# 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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