Given curve and surface, proving uniqueness of w for given partial differential equation

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Let $Gamma$ be a closed curve in xy plane and S denote the region bounded by curve $Gamma$. let $fracpartial^2 wpartial^2 x +fracpartial^2 wpartial^2 y = f(x,y) forall (x,y) in S$. If f is prescribed at each point $(x,y)$ of S and w is prescribed on the boundary $Gamma$ of S, then prove that any solution $w = w(x,y)$ satisfying the conditions is unique



My attempt:



On boundary $Gamma$:
f is not defined. So taking f(x,y) = 0 on $Gamma$



Inside Surface S, w is not defined. So taking w(x,y) = 0 in S



Now Solution of give PDE is given by



$w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 f(x,y), D = fracpartialpartial x, D' = D = fracpartialpartial y$



But w is defined on boundary $Gamma$; On boundary, f(x,y) = 0 as f is defined inside Surface S.



So $w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 0 \ w = phi_1(y+ix) + phi_2(y-ix) = phi_1(i(x-iy)) + phi_2(-i(x+iy)) = f_1(frac1z)+ f_2(z), \z = x+iy, x = r cos theta, y = r sin theta\ w = f_1(1/z)+f_2(z)$



Now for uniqueness,



Let $w(x_1,y_1)=w(x_2,y_2) \ implies f_1(1/z_1)+f_2(z_1) = f_1(1/z_2)+f_2(z_2) \ f_1(1/z_1)-f_1(1/z_2) = f_2(z_2)-f_2(z_1)$



Now to choose that $f_1 , f_2$ which satisfy above?




complex-analysis pde boundary-value-problem laplacian




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    Let $Gamma$ be a closed curve in xy plane and S denote the region bounded by curve $Gamma$. let $fracpartial^2 wpartial^2 x +fracpartial^2 wpartial^2 y = f(x,y) forall (x,y) in S$. If f is prescribed at each point $(x,y)$ of S and w is prescribed on the boundary $Gamma$ of S, then prove that any solution $w = w(x,y)$ satisfying the conditions is unique



    My attempt:



    On boundary $Gamma$:
    f is not defined. So taking f(x,y) = 0 on $Gamma$



    Inside Surface S, w is not defined. So taking w(x,y) = 0 in S



    Now Solution of give PDE is given by



    $w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 f(x,y), D = fracpartialpartial x, D' = D = fracpartialpartial y$



    But w is defined on boundary $Gamma$; On boundary, f(x,y) = 0 as f is defined inside Surface S.



    So $w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 0 \ w = phi_1(y+ix) + phi_2(y-ix) = phi_1(i(x-iy)) + phi_2(-i(x+iy)) = f_1(frac1z)+ f_2(z), \z = x+iy, x = r cos theta, y = r sin theta\ w = f_1(1/z)+f_2(z)$



    Now for uniqueness,



    Let $w(x_1,y_1)=w(x_2,y_2) \ implies f_1(1/z_1)+f_2(z_1) = f_1(1/z_2)+f_2(z_2) \ f_1(1/z_1)-f_1(1/z_2) = f_2(z_2)-f_2(z_1)$



    Now to choose that $f_1 , f_2$ which satisfy above?




    complex-analysis pde boundary-value-problem laplacian




    share|cite|improve this question





















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      Let $Gamma$ be a closed curve in xy plane and S denote the region bounded by curve $Gamma$. let $fracpartial^2 wpartial^2 x +fracpartial^2 wpartial^2 y = f(x,y) forall (x,y) in S$. If f is prescribed at each point $(x,y)$ of S and w is prescribed on the boundary $Gamma$ of S, then prove that any solution $w = w(x,y)$ satisfying the conditions is unique



      My attempt:



      On boundary $Gamma$:
      f is not defined. So taking f(x,y) = 0 on $Gamma$



      Inside Surface S, w is not defined. So taking w(x,y) = 0 in S



      Now Solution of give PDE is given by



      $w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 f(x,y), D = fracpartialpartial x, D' = D = fracpartialpartial y$



      But w is defined on boundary $Gamma$; On boundary, f(x,y) = 0 as f is defined inside Surface S.



      So $w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 0 \ w = phi_1(y+ix) + phi_2(y-ix) = phi_1(i(x-iy)) + phi_2(-i(x+iy)) = f_1(frac1z)+ f_2(z), \z = x+iy, x = r cos theta, y = r sin theta\ w = f_1(1/z)+f_2(z)$



      Now for uniqueness,



      Let $w(x_1,y_1)=w(x_2,y_2) \ implies f_1(1/z_1)+f_2(z_1) = f_1(1/z_2)+f_2(z_2) \ f_1(1/z_1)-f_1(1/z_2) = f_2(z_2)-f_2(z_1)$



      Now to choose that $f_1 , f_2$ which satisfy above?




      complex-analysis pde boundary-value-problem laplacian




      share|cite|improve this question











      Let $Gamma$ be a closed curve in xy plane and S denote the region bounded by curve $Gamma$. let $fracpartial^2 wpartial^2 x +fracpartial^2 wpartial^2 y = f(x,y) forall (x,y) in S$. If f is prescribed at each point $(x,y)$ of S and w is prescribed on the boundary $Gamma$ of S, then prove that any solution $w = w(x,y)$ satisfying the conditions is unique



      My attempt:



      On boundary $Gamma$:
      f is not defined. So taking f(x,y) = 0 on $Gamma$



      Inside Surface S, w is not defined. So taking w(x,y) = 0 in S



      Now Solution of give PDE is given by



      $w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 f(x,y), D = fracpartialpartial x, D' = D = fracpartialpartial y$



      But w is defined on boundary $Gamma$; On boundary, f(x,y) = 0 as f is defined inside Surface S.



      So $w = phi_1(y+ix) + phi_2(y-ix) + frac1D^2+D'^2 0 \ w = phi_1(y+ix) + phi_2(y-ix) = phi_1(i(x-iy)) + phi_2(-i(x+iy)) = f_1(frac1z)+ f_2(z), \z = x+iy, x = r cos theta, y = r sin theta\ w = f_1(1/z)+f_2(z)$



      Now for uniqueness,



      Let $w(x_1,y_1)=w(x_2,y_2) \ implies f_1(1/z_1)+f_2(z_1) = f_1(1/z_2)+f_2(z_2) \ f_1(1/z_1)-f_1(1/z_2) = f_2(z_2)-f_2(z_1)$



      Now to choose that $f_1 , f_2$ which satisfy above?





      complex-analysis pde boundary-value-problem laplacian





      share|cite|improve this question










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      asked Jul 16 at 15:55









      Magneto

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