Two Dimensions

In , we define the arc length of a function over the interval to be

If we define a parametric function and , we observe that .  Substituting this into the arc length formula yields

Getting a common denominator gives us

In polar coordinates, we know that and .   If and , then we observe that

Therefore,

Three Dimensions

Let us now consider two additional coordinate systems in :  the cylindrical and spherical coordinate system.

The cylindrical coordinate system is a 3-D version of the polar coordinate system in 2-D with an extra component for .  We can slightly modify our arc length equation in polar to make it apply to the cylindrical coordinate system given that , .  We start from this step:

From rectangular coordinates, the arc length of a parameterized function is

.

So from this, it follows that in cylindrical coordinates, we have arc length defined to be

To define arc length in the spherical coordinate system, we need to know first how to convert points from spherical to cylindrical coordinates (and then spherical to rectangular).   I leave without proof that , , and .  Since we know that , , and goes from cylindrical to rectangular, it follows that , , and takes points in spherical coordinates and converts them into points from the rectangular coordinate system.

Again we consider the arc length formula for a parametric curve in the rectangular coordinate system:

If we let , and , we now can find expressions for , and .  I leave the calculations of these three to the reader, but I will, for the sake of space, write the values to each one.

Since , we have

Since , we have

Since , we have

When you plug this into the rectangular arc length formula and after a ton of painful simplifying (which I have left out to spare your sanity), we see that the arc length formula in spherical coordinates is:

where is the displacement function of a vibrating string with fixed ends, is the initial position function, and is the initial velocity function.

Solution:  The best way to solve this would be to set up two boundary value problems [where each one has one nonhomogeneous condition], and then solve the problems using the technique of Separation of Variables:

Problem A:

If we let , then the PDE becomes .  Now separate the variables to get:

The two functions and agree if they both are equal to the same constant.  We now let :

What we now have is an eigenvalue problem.  We can now solve two ordinary DEs:

Since the endpoint conditions and require that if is non-trivial.  Thus, must satisfy the eigenvalue problem

If , then the eigenvalue problem becomes:

When solved, the general solution is .  When we apply the endpoint conditions, we have .  This gives us the trivial solution .  Thus, is NOT an eigenvalue.

If , we will let .  The eigenvalue problem now becomes:

When solved, the general solution is:

where

Applying the endpoint condition , we have so that .  When we apply the other endpoint , we have .  Thus, since and for .  Thus the only solution is the trivial solution .  As a result, there are no negative eigenvalues.

Now, when , we let   The eigenvalue problem then becomes:

The general solution has the form:

Applying the endpoint condition , we have so that .  But if we apply the other condition , we have .  This can occur when when is a positive multiple of :

Therefore,

Solving for , we get:

Since , we can say that:

We now can see that we have an infinte sequence of eigenvalues defined by:

The associated eigenfunction is

——-
***

The homogeneous initial condition

Implies that .  Thus, the solution that is associated with the eigenvalue must also satisfy the conditions

After some calculations [which I have omitted], we get the general solution:

To apply the condition, we need to find :

Applying the condition , we get:

Therefore, .  Thus, can be defined as the non-trivial solution

Since , we can say that

Each of these terms satisfy the wave equation and the homogeneous boundary condition given at the start of this part of the problem.  Thus, by the principle of superposition, we define as the infinte series:

All we need to do now is find such that it satisfies the nonhomogeneous condition

for .  However, this will be the Fourier Sine Series of if we choose

Therefore, the Formal Series Solution to Problem A is

where

———————————————-

Problem B

Let us start from ***:

The homogeneous initial condition

implies that .  Thus, the solution associated with the eigenvalue must also satisfy the conditions

After some calculations [which I have omitted again...], the general solution takes the form

Applying the initial condition, we get

Thus, can be defined as the non-trivial solution

Since , we can say that

Each of these terms satisfy the wave equation and the homogeneous boundary condition given at the start of this part of the problem.  Thus, by the principle of superposition, we define as the infinte series:

Find :

All we need to do now is find such that it satisfies the nonhomogeneous condition

for .  However, this will be the Fourier Sine Series of if we choose

Therefore, the Formal Series Solution to Problem B is

where

————-

Thus, the Formal Series Solution to the One Dimensional Wave Equation is