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The case when the separation constant, , is negative is the interesting case and generates a non-trivial solution. The two equations are now
 (2.29) (2.30)

As in the last section we look for a solution to (2.29) of the form

where

and so we obtain the purely imaginary solutions

Thus, the solution to (2.29) is
 (2.31)

The boundary condition at implies that
 (2.32)

and so using this in (2.31) gives

The condition at gives

If we have the trivial solution, but there is a non-trivial solution if
 (2.33)

where is an integer.

With constrained by (2.33), the solution for (2.30) follows in a similar manner and we obtain

 (2.34)

where . Thus, a solution for is
 (2.35)

Note that the constants , and can be combined to give just two constants. One multiplies the cosine term and the other multiplies the sine term. Thus, we can set without any loss of generality.

Now, we have a solution for every integer value of . Thus, to form the most general solution we must add together all the possible solutions.

Hence, we obtain

 (2.36)

The subscripts on the constants are there to signify that the constants, and , will have different values for different values of .

We now need to apply the initial conditions. Setting , (2.35) becomes

 (2.37)

since and . This must equal . Hence, we must express as a Fourier sine series and the coefficients of that Fourier sine series must equal the unknown coefficients in (2.37). Equivalently, we write (2.37) as

To determine the constants, , we multiply both sides of the equation by and integrate from to . Hence,
 (2.38)

In (2.38) we interchange the order of integration and summation to obtain
 (2.39)

Now we use the orthogonality conditions given above to obtain
 (2.40)

Replacing by we obtain
 (2.41)

The other initial condition requires the time derivative of . Thus,

Hence, at we have

Using the initial condition (2.19), we now have
 (2.42)

Repeating the same procedure as above, we multiply by and integrate from to . The orthogonality properties of the sine functions means that only one term remains on the right hand side of (2.42), namely the term when . Hence,
 (2.43)

Thus, the coefficients are given by
 (2.44)

Using the definitions of and , the final solution to the wave equation, that satisfies both the initial and boundary conditions is

Notice that, although the solution was obtained by looking for solutions that were separable in and , the final answer is NOT separable (unless all the coefficients are zero except one).

Example 2. .20Using the geometric series

we see that we can express

Each term in the sum is separable, having the form , but the left hand side cannot be expressed as a function of times a function of .

Example 2. .21We can relate our solution in (2.36) to d'Alembert's solution as follows. Using the trigonometric identities

and

These expressions are both of the form of and . The solution that was obtained in (2.36) corresponds to standing waves whereas d'Alembert's solution corresponds to travelling waves we see that standing waves can be considered as the superposition of an inifinite number of travelling waves travelling to the left and right but interacting in such a way as to produce standing waves.We now illustrate the method of separation with a few examples corresponding to different initial conditions. We choose different forms for and . In all these examples we choose and .

Example 2. .22Consider

From (2.44) we have for all and from (2.41) we have and for . We can verify this quite easily. Consider

for . Clearly,

and so

Finally,

Thus, the only term in the Fourier series that is non-zero is the first term and so .

Example 2. .23 Consider

 (2.45)

This is a triangular profile for and again we take . Since we have and are given by

To evaluate these integrals we use integration by parts. Taking the first integral on the right hand side we obtain

The second integral can be evaluated in a similar manner and we obtain

Adding the two terms together gives

Notice that when is even we have the sine of a multiple of and so the even coefficients are zero. For , so that is odd, we have

Thus,

Hence the solution to the wave equation, for these initial conditions, is

Example 2. .24Here we take the initial conditions as and so that there is an initial velocity and well as an initial displacement. However, since the functions only involve a single sine term, the solution will simply be

Thus, and so that

It is easily verified that this solution does indeed satisfy the initial conditions.

Example 2. .25This time we consider and

 (2.46)

This is similar to example (23). However, now we have

and, from (2.44), we have

Using the result of example (23), we obtain

Hence, the solution to the wave equation is

Finally, we give two examples of the method of separation applied to other partial differential equations. This time we do not include either the boundary or initial conditions.

Example 2. .26Consider the equation

Again we set

so that the equation becomes

Dividing by , instead of , we obtain

Since the left hand side is a function of alone and the right hand side is function of alone, the only possible solution is if they are both equal to a constant, say . Thus, we get two ordinary differential equations of the form

and

These are easily solved in terms of exponential functions to give

and

Thus, the general solution to the partial differnetial equation is

Example 2. .27This example shows how the method can be applied to problems for which the coefficients are functions of and . For the separation to work the coefficients do need to have particular forms. Here

Again we set

substitute into the equation and divide by to obtain

Here the left hand side is a function of alone, whereas the right hand side is a function of alone and so they must both equal a constant, . Thus, the two equations are

and

These equations are both separable first order equations and so we obtain, for the dependence

and hence

The dependence gives

so that

Thus, the general solution to the equation is

Next: Modelling: Derivation of the Up: The Method of Separation Previous:
Prof. Alan Hood
2000-10-30