I am trying to solve the PDE
\begin{align}
u_{tt} – u_{xx} + 2u &= 0
\end{align}
where $0 \leq x \leq \pi, \ t \geq 0$. The initial conditions are
$$u(x,0) = 0, \quad u_t(x,0) = \frac{x}{\pi}$$
and the boundary conditions are
$$ u_x(0,t) = 0, \quad u_x(\pi,t) = 0$$
I am trying to solve using separation of variables. Making the ansatz
$$u(x,t) = F(x)G(t)$$
after substitution I get
$$\frac{F''(x)-2F(x)}{F(x)} = \frac{\ddot{G}(t)}{G(t)}=-n^2-2$$
I get finally a solution of the form
$$ u(x,t)=\sum_{n=1}^{\infty} A_n \cos(nx) \sin(\sqrt{n^2+2}t) $$
Now I still have to comply with the initial condition for the time derivative
$$u_t(x,0) = \frac{x}{\pi}$$
This implies
$$ u(x,t)=\sum_{n=1}^{\infty} A_n \cos(nx) \sqrt{n^2+2} = \frac{x}{\pi} $$
But $x/\pi$ is an odd function, so it can not be represented by a sum of cosine functions. What did I do wrong?
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\mbox{Lets}\ \mrm{u}_{x}\pars{x,t} = \sum_{n = 1}^{\infty}a_{n}\pars{t}\sin\pars{nx}\ \mbox{which already satisfies}\ \mrm{u}_{x}\pars{0,t} = \mrm{u}_{x}\pars{\pi,t} = 0. \\[5mm] &\ \mbox{Then,}\ \mrm{u}\pars{x,t} = -\sum_{n = 1}^{\infty}a_{n}\pars{t}\,{\cos\pars{nx} \over n} + \mrm{f}\pars{t}. \\ &\ \mrm{f}\pars{t}\ \mbox{is a time dependent }\ arbitrary\ \mbox{function ( for the time being ).} \end{align}
$\ds{\mrm{u}\pars{x,t}}$ must satisfy the differential equation. Namely, \begin{align} &0 = \bracks{-\sum_{n = 1}^{\infty}\ddot{a}_{n}\pars{t}\,{\cos\pars{nx} \over n} + \ddot{\mrm{f}}\pars{t}} - \bracks{\sum_{n = 1}^{\infty}a_{n}\pars{t}n\cos\pars{nx}} \\[5mm] &\ + 2\bracks{-\sum_{n = 1}^{\infty}a_{n}\pars{t}\,{\cos\pars{nx} \over n} + \mrm{f}\pars{t}} \end{align} Integrating both sides over $\ds{\pars{0,\pi}} \implies \ddot{\mrm{f}}\pars{t} + 2\,\mrm{f}\pars{t} = 0 \implies \mrm{f}\pars{t} = A\sin\pars{\root{2}t} + B\cos\pars{\root{2}t}$. $\ds{A\ \mbox{and}\ B}$ are constants.
Similarly, integrate after multiplying both sides for a factor $\ds{\cos\pars{nx}}$ to get \begin{align} &\ddot{a}_{n}\pars{t} + \pars{n^{2} + 2}a_{n}\pars{t} = 0 \\ &\ \implies a_{n}\pars{t} = a_{n}\pars{0}\cos\pars{\root{n^{2} + 2}t} + \dot{a}_{n}\pars{0}\,{\sin\pars{\root{n^{2} + 2}t} \over \root{n^{2} + 2}} \end{align} The general solution becomes: \begin{align} \mrm{u}\pars{x,t} & = -\sum_{n = 1}^{\infty}\bracks{a_{n}\pars{0}\cos\pars{\root{n^{2} + 2}t} + \dot{a}_{n}\pars{0}\,{\sin\pars{\root{n^{2} + 2}t} \over \root{n^{2} + 2}}}\,{\cos\pars{nx} \over n} \\ & + A\sin\pars{\root{2}t} + B\cos\pars{\root{2}t} \end{align} Also, $$ 0 = \mrm{u}\pars{x,0} = -\sum_{n = 1}^{\infty}a_{n}\pars{0}\,{\cos\pars{nx} \over n} + B \implies a_{n}\pars{0} = B = 0 $$ The general solution is reduced to \begin{align} \mrm{u}\pars{x,t} & = -\sum_{n = 1}^{\infty} \dot{a}_{n}\pars{0}\,{\sin\pars{\root{n^{2} + 2}t} \over \root{n^{2} + 2}}\,{\cos\pars{nx} \over n} + A\sin\pars{\root{2}t} \end{align} In addition, \begin{align} \mrm{u}_{t}\pars{x,0} & = {x \over \pi} = -\sum_{n = 1}^{\infty} \dot{a}_{n}\pars{0}\,{\cos\pars{nx} \over n} + \root{2}A \end{align} Integrating both sides over $\ds{\pars{0,\pi} \implies {\pi \over 2} = \root{2}A\pi \implies A = {\root{2} \over 4}}$. Also, \begin{align} &\int_{0}^{\pi}{x \over \pi}\,\cos\pars{nx}\,\dd x = -\,{\pi \over 2n}\dot{a}_{n}\pars{0} \implies {\pars{-1}^{n} - 1 \over n^{2}\pi} = -\,{\pi \over 2n}\dot{a}_{n}\pars{0} \\[5mm] &\ \implies \dot{a}_{n}\pars{0} = {2 \over \pi^{2}}\,{1 - \pars{-1}^{n} \over n} \end{align} Finally, \begin{align} \mrm{u}\pars{x,t} & = -\sum_{n = 1}^{\infty} {2 \over \pi^{2}}\,{1 - \pars{-1}^{n} \over n}\,{\sin\pars{\root{n^{2} + 2}t} \over \root{n^{2} + 2}}\,{\cos\pars{nx} \over n} + {\root{2} \over 4}\,\sin\pars{\root{2}t} \\[5mm] & = \color{red}{-\,{4 \over \pi^{2}}\sum_{n = 0}^{\infty} {1 \over \pars{2n + 1}^{2}}\,{\sin\pars{\root{\bracks{2n + 1}^{2} + 2}t} \over \root{\bracks{2n + 1}^{2} + 2}}\cos\pars{\bracks{2n + 1}x}} \\[2mm] &\ \color{red}{+ {\root{2} \over 4}\,\sin\pars{\root{2}t}} \end{align}