The ODE:



$$y(0,t)=0\text{ and } y_x(L,t)=\alpha [y(L,t)+\beta]$$

So the latter is an inhomogeneous Neumann BC.

The domain:

$$0\leq x \leq L\text{ and }t\geq 0$$

An IC is also needed but not relevant to my question, right now.

I'm familiar with the method of *homogenisation* where a separate function is added to the target function so that the PDE and/or its BCs becomes homogeneous. That works very well in simple cases.

In accordance with that, for my first attempt, I assumed that:


where $y_E(x)$ is the steady-state equation (so for $y_t=0$):

$$y_t=0\Rightarrow y_E''=0$$

$$\Rightarrow y_E(x)=c_1x+c_2$$

With $y(0,t)=0$:

$$\Rightarrow c_2=0$$

$$y_E'=c_1=\alpha [c_1L+\beta]$$ $$c_1=\alpha c_1+\alpha \beta$$ $$c_1=\frac{\alpha \beta}{1-\alpha L}$$ Recapping: $$y_t(x,t)=z_t(x,t)$$ And: $$y_{xx}(x,t)=z_{xx}(x,t)$$ And: $$y_x(L,t)=\alpha [y(L,t)+\beta]$$ $$c_1 +z_x(L,t)=\alpha [c_1L+z(L,t)+\beta]$$ So homogenisation hasn't been achieved.

Any serious pointers would be much appreciated.

  • $\begingroup$ Welcome also to you from PSE to Math.SE. My sincere welcome. $\endgroup$
    – Sebastiano
    Aug 15, 2020 at 22:01
  • $\begingroup$ Errm... I've contributed to math.SE for a while now. But thanks anyway. $\endgroup$
    – Gert
    Aug 15, 2020 at 22:28
  • $\begingroup$ It's not important to me...:-) thank you always for your help. $\endgroup$
    – Sebastiano
    Aug 15, 2020 at 22:32
  • $\begingroup$ We seemed to have dropped an $L$ when solving for $c_1$ (specifically in the step after distributiong $\alpha$). It ought to be $c_1=\frac{\alpha \beta}{1-\alpha L}$, no? $\endgroup$ Aug 15, 2020 at 23:12
  • $\begingroup$ Yes, I've edited it now, Ta. $\endgroup$
    – Gert
    Aug 15, 2020 at 23:20

2 Answers 2


The steady state solution for the original problem is $y_E(x)=\frac{\alpha \beta}{1-\alpha L} x$. The transient solution is given by $z(x,t)=y(x,t)-y_E(x)$ which now solves the PDE, for $(x,t)\in (0, L)\times (0,\infty)$, $$z_t=k z_{xx},$$ and with BCs $$z(0,t)=0 \,\text{ and } \, z_x(L,t)-\alpha z(L,t)=0$$ for $t>0$ and IC $z(x,0)=g(x)-y_E(x)$ (where $g$ is the IC of the original problem, unspecified in the OP). Thus the $\beta$ term has vanished as indicated in my comment on the OP. The transient solution can then be found by separation of variables.

For then we get two ODEs one in $x$, $\phi''+\lambda^2 \phi = 0$ for $0<x<L$ with BC $\phi(0)=0$ and $\phi'(L)-\alpha \phi(L)=0$ and one in $t$, $T'+\lambda^2 k T=0$ with the IC. Solving the first ODE and imposing the first BC gives $\phi = c_2 \sin(\lambda x),$ and imposing the second BC and avoiding trivial solutions requires $\lambda$ to solve $\tan(\lambda L)=\lambda/\alpha$ which has infinite solutions $\lambda_n$ for $n\geq 1$. All together, we get $$z(x,t)=\sum_{n} b_n \sin(\lambda_n x)e^{-\lambda_n^2 k t},$$ with initial condition $$z(x,0)=\sum_n b_n \sin(\lambda_n x)=g(x)-y_E(x),$$ which leads to $$b_n=\frac{\int_0^L [g(x)-y_E(x)]\sin(\lambda_n x) dx}{\int_0^L \sin^2(\lambda_n x) dx},$$ so that finally, returning back to $y=z+y_E$, we have $$y(x,t)=\frac{\alpha \beta}{1-\alpha L} x + \sum_n b_n \sin(\lambda_n x) e^{-\lambda_n^2 k t},$$ where $\lambda_n$ and $b_n$ are defined above.

Please comment for clarifications or corrections.

  • 1
    $\begingroup$ Thank you! If I had understood from your first comment how $\beta$ disappears I would have been much closer to a solution. It's heart-warming to see that my approach with $y_E(x)$ is valid but a bit galling that I 'missed a trick' there! Live and learn. Thanks again! $\endgroup$
    – Gert
    Aug 17, 2020 at 18:05

The problem has a solution if we simplify the second BC slightly, so that $\beta=0$: $$y(0,t)=0\text{ and } y_x(L,t)=\alpha y(L,t)$$

Carry out separation of variables, this will yield, with $-m^2$ the separation constant:

$$y_n(x,t)=A_n\exp(-\alpha m^2 t)\sin(mx)$$ Insert into:

$$y_x(L,t)=\alpha y(L,t)$$ $$-mA_n\cos(mL) =A_n\alpha\sin(mL)$$ $$\Rightarrow \tan(mL)=-\frac{m}{\alpha }$$ $$\mu=mL \Rightarrow$$ $$\tan \mu=-\frac{\mu}{\alpha L}$$

This is a transcendental equation which can be solved numerically for $\mu$.

But it is not a solution to the original problem.

Then I thought to carry out a substitution: $$y(x,t)=u(x,t)-\beta$$ $$\Rightarrow u_x(L,t)=\alpha u(L,t)$$

But also:


Snookered again!

Finally, changing the nature of the first BC to:


Then with $y(x,t)=u(x,t)-\beta$:


This would yield:

$$\tan \mu=-\frac{\alpha L}{\mu}$$

But this too changes the nature of the original problem.

The key seems to be to eliminate $\beta$ while keeping the other BC homogeneous. But how?


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