Sloving The Laplace's Equation

Solutions for PDE handout #7

by Mohammad Ghomi(the grader)
March 23, 1998

Problem 1

1.1 Basic Formulas

The solution of Laplace's equation over the recatngle [0,Pi]x[0,Pi], subject to the boundary conditions u(x,0)=f1(x), u(x, Pi)=f2(x), u(0,y)=g1(y), and u(Pi,y)=g2(y), is given by

where

and the coefficients A1,..., A4 are :

1.2 Graphs

For some of the graphs you may need to increase the plot range. Also, If your machine is slow, you may decrease the value of for plot points.

1.2.1 Part (a)

If the boundary conditions are:

then the corresponding solution looks like

If you just want to see the function type

and you get your answer for part (a):
Csch[5*Pi]*Sin[5*y]*Sinh[5*x] -
Csch[2*Pi]*Sin[2*y]*Sinh[2*(-Pi + x)] +
Csch[Pi]*Sin[x]*Sinh[Pi - y] + Csch[Pi]*Sin[x]*Sinh[y]

It is also instructive to look at the graph of the four components u1,..., u4:

1.2.2 Part (b)

Here we just need to change the boundary conditions:

  Csch[Pi]*Sin[y]*Sinh[x]
- 3*Csch[4*Pi]*Sin[4*y]*Sinh[4*x]
+ Csch[2*Pi]*Sin[2*x]*Sinh[2*y]

1.2.3 Part (c)

1.2.4 Part (d)

Probelm 2

2.1 Basic Formulas

The solution to the Laplace's equation over the unit disk, subject to the boundary condition f(theta) is given by

where

2.2 Graphs

2.2.1 Part (a)

If our boundary condition is

Then the graph looks

r Sin[t]

2.2.2 Part(b)

If we change the boundary condition

then we get

2
r  Sin[2 t]

Problem 3

Yes except for parts a, and e.