# Sloving The Laplace's Equation

## Solutions for PDE handout #7

### 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

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]

### 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.

Converted by Mathematica      July 3, 2000