Poisson equation on semi-infinite strip











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Function $u(x,y)$ satisfy the equation:
$$Delta u = e^{-2y}sin x$$



in the semi-infinite strip:
$$0<x<pi, y>0$$
and the boundary condition:
$$u(0,y) = u(pi,y),text{ }u(x,0)=sin(3x),text{ } lim_{ytoinfty} u =0$$



Find $u_y(x,0)$



I've seen someone using Fourier tranform to solve similar poisson equation in the infinite strip, and separation of variables to solve Laplace in semi-infinite strip. So I'm wondering how to solve this one? (How do we usually decide when to use separation of variables and Fourier) Or the questions only asks for $u_y(x,0)$, so is it possible to find $u_y$ without solving the equation?






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    down vote

    favorite
    1












    Function $u(x,y)$ satisfy the equation:
    $$Delta u = e^{-2y}sin x$$



    in the semi-infinite strip:
    $$0<x<pi, y>0$$
    and the boundary condition:
    $$u(0,y) = u(pi,y),text{ }u(x,0)=sin(3x),text{ } lim_{ytoinfty} u =0$$



    Find $u_y(x,0)$



    I've seen someone using Fourier tranform to solve similar poisson equation in the infinite strip, and separation of variables to solve Laplace in semi-infinite strip. So I'm wondering how to solve this one? (How do we usually decide when to use separation of variables and Fourier) Or the questions only asks for $u_y(x,0)$, so is it possible to find $u_y$ without solving the equation?






    pde poissons-equation







    share|cite|improve this question
























      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1





      Function $u(x,y)$ satisfy the equation:
      $$Delta u = e^{-2y}sin x$$



      in the semi-infinite strip:
      $$0<x<pi, y>0$$
      and the boundary condition:
      $$u(0,y) = u(pi,y),text{ }u(x,0)=sin(3x),text{ } lim_{ytoinfty} u =0$$



      Find $u_y(x,0)$



      I've seen someone using Fourier tranform to solve similar poisson equation in the infinite strip, and separation of variables to solve Laplace in semi-infinite strip. So I'm wondering how to solve this one? (How do we usually decide when to use separation of variables and Fourier) Or the questions only asks for $u_y(x,0)$, so is it possible to find $u_y$ without solving the equation?






      pde poissons-equation







      share|cite|improve this question













      Function $u(x,y)$ satisfy the equation:
      $$Delta u = e^{-2y}sin x$$



      in the semi-infinite strip:
      $$0<x<pi, y>0$$
      and the boundary condition:
      $$u(0,y) = u(pi,y),text{ }u(x,0)=sin(3x),text{ } lim_{ytoinfty} u =0$$



      Find $u_y(x,0)$



      I've seen someone using Fourier tranform to solve similar poisson equation in the infinite strip, and separation of variables to solve Laplace in semi-infinite strip. So I'm wondering how to solve this one? (How do we usually decide when to use separation of variables and Fourier) Or the questions only asks for $u_y(x,0)$, so is it possible to find $u_y$ without solving the equation?






      pde poissons-equation




      pde poissons-equation






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      asked 2 days ago









      QD666

      1186




      1186






















          1 Answer
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          Let $v=u-frac{1}{3}e^{-2y}sin(x)$, where $u$ is a solution of the stated problem. Then $v$ is a solution of
          $$
          Delta v = Delta u - e^{-2y}sin(x)= 0,
          $$
          with conditions
          $$
          v(0,y)=v(pi,y) \
          v(x,0)=u(x,0)-frac{1}{3}sin(x)=sin(3x)-frac{1}{3}sin(x)
          $$
          The solutions $v$ is
          $$
          v(x,y)=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}
          $$
          So,



          begin{align}
          u(x,y)&=v(x,y)+frac{1}{3}e^{-2y}sin(x) \ &=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}+frac{1}{3}sin(x)e^{-2y}.
          end{align}
          And,
          $$
          u_y(x,0)=-3sin(3x)+frac{1}{3}sin(x)-frac{2}{3}sin(x) \
          =-3sin(3x)-frac{1}{3}sin(x).
          $$






          share|cite|improve this answer





















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            up vote
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            accepted










            Let $v=u-frac{1}{3}e^{-2y}sin(x)$, where $u$ is a solution of the stated problem. Then $v$ is a solution of
            $$
            Delta v = Delta u - e^{-2y}sin(x)= 0,
            $$
            with conditions
            $$
            v(0,y)=v(pi,y) \
            v(x,0)=u(x,0)-frac{1}{3}sin(x)=sin(3x)-frac{1}{3}sin(x)
            $$
            The solutions $v$ is
            $$
            v(x,y)=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}
            $$
            So,



            begin{align}
            u(x,y)&=v(x,y)+frac{1}{3}e^{-2y}sin(x) \ &=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}+frac{1}{3}sin(x)e^{-2y}.
            end{align}
            And,
            $$
            u_y(x,0)=-3sin(3x)+frac{1}{3}sin(x)-frac{2}{3}sin(x) \
            =-3sin(3x)-frac{1}{3}sin(x).
            $$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Let $v=u-frac{1}{3}e^{-2y}sin(x)$, where $u$ is a solution of the stated problem. Then $v$ is a solution of
              $$
              Delta v = Delta u - e^{-2y}sin(x)= 0,
              $$
              with conditions
              $$
              v(0,y)=v(pi,y) \
              v(x,0)=u(x,0)-frac{1}{3}sin(x)=sin(3x)-frac{1}{3}sin(x)
              $$
              The solutions $v$ is
              $$
              v(x,y)=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}
              $$
              So,



              begin{align}
              u(x,y)&=v(x,y)+frac{1}{3}e^{-2y}sin(x) \ &=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}+frac{1}{3}sin(x)e^{-2y}.
              end{align}
              And,
              $$
              u_y(x,0)=-3sin(3x)+frac{1}{3}sin(x)-frac{2}{3}sin(x) \
              =-3sin(3x)-frac{1}{3}sin(x).
              $$






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Let $v=u-frac{1}{3}e^{-2y}sin(x)$, where $u$ is a solution of the stated problem. Then $v$ is a solution of
                $$
                Delta v = Delta u - e^{-2y}sin(x)= 0,
                $$
                with conditions
                $$
                v(0,y)=v(pi,y) \
                v(x,0)=u(x,0)-frac{1}{3}sin(x)=sin(3x)-frac{1}{3}sin(x)
                $$
                The solutions $v$ is
                $$
                v(x,y)=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}
                $$
                So,



                begin{align}
                u(x,y)&=v(x,y)+frac{1}{3}e^{-2y}sin(x) \ &=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}+frac{1}{3}sin(x)e^{-2y}.
                end{align}
                And,
                $$
                u_y(x,0)=-3sin(3x)+frac{1}{3}sin(x)-frac{2}{3}sin(x) \
                =-3sin(3x)-frac{1}{3}sin(x).
                $$






                share|cite|improve this answer












                Let $v=u-frac{1}{3}e^{-2y}sin(x)$, where $u$ is a solution of the stated problem. Then $v$ is a solution of
                $$
                Delta v = Delta u - e^{-2y}sin(x)= 0,
                $$
                with conditions
                $$
                v(0,y)=v(pi,y) \
                v(x,0)=u(x,0)-frac{1}{3}sin(x)=sin(3x)-frac{1}{3}sin(x)
                $$
                The solutions $v$ is
                $$
                v(x,y)=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}
                $$
                So,



                begin{align}
                u(x,y)&=v(x,y)+frac{1}{3}e^{-2y}sin(x) \ &=sin(3x)e^{-3y}-frac{1}{3}sin(x)e^{-y}+frac{1}{3}sin(x)e^{-2y}.
                end{align}
                And,
                $$
                u_y(x,0)=-3sin(3x)+frac{1}{3}sin(x)-frac{2}{3}sin(x) \
                =-3sin(3x)-frac{1}{3}sin(x).
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                DisintegratingByParts

                57.6k42376




                57.6k42376






























                     

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