 1& 0\
          end{pmatrix}
          partial_x
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}
          +
          begin{pmatrix}
          0 & 1& 0\
          a & 0& 0\
          0 & 0& 0\
          end{pmatrix}
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}.
          $$
          This is the form you want.






          share|cite|improve this answer












          You need to include $y$ in your $z$ vector. Set $z^1 =y, z^2 = y_t, z^3 = y_x$. Then
          $$
          partial_t
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}
          =
          begin{pmatrix}
          y_t \ y_{xx} + a y \ y_{xt}
          end{pmatrix}
          =
          begin{pmatrix}
          0 & 0& 0\
          0 & 0& 1\
          0 & 1& 0\
          end{pmatrix}
          partial_x
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}
          +
          begin{pmatrix}
          0 & 1& 0\
          a & 0& 0\
          0 & 0& 0\
          end{pmatrix}
          begin{pmatrix}
          y \ y_t \ y_x
          end{pmatrix}.
          $$
          This is the form you want.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Glitch

          5,5431030




          5,5431030












          • Thank you @Glitch for replaying. Is there any alternative substitutions to write the above system under a $2 times 2$ hyperbolic system with $A$ is diagonal matrix? I h'ave tried but I have not secceeded.
            – Gustave
            yesterday








          • 1




            @Gustave You're welcome. I don't think it's possible to reduce to a $2 times 2$ first-order system when you have the lower order term $a y$ in your original equation. If you want to capture the PDE then $z$ must contain both $y_t$ and $y_x$, but $y$ itself cannot be obtained from these without integration. If you wanted to integrate, then you could obtain $y$, but you would break the structure of the first-order PDE system. Is there a particular reason you don't like the $3 times 3$ system? The tricks to solve a $2times 2$ should work here just as well...
            – Glitch
            yesterday










          • Thank you sir..
            – Gustave
            yesterday


















          • Thank you @Glitch for replaying. Is there any alternative substitutions to write the above system under a $2 times 2$ hyperbolic system with $A$ is diagonal matrix? I h'ave tried but I have not secceeded.
            – Gustave
            yesterday








          • 1




            @Gustave You're welcome. I don't think it's possible to reduce to a $2 times 2$ first-order system when you have the lower order term $a y$ in your original equation. If you want to capture the PDE then $z$ must contain both $y_t$ and $y_x$, but $y$ itself cannot be obtained from these without integration. If you wanted to integrate, then you could obtain $y$, but you would break the structure of the first-order PDE system. Is there a particular reason you don't like the $3 times 3$ system? The tricks to solve a $2times 2$ should work here just as well...
            – Glitch
            yesterday










          • Thank you sir..
            – Gustave
            yesterday
















          Thank you @Glitch for replaying. Is there any alternative substitutions to write the above system under a $2 times 2$ hyperbolic system with $A$ is diagonal matrix? I h'ave tried but I have not secceeded.
          – Gustave
          yesterday






          Thank you @Glitch for replaying. Is there any alternative substitutions to write the above system under a $2 times 2$ hyperbolic system with $A$ is diagonal matrix? I h'ave tried but I have not secceeded.
          – Gustave
          yesterday






          1




          1




          @Gustave You're welcome. I don't think it's possible to reduce to a $2 times 2$ first-order system when you have the lower order term $a y$ in your original equation. If you want to capture the PDE then $z$ must contain both $y_t$ and $y_x$, but $y$ itself cannot be obtained from these without integration. If you wanted to integrate, then you could obtain $y$, but you would break the structure of the first-order PDE system. Is there a particular reason you don't like the $3 times 3$ system? The tricks to solve a $2times 2$ should work here just as well...
          – Glitch
          yesterday




          @Gustave You're welcome. I don't think it's possible to reduce to a $2 times 2$ first-order system when you have the lower order term $a y$ in your original equation. If you want to capture the PDE then $z$ must contain both $y_t$ and $y_x$, but $y$ itself cannot be obtained from these without integration. If you wanted to integrate, then you could obtain $y$, but you would break the structure of the first-order PDE system. Is there a particular reason you don't like the $3 times 3$ system? The tricks to solve a $2times 2$ should work here just as well...
          – Glitch
          yesterday












          Thank you sir..
          – Gustave
          yesterday




          Thank you sir..
          – Gustave
          yesterday


















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