Fourier's Heat PDE with time dependent heat source

Question

I have a PDE for $x\in [0,L]$: $$u_t=\alpha u_{xx}+f(t)$$ BCs and IC: $$u(0,t)=u(L,t)=0\text{ and }u(x,0)=f(x)$$ My first 'instinct' is to solve the homogeneous PDE. $$\alpha u_{xx}+f(t)=0$$ $$\alpha u_E''(x)=-f(t)$$ $$u_E(x)=-\frac12 f(t)x^2+c_1 x+c_2$$ With the BCs: $$c_2=0$$ $$c_1=\frac12 f(t)L$$ Then, with the superposition principle, we look for a new function $v(x,t)$: $$u(x,t)=v(x,t)+u_E(x)$$ $$u_t=v_t$$ $$u_x=v_{xx}$$ The latter solves easily to: $$v\left( {x,t} \right) = \sum\limits_{n = 1}^\infty {{A_n}\sin \left( {\frac{{n\pi x}}{L}} \right){{\bf{e}}^{ - \alpha{{\left( {\frac{{n\pi }}{L}} \right)}^2}\,t}}}$$ At $t=0$: $$v(x,0)=f(x)-u_E(x)$$ $$v(x,0)=f(x)-\left[\frac12 f(0)\left(L-x^2\right)\right]$$ Using the Fourier series: $${A_n} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{\left( {f\left( x \right) - \left[\frac12 f(0)\left(L-x^2\right)\right]\left( x \right)} \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,dx}}\,\,\,\,\,\,\,\hspace{0.25in}n = 1,2,3, \ldots$$ However, I have a niggling doubt in that $u_E(x)$, where the heat source is of the kind $F(x)$, is the steady state (or Equilibrium) function and it is free of time $t$. That is not the case here.

So is this derivation correct or not?

Answer 1

Here is the approach outlined in the comments.

Solving the homogenous equation is straight forward and gives the solution $$u_h(x,t) = \sum a_n \sin\left(\frac{n\pi x}{L}\right)e^{-\alpha n^2\pi^2 t/L^2}$$ where $a_n$ is determined by the initial condition $u(x,0)$. Taking $u = u_h + u_p$ we see that $u_p$ satisfy the same PDE with the same boundary conditions but with initial conditions $u_p(x,0) = 0$.

Now how do we find $u_p$? To be able to not mess up the boundary conditions the basis functions for $x$ should remain the same, i.e. we want to find a solution as a series in terms of $\sin\left(\frac{n\pi x}{L}\right)$. Thus we want to find a solution on the form $$u_p(x,t) = \sum \sin\left(\frac{n\pi x}{L}\right) T_n(t)$$ for some functions $T_n$ to be determined. Inserting it into the PDE we get $$\sum \sin\left(\frac{n\pi x}{L}\right)\left[T_n'(t) + \alpha\frac{n^2\pi^2}{L^2}T_n(t)\right] = f$$ Thus we see that if we could expand $f$ in a Fourier sin series then we can extract an ODE for $T_n(t)$. To do so lets extend $f$ to be an odd function (such that it has a Fourier sin series), i.e. we take $f(x,t) = f(t)$ for $x>0$ and $-f(t)$ for $x<0$. In other words we consider the source to be $f(t)H(x)$ where $H$ is the step function. We are only interested in the source for $x>0$ so this is just a trick to be able to write down the Fourier series. This is $$H(x) = \sum c_n \sin\left(\frac{n\pi x}{L}\right)\sin\left(\frac{n\pi x}{L}\right)$$ where $c_n = \frac{2}{n\pi}(1 - (-1)^n)$. In our equation this becomes $$\sum \sin\left(\frac{n\pi x}{L}\right)\left[T_n'(t) + \alpha\frac{n^2\pi^2}{L^2}T_n(t) - c_n f(t)\right] = 0$$ which is only possible (the Fourier series is unique) if $$T_n'(t) + \alpha\frac{n^2\pi^2}{L^2}T_n(t) - c_n f(t) = 0$$ Solve this ODE with initial conditions $T_n(0) = 0$ (recall $u_p(x,0)=0$) and you will find the solution. Adding in the homogenous solution gives you the total solution.

Answer 2

So, thanks to @Winther, for the original problem: $$u_t=\alpha u_{xx}+f(t)$$ $$u(0,t)=u(L,t)=0\text{ and }u(x,0)=f(x)$$ Assume: $$u(x,t)=v(x,t)+F(t)$$ where: $$F'(t)=f(t)$$ Thus: $$u_t=v_t+f(x)$$ $$u_{xx}=v_{xx}$$ $$\Rightarrow v_t=\alpha v_{xx}\tag{1}$$ The originally homogeneous BCs are now time-dependent: $$u(0,t)=v(0,t)+F(t)\Rightarrow v(0,t)=-F(t)$$ $$u(L,t)=v(L,t)+F(t)\Rightarrow v(L,t)=-F(t)$$

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I'll now complete the answer using the method linked to be @Winther.

The homogeneous PDE $(1)$ solves to:

$$u(x,t)=\sum_{n=1}^\infty u_n(t)\sin(\sqrt{\lambda_n}\,x)\quad\text{where}\quad u_n(t)={2\over L}\int_0^L u(x,t)\sin(\sqrt{\lambda_n}\,x)\,dx$$ $$\lambda_n=(n\pi/L)^2\text{ for }n=1,2,3,...$$ $$G(t)=-\lambda_nu_n(t)-\underbrace{{2\sqrt{\lambda_n}\over L}F(t)\left[1+(-1)^{n+1}\right]}_{G(t)}$$

So that: \begin{align} {du_n\over dt}+\alpha \lambda_nu_n(t)&=-\alpha G(t),\\ u_n(0)&={2\over L}\int_0^L f(x)\sin(\sqrt{\lambda_n}\,x)\,dx, \end{align}