"Damped" wave equation with Fourier method

Question

The problem I was trying to solve is the following PDE problem

$$\begin{cases} \partial_{tt}^2 u = \partial_{xx}^2 u -\gamma\partial_t u \\[5 pt] u(0,t)= u(\pi, t) = 0 \\[5 pt] u(x,0) = (\sin2x)^4 -{1\over 5}\sin 10x \\[5 pt] \partial_t u(x,t)|_{t=0}=0 \end{cases}\tag 1$$

with the Fourier series method. But I got stuck on the calculations.

What I've done is, first thing first, to evaluate the initial condition to eliminate that fourth power, which can be easily done, and I've got

$$ u(x,0) = {3\over 8}-{1\over 2}\cos4x +{1\over8}\cos8x-{1\over 5}\sin10x $$

This says to me that the solution ought to be of the form

$$ u(x,t) = \sum_n a_n(t)\sin(nx)+b_n(t)\cos(nx) $$

or simply by using the complex exponential, which doesn't change much. I then used the ansatz in the PDE to get two ODE's for the coefficients $a_n(t), b_n(t)$

$$ \sum_n(a''_n(t)\sin(nx)+b''_n(t)\cos(nx))= \\ =-\sum_n(a_n(t)\sin(nx)+b_n(t)\cos(nx))-\gamma\sum_n(a'_n(t)\sin(nx)+b'_n(t)\cos(nx)) $$

and got, equating the coefficients

$$ a''_n(t)= -\gamma a'_n(t)-a_n(t) \\ b''_n(t)= -\gamma b'_n(t)-b_n(t) $$

which are the same equations: the equation of a damped harmonic oscillator. To find the solution we search for the solutions of the polynomial equation

$$ \lambda^2 +\gamma\lambda +1 = 0 $$

which are

$$ \lambda_1 = -{1\over 2}\left(\gamma+\sqrt{\gamma^2-4}\right)\;\;\;\;\; \lambda_2 = -{1\over 2}\left(\gamma-\sqrt{\gamma^2-4}\right) $$

Clearly the solution for the ODE's depends on the value of the "damping coefficient" gamma

$$ \gamma^2-4 \gt 0 \implies \color{red}{a_n(t) = c^a_1e^{\lambda_1 t}+c^a_2e^{\lambda_2 t}}\\ \gamma_2-4\lt 0 \implies \lambda_{1/2} = \mu\pm i\nu \implies \color{orange}{a_n(t) = c^a_1 e^{(\mu+i\nu)t}+c^a_2 e^{(\mu-i\nu)t}} \\ \gamma^2-4=0\implies \lambda_1=\lambda_2=\lambda \implies \color{green}{a_n(t)=c^a_1 e^{\lambda t}+c^a_2 t e^{\lambda t}} $$

and the same goes for $b_n(t)$. But then jumped to my mind that the solution would become very ugly! Knowing my professor I think that there could be a easier way to solving this.

Question 1: In my solution, am I headed in the right way?

Question 2: Is there a simpler method to solve this problem?

Question 3: I thought about using Laplace transform but the initial condition make matter worse: could this be a viable way?

Answer 1

Putting it short, the solution is "ugly", I might guess your professor meant for a nicer initial condition to force many fourier coefficients to zero out..

Anyways, most of your solution so far is fine. You should get that, through SepVar assuming $u(x,t) = X(x)T(t)$ (or your eigenvalue technique, similarly)

$$\begin{align*} X(x) &= c_n \sin{(nx)}\\ T(t) &= B e^{-\frac{t}{2}}\left(\gamma_n \cos{\left(\frac{t}{2}\gamma_n \right)}+\sin{\left(\frac{t}{2}\gamma_n \right)}\right) \end{align*}$$

Where $\gamma_n = \sqrt{4n^2 - 1}$, $n = 1, 2, ...$

Cool, our solution to our problem is

$$u(x, t) = \sum_{n=1}^{\infty}c_n \sin{(nx)}e^{-\frac{t}{2}}\left(\gamma_n \cos{\left(\frac{t}{2}\gamma_n \right)}+\sin{\left(\frac{t}{2}\gamma_n \right)}\right)$$

The not so cool thing is our orthogonality relation gives us a pretty disgusting integral to solve. The formula for our $c_n$'s follows

$$c_n = \frac{2}{\gamma_n \pi}\int_{0}^{\pi}\left(\sin^4{2x}-\frac1{5}\sin{10x}\right)\sin{(nx)}\, dx = -\frac{2}{\pi \gamma_n} \left[\frac{384((-1)^n - 1)}{n(n^4 - 80n^2 + 1024)} \right]$$.

Note for $n = 4, n = 8$, that $c_n = 0$, and whene $n = 10$, $c_n = -\frac{1}{5\sqrt{399}}$.

Hence

$$u(x, t) = e^{-\frac{t}{2}}\sum_{n=1}^{\infty}c_n \sin{(nx)}\left(\gamma_n \cos{\left(\frac{t}{2}\gamma_n \right)}+\sin{\left(\frac{t}{2}\gamma_n \right)}\right)$$

with the above formula for the coefficients.

Answer 2

This problem is the same as this one, but with a specification of the initial data $$ u(x,0) = \sin^4 2x - \tfrac{1}{5}\sin 10x = f(x)\, , $$ which is consistent with the boundary conditions at $x=0$ and $x=\pi$. As such, it is also a sub-case of Exercise 4.4.3 p. 142 of (1). Hence, we have $$ u(x,t) = e^{-\gamma t/2} \sum_{n = 1}^{+\infty} A_n \left(\cos \beta_n t + \frac{\gamma}{2\beta_n} \sin \beta_n t \right) \sin nx \, , \qquad \beta_n = \sqrt{n^2 - \tfrac14 \gamma^2} $$ with $A_n = \frac{2}{\pi}\int_0^{\pi} f(x) \sin nx\,\text dx$, for weak damping $0\leq \gamma < 2$. Using trigonometry, one writes $$ f(x) = \tfrac{3}{8} - \tfrac{1}{2} \cos 4x + \tfrac{1}{8} \cos 8x - \tfrac{1}{5}\sin 10x \, , $$ from which the coefficients $A_n$ can be obtained by linearity. Finally, the only nonzero coefficients are $A_{10}=-\frac15$, and $A_n = \frac{2}{\pi} \frac{768}{n (n^2 - 64) (n^2 - 16)}$ for odd natural integers $n$.

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Note: The above solution seems consistent with the one in @DaveNine 's answer where apparently $\gamma=1$ is implicitly assumed.

Note: this equation is also known as telegraphers' equation or simply telegraph equation.

(1) R. Habermann, Applied Partial Differential Equations; with Fourier Series and Boundary Value Problems, 5th ed., Pearson Education Inc., 2013.