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Let $v$ a solution of he heat equation, given by $\frac{\partial v}{\partial t}(t,x)=\frac{\partial^2v}{\partial x^2}(t,x)$ for $t>0,x\in\mathbb R$ with the following properties

$v(0,x)=u_0(0) \forall x\in[0,1]$

$v(t,x)=-v(t,-x)$

$v(t,x+2)=v(t,x)$

Now I evolve the periodic function $v$ in a fourier series $v(t,x)=\sum_{k=-\infty}^{\infty}a_k(t)e^{\pi ikx}$

Question: Why does $a_k$ has to satify the differential equations $a_k'(t)=-\pi^2k^2a_k(t)$ ?

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2 Answers 2

up vote 3 down vote accepted

Let $v(t,x)=\sum\limits_{k=1}^\infty a(k,t)\sin k\pi x+\sum\limits_{k=0}^\infty b(k,t)\cos k\pi x$ so that it automatically satisfies $v(t,x+2)=v(t,x)$ ,

Then $\sum\limits_{k=1}^\infty a_t(k,t)\sin k\pi x+\sum\limits_{k=0}^\infty b_t(k,t)\cos k\pi x=-\sum\limits_{k=1}^\infty k^2\pi^2a(k,t)\sin k\pi x-\sum\limits_{k=0}^\infty k^2\pi^2b(k,t)\cos k\pi x$

$\therefore a_t(k,t)=-k^2\pi^2a(k,t)$ and $b_t(k,t)=-k^2\pi^2b(k,t)$

$a(k,t)=A(k)e^{-k^2\pi^2t}$ and $b(k,t)=B(k)e^{-k^2\pi^2t}$

$\therefore v(t,x)=\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin k\pi x+\sum\limits_{k=0}^\infty B(k)e^{-k^2\pi^2t}\cos k\pi x$

$v(t,x)=-v(t,-x)$

$\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin k\pi x+\sum\limits_{k=0}^\infty B(k)e^{-k^2\pi^2t}\cos k\pi x=-\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin(k\pi(-x))-\sum\limits_{k=0}^\infty B(k)e^{-k^2\pi^2t}\cos(k\pi(-x))$

$\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin k\pi x+\sum\limits_{k=0}^\infty B(k)e^{-k^2\pi^2t}\cos k\pi x=\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin k\pi x-\sum\limits_{k=0}^\infty B(k)e^{-k^2\pi^2t}\cos k\pi x$

$\sum\limits_{k=0}^\infty2B(k)e^{-k^2\pi^2t}\cos k\pi x=0$

$B(k)=0$

$\therefore v(t,x)=\sum\limits_{k=1}^\infty A(k)e^{-k^2\pi^2t}\sin k\pi x$

$v(0,x)=u_0(x)~\forall x\in[0,1]$ :

$\sum\limits_{k=1}^\infty A(k)\sin k\pi x=u_0(x)~\forall x\in[0,1]$

$A(k)=2\int_0^1u_0(x)\sin k\pi x~dx~\forall x\in[0,1]$

$\therefore v(t,x)=\sum\limits_{k=1}^\infty2\int_0^1u_0(x)\sin k\pi x~dx~e^{-k^2\pi^2t}\sin k\pi x~\forall x\in[0,1]$

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It took me some time to get through, but that seems legit :) –  bonext Dec 7 '12 at 20:11

Quick hint: substitute your series for $v(t,x)$ to heat equation.

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Thanks, I have one more question concerning the fourier coefficients $a_k(t)$. Why are they given by $a_k(t)=\frac{1}{2}\int_{-1}^{1}v(0,y)e^{-\pi iky}dy=\frac{1}{i}\int_{0}^{1}sin(\pi ky)dy$ –  Montaigne Dec 7 '12 at 16:16
    
You got something wrong in the above formula and also in your question. Your initial condition should be $v(0,x)=u_0(x)$, and here I assume you wanted to write $a_k(0)$? If that's the case, then here's why you do that: your fourier series for $v(t,x)$ led to a number of ODEs to find $a_k(t)$, so in order to solve these ODEs you have to find initial values ($a_k(0)$). This is done by simply expanding your initial condition $v(0,x)$ into fourier series and taking coefs. for same k's. Does it help? –  bonext Dec 7 '12 at 16:30
    
Yes I want to calculate $a_k(0)$. I know that $v(t,x)=\sum_{-\infty}^{\infty}a_k(0)e^{-\pi^2 k^2 t}e^{\pi ikx}$. Now I set t=0. $v(0,x)=u_0(x)=\sum_{-\infty}^{\infty}a_k(0)e^{\pi ikx}$, but how to derive $a_k(0)$ now? –  Montaigne Dec 7 '12 at 16:50
    
Like any other fourier coef, by integration: $a_k(0)=\int_0^1 u_0(x) e^{i\pi k x}dx$ –  bonext Dec 7 '12 at 20:23

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