Evaluating a limit of a Fourier series

Question

I have a Fourier series representation of a solution to the heat equation, given by

$\displaystyle u(t,x) = \\ \displaystyle\sin \omega t + \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\omega\left\lbrack \frac{\omega\sin \omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\left( \cos \omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack\cos\left\lbrack \left( n - \frac{1}{2} \right)\pi x \right\rbrack}.$

How would one go about computing

$\displaystyle \lim_{\omega\rightarrow\infty}{u(t,0)}$

explicitly?

Answer 1

A quicker method to obtain the same result is as follows. Firstly, merge the $\sin\omega t$ term with the rest of the Fourier series. This is achieved by noting that the coefficient of $\sin\omega t$ is unity, and so the Fourier series of unity in terms of the spatial variable $x$ may be employed:

\begin{align*} \sin\omega t &= \sum_{n=1}^{\infty}{\frac{4\left( -1 \right)^{n+1}}{(2n-1)\pi}\sin\omega t\cos\left[\left( n - \frac{1}{2}\right)\pi x\right]} \\ &= - \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\sin\omega t\cos\left\lbrack \left( n - \frac{1}{2} \right)\pi x \right\rbrack}. \end{align*}

Thus

\begin{align*} u(t,x) &= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\omega\left\lbrack \frac{\omega\sin\omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} - \frac{\sin\omega t}{\omega} \right\rbrack\cos\left\lbrack \left( n - \frac{1}{2} \right)\pi x \right\rbrack}. \end{align*}

Hence,

\begin{align*} u(t,0) &= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\left\lbrack \frac{\omega^{2}\sin\omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} - \sin\omega t \right\rbrack} \\ &= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\frac{\omega^{2}\sin\omega t + \left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right) - \left( \omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4} \right)\sin\omega t}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}}} \\ &= \sum_{n = 1}^{\infty}{\frac{4( - 1)^{n}}{(2n - 1)\pi}\frac{\left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right) - \left( n - \frac{1}{2} \right)^{4}\pi^{4}\sin\omega t}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}}}. \end{align*}

To determine the limit as $\omega \rightarrow \infty$ of $u(t,0)$, consider each term separately:

\begin{align*} \text{Limit 1:} && \lim_{\omega \rightarrow \infty}\left\lbrack \frac{\left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( \cos\omega t \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack, \\ \text{Limit 2:} && \lim_{\omega \rightarrow \infty}\left\lbrack \frac{\left( n - \frac{1}{2} \right)^{2}\pi^{2}\omega\left( - e^{- \left( n - \frac{1}{2} \right)^{2}\pi^{2}t} \right)}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack, \\ \text{Limit 3:} && \lim_{\omega \rightarrow \infty}\left\lbrack \frac{- \left( n - \frac{1}{2} \right)^{4}\pi^{4}\sin\omega t}{\omega^{2} + \left( n - \frac{1}{2} \right)^{4}\pi^{4}} \right\rbrack, \end{align*}

which all tend to zero, giving

\begin{equation*} \lim_{\omega\rightarrow\infty}{u(t,0)} = 0 \end{equation*}

as required.

Answer 2

Funny enough, but there is no problem with $\sin\omega t$ at $\omega\to\infty:\boxed{\,\,\, \lim_{\omega\to\infty}u(t;0)=0\,\,}$

To show this, let's consider $$u(t;0)=\sin\omega t+\sum_{n = 1}^\infty\frac{4(-1)^{n}}{(2n - 1)\pi}\omega \frac{\omega\sin \omega t + ( n - \frac{1}{2} )^2\pi^2( \cos \omega t - e^{- ( n - \frac{1}{2})^2\pi^2t})}{\omega^2 +( n - \frac{1}{2})^4\pi^4} $$ $$=\sin\omega t+\frac{2\omega^2\sin\omega t}{\pi}\sum_{n = 1}^\infty\frac{(-1)^{n}}{(n - \frac{1}{2})\big(\omega^2 +( n - \frac{1}{2})^4\pi^4\big)}$$ $$+2\pi\omega\cos\omega t\sum_{n = 1}^\infty\frac{(-1)^{n}(n - \frac{1}{2})}{\big(\omega^2 +( n - \frac{1}{2})^4\pi^4\big)}+2\pi\omega\cos\omega t\sum_{n = 1}^\infty\frac{(-1)^{n}e^{- ( n - \frac{1}{2})^2\pi^2t}}{(n - \frac{1}{2})\big(\omega^2 +( n - \frac{1}{2})^4\pi^4\big)}\qquad(1)$$ Evaluating the last term: $$\sum_{n = 1}^\infty\frac{\omega(-1)^{n}e^{- ( n - \frac{1}{2})^2\pi^2t}}{(n - \frac{1}{2})\big(\omega^2 +( n - \frac{1}{2})^4\pi^4\big)}<2\sum_{n = 1}^\infty\frac{\omega e^{- ( n - \frac{1}{2})^2\pi^2t}}{\omega^2 }<\frac{\operatorname{Const}}{\omega}\to0\,\,\text{at}\,\,\omega\to 0$$ To evaluate the third term we notice that $$\frac{1}{2}S=\sum_{n = 1}^\infty\frac{(-1)^{n}(n - \frac{1}{2})}{\big(\omega^2 +( n - \frac{1}{2})^4\pi^4\big)}=\frac{1}{2}\sum_{n = -\infty}^\infty\frac{(-1)^{n}(n - \frac{1}{2})}{\omega^2 +( n - \frac{1}{2})^4\pi^4}$$ To evaluate the sum $S$ we use a standart approach: integrating in the complex plane along a big circle of the radius $R\to\infty\,$ the function $f(z)=\frac{(z - \frac{1}{2})}{\omega^2 +( z - \frac{1}{2})^4\pi^4}\,\frac{\pi}{\sin\pi z}$: $$\oint\frac{(z - \frac{1}{2})}{\omega^2 +( z - \frac{1}{2})^4\pi^4}\,\frac{\pi}{\sin\pi z}dz$$ The integrand declines rapidly enough at $R\to\infty$, so $\oint\to0$. On the other hand $$\oint\frac{(z - \frac{1}{2})}{\omega^2 +(z - \frac{1}{2})^4\pi^4}\,\frac{\pi}{\sin\pi z}dz=2\pi i\sum\operatorname{Res}\frac{(z - \frac{1}{2})}{\omega^2 +(z - \frac{1}{2})^4\pi^4}\,\frac{\pi}{\sin\pi z}=0\qquad(2)$$ where we take all residues lying inside the circle. We have simple poles of $\frac{\pi}{\sin \pi z}$ at $z=\pi n, n=0,\pm1,\pm2...$ with residues $(-1)^n$ (and evaluating these residues we get the sum $S$). We also have four poles at $\omega^2 +( z - \frac{1}{2})^4\pi^4\,\Rightarrow\, z_k=\frac{\sqrt\omega}{\pi}e^{\pi i/4+\pi i k/2}+\frac{1}{2}, \,k=0,1,2,3$. What is important for us is that $z_k$ contain non-zero imaginary parts: $z_k = i\sqrt\omega\alpha_k+\sqrt\omega\beta_k; \,\alpha_k,\,\beta_k\neq 0$ - some real constants.
From $(2)$ we get: $$S=-\sum_{k=0,1,2,3}\operatorname{Res}_{z=z_k}\frac{(z - \frac{1}{2})}{\omega^2 +(z - \frac{1}{2})^4\pi^4}\,\frac{\pi}{\sin\pi z}$$ But $$\frac{\pi}{\sin z_k}=\frac{2\pi i}{e^{-\alpha_k\sqrt\omega+i\beta_k\sqrt\omega}-e^{\alpha_k\sqrt\omega-i\beta_k\sqrt\omega}}$$ so, it is exponentially small ($\sim e^{-|\alpha_k|\sqrt\omega}$) at $\omega\to\infty$. Therefore, $S\to0$, and the third term tends to zero as well.

To evaluate the second term in $(1)$, we consider the integral $$\oint\frac{1}{(z - \frac{1}{2})\big(\omega^2 +(z-\frac{1}{2})^4\pi^4\big)}\,\frac{\pi}{\sin\pi z}dz=0$$ The evaluation is similar to the evaluation of the third term, but in this case there is also a simple pole at $z=\frac{1}{2}$. The residue in this pole is $$\operatorname{Rez}_{z=\frac{1}{2}}\frac{1}{(z - \frac{1}{2})\big(\omega^2 +(z-\frac{1}{2})^4\pi^4\big)}\,\frac{\pi}{\sin\pi z}=\frac{\pi}{\omega^2\sin\frac{\pi}{2}}=\frac{\pi}{\omega^2}$$ Given that other four terms (residues) are exponentially small, we get for the second term in $(1)$ $$\frac{2\omega^2\sin\omega t}{\pi}\sum_{n = 1}^\infty\frac{(-1)^{n}}{(n - \frac{1}{2})\big(\omega^2 +( n - \frac{1}{2})^4\pi^4\big)}\to \frac{2\omega^2\sin\omega t}{\pi}\cdot\Big(-\frac{1}{2}\frac{\pi}{\omega^2}\Big)=-\sin\omega t$$ And this term cancels the first term, providing $\lim_{\omega\to\infty}u(t;0)=0$.

The numeric check at WolframAlpha confirms this conclusion.