Use conformal mapping to solve the Laplace equation ∇²φ = 0 in the upper half-plane with boundary conditions φ(x,0) = 1 for -1<x<1 and 0 elsewhere.
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Step-by-Step Solution
Step 1: Identify the problem and boundary conditions
We are asked to solve Laplace's equation in the upper half-plane, which means we are looking for a harmonic function ϕ(x,y) for y>0. The boundary conditions are given on the x -axis: ϕ(x,0) is 1 between x=−1 and x=1, and 0 everywhere else on the x -axis.
Step 2: Choose a suitable conformal mapping
A common strategy for problems with piecewise constant boundary conditions on the real axis is to use a conformal mapping that transforms the real axis into a simpler boundary. The mapping w=z+1z−1 transforms the real axis in the z -plane to the real axis in the w -plane, and specifically maps the interval (−1,1) to the negative real axis (−∞,0) in the w -plane. The upper half-plane in z maps to the upper half-plane in w.
Step 3: Transform boundary conditions to the w -plane
Under the mapping w=z+1z−1: When z=x (real axis), w=x+1x−1. For −1<x<1, x−1 is negative and x+1 is positive, so w is negative. As x→−1+, w→−∞. As x→1−, w→0−. Thus, the segment (−1,1) maps to (−∞,0) on the u -axis. For x>1, x−1 and x+1 are both positive, so w is positive. For x<−1, x−1 and x+1 are both negative, so w is positive. Therefore, the boundary condition becomes ϕ(u,0)=1 for u<0 and ϕ(u,0)=0 for u>0.
Step 4: Solve Laplace's equation in the w -plane
In the w -plane, we have a standard problem: ϕ(u,0)=1 for u<0 and ϕ(u,0)=0 for u>0. The solution to this problem in the upper half-plane is given by ϕ(u,v)=π1arg(w), where arg(w) is the principal argument of w, ranging from 0 to π in the upper half-plane. This function is harmonic and satisfies the boundary conditions.
Step 5: Substitute back to the z -plane
Now we substitute w=z+1z−1 back into the solution for ϕ(u,v) to get the solution in terms of z=x+iy. The argument of a complex number w=u+iv can be expressed as arctan(uv) for u>0 and π+arctan(uv) for u<0, or more generally using the `atan2` function. This gives the final solution for ϕ(x,y).