Suppose I want to solve [imath]u_{xy} = xy[/imath] via Green's Function. This will correspond to the associated PDE [imath]G_{xy} = \delta(x - x_G,\ y - y_G)[/imath], and I want my boundary conditions for this Green problem to be [imath]G(0,\ y) = 0,\ G_x(x,\ 0) = 0[/imath].
The first step logically would actually be to find the adjoint of the differential operator and boundary conditions for [imath]u[/imath] to arrive at the boundary value problem for [imath]G[/imath], but I'll just assume everything is self-adjoint for now, and return to that step at the end. I solve [imath]G_{xy} = \delta(x - x_G,\ y - y_G),\ G(0,\ y) = 0,\ G_x(x,\ 0) = 0[/imath] using a double Laplace Transform. My double-transformed equation is [imath]\hat{\hat G} = \frac{1}{rs}e^{-y_G s - x_G r}[/imath], and unwinding the transforms, my Green's Function turns out to be [imath]G = H[x - x_G]H[y - y_G][/imath], where [imath]H[\ ][/imath] is the Heaviside Function.
The solution to the original PDE thus should be [imath]u = \int_0^L \int_0^W x_G y_G H[x - x_G]H[y - y_G]\ dy_G\ dx_G[/imath]. Making use of the identity [imath]H[a - b] = 1 - H[b - a][/imath], I can eliminate the Heaviside Functions by modifying the integration limits, and then straightforward calc computations yield [imath]u = \frac{x^2 y^2}{4}[/imath] (the domain dimensions [imath]L[/imath] and [imath]W[/imath] cancel).
If the operator/boundary conditions are indeed self-adjoint, and if I did everything correctly, then [imath]u_{xy} = xy,\ u(0,\ y) = 0,\ u_x(x,\ 0) = 0[/imath] should yield [imath]u = \frac{x^2 y^2}{4}[/imath]. Solving this problem by simple integration and application of the boundary conditions confirms this answer, which in turn strongly confirms the self-adjointness hypothesis as well as my computations so far.
Returning to the beginning, all that remains is to find the adjoint to verify this. I have [imath]u_{xy} = xy[/imath], and I take the inner product of both sides with [imath]G[/imath]. So I have to use integration by parts on [imath]\int_0^L \int_0^W u_{xy}G\ dy\ dx[/imath] with respect to [imath]y[/imath]. This gives [imath]\int_0^L (u_x G |_{y = 0}^{y = W} - \int_0^W u_x G_y\ dy) dx[/imath]. After distributing the outer integral, swapping the order of integration in the double integral, and applying integration by parts again, this time with respect to [imath]x[/imath], I get [imath]\int_0^L u_x G |_{y = 0}^{y = W}\ dx - \int_0^W u G_y |_{x = 0}^{x = L}\ dy + \int_0^L \int_0^W uG_{xy}\ dy\ dx[/imath].
The final term is indeed the original inner product with the differential operator moved from [imath]u[/imath] to [imath]G[/imath]. I just have to check the boundary conditions now; I plug [imath]u(0,\ y) = 0[/imath] and [imath]u_x(x,\ 0) = 0[/imath] into [imath]\int_0^L u_x G |_{y = 0}^{y = W}\ dx - \int_0^W u G_y |_{x = 0}^{x = L}\ dy[/imath], and am left with [imath]\int_0^L u(x,\ W)G(x,\ W)\ dx - \int_0^W u(L,\ y)G(L,\ y)\ dy[/imath].
However, my understanding is that I now have to pick two boundary conditions in [imath]G[/imath] that make this expression [imath]0[/imath], and all I have left to use are [imath]G(x,\ W)[/imath] and [imath]G(L,\ y)[/imath]. Setting these to [imath]0[/imath] won't give me the [imath]G(0,\ y) = 0[/imath] and [imath]G_x(x,\ 0) = 0[/imath] boundary conditions I expected. This seems to suggest that the operator [imath]\frac{\partial}{\partial y \partial x}[/imath] is actually not self-adjoint. But this doesn't make sense either, because when I assumed self-adjointness, I got the correct answer, and when I use simple integration to solve [imath]u_{xy} = xy[/imath] and apply the boundary conditions in [imath]u[/imath] which would produce the needed boundary conditions in [imath]G[/imath] (they are [imath]u(x,\ W) = 0[/imath] and [imath]u_y(L,\ y) = 0[/imath]), I no longer get the solution [imath]u = \frac{x^2 y^2}{4}[/imath] to match my Green's Function solution.
What am I doing wrong here? Is there some simple mistake in my computations, or do I understand the procedure of finding the adjoint operator/boundary conditions incorrectly?
The first step logically would actually be to find the adjoint of the differential operator and boundary conditions for [imath]u[/imath] to arrive at the boundary value problem for [imath]G[/imath], but I'll just assume everything is self-adjoint for now, and return to that step at the end. I solve [imath]G_{xy} = \delta(x - x_G,\ y - y_G),\ G(0,\ y) = 0,\ G_x(x,\ 0) = 0[/imath] using a double Laplace Transform. My double-transformed equation is [imath]\hat{\hat G} = \frac{1}{rs}e^{-y_G s - x_G r}[/imath], and unwinding the transforms, my Green's Function turns out to be [imath]G = H[x - x_G]H[y - y_G][/imath], where [imath]H[\ ][/imath] is the Heaviside Function.
The solution to the original PDE thus should be [imath]u = \int_0^L \int_0^W x_G y_G H[x - x_G]H[y - y_G]\ dy_G\ dx_G[/imath]. Making use of the identity [imath]H[a - b] = 1 - H[b - a][/imath], I can eliminate the Heaviside Functions by modifying the integration limits, and then straightforward calc computations yield [imath]u = \frac{x^2 y^2}{4}[/imath] (the domain dimensions [imath]L[/imath] and [imath]W[/imath] cancel).
If the operator/boundary conditions are indeed self-adjoint, and if I did everything correctly, then [imath]u_{xy} = xy,\ u(0,\ y) = 0,\ u_x(x,\ 0) = 0[/imath] should yield [imath]u = \frac{x^2 y^2}{4}[/imath]. Solving this problem by simple integration and application of the boundary conditions confirms this answer, which in turn strongly confirms the self-adjointness hypothesis as well as my computations so far.
Returning to the beginning, all that remains is to find the adjoint to verify this. I have [imath]u_{xy} = xy[/imath], and I take the inner product of both sides with [imath]G[/imath]. So I have to use integration by parts on [imath]\int_0^L \int_0^W u_{xy}G\ dy\ dx[/imath] with respect to [imath]y[/imath]. This gives [imath]\int_0^L (u_x G |_{y = 0}^{y = W} - \int_0^W u_x G_y\ dy) dx[/imath]. After distributing the outer integral, swapping the order of integration in the double integral, and applying integration by parts again, this time with respect to [imath]x[/imath], I get [imath]\int_0^L u_x G |_{y = 0}^{y = W}\ dx - \int_0^W u G_y |_{x = 0}^{x = L}\ dy + \int_0^L \int_0^W uG_{xy}\ dy\ dx[/imath].
The final term is indeed the original inner product with the differential operator moved from [imath]u[/imath] to [imath]G[/imath]. I just have to check the boundary conditions now; I plug [imath]u(0,\ y) = 0[/imath] and [imath]u_x(x,\ 0) = 0[/imath] into [imath]\int_0^L u_x G |_{y = 0}^{y = W}\ dx - \int_0^W u G_y |_{x = 0}^{x = L}\ dy[/imath], and am left with [imath]\int_0^L u(x,\ W)G(x,\ W)\ dx - \int_0^W u(L,\ y)G(L,\ y)\ dy[/imath].
However, my understanding is that I now have to pick two boundary conditions in [imath]G[/imath] that make this expression [imath]0[/imath], and all I have left to use are [imath]G(x,\ W)[/imath] and [imath]G(L,\ y)[/imath]. Setting these to [imath]0[/imath] won't give me the [imath]G(0,\ y) = 0[/imath] and [imath]G_x(x,\ 0) = 0[/imath] boundary conditions I expected. This seems to suggest that the operator [imath]\frac{\partial}{\partial y \partial x}[/imath] is actually not self-adjoint. But this doesn't make sense either, because when I assumed self-adjointness, I got the correct answer, and when I use simple integration to solve [imath]u_{xy} = xy[/imath] and apply the boundary conditions in [imath]u[/imath] which would produce the needed boundary conditions in [imath]G[/imath] (they are [imath]u(x,\ W) = 0[/imath] and [imath]u_y(L,\ y) = 0[/imath]), I no longer get the solution [imath]u = \frac{x^2 y^2}{4}[/imath] to match my Green's Function solution.
What am I doing wrong here? Is there some simple mistake in my computations, or do I understand the procedure of finding the adjoint operator/boundary conditions incorrectly?