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The Fourier PDE for a uniform $1D$ rod with convection loss all along the length $L$ is given by: $$u_t=\alpha u_{xx}-\beta u$$

Where both $\alpha$ and $\beta$ are Real and positive numbers.

Domain: $[0,L]$

Boundary conditions (BCs): $$u(0,t)=0\text{ and }u_x(L,t)=0$$ Initial condition: $$u(x,0)=T_0$$

  1. Steady State solution ($u_t=0$):

$$u''(x)-\frac{\beta}{\alpha}u(x)=0$$

If:

$$\lambda^2=\frac{\beta}{\alpha}$$ Then:

$$u(x)=c_1e^{\lambda x}+c_2e^{-\lambda x}$$

The integration constants $c_1$ and $c_2$ are obtained by applying the BCs. The obtained solution is in agreement with published solutions for cooling fins (for example).

  1. Separation of values:

Ansatz:

$$u(x,t)=X(x)\Theta(t)$$

Separation:

$$X\Theta'=\alpha \Theta X''+\beta X\Theta$$ $$\frac{\Theta'}{\Theta}=\alpha \frac{X''}{X} +\beta$$ $$\frac{\Theta'}{\Theta}-\beta=\alpha \frac{X''}{X}$$ $$\frac{\Theta'}{\alpha \Theta}-\frac{\beta}{\alpha}= \frac{X''}{X}=-m^2\tag{1}$$ $$\frac{\Theta'}{\alpha \Theta}-\frac{\beta}{\alpha}=-m^2$$ $$\frac{\Theta'}{ \Theta}-\beta=-m^2\alpha$$ $$\frac{\Theta'}{ \Theta}=-m^2\alpha+\beta$$ $$\Theta=\exp[(-m^2\alpha+\beta)t]$$ As the rod's overall temperature decreases with time ($T_0>0$):

$$-m^2\alpha+\beta<0 \Rightarrow -m^2<0$$ From $(1)$: $$X''(x)+m^2X(x)=0$$ With the BCs: $$X_n(x)=A_n\sin mx$$ $$m=\frac{n\pi}{2L}\text{ with }n=1,3,5,...$$ So that:

$$u_n(x,t)=A_n\exp[(-m^2\alpha+\beta)t]\sin mx$$ So the steady state solution is:

$$u_n(x,+\infty)=A_n\sin mx$$

which is very different from the solution found under $1.$

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2 Answers 2

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With $\ds{k_{n} \equiv {\pars{2n + 1}\pi \over 2L}}$, lets $\ds{\mrm{u}\pars{x,t} = \sum_{n = 0}^{\infty}a_{n}\pars{t}\sin\pars{k_{n}x}}$ which already satisfies the boundary condition $\ds{\mrm{u}\pars{0,t} = \mrm{u}_{x}\pars{L,t} = 0}$. However, $\ds{\mrm{u}\pars{x,t}}$ must satisfies the above differential equation: \begin{align} &\sum_{n = }^{\infty}\dot{a}_{n}\pars{t}\sin\pars{k_{n}x} = \alpha\sum_{n = 0}^{\infty}a_{n}\pars{t}\pars{-k_{n}^{2}}\sin\pars{k_{n}x} - \beta\sum_{n = 0}^{\infty}a_{n}\pars{t}\sin\pars{k_{n}x} \end{align} Multiply both members by $\ds{\pars{2/L}\sin\pars{k_{n}x}}$ and integrate over $\ds{x \in \pars{0,L}}$ which leads to \begin{align} &\dot{a}_{n}\pars{t} + \pars{\alpha k_{n}^{2} + \beta}a_{n}\pars{t} = 0 \implies a_{n}\pars{t} = a_{n}\pars{0}\expo{-\pars{\alpha k_{n}^{2} + \beta}t} \end{align} The solution is reduced to \begin{align} \mrm{u}\pars{x,t} & = \expo{-\beta t}\sum_{n = 0}a_{n}\pars{0}\exp\pars{-\alpha k_{n}^{2}t} \sin\pars{k_{n}x} \end{align} However, \begin{align} & T_{0} = \mrm{u}\pars{x,0} = \sum_{n = 0}a_{n}\pars{0}\sin\pars{k_{n}x} \end{align} Multiply both members by $\ds{\pars{2/L}\sin\pars{k_{n}x}}$ and integrate over $\ds{x \in \pars{0,L}}$: \begin{align} & {1 \over 2n + 1}\,{4 \over \pi}\,T_{0} = a_{n}\pars{0} \end{align} \begin{align} \mrm{u}\pars{x,t} & = \bbx{{4T_{0} \over \pi}\expo{-\beta t} \sum_{n = 0}{\exp\pars{-\alpha k_{n}^{2}t}\sin\pars{k_{n}x} \over 2n + 1}} \,,\qquad k_{n} \equiv {\pars{2n + 1}\pi \over 2L} \\ & \end{align}

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  • $\begingroup$ Slightly different from the usual SoV, right? $\endgroup$
    – Gert
    Aug 29, 2020 at 23:41
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Consider your initial steady state solution $u(x)=c_1 e^{\lambda x} + c_2 e^{-\lambda x}$.

Substitute $\lambda=im$ to find: $$u(x)=c_1 e^{\lambda x} + c_2 e^{-\lambda x}=c_1 e^{imx} + c_2 e^{-im x} \\ = c_1(\cos(mx)+i\sin(mx)) + c_2(\cos(-mx) +i\sin(-mx))\\ = A_m \sin(mx) + B_m\cos(mx)$$

Apply the first boundary condition to find $u(x)=A_m \sin(mx)$, which agrees with the steady state solution you found in 2.

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  • $\begingroup$ Thanks very much. It had to be something simple! $\endgroup$
    – Gert
    Aug 29, 2020 at 18:19

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