# Bar problem and heat conduction equation

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!

• 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 – phdmba7of12 Jul 7 at 20:42

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 $$\lambda_k$$, $$k=0,1,2,\ldots$$. 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.
• 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. – DisintegratingByParts Jul 17 at 4:21