Using the method of separation of variables and writing $u(x,t)=M(x)N(t)$, we can solve the equation $$u_{tt}-\gamma^2 u_{xx} + 2\alpha u_t=0$$ $$0<x<l$$ $$t\ge0, \alpha>0 \text{ (}{\alpha} \text{ is small}) $$ $$u(x,0)=f(x), u_t(x,0)=g(x)$$ $$u(0,t)=u(l,t)=0$$

When we separate the variables, we get two ODEs: $$N''(t)+2\alpha N'(t) +\lambda N(t)=0\text{ [eq. (1)]}$$ $$M''(x)+\frac{-\lambda}{\gamma^2}M(x)=0$$

Here I consider the case when $\lambda > 0$. If I'm correct then what we get for solutions of the above ODEs is $$M_n(x)=\underbrace{\sqrt{\frac{2}{l}}}_{\text{normalization condition}}\sin\left(\frac{n\pi x}{l} \right)$$

$$N_n(t) = e^{-t} \left[a_n \cos\left(\sqrt{\alpha^2-\left(\frac{n\pi\alpha}{l}\right)^2}t\right)+b_n\sin\left(\sqrt{\alpha^2-\left(\frac{n\pi\alpha}{l}\right)^2}t\right)+a_0\right]$$

Thus $$u(x,t)=e^{-t} \sum\limits_{n=1}^\infty{ \left[a_n \cos\left(\sqrt{\alpha^2-\left(\frac{n\pi\alpha}{l}\right)^2}t\right)+b_n\sin\left(\sqrt{\alpha^2-\left(\frac{n\pi\alpha}{l}\right)^2}t\right)+a_0\right]\sqrt{\frac{2}{l}}\sin\left(\frac{n\pi x}{l} \right)}$$

Now the part I find might be indicative of some errors I made. In this part we find the Fourier coefficients $a_n$ and $b_n$.

After applying the initial conditions, we get that $$a_n = \sqrt{\frac{l}{2}} f(x) \sin\left(\frac{n\pi x}{l}\right)$$

$$b_n = \frac{\left(g(x)+\sum\limits_{n=1}^\infty a_n\sqrt{\frac{l}{2}}\sin\left(\frac{n\pi x}{l}\right)\right)}{\sqrt{\alpha^2 - \left(\frac{m\pi\alpha}{2}\right)^2}\sqrt{\frac{l}{2}}}$$

The expressions for $a_n$ and $b_n$ seem too tedious. I feel like I've made an error some where, or maybe more than just one error. I would appreciate it very much if I could be given a hint of where I might be mistaken.

I also think that $a_0$ should be solved for as well, but that's an easy step.

  • $\begingroup$ @Rodrigo de Azevedo : in the wording of the question $t>0$ is specified. So, what is the meaning of $u(x,0)=f(x)$ ? Isn't it $t\geq 0$ ? $\endgroup$ – JJacquelin Jul 3 '16 at 13:22
  • $\begingroup$ @JJacquelin I am not the OP. $\endgroup$ – Rodrigo de Azevedo Jul 3 '16 at 13:25
  • $\begingroup$ @Rodrigo de Azevedo : Sorry for the confusion. $\endgroup$ – JJacquelin Jul 3 '16 at 13:29
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    $\begingroup$ @sequence : in the wording of the question $ t>0$ is specified. So, what is the meaning of $u(x,0)=f(x)$ ? Isn't it $t\geq 0$ ? $\endgroup$ – JJacquelin Jul 3 '16 at 13:31
  • $\begingroup$ Thanks, that has been corrected. $\endgroup$ – sequence Jul 3 '16 at 18:21

You've made a couple mistakes here. The first is in the $M$ equation, which should read $M'' = -(\lambda/\gamma^2) M$. This means that your solutions for $\lambda$ are off by a factor involving $\gamma$.

Next, your general solution for $N$ is off. You should double check the algebra you used to find the general solutions. Be careful with how $\lambda$ and $\alpha$ relate to each other.

After that you're missing integrals in your computation of $a_n$ and $b_n$. These are numbers, not functions. You are right, though, that they can be computed in terms of the initial data once you've sorted out the correct form of $M$ and $N$.

  • $\begingroup$ Thanks, I'll re-check my algebra. $f(x)$ and $g(x)$ aren't given explicitly though, so I don't think that $a_n$ and $b_n$ will be numbers in this case. $\endgroup$ – sequence Jul 2 '16 at 17:38
  • $\begingroup$ I initially equated the two equations for $M(x)$ and $N(t)$ to $-\lambda$, that's where the minus sign is coming from in the equation for $M(x)$. $\endgroup$ – sequence Jul 2 '16 at 17:44

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