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# 5.5: More Complex Initial/Boundary Conditions

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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.

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 $$t=0$$s is one meter and the initial velocity is $$x/t_0~\rm m/s$$.

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

\begin{aligned} \dfrac{\partial^2}{\partial x^2} u &= \frac{1}{c^2} \dfrac{\partial^2}{\partial t^2} u ,\nonumber\\ \dfrac{\partial}{\partial x} u(0,t) &= \dfrac{\partial}{\partial x} u(4,t) = 0, \;t>0, \nonumber\\ u(x,0) & = 1, \nonumber\\ \dfrac{\partial}{\partial t} u(x,0) & = x /t_0.\end{aligned}

Exercise $$\PageIndex{1}$$

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{aligned} v(x,0) =0 &,& \dfrac{\partial}{\partial t} v(x,0) = x /t_0, \nonumber\\ w(x,0) =1 &,& \dfrac{\partial}{\partial t} w(x,0) = 0?\end{aligned}

$$u$$=$$v+w$$, 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 $$X(x)$$, $X'(0)=X'(4) = 0,$ but we have no such luck for $$(t)$$. As before we solve the eigenvalue equation for $$X$$, and find solutions for $$\lambda_n=\frac{n^2\pi^2}{16}$$, $$n=0,1,...$$, and $$X_n=\cos(\frac{n\pi}{4}x)$$. Since we have no boundary conditions for $$T(t)$$, we have to take the full solution,
\begin{aligned} T_0(t) &= A_0 + B_0 t, \nonumber\\ T_n(t) &= A_n \cos \frac{n\pi}{4} ct + B_n \sin \frac{n\pi}{4} ct,\end{aligned} and thus $u(x,t) = \dfrac{1}{2}(A_0 + B_0 t ) + \sum_{n=1}^\infty \left(A_n \cos \frac{n\pi}{4} ct + B_n \sin \frac{n\pi}{4} ct\right) \cos \frac{n\pi}{4}x.$
• $u(x,0) = 1 = \dfrac{1}{2} A_0 + \sum_{n=1}^\infty A_n \cos \frac{n\pi}{4}x,$ which implies $$A_0=2$$, $$A_n=0, n>0$$.
• $\dfrac{\partial}{\partial t} u(x,0) = x/t_0 = \dfrac{1}{2} B_0 + \sum_{n=1}^\infty \frac{n\pi c}{4} B_n \cos \frac{n\pi}{4}x.$ This is the Fourier sine-series of $$x$$, which we have encountered before, and leads to the coefficients $$B_0=4$$ and $$B_n= -\frac{64}{n^3\pi^3c}$$ if $$n$$ is odd and zero otherwise.
So finally $u(x,t) = (1+2t) -\frac{64}{\pi^3} \sum_{n~\rm odd} \frac{1}{n^3} \sin \frac{n\pi ct}{4 } \cos \frac{n\pi x}{4}.$