PDEs, Fourier Series

Question

I've been having some trouble solving the following exercise: I need to find the solution $f(x,t)$ of the following PDE: $$\partial_tf(x,t) = -\partial_x f(x,t)-f(x,t) + \delta(t^2+t)(\sin2x+2\cos3x-\sin5x+1)\tag{1}$$ where $x \in ]-\pi,\pi[$, we have periodic boundary conditions for $f$ and $\partial_xf$ and initial condition $f(x,-1/2) = 0$ My approach was a bit different than usual, where I would use Fourier transform and so on. Nevertheless, I opted for a Fourier series expansion of $f(x)$, in particular the complex form, since the $g(x) = (\sin2x+2\cos3x-\sin5x+1)$ presents both sines and cosines. Now I have basically zero experience with the Fourier series and its applications to PDEs, but I gave it a try anyway: For $t > 0$ the equation reduces to: $$ \partial_tf(x,t) = -\partial_x f(x,t)-f(x,t)\tag{2}$$ Moreover, following the procedure explained to me in one of my previous questions, I can easily find the initial condition at $t = 0$ by integrating $(1)$ between $t \in [-\epsilon, \epsilon], \epsilon \to 0^+$, thus finding $f(x,0) = g(x)$. Now, we can go on writing: $$ f(x,t) = \sum_{n = -\infty}^{+\infty}c_n(t)e^{inx}\tag{3}$$ substituting in $(2)$ and by comparison we get: $$ \dot{c_n}(t) = -c_n(t)(in+1)$$ thus we get: $$ c_n(t) = c_n(0)e^{-t(1+in)}$$ And finally resubstituting in $(3)$: $$f(x,t) = \sum_{n = -\infty}^{+\infty}c_n(0)e^{in(x-t)}e^{-t}$$ Now, from the form of our initial condition, I can see that the only indices I need are: $n=[0,2,3,5]$. The problem is that I do not know how to determine the coefficients $c_n(0)$ for those particular values of $n$. I thought that maybe I should've considered the equation at $t = 0$ (or more precisely at $t \in [-\epsilon, +\epsilon], \epsilon \to 0^+$) \begin{align} \partial_tf(x,t) &= -\partial_x f(x,t)-f(x,t) + \int_{-\epsilon}^{+\epsilon}\delta(t)(\sin2x+2\cos3x-\sin5x+1)dt \\ &= -\partial_x f(x,t)-f(x,t) + g(x) \end{align} Then express $g(x),f(x)$ as: $$ f(x) = \sum_{n = -\infty}^{+\infty}c_n(t)e^{inx},\qquad g(x) = \sum_{n = -\infty}^{+\infty}p_n(t)e^{inx}$$ and then somehow solve 4 ODEs of the form: $$ \dot{c}_n(t) = -c_n(t)(1+in) + p_n$$ But I don't think this procedure is the one to follow, since it does seem to be pretty long and time is not much during an exam (and, to be honest, I don't think it is mathematically justified in the slightest...). If anyone would like to help, I'd be very grateful.

Answer 1

After you have found the general solution to the PDE for $t>0$, $$ f(x,t) = \sum_{n = -\infty}^{+\infty}c_n(0)e^{in(x-t)}e^{-t}, \tag{1} $$ the next step is to apply the condition $f(x,0)=g(x)$, or \begin{align} \sum_{n = -\infty}^{+\infty}c_n(0)e^{inx} &= \sin2x+2\cos3x-\sin5x+1 \\ &=\frac{e^{2ix}-e^{-2ix}}{2i}+e^{3ix}+e^{-3ix}-\frac{e^{5ix}-e^{-5ix}}{2i}+1. \tag{2} \end{align} Comparing the LHS with the RHS of $(2)$, one obtains $c_0(0)=1$, $c_{\pm 2}(0)=\pm\frac{1}{2i}$, $c_{\pm 3}(0)=1$, $c_{\pm 5}(0)=\mp\frac{1}{2i}$, and $c_n(0)=0$ for $n\in\mathbb{Z}\setminus\{0,\pm 2, \pm 3, \pm 5\}$. Substituting this result in $(1)$, one finally obtains, after some simplification, $$ f(x,t)=e^{-t}[\sin 2(x-t) + 2\cos 3(x-t) - \sin 5(x-t) + 1]\qquad(t>0). \tag{3} $$