I want to calculate Green's Function to solve [imath]\triangle u = f(x,\ y)[/imath], 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 [imath]G_{xx} + G_{yy} = \delta(x - x_G,\ y - y_G)[/imath].
Laplace Transforming [imath]y[/imath] to [imath]s[/imath] and [imath]G[/imath] to [imath]\hat G[/imath], I get [imath]\hat G_{xx} + s^2 \hat G - sG(x,\ 0) - G_y(x,\ 0) = e^{-y_G s} \delta(x - x_G)[/imath]. Thus to simplify things, I'll go back and include [imath]u(x,\ 0) = 0,\ u_y(x,\ 0) = 0[/imath] in the problem statement, so that the terms with [imath]G(x,\ 0)[/imath] and [imath]G_y(x,\ 0)[/imath] drop out.
Laplace Transforming [imath]x[/imath] to [imath]r[/imath] and [imath]\hat G[/imath] to [imath]\hat{\hat G}[/imath], I get [imath]r^2 \hat{\hat G} - r \hat G(0,\ s) - \hat G_x(0,\ s) + s^2 \hat{\hat G} = e^{-y_G s}e^{-x_G r}[/imath]. Again to simplify things, I'll go back and include [imath]u(0,\ y) = 0,\ u_x(0,\ y) = 0[/imath] in the problem statement. Now I have [imath]\hat{\hat G} = e^{-y_G s}e^{-x_G r}\frac{1}{r^2 + s^2}[/imath].
Inverting from [imath]r[/imath] back to [imath]x[/imath], by applying the shifting theorem, I think I should get [imath]\hat G = e^{-y_G s}\frac{\sin(s(x - x_G))}{s}H[x - x_G][/imath], where [imath]H[\ ][/imath] is the Heaviside step function.
Inverting from [imath]s[/imath] back to [imath]y[/imath], the first factor induces another shift, and the last factor is constant with respect to [imath]s[/imath], so the challenge is to find the inverse transform of [imath]\frac{\sin(s(x - x_G))}{s}[/imath]. 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 [imath]\frac{e^{\_\_\_s}}{s} \iff H[y + \_\_\_][/imath] for each exponential. After some algebra, I arrive at [math]G = \frac{H[y - y_G - xi + x_G i]H[x - x_G]H[y - y_G]i - H[y - y_G + xi - x_G i]H[x - x_G]H[y - y_G]i}{2}[/math]
The Boundary Value Problem has been refined into [imath]\triangle u = f(x,\ y),\ u(x,\ 0) = 0,\ u_y(x,\ 0) = 0,\ u(0,\ y) = 0,\ u_x(0,\ y) = 0[/imath]. 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 [imath]i[/imath]'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 [imath]G_{xx} + G_{yy} = \delta(x - x_G,\ y - y_G)[/imath].
Laplace Transforming [imath]y[/imath] to [imath]s[/imath] and [imath]G[/imath] to [imath]\hat G[/imath], I get [imath]\hat G_{xx} + s^2 \hat G - sG(x,\ 0) - G_y(x,\ 0) = e^{-y_G s} \delta(x - x_G)[/imath]. Thus to simplify things, I'll go back and include [imath]u(x,\ 0) = 0,\ u_y(x,\ 0) = 0[/imath] in the problem statement, so that the terms with [imath]G(x,\ 0)[/imath] and [imath]G_y(x,\ 0)[/imath] drop out.
Laplace Transforming [imath]x[/imath] to [imath]r[/imath] and [imath]\hat G[/imath] to [imath]\hat{\hat G}[/imath], I get [imath]r^2 \hat{\hat G} - r \hat G(0,\ s) - \hat G_x(0,\ s) + s^2 \hat{\hat G} = e^{-y_G s}e^{-x_G r}[/imath]. Again to simplify things, I'll go back and include [imath]u(0,\ y) = 0,\ u_x(0,\ y) = 0[/imath] in the problem statement. Now I have [imath]\hat{\hat G} = e^{-y_G s}e^{-x_G r}\frac{1}{r^2 + s^2}[/imath].
Inverting from [imath]r[/imath] back to [imath]x[/imath], by applying the shifting theorem, I think I should get [imath]\hat G = e^{-y_G s}\frac{\sin(s(x - x_G))}{s}H[x - x_G][/imath], where [imath]H[\ ][/imath] is the Heaviside step function.
Inverting from [imath]s[/imath] back to [imath]y[/imath], the first factor induces another shift, and the last factor is constant with respect to [imath]s[/imath], so the challenge is to find the inverse transform of [imath]\frac{\sin(s(x - x_G))}{s}[/imath]. 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 [imath]\frac{e^{\_\_\_s}}{s} \iff H[y + \_\_\_][/imath] for each exponential. After some algebra, I arrive at [math]G = \frac{H[y - y_G - xi + x_G i]H[x - x_G]H[y - y_G]i - H[y - y_G + xi - x_G i]H[x - x_G]H[y - y_G]i}{2}[/math]
The Boundary Value Problem has been refined into [imath]\triangle u = f(x,\ y),\ u(x,\ 0) = 0,\ u_y(x,\ 0) = 0,\ u(0,\ y) = 0,\ u_x(0,\ y) = 0[/imath]. 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 [imath]i[/imath]'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?