# 1D Wave Equation Problem Separation of Variables

I need to solve the following 1D Wave Equation problem using Separation of Variables, but I cannot figure it out. \begin{align} u_{tt} &= u_{xx}\\ u_x(t,0) &= u_x(t,1) = 0\\ u(0,x) &= x(1-x)\\ u_t(0,x) &= 0 \end{align} I have done the following work to solve it, but do not know how to find the values of $$D$$, $$A$$, $$B$$, $$d_k$$, or $$b_k$$.

Assume $$u(t,x) = w(t)v(x) \longrightarrow w''(t) = \lambda w(t), v''(x) =\lambda v(x)$$

Case 1: $$\lambda = 0$$ \begin{align} v(x) &= Cx+D\\ v'(x) &= C\\ v'(0) &= v'(1) = C = 0 \end{align} $$D$$ can be anything.

\begin{align} u(t,x) &= D(At+B) +\sum_{k=1}^{\infty}{d_k\cos(k\pi t)\sin(k\pi x) + b_k \sin(k\pi t)\sin(k\pi x)}\\ u(0,x) &= x(1-x) = DB+\sum_{k=1}^{\infty}{d_k\sin(k\pi x)}\\ u_t(0,x) &= 0 = DA+\sum_{k=1}^{\infty}{b_kk\pi\sin(k\pi x)} \end{align} This is as far as I have gotten and I cannot seem to be able to find $$D$$, $$A$$, $$B$$, $$d_k$$, or $$b_k$$. Any help is greatly appreciated.

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$

• $$\ds{\on{u}_{x}\pars{t,x} = -\pi\sum_{n = 1}^{\infty}\on{a}_{n}\pars{t}\,n\sin\pars{n\pi x}}$$ already satisfies the boundary conditions.

\begin{align} &\implies \on{u}\pars{t,x} = \sum_{n = 1}^{\infty}\on{a}_{n}\pars{t}\cos\pars{n\pi x} + \on{f}\pars{t}\label{1}\tag{1} \\[5mm] & \mbox{Note that} \\ &\ \left.\begin{array}{rcl} \ds{\int_{0}^{1}\cos\pars{m\pi x}\cos\pars{n\pi x} \,\dd x} & \ds{=} & \ds{{1 \over 2}\,\delta_{mn}} \\ \ds{\int_{0}^{1}\cos\pars{n\pi x}\,\dd x} & \ds{=} & \ds{\delta_{n0}} \end{array}\right\}\,,\qquad m, n \in \mathbb{N}_{\,\geq\ 0} \label{2}\tag{2} \end{align}

• $$\ds{\on{u}\pars{t,x}}$$ must satisfies the above differential equation: \begin{align} &\sum_{n = 1}^{\infty} \ddot{\on{a}}_{n}\pars{t}\cos\pars{n\pi x} + \ddot{\on{f}}\pars{t} \\[2mm] = &\ \sum_{n = 1}^{\infty}\on{a}_{n}\pars{t} \pars{-n^{2}\pi^{2}}\cos\pars{n\pi x} \label{3}\tag{3} \end{align}
• Integrate -see (\ref{2})- both members of (\ref{3}) over $$\ds{x \in \pars{0,1}}$$: $$\implies\ddot{\on{f}}\pars{t} = 0\implies \on{f}\pars{t} = bt + c\,,\quad b, c = \mbox{constants}$$
• With (\ref{2}) and (\ref{3}): \begin{align} &\ddot{\on{a}}_{n}\pars{t} = -n^{2}\pi^{2}\on{a}\pars{t} \\[2mm] & \implies \on{a}_{n}\pars{t} = \on{a}_{n}\pars{0}\cos\pars{n\pi t} + \dot{\on{a}}_{n}\pars{0}\,{\sin\pars{n\pi t} \over n\pi} \end{align}
• $$\ds{\on{u}\pars{t,x}}$$ is reduced to \begin{align} \on{u}\pars{t,x} & = \sum_{n = 1}^{\infty}\bracks{\on{a}_{n}\pars{0}\cos\pars{n\pi t} + \dot{\on{a}}_{n}\pars{0}\,{\sin\pars{n\pi t} \over n\pi}} \\[2mm] & \cos\pars{n\pi x} + bt + c \\[5mm] \on{u}_{t}\pars{t,x} & = \sum_{n = 1}^{\infty}\bracks{% -n\pi\on{a}_{n}\pars{0}\sin\pars{n\pi t} + \dot{\on{a}}_{n}\pars{0}\cos\pars{n\pi t}} \\[2mm] & \cos\pars{n\pi x} + b \end{align}

\begin{align} x\pars{1 - x} = \on{u}\pars{0,x} & = \sum_{n = 1}^{\infty}\on{a}_{n}\pars{0} \cos\pars{n\pi x} + c\label{4}\tag{4} \\[5mm] 0 = \on{u}_{t}\pars{0,x} & = \sum_{n = 1}^{\infty}\dot{\on{a}}_{n}\pars{0}\cos\pars{n\pi x} + b \end{align}

• With the last equation and (\ref{2}), I found $$\ds{\dot{\on{a}}_{n}\pars{0} = 0}$$ and $$\ds{b = 0}$$.
• Integrate both members of (\ref{4}) over $$\ds{x \in \pars{0,1}}$$: $$\int_{0}^{1}x\pars{1 - x}\dd x = \int_{0}^{1}c\,\dd x \implies c = {1 \over 6}$$
• With (\ref{2}) and (\ref{4}): $$\underbrace{2\int_{0}^{1}x\pars{1 - x}\cos\pars{n\pi x} \,\dd x} _{\ds{-\,\bracks{1 + \pars{-1}^{n}}{2 \over n^{2}\pi^{2}}}} = \on{a}_{n}\pars{0}$$
• Finally, $$\bbx{\on{u}\pars{t,x} = {1 \over 6} - {1 \over \pi^{2}} \sum_{n = 1}^{\infty} {\cos\pars{2n\pi t}\cos\pars{2n\pi x} \over n^{2}}} \\$$