Applying boundary conditions

Question

I wish to solve $u_{tt} = u_{xx}, \; 0 < x < 1$ with the initial boundary conditions $u_x(0,t) = 0$, $(0 < t< \infty)$, $u_x(1,t) = 0$ and $u(x,0) = 0$, $0 \leq x \leq 1$, $u_t(x,0) = \cos(\pi x) + \cos(3\pi x)$.

So we assume $u(x,y) = X(x)T(t)$ and then get $u_x = X' T$, $u_{xx} = X''T$. Similarly, $u_t = X T'$ and $u_{tt} = X T''$.

So $\frac{X''}{X} = \frac{T''}{T} = \gamma$ for some constant $\gamma$. Now I form two ODEs $T'' - \gamma T = 0$ and $X'' - \gamma X = 0$. In solving the second ODE, the char. eq. is $\lambda^2 = \gamma$. Hence, $X(x) = Ax + B$, $(\gamma = 0)$, two real equal roots. And $X(x) = Ae^{\sqrt{\gamma}x} + Be^{-\sqrt{\gamma}x}$, $(\gamma > 0)$, real but unequal roots. And $X(x) = \cos(\sqrt{-\gamma}x) + B\sin(\sqrt{-\gamma}x)$, $(\gamma < 0)$, complex roots. Here $A,B$ are constants.

Now I want to solve for the constants $A,B$ by applying the boundary conditions. But how exactly is where I am stuck. Can anyone please assist with this?

Answer 1

Hint: Use the facts that \begin{align} u_x(0, t) =& X'(0)T(t)=0 \ \ \implies \ \ \ X'(0) =0,\\ u_x(1, t) =& X'(1)T(t)= 0 \ \ \implies \ \ \ X'(1) =0 \end{align} otherwise, your answer will be trivial. Use this to get constraints on $\gamma$.

Spoiler Hint:

$\gamma=-n^2\pi^2$ for any $n \in \mathbb{Z}$ which implies $X(x)=A\cos(\pi n x)$

Finally Hint:

Since $T(t)= A\cos(n\pi t)+B\sin(n\pi t)$ and $T(0) = 0$, then $T(t) = B\sin(n\pi t)$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ It's interesting to start from the 'wave equation' general solution

$\ds{\mrm{u}\pars{x,t} = \mrm{f}\pars{x + t} + \mrm{g}\pars{x - t}.\qquad \require{cancel} \left\{\begin{array}{rcl} \ds{\mrm{u}\pars{x,0}} & \ds{=} & \ds{0} \\[1mm] \ds{\mrm{u}_{t}\pars{x,0}} &\ds{=} & \ds{\cos\pars{\pi x} + \cos\pars{3\pi x}} \end{array}\right.}$

$$ \bbx{\mbox{The boundary condition}\ \mrm{u}_{x}\pars{0,t} =\mrm{u}_{x}\pars{1,t} = 0\ \mbox{is}\ \color{red}{\textsf{redundant}}.} $$

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$$ \left\{\begin{array}{rcccl} \ds{0} & \ds{=} & \ds{\mrm{u}\pars{x,0}} & \ds{=} & \ds{\mrm{f}\pars{x} + \mrm{g}\pars{x}} \\[2mm] \ds{\cos\pars{\pi x} + \cos\pars{3\pi x}} & \ds{=} & \ds{\mrm{u}_{t}\pars{x,0}} & \ds{=} & \ds{\mrm{f}'\pars{x} - \mrm{g}'\pars{x}} \end{array}\right. $$

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$$ \left\{\begin{array}{rcrcl} \ds{\mrm{f}\pars{x}} & \ds{+} & \ds{\mrm{g}\pars{x}} & \ds{=} & \ds{0} \\[2mm] \ds{\mrm{f}\pars{x}} & \ds{-} & \ds{\mrm{g}\pars{x}} & \ds{=} & \ds{{\sin\pars{\pi x} \over \pi} + {\sin\pars{3\pi x} \over 3\pi} + a} \end{array}\right. $$

where $\ds{a}$ is a constant.

Then, $$ \left\{\begin{array}{rcl} \ds{\mrm{f}\pars{x}} & \ds{=} & \ds{{1 \over 6\pi} \bracks{3\sin\pars{\pi x} + \sin\pars{3\pi x} + 3\pi a}} \\[2mm] \ds{\mrm{g}\pars{x}} & \ds{=} & \ds{{1 \over 6\pi} \bracks{-3\sin\pars{\pi x} - \sin\pars{3\pi x} - 3\pi a}} \end{array}\right. $$

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\begin{align} \mrm{u}\pars{x,t} & = {3\sin\pars{\pi\bracks{x + t}} + \sin\pars{3\pi\bracks{x + t}} - 3\sin\pars{\pi\bracks{x - t}} - \sin\pars{3\pi\bracks{x - t}}\over 6\pi} \\[5mm] & = \bbx{{3\cos\pars{\pi x}\sin\pars{\pi t} + \cos\pars{3\pi x}\sin\pars{3\pi t} \over 3\pi}} \end{align}