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# Dirichlet Problems

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1. Consider the Dirichlet Problem where the temperature within a rectangular plate R is steady-state and does not change with respect to time. Find the temperature u(x,y) within the plate for the boundary conditions below and where (see attached).

2. Solve the Dirichlet problem for steady-state (constant with respect to time) temperature within the wedge shown below. Find the temperature (see attached) within the wedge. The temperature along the arc is (see attached).

Hint: For the solution in ρ, which is the radius, the coefficient for one of the two independent solutions must be zero, since the solution must be finite (can't go to infinity) when ρ=0.

https://brainmass.com/math/calculus-and-analysis/dirichlet-temperature-boundary-conditions-530410

## SOLUTION This solution is FREE courtesy of BrainMass!

The solution is attached below in two files. the files are identical in content, only differ in format. The first is in MS Word format, while the other is in Adobe pdf format. Therefore you can choose the format that is most suitable to you.

1.
The two-dimensional heat equation is:
(1.1)
Under steady state conditions and the problem turns into Laplace equation:
(1.2)
In the case of the rectangle the boundary conditions are:
(1.3)
And:
(1.4)
Note that (1.4) are periodic boundary conditions that tells us w should find the eigenfunctions in the form of .
We start with separation of variables. We write the temperature function as a product of two single-variable independent functions:
(1.5)
Therefore:
(1.6)

Plugging it back into the equation we obtain:

(1.7)
So each side of equation (1.7) is completely independent of the other, and since it is true for any both sides must equal the same constant:
(1.8)
The conditions for are given in (1.4):
(1.9)
Now we need to solve equation (1.8) for :
(1.10)

Case 1:
The equation becomes:
(1.11)
Its solution is:
(1.12)
Applying boundary conditions (recall that )

(1.13)
We get the trivial solution.

Case 2:
The equation becomes:
(1.14)
Its solution is
(1.15)
Applying boundary conditions:

So again we get the trivial solution
(1.16)

Case 3:
The equation (1.10) is now:
(1.17)
This is the harmonic equation. Its solution is:
(1.18)
Applying boundary conditions:

The eigenfunctions are:
(1.19)
And the eigenvalues are
(1.20)

(1.21)
The solution is:
(1.22)
Therefore the general solution of the Laplace equation is a linear combination of all possible solutions:

(1.23)
From the boundary condition we get:

Thus:
(1.24)

From the last boundary condition we get:
(1.25)
If we define:
(1.26)
Then (1.25) can be written as
(1.27)
To find the coefficients we use the orthogonality of the eigenfunctions for m,n integers:
(1.28)
Multiplying (1.27) by we get:
(1.29)

Since the summation is over n and not m, we can bring it into the sum:
(1.30)
Integrating:
(1.31)
Integral of a sum is the same as a sum of integrals:
(1.32)
Due to the orthogonality, the only non-zero term in the sum is that where m=n, so according to (1.28):
(1.33)
And
(1.34)
In our case
(1.35)

Using integration by parts:

(1.36)
Therefore:

(1.37)
Plugging this into (1.34):
(1.38)
So the solution is:

(1.39)
2.

Again, the heat equation in steady state becomes Laplace equation:
(2.1)
In polar coordinates the two dimensional Laplacian is:
(2.2)
So our equation is:
(2.3)
Or:
(2.4)
With the boundary conditions:
(2.5)
And
(2.6)
An implicit condition is that the solution is physical, that is, it is finite everywhere.
As before we use separation of variables:
(2.7)
Note that (2.5) are periodic boundary conditions, so our eigenfunctions are
Then:
(2.8)
Equation (1.24) becomes:

(2.9)
This is now a separated equation and as before it can be true only if both sides equal the same constant
(2.10)

The angular equation is:
(2.11)
The boundary conditions:
(2.12)
Case 1:
The equation becomes:
(2.13)
Its solution is:
(2.14)
Applying boundary conditions:

We get the trivial solution
(2.15)

Case 2:
The equation becomes:
(2.16)

Its solution is:
(2.17)
Applying boundary conditions:

Again, we obtain the trivial solution
(2.18)
Case 3:
The equation becomes:
(2.19)
The harmonic solutions are:
(2.20)
Applying boundary conditions:

The eigenfunctions are:
(2.21)
The eigenvalues are:
(2.22)
With this we can solve the radial part of equation (2.10):
(2.23)
Or:
(2.24)
This is Cauchy equation.
We "guess" a solution in the form:
(2.25)
Then:

(2.26)
(2.27)
Or:
(2.28)
Now come the implicit condition that we require to be finite everywhere in the domain, especially at the origin
This immediately sets B=0 and we are left with:

(2.29)

So, the general solution for equation (2.4) is:

(2.30)
We are left with the last boundary condition:
(2.31)

We use the orthogonality of the eigenfunctions the same way we did in problem (1) to write:
(2.32)
In our case:
(2.33)
This is practically the same function as in problem (1) equation (1.37), so we get:

(2.34)
And the solution is:

(2.35).

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