How do I inverse sine transform U(k,y) to get u(x,y)?

[urista]

I'm trying to solve the Laplacian in 2D:
uxx+uyy=0 in the quarter plane x>0, y>0
using a Fourier sine transform
Boundary Conditions:
u(x,0)=DiracDelta(x-a) , 0< x < infinity and 0< a< infinity
u(0,y)=0, 0< y < infnity

I transformed the PDE in x using the definition of the transform with squareroot(2/Pi) in front of the integral transform and got:
Uyy-k^2*U=0
hence,
U=SquareRoot(2/Pi)*Sin(k*a)*Exp(-k*y) Is this correct?
and how do I inverse sine transform this U(k,y) to get u(x,y)

Any help please?

 

[M98Ranger]

To inverse sine transform U(k,y) and get u(x,y), you can use the inverse Fourier sine transform defined as:

u(x,y) = (2/Pi) * Integral from 0 to infinity of U(k,y) * sin(kx) dk

In this case, since U(k,y) = SquareRoot(2/Pi) * sin(ka) * exp(-ky), we can substitute it into the above formula to get:

u(x,y) = (2/Pi) * Integral from 0 to infinity of SquareRoot(2/Pi) * sin(ka) * exp(-ky) * sin(kx) dk

Using trigonometric identities, we can simplify this to:

u(x,y) = (1/Pi) * Integral from 0 to infinity of sin(ka) * sin(kx) * exp(-ky) dk

Now, we can use the identity sin(a) * sin(b) = (1/2) * (cos(a-b) - cos(a+b)) to further simplify the integral to:

u(x,y) = (1/2Pi) * Integral from 0 to infinity of (cos(ka-kx) - cos(ka+kx)) * exp(-ky) dk

Finally, we can use the inverse Fourier cosine transform to evaluate the integral and get the final solution:

u(x,y) = (1/2Pi) * (exp(-ky) * (delta(x-a) - delta(x+a)))

Where delta(x) is the Dirac delta function. This satisfies the given boundary conditions and solves the Laplacian in the quarter plane x>0, y>0.