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I'd like to solve this equation:

$$ u_t = u_{xx} + u_x $$ for $t>0$ with initial condition: $$ u(0,x) = \cos (2 \pi x) $$ And $1$-periodic, i.e we have that $u(t,x) = u(t,x+1).$ I was going to solve it using separation of variables, but I was thinking another way, namely solving it using a fourier series. I.e, first I assume that $u$ has fourier series expansion: $$ u(t,x) = \sum_{k \in \mathbb{Z}} a_k(t)e^{2\pi ikx} $$ Now, I would like to differentiate the coefficients in the series and set them equal to each other. Something akin to: $$ a_k'(t) = 2\pi ike^{2\pi i k x} ak(t) + (2 \pi ik )^2 ak(t) e^{2 \pi i k x} $$ And then solve for $ak(t)$. Two questions: 1) is this method valid? I.e is it true that the series are equal if and only if the coefficients are? 2) If I can use this method, how do I solve the ODE for $a_k(t)$? Apologies if this is particularly obvious, I have never taken an ODEs class

rubikscube09
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    Your method works, but your governing equation seems to observe some typos. It should be $a_k'e^{2\pi ikx}=2\pi ika_ke^{2\pi ikx}+\left(2\pi ik\right)^2a_ke^{2\pi ikx}$. Since $\left{e^{2\pi ikx}\right}_{k\in\mathbb{Z}}$ forms an orthonormal basis, this equation implies that the coefficient in front of each $e^{2\pi ikx}$ must vanish, which yields $a_k'=2\pi ika_k+\left(2\pi ik\right)^2a_k$. This is no more than an ODE now. – hypernova Apr 18 '18 at 20:07
  • Yes this method should work - in fact I suspect it is the same solution you'd get if you did do separation of variables - you'd get a term in $t$ and a term in $x$ which looks like that fourier term (since it must be $1$ periodic), and then to get a general solution, a general linear combination would be taken (the sum) – John Doe Apr 18 '18 at 20:08
  • @hypernova Thank you for your explanation. Just to clarify, the coefficients vanish because if we have $\sum (a_k - b_k)e_k = 0$ where $e_k$ forms a (Schauder?) basis (for what space? sufficiently smooth periodic functions) then it follows by linear independence that $a_k = b_k$. – rubikscube09 Apr 18 '18 at 20:14
  • @qbert my professor gave this problem on a review sheet for a midterm. He said it contained questions from prior exams. Perhaps you took the same class with the same professor? – rubikscube09 Apr 18 '18 at 20:30
  • @rubikscube09 looking at your previous questions, my suspicion is yes :) – operatorerror Apr 18 '18 at 20:33
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    @rubikscube09: It is a Schauder basis for general $L^p([0,1])$ ($p>1$, typically $p=2$ for a Hilbert space). You may let $x\to x/L$ to get it to $L^p([0,L])$ as well. This is why, for mathematical rigor, you need to check what you get from the equations is indeed a solution (e.g., using dominate convergence theorem to show that $\partial_t$, $\partial_x^m$ and $\sum_{j\in\mathbb{Z}}$ are commutative, using absolute convergence to show that terms in this series could be rearranged so that your governing equation makes sense, etc.). – hypernova Apr 19 '18 at 00:21

1 Answers1

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  1. Yes this method is valid (under the same hypothesis as for the usual heat equation).
  2. Your ODE for $a_k$ should be: $$a_k'(t)=2i \pi k a_k(t)+(2i \pi k)^2 a_k(t)=(-4 \pi k^2+2i \pi k)a_k(t)$$ so: $$a_k(t)=e^{(-4 \pi k^2+2i \pi k) t} a_k(0)$$
Delta-u
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  • Thank you. The above solution to the ODE is simply solving the ODE $f ' = f$ right? Or some derivative of it? Apologies for the rather elementary questions, although I am taking PDE I have not taken ODE. – rubikscube09 Apr 18 '18 at 20:15
  • Yes it is simply an ODE such as $f'=\lambda f$. – Delta-u Apr 18 '18 at 20:19