Solve the following Laplace’s equations in polar coordinates. (a) On a disk of r

Question

Solve the following Laplace’s equations in polar coordinates.

(a) On a disk of radius 1:

uxx + uyy = 0; u(x, y) = x2y when x2 + y2 = 1.

(Note: You’ll need to convert this problem to polar coordinates. The boundary condition, for example, is: u=cos2sin when r = 1 (since x=rcos and y=rsin ). When you’ve got your final answer in polar coordinates, convert back to x, y coordinates, if you can.)

(b) On a semicircle of radius 1, x2 + y2 1, y 0:

uxx + uyy = 0; u = 0 on the diameter (i.e., when y = 0), and u = 1 on the arc (i.e., when x2 + y2 = 1).

Explanation / Answer

a)

This is Laplace's equation on a circular disk of unit radius.It is easier to convert to polar coordinates using the standardformula:

      Laplacian = (1/r)(d(rdu/dr)/dr) +(1/r2)(d2u/d2)

Using separation of variables we can write the solution as u =R(r)(). Substitute this for u in the partialdifferential equation, divide by R(r)(), and multipleby r2 and we obtain:

      (r/R)(d(r dR/dr)/dr) +(1/)(d2/d2) = 0

The second term only involves the variable and theother term is independent of . Therefore, the second term ofthe this modified partial differential equation must be a constantwhich we'll call -n2 (chosen to be negative so that thesolution with respect to will be periodic and notexponential).

The second term's partial differential equation is(1/[])''[] = -n2. The solutionto this is just c1 cos(n ) + c2 sin(n). Here n must be an integer so that +2n willalways produce the same value as we move around the unitdisk.

The other partial differential equation is then:

      (r/R[r])(r R''[r] + R'[r])= n2. This has the solution c3 cosh(n log[r])+ c4 sinh(n log[r])

The complete solution is the product of these two solutions.We now need to determine the constants that satisfy the boundarycondition when r = 1. Note that when r = 1 the log[r]=0. Alsocosh(0) = 1 but sinh(0) = 0. In polar coordinates the expressionxy2 = r3cos sin2 =(r3/4)(cos -cos3), and r =1 on the boundary.Therefore, the solution must be:

      (1/4)(coscosh(log[r]) -cos3 cosh(3 log[r])) = (1+ r2)/(8r)cos - (1/4)(cos3 cosh(3 log[r]))

This can be written in some other forms if you like. You cancheck that it satisfies the differential equation and works on the boundary.

post b) part again

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