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A thin bar, defined through $ x\in [0,l] $ has a temparature distribution $ \theta (x,t) $

and has at the point $ x=0 $ a temperature of $0$ . At the other end there is a heat emission to another medium of temperature $0$. Here holds $$ \frac{ \partial \theta }{ \partial x} (l,t)+ \sigma \theta (l,t)= 0 $$ for all $ t \geq 0 $ At the timepoint $ t_o =0 $ the bar has a temperature destribution $ x \mapsto f(x) $

And it holds the heat conduction equation $$ \frac{ \partial \theta }{ \partial t} = a^2 \frac{ \partial^2 \theta }{ \partial x^2} $$ where $ \sigma \in \mathbb{ R}^+, a \in \mathbb{R} \backslash \{0\} $ are constants.

Firstly..how can I show with the seperation method , so plugging in $ \theta (x,t)= u(x)v(t) $ I am confused, at which equation do I have to use the separation method?

that it will become the Sturm-Liouville Eigenvalueproblem $$ u''+ \lambda u = 0 , u(0)=0$$ $$ \sigma u(l)+ u'(l)=0 $$ where $ \lambda \in \mathbb{R} $ is a constant.

And how can I solve the problem and determine the Eigenvalues? many thanks in advance!

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  • $\begingroup$ each eigenmode will decay exponentially with its own time constant. take the initial spatial function and express it as a linear combination of the eigenmodes (the eigenvalues pertain to the time constant). then time evolve the sum with each term relaxing at its characteristic rate $\endgroup$ – phdmba7of12 Jul 7 at 20:42
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Let us consider a solution of type $\theta(x,t)=u(x)v(t)$, then substituting into the equation we have $$ u(x)v'(t)=a^2u''(x)v(t)\quad\implies\quad a^2\frac{u''(x)}{u(x)}=\frac{v'(t)}{v(t)}=-\lambda^2, $$ so that, for $\lambda>0,$ \begin{align} u(x) &= c_1\cos\left(\frac{\lambda}{a}x\right)+c_2\sin\left(\frac{\lambda}{a}x\right),\\ v(t) &= d_1e^{-\lambda^2t}, \end{align} while for $\lambda=0$ we have \begin{align} u(x) &= c_1+c_2x,\\ v(t) &= d_1, \end{align}

The condition $\theta(0,t)=0$ gives $u(0)=0$ so that $c_1=0$, while the condition $\partial\theta/\partial x+\sigma\theta|_{x=l}=0$ gives $$ \frac{\lambda}{a}\cos\left(\frac{\lambda}{a}l\right)+\sigma\sin\left(\frac{\lambda}{a}l\right)=0 $$ if we set $x=\lambda l/a$, this can be recast as $\tan x=-\mu x$, where $\mu=l/\sigma$, whose solution cannot be obtained exactly (see graph), but provides an infinite number of solutions $\lambda_k$, $k=0,1,2,\ldots$.

enter image description here

So the general solution is $$ \theta(x,t)=c_0x+\sum_{k=1}^\infty c_k\sin\left(\frac{\lambda_k}{a}x\right)e^{-\lambda_k^2t} $$

It remains the initial condition $$ \theta(x,0)=c_0x+\sum_{k=1}^\infty c_k\sin\left(\frac{\lambda_k}{a}x\right)=f(x),\quad x\in[0,l]. $$ When the $\lambda_k$ are multiple of integers, one can use Fourier series to obtain the $c_k$, but in this case I don't know how to proceed.

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    $\begingroup$ There is a special case when $\lambda=0$, where the solutions of the ODE are $A+Bx$. That has to be handled separately. The other cases are generic. $\endgroup$ – DisintegratingByParts Jul 17 at 4:21
  • $\begingroup$ @DisintegratingByParts thank you, I made some correction $\endgroup$ – enzotib Jul 17 at 10:02

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