More complex initial/boundary conditions

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5.5 More complex initial/boundary conditions

It is not always possible on separation of variables to separate initial or boundary conditions in a condition on one of the two functions. We can either map the problem into simpler ones by using superposition of boundary conditions, a way discussed below, or we can carry around additional integration constants.

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Let me give an example of these procedures. Consider a vibrating string attached to two air bearings, gliding along rods 4m apart. You are asked to find the displacement for all times, if the initial displacement, i.e. at t = 0s is one meter and the initial velocity is x∕{t}_{0}\ m∕s.

The differential equation and its boundary conditions are easily written down,

\begin{eqnarray} {{∂}^{2}u\over ∂{x}^{2}}& =& {1\over { c}^{2}} {{∂}^{2}u\over ∂{t}^{2}} , %& \\ {∂u\over ∂x}(0,t)& =& {∂u\over ∂x}(4,t) = 0,\kern 2.77695pt t > 0,%& \\ u(x,0)& =& 1, %& \\ {∂u\over ∂t} (x,0)& =& x∕{t}_{0}. %&(5.40) \\ \end{eqnarray}

Question: What happens if I add two solutions v and w of the differential equation that satisfy the same BC’s as above but different IC’s,

\begin{eqnarray} v(x,0) = 0& ,& {∂v\over ∂t} (x,0) = x∕{t}_{0},%& \\ w(x,0) = 1& ,& {∂w\over ∂t} (x,0) = 0? %&(5.41) \\ \end{eqnarray}

Answer: u=v + w, we can add the BC’s.

If we separate variables, u(x,t) = X(x)T(t), we find that we obtain easy boundary conditions for X(x),

X'(0) = X'(4) = 0,

(5.42)

but we have no such luck for (t). As before we solve the eigenvalue equation for X, and find solutions for {λ}_{n} = {{n}^{2}{π}^{2}\over 16} , n = 0,1,..., and {X}_{n} =\mathop{ cos}\nolimits ({nπ\over 4} x). Since we have no boundary conditions for T(t), we have to take the full solution,

\begin{eqnarray} {T}_{0}(t)& =& {A}_{0} + {B}_{0}t, %& \\ {T}_{n}(t)& =& {A}_{n}\mathop{ cos}\nolimits {nπ\over 4} ct + {B}_{n}\mathop{ sin}\nolimits {nπ\over 4} ct,%&(5.43) \\ \end{eqnarray}

and thus

u(x,t) = {1\over 2}({A}_{0} + {B}_{0}t) +{ \mathop{∑ }}_{n=1}^{∞}\left ({A}_{ n}\mathop{ cos}\nolimits {nπ\over 4} ct + {B}_{n}\mathop{ sin}\nolimits {nπ\over 4} ct\right )\mathop{cos}\nolimits {nπ\over 4} x.

(5.44)

Now impose the initial conditions

u(x,0) = 1 = {1\over 2}{A}_{0} +{ \mathop{∑ }}_{n=1}^{∞}{A}_{ n}\mathop{ cos}\nolimits {nπ\over 4} x,
(5.45)

which implies {A}_{0} = 2, {A}_{n} = 0,n > 0.

{∂u\over ∂t} (x,0) = x∕{t}_{0} = {1\over 2}{B}_{0} +{ \mathop{∑ }}_{n=1}^{∞}{nπc\over 4} {B}_{n}\mathop{ cos}\nolimits {nπ\over 4} x.

(5.46)

This is the Fourier sine-series of x, which we have encountered before, and leads to the coefficients {B}_{0} = 4 and {B}_{n} = − {64\over { n}^{3}{π}^{3}c} if n is odd and zero otherwise.

So finally

u(x,t) = (1 + 2t) −{64\over { π}^{3}}{ \mathop{∑ }}_{n\ odd} {1\over { n}^{3}}\mathop{ sin}\nolimits {nπct\over 4} \mathop{cos}\nolimits {nπx\over 4} .

(5.47)

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