PDE using separation of variables

Question

Use the method of separation of variables to solve the equation $\frac{∂w}{∂t} −\frac{∂^2w}{∂x^2}= 0, 0 ≤ x ≤ π,$ subject to the conditions $w(x, 0) = 0, w_x(0, t) = −1, w_x(π, t) = −1.$

Would I be right in saying that the BCs are inhomogeneous and I would have to transform the original equation? I tried this but am struggling to get an answer in the form of a series.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{% \begin{array}{l} \ds{\partiald{\mrm{w}\pars{x,t}}{t} - \partiald[2]{\mrm{w}\pars{x,t}}{x} = 0} \\[2mm] \ds{x \in \pars{0,\pi}} \end{array}\,,\qquad\qquad \left\{\begin{array}{rclcl} \ds{\mrm{w}\pars{x,0}} & \ds{=} & \ds{0}&& \\[3mm] \ds{\left.\partiald{\mrm{w}\pars{x,t}}{x}\right\vert_{\ x\ =\ 0}} & \ds{=} & \ds{\left.\partiald{\mrm{w}\pars{x,t}}{x}\right\vert_{\ x\ =\ \pi}} & = &\ds{-1} \end{array}\right.}$

\begin{align} &\partiald{\mrm{w}\pars{x,t}}{x} = -1 + \sum_{n = 1}^{\infty}\mrm{A}_{n}\pars{t}\sin\pars{nx} \\ &\mrm{w}\pars{x,t} = \mrm{f}\pars{t} - x - \sum_{n = 1}^{\infty}{\mrm{A}_{n}\pars{t} \over n}\,\cos\pars{nx} \\ &0 = \dot{\mrm{f}}\pars{t} - \sum_{n = 1}^{\infty}{1 \over n}\bracks{\dot{\mrm{A}}_{n} + n^{2}\,\mrm{A}_{n}\pars{t}}\,\cos\pars{nx} \\ &\mrm{f}\pars{t} = a = \mbox{constant}\,,\qquad \mrm{A}_{n}\pars{t} = A_{n}\pars{0}\expo{-n^{2}t} \end{align}

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$$ \bbx{\ds{\mrm{w}\pars{x,t} = a - x - \sum_{n = 1}^{\infty}{\mrm{A}_{n}\pars{0} \over n}\,\cos\pars{nx} \expo{-n^{2}t}}} $$

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$$ 0 = \mrm{w}\pars{x,0} = a - x - \sum_{n = 1}^{\infty}{\mrm{A}_{n}\pars{0} \over n}\,\cos\pars{nx} $$

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$$ 0 = \int_{0}^{\pi}\bracks{% a - x - \sum_{n = 1}^{\infty}{\mrm{A}_{n}\pars{0} \over n}\,\cos\pars{nx}}\dd x \implies a = {\pi \over 2} $$

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$$ 0 = \int_{0}^{\pi}\bracks{% {\pi \over 2} - x - \sum_{m = 1}^{\infty} {\mrm{A}_{m}\pars{0} \over m}\,\cos\pars{mx}}\cos\pars{nx}\,\dd x $$

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$$ 0 = {1 - \pars{-1}^{n} \over n^{2}} - {\pi \over 2}\, {\mrm{A}_{n}\pars{0} \over n} \implies {\mrm{A}_{n}\pars{0} \over n} = {1 - \pars{-1}^{n} \over n^{2}}\,{2 \over \pi} $$

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$$\bbox[#ffe,15px,border:1px dotted navy]{\ds{% \mrm{w}\pars{x,t} = {\pi \over 2} - x - {4 \over \pi} \sum_{n = 0}^{\infty}{\cos\pars{\bracks{2n + 1}x} \over \pars{2n + 1}^{2}}\, \expo{-\pars{2n + 1}^{2}t}}} $$

Answer 2

Yes, you would be right. Look for a (simple) function $\phi(x)$ satisfying the boundary conditions and define a new unknown $u$ as $u(x,t)=w(x,t)-\phi(x)$.