Solving the PDE $\frac{\partial T}{\partial t} = K\frac{\partial^{2} T}{\partial x^{2}}$

Question

I am trying to find solutions to the following PDE:

$$\frac{\partial T}{\partial t} = K\frac{\partial^{2} T}{\partial x^{2}},$$ With boundary condition: $T(0,t)=T_{0}\cos(\omega t)$, where $\omega = \frac{2K}{a^{2}}$.

I am trying to solve this by separation of variables into $T(x,t) = X(x)F(t)$ so we have:

$$X(x)\frac{\partial F}{\partial t}=KF(t)\frac{\partial^{2} X}{\partial x^{2}}$$

Thus we have the following two conditions:

$$\frac{\partial F}{\partial t}=-\lambda K F(t) \land \frac{\partial ^{2} X}{\partial x^{2}}=-\lambda X(x)$$

Solving these (and taking into account that $\lambda$ can be arbitrary), we have:

$$F(t)=\sum_{\lambda} \alpha_{\lambda} \exp(-\lambda K t) \land X(x)=\sum_{\lambda}\left(A_{\lambda}\cos(\sqrt{\lambda}x)+B_{\lambda}\sin(\sqrt{\lambda}x)\right)$$

We have boundary condition:

$$T(0,t)=T_{0}\cos(\omega t)$$

However, I am having trouble now computing the coefficients $\alpha_{\lambda}$, $A_{\lambda}$ and $B_{\lambda}$.

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The question in it's full form is given by:

The temperature $T$ in a one-dimensional bar whose sides are perfectly insulated obeys the heat flow equation: $$\frac{\partial T(x,t)}{\partial t}=K\frac{\partial^{2}T(x,t)}{\partial x^{2}}$$ where $K$ is a constant. The bar extends from $x=0$ to $x=\infty$. The temperature at the end $x=0$ oscillates in time according to $T(x=0,t)=T_{0}\cos(\omega t)$. By looking for solutions that are separated in $x$ and $t$ ($T = X(x)F(t)$) find the solution for all $x \geq 0$ and $t$, which matches the boundary condition at $x=0$. Sketch $T$ versus $x$ for $\omega t = \pi/2$, given that $\frac{\omega}{2K}=\frac{1}{a^{2}}$.

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Look for solutions of the form $\ds{\,{\rm T}\pars{x,t} \equiv \bracks{% \,{\rm s}\pars{x}\sin\pars{\omega t} + \,{\rm c}\pars{x}\cos\pars{\omega t}}}$ which satisfies $\ds{\partiald{\,{\rm T}\pars{x,t}}{t} = K\,\partiald[2]{\,{\rm T}\pars{x,t}}{x}}$ with $\ds{\,{\rm s}\pars{0} = 0}$ and $\ds{\,{\rm c}\pars{0} = T_{0}}$.

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\begin{align}&\omega\,{\rm s}\pars{x}\cos\pars{\omega t} -\omega\,{\rm c}\pars{x}\sin\pars{\omega t} =K\,{\rm s}''\pars{x}\sin\pars{\omega t} + K\,{\rm c}''\pars{x}\cos\pars{\omega t} \\[5mm]& \imp\quad\,{\rm c}''\pars{x} - {\omega \over K}\,{\rm s}\pars{x} = 0\,,\qquad \,{\rm s}''\pars{x} + {\omega \over K}\,{\rm c}\pars{x} = 0 \\[5mm]&\imp\quad\bracks{\,{\rm c}''\pars{x} + \ic\,{\rm s}''\pars{x}} +\ic\,{\omega \over K}\bracks{\,{\rm c}\pars{x} + \ic\,{\rm s}\pars{x}}=0 \end{align}

Solutions are linear combinations of

\begin{align}&\dsc{\exp\pars{\pm\root{\ic\,{\verts{w} \over K}}x}} =\exp\pars{\pm\pars{1 + \ic}\root{{\verts{w} \over 2K}}x} \\[5mm]&=\exp\pars{\pm\,{x \over a}} \exp\pars{\pm\ic\,{x \over a}} \end{align}

Since the equation is the second order in $\ds{x}$, it will require another condition. Lets require, $\dsc{\mbox{for example}}$, that $\ds{\lim_{x\ \to\ \infty}\,{\rm T}\pars{x,t} = 0}$. In that case the solution becomes: $$ \,{\rm c}\pars{x} + \ic\,{\rm s}\pars{x}=T_{0}\expo{-x/a}\cos\pars{x \over a} \quad\imp\quad\left\{\begin{array}{rcl} \,{\rm c}\pars{x} & = & T_{0}\expo{-x/a}\cos\pars{x \over a} \\[2mm] \,{\rm s}\pars{x} & = & 0 \end{array}\right. $$

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$$ \color{#66f}{\large\,{\rm T}\pars{x,t}} =\color{#66f}{\large T_{0}\expo{-x/a}\cos\pars{x \over a}\cos\pars{\omega t}} $$