[Fourier transform and its application study notes] five. Fourier series continuity discussion, thermal equation

This is my study notes, the course for NetEase Open Class Stanford University Open Class: Fourier transform and its application.

Thermal equation Follow-up

The Fourier coefficients of the thermal equation are deduced from the previous lesson:

$C _k (t) = c_k (0) e^{-2\pi ^2 k^2t}$

So what is $c_k (0) $?

The previous lesson mentioned that temperature has the following relationship:

$U (x,t) = \displaystyle{\sum_{k=-\infty}^{\infty}c_k (t) e^{2\pi ikx}}$

When $t=0$, represents the temperature distribution on the initial time ring

$f (x) = U (x,0) = \displaystyle{\sum_{k=-\infty}^{\infty}c_k (0) e^{2\pi Ikx}}$

Then, $C _k (0) $ for $f (x) $ for the Fourier coefficients

$C _k (0) = \hat{f} (k) $

Therefore, the temperature distribution formula (thermal equation) is as follows:

$U (x,t) = \displaystyle{\sum_{k=-\infty}^{\infty}\hat{f} (k) E^{-2\pi^2k^2t}e^{2\pi Ikx}}$

The relationship between temperature $u$ and time $t$ is as follows: When $t \to \infty$,$-2\pi^2k^2t \to–\infty$, $e ^{-2\pi^2k^2t} \to 0$, $U \to 0$. Therefore, the temperature of the ring will eventually change to 0.

Further derivation of the thermal equation, the introduction of convolution

We can further decompose the $\hat{f} (k) $ in the thermal equation

$\hat{f} (k) = \displaystyle{\int_0^1 e^{-2\pi iky}f (y) dy}$

Considering the initial time of the temperature distribution $f (x) $ with the heat equation $u (x,t) $ in the position variable $x$ may take a different value, we here put $f (x) $ into $f (y) $.

After the $\hat{f} (k) $ is put into the heat equation,

$\begin{align*}
U (x,t) &=\displaystyle{\sum_{k=-\infty}^{\infty} (\int_0^1 e^{-2\pi iky}f (y) dy) e^{-2\pi^2k^2t}e^{2\pi ikx}} \ \
&=\displaystyle{\int_0^1 (\sum_{k=-\infty}^{\infty}e^{-2\pi iky}e^{2\pi ikx}e^{-2\pi^2k^2t}) F (y) dy} \ \
&=\displaystyle{\int_0^1 (\sum_{k=-\infty}^{\infty}e^{2\pi ik (x-y)}e^{-2\pi^2k^2t}) f (y) dy}
\end{align*}$

Make

$g (x,t) = \displaystyle{\sum_{k=-\infty}^{\infty}e^{2\pi ikx}e^{-2\pi^2k^2t}}$

The above equation is called the thermonuclear equation (heat kernel), then

$U (x,t) = \displaystyle{\int_0^1g (x-y,t) f (y) dy}$

As the equation above, the heat equation is converted into a convolution representation.

From Fourier series to Fourier transform

Fourier series to Fourier transform is a transition from periodic phenomenon to non-periodic phenomenon, we can consider the Aperiodic function as a special case of periodic function: The period tends to infinity.

For a function with a period of 1

$C _k = \displaystyle{\hat{f} (k) = \int_0^1e^{-2\pi ikt}f (t) DT}$

$f (t) = \displaystyle{\sum_{k=-\infty}^{\infty}\hat{f} (k) E^{2\pi Ikt}}$

The spectrum diagram is as follows

Since $c_k$ is a plural form, we cannot draw it on the graph, so we can only draw $\left | C_k \right | = \left | A + bi \right | = \sqrt{a^2 + b^2}$. In the second lesson, we also learned that $C _k$ is symmetrical in the Y axis.

For a function with a period of T

$\begin{align*}
C_k = \displaystyle{\hat{f} (k)} &= \displaystyle{\frac{1}{t}\int_0^1e^{-\frac{2\pi}{T}ikt}f (T) dt} \ \
&= \displaystyle{\frac{1}{t}\int_{-\frac{t}{2}}^{\frac{t}{2}}e^{-2\pi i\frac{k}{t}t}f (T) DT}
\end{align*}.$

$f (t) = \displaystyle{\sum_{k=-\infty}^{\infty}\hat{f} (k) E^{2\pi i\frac{k}{t}t}}$

The spectrum diagram is as follows

Since the cycle is $t$, the frequency is $\frac{1}{t}$. When $t \to \infty$,$\frac{1}{t} \to 0$, the spectrum becomes contiguous.

$T \to \infty$

But is it possible to get the Fourier transform just by $t \to \infty$? The answer is no, let's take a look at an example

There is a function f (t) such as

The Fourier coefficients of the function are solved as follows

$\begin{align*}
C_k = \displaystyle{\hat{f} (k)}
&= \displaystyle{\frac{1 }{t}\int_{-\frac{t}{2}}^{\frac{t}{2}}e^{-2\pi i\frac{k}{t}t}f (T) dt} \ \
&= \displaystyle{\frac{1}{t}\int_a^ b e^{-2\pi i\frac{k}{t}t}f (T) dt} \ \
&\leqslant \displaystyle{\frac{1}{t}\int_a^b \left | e^{-2\pi i\frac{k}{T}t }\right | \left |f (t) \right |dt} \ \
&= \displaystyle{\frac{1}{t}\int_a^b Mod (e^{-2\pi i\frac{k}{t}t}) \left |f (t) \right |d T} \ \
&= \displaystyle{\frac{1}{t}\int_a^b Mod (cos ( -2\pi \frac{k}{t}t) + isin ( -2\pi \frac{k}{t}t)) \left |f (t) \ r ight |dt} \quad spread \ with \ eular \ Formula \
&= \displaystyle{\frac{1}{t}\int_a^b \sqrt{cos^2 ( -2\pi \frac{k} {t}t) + sin^2 ( -2\pi \frac{k}{t}t)} \left |f (t) \right |dt} \ \
&= \displaystyle{\frac{1}{t}\int_a^b 1\left |f (t) \ r ight |dt} \ \
&= \displaystyle{\frac{1}{t}\int_a^b \left |f (T) \right |dt} \ \
&= \frac{m}{t}
\end{align *}.$

That is, for all $c_k$ there are $c_k \leqslant \frac{m}{t}$.

$M $ is the integral of the absolute value of the function, which is finite, if $t \to \infty$, then all $c_k \to 0$. All Fourier coefficients are 0 and the Fourier transform is meaningless.

[Fourier transform and its application study notes] five. Fourier series continuity discussion, thermal equation