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Next: Modelling: Derivation of the Up: The Method of Separation Previous:

$C = - p^{2}$

The case when the separation constant, $C$, is negative is the interesting case and generates a non-trivial solution. The two equations are now
$\displaystyle X^{\prime\prime} + p^{2} X$ $\textstyle =$ $\displaystyle 0,$ (2.29)
$\displaystyle \ddot{T} + c^{2}p^{2} T$ $\textstyle =$ $\displaystyle 0.$ (2.30)

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

\begin{displaymath}
X(x) = e^{\lambda x},
\end{displaymath}

where

\begin{displaymath}
\lambda^{2} = - p^{2},
\end{displaymath}

and so we obtain the purely imaginary solutions

\begin{displaymath}
\lambda = \pm i p.
\end{displaymath}

Thus, the solution to (2.29) is
\begin{displaymath}
X(x) = A \cos px + B \sin px .
\end{displaymath} (2.31)

The boundary condition at $x=0$ implies that
\begin{displaymath}
u(0,t) = X(0)T(t) = 0, \qquad \Rightarrow \qquad X(0) = 0,
\end{displaymath} (2.32)

and so using this in (2.31) gives

\begin{displaymath}
A = 0.
\end{displaymath}

The condition at $x=L$ gives

\begin{displaymath}
X(L) = B \sin pL = 0.
\end{displaymath}

If $B=0$ we have the trivial solution, but there is a non-trivial solution if
\begin{displaymath}
\sin pL = 0, \qquad \Rightarrow \qquad pL = n\pi,
\end{displaymath} (2.33)

where $n$ is an integer.

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

\begin{displaymath}
T(t) = D \cos pct + E \sin pct,
\end{displaymath} (2.34)

where $p = n\pi /L$. Thus, a solution for $u(x,t)$ is
\begin{displaymath}
u(x,t) = A \sin {n\pi\over L}x \left (D \cos {n\pi c\over L}t + E
\sin {n\pi c\over L}t \right ).
\end{displaymath} (2.35)

Note that the constants $A$, $D$ and $E$ can be combined to give just two constants. One multiplies the cosine term and the other multiplies the sine term. Thus, we can set $A=1$ without any loss of generality.

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

Hence, we obtain

\begin{displaymath}
u(x,t) = \sum_{n=1}^{\infty} \sin {n\pi\over L}x \left (D_{n}\cos
{n\pi c\over L}t + E_{n}\sin {n\pi c\over L}t \right ).
\end{displaymath} (2.36)

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

We now need to apply the initial conditions. Setting $t=0$, (2.35) becomes

\begin{displaymath}
u(x,0) = \sum_{n=1}^{\infty} D_{n}\sin {n\pi\over L}x ,
\end{displaymath} (2.37)

since $\sin (0) = 0$ and $\cos (0) = 1$. This must equal $f(x)$. Hence, we must express $f(x)$ 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

\begin{displaymath}
f(x) = \sum_{n=1}^{\infty} D_{n} \sin {n\pi\over L}x .
\end{displaymath}

To determine the constants, $D_{n}$, we multiply both sides of the equation by $\sin m\pi x/L$ and integrate from $x=0$ to $x=L$. Hence,
\begin{displaymath}
\int_{0}^{L}f(x) \sin {m\pi\over L}x dx =
\int_{0}^{L}\le...
... D_{n} \sin {n\pi\over L}x \sin
{m\pi \over L}x \right ) dx.
\end{displaymath} (2.38)

In (2.38) we interchange the order of integration and summation to obtain
\begin{displaymath}
\int_{0}^{L}f(x) \sin {m\pi\over L}x dx = \sum_{n=1}^{\inft...
... D_{n} \sin {n\pi\over L}x \sin {m\pi \over L}x dx
\right ).
\end{displaymath} (2.39)

Now we use the orthogonality conditions given above to obtain
\begin{displaymath}
\int_{0}^{L}f(x) \sin {m\pi\over L}x dx = D_{m} {L\over 2}.
\end{displaymath} (2.40)

Replacing $m$ by $n$ we obtain
\begin{displaymath}
D_{n} = {2\over L} \int_{0}^{L}f(x) \sin {n\pi\over L}x dx.
\end{displaymath} (2.41)

The other initial condition requires the time derivative of $u(x,t)$. Thus,

\begin{displaymath}
{\partial u \over \partial t} = \sum_{n=1}^{\infty} {n\pi c...
...cos {n\pi c\over L}t - D_{n}
\sin {n\pi c\over L}t \right ).
\end{displaymath}

Hence, at $t=0$ we have

\begin{displaymath}
{\partial u \over \partial t}(x,0) = \sum_{n=1}^{\infty} {n\pi c \over
L} E_{n} \sin {n \pi \over L}x.
\end{displaymath}

Using the initial condition (2.19), we now have
\begin{displaymath}
g(x) = \sum_{n=1}^{\infty} {n\pi c \over
L} E_{n} \sin {n \pi \over L}x.
\end{displaymath} (2.42)

Repeating the same procedure as above, we multiply by $\sin m\pi x/L$ and integrate from $x=0$ to $x=L$. 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 $n=m$. Hence,
\begin{displaymath}
\int_{0}^{L}g(x) \sin {m\pi\over L}x dx = {m\pi c\over L}E_{m}{L\over 2}.
\end{displaymath} (2.43)

Thus, the coefficients $E_{n}$ are given by
\begin{displaymath}
E_{n} = {2\over n\pi c} \int_{0}^{L}g(x) \sin {n\pi\over L}x dx.
\end{displaymath} (2.44)

Using the definitions of $D_{n}$ and $E_{n}$, the final solution to the wave equation, that satisfies both the initial and boundary conditions is

\begin{displaymath}
u(x,t) = \sum_{n=1}^{\infty} \sin {n\pi\over L}x \left (D_{n}\cos
{n\pi c\over L}t + E_{n}\sin {n\pi c\over L}t \right ).
\end{displaymath}

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

Example 2. .20Using the geometric series

\begin{displaymath}
{1\over 1-x} = 1 + x + x^{2} + x^{3} + \cdots,
\end{displaymath}

we see that we can express

\begin{displaymath}
{1\over 1 - {1\over 2}\sin x \cos t} = \sum_{n=0}^{\infty} {1\over
2^{n}}\sin^{n}x \cos^{n}t.
\end{displaymath}

Each term in the sum is separable, having the form $\sin^{n}x
\cos^{n}t$, but the left hand side cannot be expressed as a function of $x$ times a function of $t$.

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

\begin{displaymath}
\sin {n\pi \over L}x \cos {n\pi c\over L}t = {1\over 2}\lef...
...n \pi \over L}(x - ct) + \sin {n\pi \over L}(x + ct) \right ),
\end{displaymath}

and

\begin{displaymath}
\sin {n\pi \over L}x \sin {n\pi c\over L}t = {1\over 2}\lef...
...n \pi \over L}(x - ct) - \cos {n\pi \over L}(x + ct) \right ).
\end{displaymath}

These expressions are both of the form of $F(x - ct)$ and $G(x+ct)$. 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 $f(x)$ and $g(x)$. In all these examples we choose $L= \pi$ and $ c = 1$.

Example 2. .22Consider

\begin{displaymath}
u(x,0) = f(x) = \sin x, \qquad \hbox{and} \qquad {\partial u \over \partial
t}(x,0) = g(x) = 0.
\end{displaymath}

From (2.44) we have $E_{n}= 0$ for all $n$ and from (2.41) we have $D_{1} = 1$ and $D_{n} = 0$ for $n > 1$. We can verify this quite easily. Consider

\begin{displaymath}
\sin x\sin nx = {1\over 2}\left ( \cos (1-n )x - \cos (1 + n)x
\right ),
\end{displaymath}

for $n \ne 1$. Clearly,

\begin{displaymath}
\int_{0}^{\pi}\sin x \sin nx dx = {1\over 2} \left [ -{\sin
(1-n)x\over 1-n} + {\sin (1+n)x\over 1+n} \right ]_{0}^{\pi},
\end{displaymath}

and so

\begin{displaymath}
\int_{0}^{\pi}\sin x \sin nx dx = {1\over 2} \left [\left (...
...(1+n)\pi\over 1+n} \right ) - \left (0 \right )\right ] =
0.
\end{displaymath}

Finally,

\begin{displaymath}
\int_{0}^{\pi}\sin^{2} x dx = {1\over 2}\int_{0}^{\pi}\left ( 1 -
\cos 2x \right ) dx = {\pi \over 2}.
\end{displaymath}

Thus, the only term in the Fourier series that is non-zero is the first term and so $D_{1} = 1$.

Example 2. .23 Consider

\begin{displaymath}
f(x) = \left \{
\begin{array}{cc}
{2 x\over \pi} & \hbox{...
... x) & \hbox{if } {\pi \over 2} < x < \pi
\end{array} \right .
\end{displaymath} (2.45)

This is a triangular profile for $u(x,0) = f(x)$ and again we take $\partial u /\partial t (x,0) = g(x) = 0$. Since $\partial u /\partial t = 0$ we have $E_{n}= 0$ and $D_{n}$ are given by

\begin{displaymath}
D_{n} = {2\over \pi}{2\over \pi}\left ( \int_{0}^{\pi/2}x \sin nx
dx + \int_{\pi/2}^{\pi}(\pi - x) \sin nx dx \right ).
\end{displaymath}

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

\begin{eqnarray*}
\int_{0}^{\pi/2}x \sin nx dx & = & \left [ -{x\over n}\cos nx...
...}\right ) + {1\over
n^{2}} \sin \left ({n\pi \over 2}\right ).
\end{eqnarray*}



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

\begin{eqnarray*}
\int_{\pi /2}^{\pi}(\pi - x )\sin nx dx & = & \left [ -{(\pi ...
...}\right ) + {1\over
n^{2}} \sin \left ({n\pi \over 2}\right ).
\end{eqnarray*}



Adding the two terms together gives

\begin{displaymath}
D_{n} = {4\over \pi^{2}}\left ( {2\over n^{2}} \sin \left ({n\pi
\over 2}\right )\right ).
\end{displaymath}

Notice that when $n$ is even we have the sine of a multiple of $\pi$ and so the even coefficients are zero. For $n= 2m +1$, so that $n$ is odd, we have

\begin{displaymath}
\sin \left (m + {1\over 2}\right )\pi = (-1)^{m}.
\end{displaymath}

Thus,

\begin{displaymath}
D_{2m+1} = (-1)^{m}{8\over (2m+1)^{2}\pi^{2}}.
\end{displaymath}

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

\begin{displaymath}
u(x,t) = {8\over \pi^{2}}\sum_{m=0}^{\infty}{(-1)^{m}\over
(2m+1)^{2}}\sin (2m+1)x \cos (2m+1)t.
\end{displaymath}

Example 2. .24Here we take the initial conditions as $u(x,0) = f(x) = \sin x$ and $\partial
u /\partial t = g(x) = - \sin
x$ 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

\begin{displaymath}
u(x,t) = D \sin x \cos t + E \sin x \sin t.
\end{displaymath}

Thus, $D=1$ and $E=-1$ so that

\begin{displaymath}
u(x,t) = \sin x \cos t - \sin x \sin t.
\end{displaymath}

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

Example 2. .25This time we consider $u(x,0) = f(x) = 0$ and

\begin{displaymath}
{\partial u \over \partial t}(x,0) = g(x) = \left \{
\begi...
... x) & \hbox{if } {\pi \over 2} < x < \pi
\end{array} \right .
\end{displaymath} (2.46)

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

\begin{displaymath}
D_{n} = 0
\end{displaymath}

and, from (2.44), we have

\begin{displaymath}
E_{n} = {2\over n\pi }{2\over \pi}\left ( \int_{0}^{\pi/2}x \sin nx
dx + \int_{\pi/2}^{\pi}(\pi - x) \sin nx dx \right ).
\end{displaymath}

Using the result of example (23), we obtain

\begin{displaymath}
E_{n} = {8\over n^{3}\pi^{2}}\sin \left ({n\pi \over 2}\right ).
\end{displaymath}

Hence, the solution to the wave equation is

\begin{displaymath}
u(x,t) = {8\over \pi^{2}}\sum_{n=0}^{\infty}{(-1)^{n}\over
(2n+1)^{3}} \sin (2n+1)x \sin (2n+1)t.
\end{displaymath}

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

\begin{displaymath}
{\partial^{2} u\over \partial y \partial x} - u = 0.
\end{displaymath}

Again we set

\begin{displaymath}
u(x,y) = X(x)Y(y).
\end{displaymath}

so that the equation becomes

\begin{displaymath}
{dX\over dx}{dY\over dy} = XY.
\end{displaymath}

Dividing by $(dX/dx) Y$, instead of $XY$, we obtain

\begin{displaymath}
{1\over Y}{dY\over dy} = X {1\over dX/dx}.
\end{displaymath}

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

\begin{displaymath}
{dY\over dy} = k Y,
\end{displaymath}

and

\begin{displaymath}
{dX\over dx} = {1\over k} X.
\end{displaymath}

These are easily solved in terms of exponential functions to give

\begin{displaymath}
Y(y) = A e^{ky},
\end{displaymath}

and

\begin{displaymath}
X(x) = B e^{x/k}.
\end{displaymath}

Thus, the general solution to the partial differnetial equation is

\begin{displaymath}
u(x,y) = C e^{ky + x/k}.
\end{displaymath}

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

\begin{displaymath}
x^{2}{\partial^{2}u\over \partial y \partial x} + 3y^{2}u = 0.
\end{displaymath}

Again we set

\begin{displaymath}
u(x,y) = X(x)Y(y),
\end{displaymath}

substitute into the equation and divide by $X(dY/dy)$ to obtain

\begin{displaymath}
x^{2}{dX\over dx}{1\over X} = - 3y^{2}Y{1\over dY/dy}.
\end{displaymath}

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

\begin{displaymath}
x^{2}{dX\over dx} = k X,
\end{displaymath}

and

\begin{displaymath}
{dY\over dy} = - 3{y^{2}\over k}Y.
\end{displaymath}

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

\begin{displaymath}
{dX\over X} = {k\over x^{2}} dx \qquad \Rightarrow \qquad \log X =
-{k\over x} + c,
\end{displaymath}

and hence

\begin{displaymath}
X(x) = A e^{-k/x}.
\end{displaymath}

The $y$ dependence gives

\begin{displaymath}
{dY\over Y} = - 3 {y^{2}\over k} Y \qquad \Rightarrow \qquad \log
Y = - {y^{3}\over k} + c,
\end{displaymath}

so that

\begin{displaymath}
Y(y) = B e^{-y^{3}/k}.
\end{displaymath}

Thus, the general solution to the equation is

\begin{displaymath}
u(x,y) = C e^{-k/x - y^{3}/k}.
\end{displaymath}


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