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Two-Dimensional Wave Equation

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1. Find the solution to the two-dimensional wave equation
[see the attachment for the full equation]

2. Solve the two-dimensional wave equation for a quarter-circular membrane
[see the attachment for the full equation]
The boundary condition is such that u=0 on the entire boundary.

3. Consider Laplace's equation
[see the attachment for the full equation]
a. Give a brief physical interpretation of this equation.
b. Suppose that u(x,y,t)=f(x)g(y)h(t)
What ordinary differential equations are satisfied by f, g, and h?

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Solution Summary

The expert finds the solution to the two-dimensional wave equations. Laplace's equations are considered.

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1.

The two dimensional wave equation on the square membrane is:
(1.1)
with boundary conditions:
(1.2)
And initial conditions:
(1.3)
We start with assuming we can write the solution as a product of three completely independent functions:
(1.4)
Therefore the partial derivative become full derivatives, for example:
(1.5)
Plugging this into (1.1) we get:

(1.6)
On the left hand side the expression depends only on t, while the right hand side is a function of x and y.
Since this equation must hold for any (x, y, t), the only way this can be satisfied is if both sides equal the same (arbitrary) constant:
(1.7)
The time dependent equation is:
(1.8)
While the spatial equation is:
(1.9)
Here we see that the left hand side is a function of x while the right hand side depends on y. Again, this can occur for any (x,y) if and only if both sides equal the same constant:
(1.10)
The x dependent equation is:
(1.11)
While the y dependent equation is:
(1.12)
Where we define the constant:
(1.13)
From the boundary conditions we get conditions for the spatial equations namely:
(1.14)
And:
(1.15)
Note that equation (1.11) and equation (1.12) are identical (up to a constant) and have the same boundary conditions, hence they will have identical solutions.
We begin with noting the three cases.

Case 1:
The equation becomes:
(1.16)
Its solution is:
(1.17)
Applying conditions (1.14):

We obtain the trivial solution
(1.18)
Case 2:
The equation becomes:
(1.19)
Its solution is:
(1.20)

Applying conditions (1.14):

We obtain the trivial solution
(1.21)

Case 3:
The equation becomes a simple harmonic:
(1.22)
Its solution is:
(1.23)
Applying conditions (1.14):

We obtain the non-trivial solution
(1.24)
And the eigenvalue is:
(1.25)
By the same process we see that the solution for the y-dependent equation (1.12) is:

(1.26)
And its eigenvalues are:
(1.27)
Thus, from equation (1.13):
(1.28)
Plugging this in the temporal equation (1.8) we obtain:

(1.29)

The general solution of equation (1.1) is a linear combination of all possible solutions:
...

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