Laplace's Equation Boundary Problem

sqrt(-1)
Messages
17
Reaction score
0

Homework Statement



I have a two part question, the first part involves solving Laplace's equation

<br /> u_{xx} + u_{yy} = 0<br />

for the boundary conditions
<br /> u_x(0,y) = u_x(2,y) = 0<br />

<br /> u(x,0) = 0<br />

<br /> u(x,1) = \sin(\pi x)<br />
for
0 &lt; x &lt; 2, 0 &lt; y &lt; 1.

The second part now states a new boundary problem for the same equation, involving the square area defined on 0 &lt; x &lt; 1, 0 &lt; y &lt; 1. This time we have
<br /> u_x(0,y) = u(1,y) = 0<br />

<br /> u(x,0) = 0<br />

<br /> u(x,1) = 2\sin(\pi x)<br />
The question asks me to use the solution from the first boundary problem to solve this problem directly (using a theorem/principle).

Homework Equations


The Attempt at a Solution



I have solved the first part using the standard method of separation of variables but I'm rather puzzled as to what this mystery theorem/principle is that can allow me to take my solution from the first problem and apply it directly to the second boundary problem :frown:
 
Physics news on Phys.org
Since the first problem has x between 0 and 2 and the second has x between 0 and 1, what would happen if you replaced the "x" in the first solution by "2x"?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top