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I was trying to solve a 1-dimensional heat equation in a confined region, with time-dependent Dirichlet boundary conditions. After some Googling, I found this wiki page that seems to have a somewhat complete method for solving the 1d heat eq.

But I think there is a potential problem lying in this method: assume that there are no heat sources in the solution domain, so that $h(x,t)=0$, and also the boundary conditions are of the first type (Dirichlet), so the $\beta$ coefficients are zero too.

In this case, the ($X_n$)s in Step 3.1 are of the form ($\sin(\frac{n \pi x}{L})$). Then in Step 3.2, when we try to write $q(x,t)$, which involves the linear function $s_t(x,t)$ obtained in Step 2, as a series in terms of these $X_n$ functions, we'll always arrive at a zero value for $q(x=L,t)$, regardless of the time-dependent coefficients of the series.

I guess that this is somehow related to the Gibbs phenomenon, and the linear function is implicitly extended beyond $[0,L]$, for e.g. to $[-L,L]$, and that's causing the problem.

Are there any ways to deal with this? Are there any better methods for solving the time-dependent heat equation?

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  • $\begingroup$ It would help if you would just state the problem, including the boundary conditions. I'd love to have bash at this but feel I'd be working in a 'vacuum'. Thanks. $\endgroup$ – Gert Sep 24 '16 at 0:57
  • $\begingroup$ E.g. would $u_t=ku_{xx}$, $u(x,0)=f(x)$, $u(0,t)=b_1(t)$ and $u(L,t)=b_2(t)$ describe your problem? $\endgroup$ – Gert Sep 24 '16 at 1:06
  • $\begingroup$ @Gert yes, that's what I have in my mind. $\endgroup$ – dedekindCuttage Sep 24 '16 at 13:03
  • $\begingroup$ @SaeedD - Ok, I'm going to give this a shot, a bit later on. $\endgroup$ – Gert Sep 24 '16 at 14:41
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I'm not finding solving this enormously difficult but it is both tedious and very lengthy. For that reason I'll take the unusual approach to publish the solution in 3 parts:

Part 1: up to solving the homogeneous PDE.

Part 2: solving the non-homogeneous PDE.

Part 3: testing the solution with a simple example.

Problem statement: $$u_t=ku_{xx}$$ With boundary and in initial conditions: $$u(x,0)=f(x)$$ $$u(0,t)=b_1(t),u(L,t)=b_2(t)$$ So we're looking to solve the heat equation in one dimension, without heat sinks or sources but with time-dependent boundary conditions.

Part 1:

We assume the solution to be of the partitioned form: $$u(x,t)=s(x,t)+v(x,t)$$ So that: $$s_t+v_t=ks_{xx}+kv_{xx}...(i)$$ Translating the boundary/initial conditions: $$s(x,0)+v(x,0)=f(x)$$ $$s(0,t)+v(0,t)=b_1(t)$$ $$s(L,t)+v(L,t)=b_2(t)$$ We assume $s(x,t)$ to be linear in $x$ but time-dependent, so the general form of $s(x,t)$ is: $$s(x,t)=a(t)x+b(t)$$ So: $$s_{xx}=0$$ So with $s(0,t)$ and $s(L,t)$ we get: $$b(t)=b_1(t)$$ $$a(t)L+b(t)=b_2(t)$$ $$a(t)=\frac{b_2(t)-b_1(t)}{L}$$ So: $$s(x,t)=\frac{b_2(t)-b_1(t)}{L}x+b_1(t)$$ With $(i)$ and $s_{xx}=0$, we get: $$v_t=kv_{xx}-s_t$$ We solve the homogeneous equation first: $$v_t=kv_{xx}$$ With boundary conditions: $$v(0,t)=0,v(L,t)=0...(ii)$$ Using the Ansatz: $$v(x,t)=X(x)T(t)$$ Obtaining separation of variables: $$\frac1k \frac{T'}{T}=\frac{X''}{X}=-m^2$$ ODE for $X(x)$: $$X''+m^2X=0$$ $$X=c_1\cos mx+c_2\sin mx$$ With boundary conditions $(ii)$: $$c_1=0$$ $$m=\frac{n\pi}{L}, n=1,2,3....$$ $$X_n(x)=\sin\Big(\frac{n\pi x}{L}\Big)$$

Part 2:

Now for the non-homogeneous PDE: $$v_t=kv_{xx}-s_t$$ With boundary/initial conditions: $$v(0,t)=0,v(L,t)=0$$ $$v(x,0)=f(x)-s(x,0)$$ Assume the solution to be of the form: $$v(x,t)=\displaystyle \sum_{n=1}^{+\infty}T_n(t)X_n(x)$$ $$s_t=\frac{b'_2(t)-b'_1(t)}{L}x+b'_1(t)=-\displaystyle \sum_{n=1}^{+\infty}Q_n(t)X_n(x)$$ $$Q_n(t)=-\frac{\int_0^Ls_tX_n(x)dx}{\int_0^LX_n(x)X_n(x)dx}$$ Our solution must of course obey the original equation, so: $$\frac{\partial}{\partial t}\displaystyle \sum_{n=1}^{+\infty}T_n(t)X_n(x)=k\frac{\partial^2}{\partial t^2}\Bigg[\displaystyle \sum_{n=1}^{+\infty}T_n(t)X_n(x)\Bigg]+\displaystyle \sum_{n=1}^{+\infty}Q_n(t)X_n(x)$$ We know that: $$X''_n(x)=-m^2X_n(x)$$ So: $$\displaystyle \sum_{n=1}^{+\infty}T'_n(t)X_n(x)=\displaystyle \sum_{n=1}^{+\infty}-km^2T_n(t)X_n(x)+\displaystyle \sum_{n=1}^{+\infty}Q_n(t)X_n(x)$$ $$\displaystyle \sum_{n=1}^{+\infty}[T'_n(t)+km^2T_n(t)]X_n(x)]=\displaystyle \sum_{n=1}^{+\infty}Q_n(t)X_n(x)$$ So that: $$T'_n(t)+km^2T_n(t)=Q_n(t)$$ Solve with an integration factor to: $$T_n(t)=e^{-km^2t}\int_0^te^{km^2t}Q_n(t)dt+C_ne^{-km^2t}$$ Using the initial condition: $$v(x,0)=f(x)-s(x,0)=\displaystyle \sum_{n=1}^{+\infty}T_n(0)X_n(x)=\displaystyle \sum_{n=1}^{+\infty}C_nX_n(x)$$ So: $$C_n=\frac{\int_0^L[f(x)-s(x,0)]X_n(x)dx}{\int_0^LX_n(x)X_n(x)dx}$$

Part 3:

In Part 3 I'll test drive the solution for the simple case:

$$u(x,0)=u_0$$ $$u(0,t)=u_0-at$$ $$u(L,t)=u_0+bt$$

Bits and pieces we need: $$s(x,t)=\frac{b_2(t)-b_1(t)}{L}x+b_1(t)=u_0+\frac{a+b}{L}xt-at$$ $$s_t=\frac{b'_2(t)-b'_1(t)}{L}x+b'_1(t)=\frac{a+b}{L}x-a$$ $$s(x,0)=u_0$$ $$Q_n(t)=-\frac{\int_0^Ls_tX_n(x)dx}{\int_0^LX_n(x)X_n(x)dx}$$ Because the boundary conditions are only linear in $t$ the $Q_n$ turn out time independent and compute to: $$Q_n=2\Big[\frac{b}{n\pi}\cos n\pi+\frac{a}{n\pi}\Big]$$ Because: $$s(x,0)=u_0=f(x)\implies C_n=0$$ So slightly reworked we get: $$T_n(t)=\frac{2L^2}{kn^3\pi^3}\Big[(-1)^nb+a\Big]\Big[1-e^{-k\Big(\frac{n\pi}{L}\Big)^2t}\Big]$$

Put it altogether with: $$u(x,t)=s(x,t)+\displaystyle \sum_{n=1}^{+\infty}T_n(t)X_n(x)$$ Where $X_n(x)=\sin\Big(\frac{n\pi x}{L}\Big)$

Let's evaluate for $u_0=100$, $L=1$, $k=1$, $a=1$ and $b=2$.

Firstly let's look at $s(x,t)$. The significance of $s(x,t)$ is as follows. If we allow temperature evolution of the boundaries for some time $t$, then stop heating/cooling and maintain the boundary temperatures until thermal equilibrium is achieved, then $s(x,t)$ describes the temperature profile, e.g.:

s(x,t)

Here's $u(x,t)$ with the first three terms computed:

u(x,t)

And here's $v(x,t)$ with the first three terms computed:

v(x,t)

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  • $\begingroup$ Thanks. As we can see, this procedure transforms a problem with time-dependent boundary conditions and without any heat sources (the eq. for u), to one with constant boundary conditions and a spatially uniform, time-dependent source, $s_t$ (the eq. for v). But then trying to express the source in terms of X's always gives zero for the endpoints of the domain, which is not the case in general. $\endgroup$ – dedekindCuttage Sep 25 '16 at 6:30
  • $\begingroup$ I'm nearly done test driving the solution. With luck it will yield some nice $x,t$ plots. $\endgroup$ – Gert Sep 25 '16 at 11:51
  • $\begingroup$ Phew, that was a lot of computing! $\endgroup$ – Gert Sep 25 '16 at 20:56
  • $\begingroup$ Nice, but the problem that I stated before remains. The answer you've come up with in step3, solves the original equation (i.e. $u_t = k u_{xx}$) everywhere except in the endpoints of the interval, as it could be easily checked. $\endgroup$ – dedekindCuttage Sep 25 '16 at 23:00
  • $\begingroup$ everywhere except in the endpoints of the interval, as it could be easily checked. I'm not sure I get your point, that's what boundary conditions do, after all: they fix $u$ in the end points of the domain. In the case used boundary-$u$ are time-dependent. Here's a very similar case but with simple Dirichlet BCs:tutorial.math.lamar.edu/Classes/DE/HeatEqnNonZero.aspx $\endgroup$ – Gert Sep 25 '16 at 23:54

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