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# Heat equation on a rectangular domain

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2-1 a, b

Consider the heat equation for a rectangular region, 0 < x < a, 0 < y < b, t > 0

ut = k(uxx + uyy) , 0 < x < a, 0 < y < b, t > 0

subject to the initial conditions: u(x,y) = f(x,y)

a) ux (0, y, t) = 0, ux (a, y, t) = 0, 0 < y < b, t > 0
uy (x, 0, t) = 0, uy (x, b, t) = 0, 0 < x < a, t > 0

b) u (0, y, t) = 0, ux (a, y, t) = 0, 0 < y < b, t > 0
uy (x, 0, t) = 0, uy (x, b, t) = 0, 0 < x < a, t > 0

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https://brainmass.com/math/fourier-analysis/heat-equation-rectangular-domain-626271

#### Solution Preview

Hi C
Here it is.
The difference between part (a) and (b) is only in the x-dependent eigenfunction, so most of the analysis of part (a) is valid to part (b).

Part (a)
The equation is:
(1.1)
With Neumann boundary conditions:
(1.2)
And
(1.3)
While the initial condition is:
(1.4)
We use separation of variables:
(1.5)
The boundary conditions become:

(1.6)
And:

(1.7)
Substituting (1.5) into (1.1) we get:

(1.8)
The left hand side is a function of t while the right hand sides is s function of x and y.
Since it must be true for any both sides must be equal a constant:
(1.9)
Furthermore,

(1.10)
The left hand side of (1.10) is a function of x while the right hand side is a function of y, therefore both sides must be equal a constant:
(1.11)

And by the same token we can write:

(1.12)
so again the equation is separated and both sides equal a different constant:
(1.13)
And from (1.9) we see that :

(1.14)
(1.15)
with the boundary conditions
(1.16)
We distinguish between three cases:
Case 1:
The equation is and its solution is
(1.17)

Applying the boundary conditions:

We get the trivial solution
Case ...

#### Solution Summary

The 17-pages solution shows in great detail how to apply the method of separation of variables to the heat equation on a rectangle with mixed boundary conditions, and how to obtain the expansion coefficients using Fourier analysis.

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