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Solving second order inhomogenous PDE by separation of variables requires homogenization of the boundary conditions. Let's say we are looking at 1D heat equation. From intuition, if we have fixed temperature on both sides(inhomogeneous Dirichlet-Dirichlet boundary conditions), there is no heat coming in or out of the 1D bar, meaning as time goes to infinity, the bar will reach an equilibrium state where the temperature would no longer depend on time meaning. So when times go to infinity the solution would be a function u(x) (so-called homogenization function), meaning the heat equation is: $$d^2u/dx^2=0$$ with the Dirichlet boundary conditions. The solution to this is $$u=c1*x+c2$$ and by applying the the conditions we can find c1 and c2. Same intuition works to generate the homogenization function if we have boundary condition Dirichlet-Neumann/Neumann-Dirichlet conditions ($u(0,t)=f(t), du(l,t)/dx=g(t) / du(o,t)/dx=f(t), u(l,t)=g(t)$). Unfortunately, when we have Neumann-Neumann conditions ($du(0,t)/dx=f(t),du(l,t)/dx=g(t)$), this intuition is not applicable and I cannot get the homogenization function in it's proper form which is $$u(x,t)= (f(t)+g(t))*x^2/l - f(t)x$$. My guess is that when there is prescribed flux on both sides there cannot be equilibrium temperature as time goes to infinity. I can see that if $d^2u/dx^2=constant$ I will reach the answer but what is the physical intuition to this statement?

P.S. Sorry if some of the terminologies are not correct in English. Also sorry for the unedited equations, I do not know how to write them out.

Edit: Basically I am trying to solve the following problem: Having the heat equation $$\partial u(x,t)/\partial t=-\epsilon \partial^2u(x,t)/\partial x^2$$ $$ \partial u(0,t)/\partial x=f(t)$$ $$ \partial u(l,t)/\partial x=g(t)$$ $$ u(x,0)=v(x) $$ So to apply separation of variables we need to make the BC homogenous. So we make the assumption that the solution is in the form: $$u(x,t)=u_0(x,t)+\tilde u(x,t) $$ which basically converts the equation to: $$\partial u_0(x,t)/\partial t-\epsilon\partial^2u_0(x,t)/\partial x^2=-\partial \tilde u(x,t)/\partial t+\epsilon\partial^2\tilde u(x,t)/\partial x^2=\tilde s(x,t)$$
and the BC transform to: $$\partial u_0(0,t)/\partial x=0$$ $$ \partial u_0(l,t)/\partial x=0$$ $$ u_0(x,0)=v(x)-\tilde u(x,0)$$ which is a homogenous problem with a source and modified initial conditions. The only thing we need is to find the explicit form of $\tilde u$ which I know is in the form of $\tilde u(x,t)= (f(t)+g(t))*x^2/l - f(t)x$ but I do not know how to get to it.

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  • $\begingroup$ This answers only the D-D boundary conditions. $\endgroup$ Jun 18 '20 at 8:22
  • $\begingroup$ Sorry. Must learn to read one fine day! ;-) $\endgroup$
    – Gert
    Jun 18 '20 at 10:31
  • $\begingroup$ "homogenization function" Did you mean solution to the homogeneous PDE? $\endgroup$
    – Gert
    Jun 18 '20 at 10:32
  • $\begingroup$ And where does $u(x,t)= \frac{(f(t)+g(t))x^2}{l} - f(t)x$ come from? $\endgroup$
    – Gert
    Jun 18 '20 at 10:36
  • $\begingroup$ Also sorry for the unedited equations, I do not know how to write them out. Just right-click on any formula to have the Latex revealed. Cut and paste! $\endgroup$
    – Gert
    Jun 18 '20 at 10:55
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There is no steady state condition for Neumann-Neumann BCs. So you must solve your problem by going through a source function like you are showing with $\bar{s}(x,t)$. The "easiest" source function would be the one you use. Although you can use other functions that can make your equation solving easier. The book author uses something close to a linear interpolation between $f(t)$ and $g(t)$ for $\bar{u}'(x,t)$: $$\bar{u}'(x,t)=f(t)+(g(t)-f(t))\frac{x}{l}$$ when integrated with respect to $x$ from $0$: $$\bar{u}(x,t)=f(t)x+(g(t)-f(t))\frac{x^2}{2l}$$ But I don't know why there are differences. Your formula doesn't give $\bar{u}'(0,t)=f(t)$ and $\bar{u}'(l,t)=g(t)$.

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I can see that if $\frac{\text{d}^2 u}{\text{d}x^2}=0$ I will reach the answer but what is the physical intuition to this statement?

Of course this is:

$$u_t=\alpha u_{xx}$$

but for $u_t=0$, i.e. steady state conditions. That's the 'physical intuition', no more, no less.

Twice integrating gives:

$$u(x)=c_1+c_2 x$$

The boundary conditions are:

$$u(0,t)=f(t), \frac{\partial u(l,t)}{\partial x}=g(t), \frac{\partial u(0,t)}{\partial x}=f(t), \frac{\partial u(l,t)}{\partial x}=g(t)$$

Assume:

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

My guess is that when there is prescribed flux on both sides there cannot be equilibrium temperature as time goes to infinity.

I think you've guessed correctly, because:

$$\frac{\partial u(l,t)}{\partial x}=\Theta(t)X(l)'=g(t)$$ and:

$$\frac{\partial u(0,t)}{\partial x}=\Theta(t)X(0)'=f(t)$$

Assuming that $\Theta(t) \neq 0$, these BCs vary in time and so the flux flowing from $x=0$ to $x=l$ will always change in time. So there can no steady state. Not even at $t=+\infty$.

But I spot another problem, (which may be due to a typo)

$$\frac{\partial u(0,t)}{\partial x}=u(0,t)=f(t)$$ means: $$\Theta(t)X(0)'=\Theta(t)X(0)=f(t)$$ Which can only be true if:

$$X(0)'=X(0)$$

(for example $X(x)=e^x$, or $X(x)=0$)

The same is true for the $x=l$ BCs.

Did you mean to deploy four functions and not two, $f(t)$ and $g(t)$? You really 'only' need two BCs and an initial condition, for a second order $(x,t)$ PDE.


I must say you've chosen a particularly hard problem to crack.
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