# Heat PDE with unusual, inhomogeneous Neumann boundary condition

The ODE:

$$y_t=ky_{xx}$$

BCs:

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

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:

$$y(x,t)=y_E(x)+z(x,t)$$

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.

• Welcome also to you from PSE to Math.SE. My sincere welcome. Aug 15, 2020 at 22:01
• Errm... I've contributed to math.SE for a while now. But thanks anyway.
– Gert
Aug 15, 2020 at 22:28
• It's not important to me...:-) thank you always for your help. Aug 15, 2020 at 22:32
• 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? Aug 15, 2020 at 23:12
• Yes, I've edited it now, Ta.
– Gert
Aug 15, 2020 at 23:20

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 $$0 with BC $$\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.

• 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!
– 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:

$$u(0,t)=\beta$$

Snookered again!

Finally, changing the nature of the first BC to:

$$y_x(x,t)=0$$

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

$$y_x(x,t)=u_x(x,t)=0$$

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?