I want to calculate Green's Function to solve △u=f(x, y), using Laplace Transforms. My plan was to tailor boundary conditions to the problem which simplify the computation.
Because the Laplacian is self-adjoint, my associated Green's Function equation can immediately be written as Gxx+Gyy=δ(x−xG, y−yG).
Laplace Transforming y to s and G to G^, I get G^xx+s2G^−sG(x, 0)−Gy(x, 0)=e−yGsδ(x−xG). Thus to simplify things, I'll go back and include u(x, 0)=0, uy(x, 0)=0 in the problem statement, so that the terms with G(x, 0) and Gy(x, 0) drop out.
Laplace Transforming x to r and G^ to G^^, I get r2G^^−rG^(0, s)−G^x(0, s)+s2G^^=e−yGse−xGr. Again to simplify things, I'll go back and include u(0, y)=0, ux(0, y)=0 in the problem statement. Now I have G^^=e−yGse−xGrr2+s21.
Inverting from r back to x, by applying the shifting theorem, I think I should get G^=e−yGsssin(s(x−xG))H[x−xG], where H[ ] is the Heaviside step function.
Inverting from s back to y, the first factor induces another shift, and the last factor is constant with respect to s, so the challenge is to find the inverse transform of ssin(s(x−xG)). This does not seem easy to find directly (using tables or software; I don't know how to do it with Complex Analysis), so I break the sine function into exponentials and use se___s⟺H[y+___] for each exponential. After some algebra, I arrive at G=2H[y−yG−xi+xGi]H[x−xG]H[y−yG]i−H[y−yG+xi−xGi]H[x−xG]H[y−yG]i
The Boundary Value Problem has been refined into △u=f(x, y), u(x, 0)=0, uy(x, 0)=0, u(0, y)=0, ux(0, y)=0. Is this the correct Green's Function for this BVP? If so, then how can I proceed to clean it up into a real, and more useful, expression and use it to solve the BVP for some particular forcing function? If not, then what went wrong? If the i's cannot be eliminated, is this an indication that I should use the Fourier Transform instead of the Laplace Transform for one or both steps? Would a better choice of boundary conditions produce a more palatable Green's Function?
Because the Laplacian is self-adjoint, my associated Green's Function equation can immediately be written as Gxx+Gyy=δ(x−xG, y−yG).
Laplace Transforming y to s and G to G^, I get G^xx+s2G^−sG(x, 0)−Gy(x, 0)=e−yGsδ(x−xG). Thus to simplify things, I'll go back and include u(x, 0)=0, uy(x, 0)=0 in the problem statement, so that the terms with G(x, 0) and Gy(x, 0) drop out.
Laplace Transforming x to r and G^ to G^^, I get r2G^^−rG^(0, s)−G^x(0, s)+s2G^^=e−yGse−xGr. Again to simplify things, I'll go back and include u(0, y)=0, ux(0, y)=0 in the problem statement. Now I have G^^=e−yGse−xGrr2+s21.
Inverting from r back to x, by applying the shifting theorem, I think I should get G^=e−yGsssin(s(x−xG))H[x−xG], where H[ ] is the Heaviside step function.
Inverting from s back to y, the first factor induces another shift, and the last factor is constant with respect to s, so the challenge is to find the inverse transform of ssin(s(x−xG)). This does not seem easy to find directly (using tables or software; I don't know how to do it with Complex Analysis), so I break the sine function into exponentials and use se___s⟺H[y+___] for each exponential. After some algebra, I arrive at G=2H[y−yG−xi+xGi]H[x−xG]H[y−yG]i−H[y−yG+xi−xGi]H[x−xG]H[y−yG]i
The Boundary Value Problem has been refined into △u=f(x, y), u(x, 0)=0, uy(x, 0)=0, u(0, y)=0, ux(0, y)=0. Is this the correct Green's Function for this BVP? If so, then how can I proceed to clean it up into a real, and more useful, expression and use it to solve the BVP for some particular forcing function? If not, then what went wrong? If the i's cannot be eliminated, is this an indication that I should use the Fourier Transform instead of the Laplace Transform for one or both steps? Would a better choice of boundary conditions produce a more palatable Green's Function?