Reaction-Diffusion Systems - Revision history

Levi: /* The discrete Fourier transform */

2025-11-06T10:37:46Z

The discrete Fourier transform

← Older revision		Revision as of 10:37, 6 November 2025
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	\end{matrix}		\end{matrix}
	\right. </math> we have:		\right. </math> we have:
	<center><math>\mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\}=\dfrac{2L}{N\sqrt{4\pi DT}}\sum_{l=-\infty}^\infty \operatorname{exp}\left[ -\dfrac{(x_m+(2l+1)L)^2}{4Dt} \right]\approx\dfrac{2L}{N\sqrt{4\pi DT}}\left( \operatorname{exp}\left[ -\dfrac{(x_m+L)^2}{4Dt} \right] + \operatorname{exp}\left[ -\dfrac{(x_m-L)^2}{4Dt} \right] \right)</math></center>		<center><math>\mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\}_m=\dfrac{2L}{N\sqrt{4\pi DT}}\sum_{l=-\infty}^\infty \operatorname{exp}\left[ -\dfrac{(x_m+(2l+1)L)^2}{4Dt} \right]\approx\dfrac{2L}{N\sqrt{4\pi DT}}\left( \operatorname{exp}\left[ -\dfrac{(x_m+L)^2}{4Dt} \right] + \operatorname{exp}\left[ -\dfrac{(x_m-L)^2}{4Dt} \right] \right)</math></center>

	The real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]		The real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]

Levi: /* The discrete Fourier transform */ fix

2025-11-04T18:12:32Z

The discrete Fourier transform: fix

← Older revision		Revision as of 18:12, 4 November 2025
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	</center>		</center>

	Therefore we can write again, similar to the continuous transform case ~~(with the only difficulty that we need to define carefully the values of~~		Therefore we can write again, similar to the continuous transform case:
	~~<math>k_n</math>)~~:
	<center>		<center>
	<math>		<math>
	c(x_m,t) = \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \mathcal{F} \left\{ c_0(x_m) \right\} \right\} = c_0(x_m) * \~~frac~~{1}{\~~sqrt~~{~~4\pi D t}}~~ e^{-~~x_m~~^2~~/4Dt~~} = \frac{1}{\sqrt{4\pi D t}} \sum_{n=0}^{N-1} c_0(x_{(m-n)\bmod N}) e^{-~~x_{n}~~^2~~/4Dt~~}		c(x_m,t) = \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \mathcal{F} \left\{ c_0(x_m) \right\} \right\} = c_0(x_m) * \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\} = \frac{1}{\sqrt{4\pi D t}} \sum_{l=0}^{N-1} c_0(x_{(m-l)\bmod N}) \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\}_l
	</math>		</math>
	</center>		</center>
			Unlike it, though, now <math>\mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\}\ne\frac{1}{\sqrt{4\pi D t}} e^{-x_m^2/4Dt}</math>. It can be proven that for our previously defined <math>x_m = -L + 2Lm/N</math> and <math>k_{n}=\left\{
			\begin{matrix}
			n\pi/L,\ \ 0\leq n\leq \frac{N}{2} \\
			\left( n-N\right)\pi/L,\ \ \frac{N}{2}+1\leq n\leq N-1
			\end{matrix}
			\right. </math> we have:
			<center><math>\mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \right\}=\dfrac{2L}{N\sqrt{4\pi DT}}\sum_{l=-\infty}^\infty \operatorname{exp}\left[ -\dfrac{(x_m+(2l+1)L)^2}{4Dt} \right]\approx\dfrac{2L}{N\sqrt{4\pi DT}}\left( \operatorname{exp}\left[ -\dfrac{(x_m+L)^2}{4Dt} \right] + \operatorname{exp}\left[ -\dfrac{(x_m-L)^2}{4Dt} \right] \right)</math></center>

	The real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]		The real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]

Levi: /* Diffusion */ Fixes + discrete convolution

2025-11-04T07:23:12Z

Diffusion: Fixes + discrete convolution

← Older revision		Revision as of 07:23, 4 November 2025
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	<center>		<center>
	<math>		<math>
	c(x,t) = \frac{1}{\sqrt{4\pi D t}} \int_{-\infty}^{\infty} c_0(x) e^{(x-s)^2/4Dt}\mathrm{d}s		c(x,t) = \mathcal{F}^{-1} \left\{ e^{-k^2 D t} \mathcal{F} \left\{ c_0(x) \right\} \right\} = c_0(x) * \frac{1}{\sqrt{4\pi D t}} e^{-x^2/4Dt} = \frac{1}{\sqrt{4\pi D t}} \int_{-\infty}^{\infty} c_0(x) e^{-(x-s)^2/4Dt}\mathrm{d}s
	</math>		</math>
	</center>		</center>
Line 149:		Line 149:
	<center>		<center>
	<math>		<math>
	c(x_m,t) = \sum_{n=~~-\infty~~}^{~~\infty~~} \hat{c}_n(0) e^{-k_n^2 D t} e^{\mathrm{i} k_n x_m}		c(x_m,t) = \sum_{n=0}^{N-1} \hat{c}_n(0) e^{-k_n^2 D t} e^{\mathrm{i} k_n x_m}
	</math>		</math>
	</center>		</center>
	<center>		<center>
	<math>		<math>
	= \sum_{n=~~-\infty~~}^{~~\infty~~} \hat{c}_n(0) e^{-k_n^2 D t} e^{2\mathrm{i} \pi nm/N} e^{-\mathrm{i} \pi n}		= \sum_{n=0}^{N-1} \hat{c}_n(0) e^{-k_n^2 D t} e^{2\mathrm{i} \pi nm/N} e^{-\mathrm{i} \pi n}
	</math>		</math>
	</center>		</center>
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	<math>		<math>
	\hat{c}_n(0) e^{-\mathrm{i} \pi n} = \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} c_0(x_m)		\hat{c}_n(0) e^{-\mathrm{i} \pi n} = \frac{1}{N} \sum_{m=0}^{N-1} e^{-2\mathrm{i} \pi nm/N} c_0(x_m)
			</math>
			</center>
			So
			<center>
			<math>
			c(x_m,t) = \frac{1}{N} \sum_{n=0}^{N-1} \sum_{l=0}^{N-1} c_0(x_l) e^{-2\mathrm{i} \pi nl/N} e^{-k_n^2 D t} e^{2\mathrm{i} \pi nm/N}
	</math>		</math>
	</center>		</center>
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	The		The
	[http://en.wikipedia.org/wiki/Discrete_Fourier_transform discrete Fourier transform]		[http://en.wikipedia.org/wiki/Discrete_Fourier_transform discrete Fourier transform]
	of a sequence of ''2N'' complex numbers ''c''<sub>0</sub>, ..., ''c''<sub>''2N''−1</sub> is transformed into the sequence of ''N'' complex numbers <math>\hat{c}</math><sub>0</sub>, ..., <math>\hat{c}</math><sub>''N''−1</sub> by the DFT according to the formula:		of a sequence of ''N'' complex numbers ''c''<sub>0</sub>, ..., ''c''<sub>''N''−1</sub> is transformed into the sequence of ''N'' complex numbers <math>\hat{c}</math><sub>0</sub>, ..., <math>\hat{c}</math><sub>''N''−1</sub> by the DFT according to the formula:
	<center>		<center>
	<math>\hat{c}_m = \sum_{n=0}^{N-1} ~~c_n~~ e^{-2\pi \mathrm{i}mn/N} \quad \quad m = 0, \dots, N-1</math>		<math>\hat{c}_n = \sum_{m=0}^{N-1} c_m e^{-2\pi \mathrm{i}mn/N} \quad \quad n = 0, \dots, N-1</math>
	</center>		</center>

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	The '''inverse discrete Fourier transform (IDFT)''' is given by		The '''inverse discrete Fourier transform (IDFT)''' is given by
	<center>		<center>
	<math>~~c_n~~ = \frac{1}{N} \sum_{m=0}^{N-1} \hat{c}_m e^{2\pi \mathrm{i}mn/N} \quad \quad n = 0,\dots,N-1.</math>		<math>c_m = \frac{1}{N} \sum_{n=0}^{N-1} \hat{c}_n e^{2\pi \mathrm{i}mn/N} \quad \quad m = 0,\dots,N-1.</math>
	</center>		</center>

	Therefore we can write		Therefore we can write again, similar to the continuous transform case (with the only difficulty that we need to define carefully the values of
			<math>k_n</math>):
	<center>		<center>
	<math>		<math>
	c(x_m,t) = \mathcal{F} \left\{ e^{-k_n^2 D t} \mathcal{F~~}^{-1~~} \left\{ c_0(x_m) \right\}		c(x_m,t) = \mathcal{F}^{-1} \left\{ e^{-k_n^2 D t} \mathcal{F} \left\{ c_0(x_m) \right\} \right\} = c_0(x_m) * \frac{1}{\sqrt{4\pi D t}} e^{-x_m^2/4Dt} = \frac{1}{\sqrt{4\pi D t}} \sum_{n=0}^{N-1} c_0(x_{(m-n)\bmod N}) e^{-x_{n}^2/4Dt}
	\right\}
	</math>		</math>
	</center>		</center>
	~~Note that the choice of where to put the <math>1/N</math> is arbitrary in the definition of FFT and IFFT and~~
	~~does not exactly match here. Of course since it appears once the formula above is correct regardless.~~
	~~The only difficulty is that we need to define carefully the values of~~
	~~<math>k_n</math>~~

	The real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]		The real power of this method lies with the [http://en.wikipedia.org/wiki/FFT Fast Fourier Transform]

Levi: /* Reaction Diffusion Equations */

2025-10-31T11:34:54Z

Reaction Diffusion Equations

← Older revision		Revision as of 11:34, 31 October 2025
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	<center>		<center>
	<math>		<math>
	\frac{1}{t^}\partial_\tau \alpha = \frac{a_0}{(x^)^2}D\partial_z^2 \alpha		\frac{a_0}{t^}\partial_\tau \alpha = \frac{a_0}{(x^)^2}D\partial_z^2 \alpha
	- k a_0^2 \alpha\beta		- k a_0^2 \alpha\beta
	</math>		</math>
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	<center>		<center>
	<math>		<math>
	\frac{1}{t^}\partial_\tau \beta = \frac{a_0}{(x^)^2}D\partial_z^2 \beta		\frac{a_0}{t^}\partial_\tau \beta = \frac{a_0}{(x^)^2}D\partial_z^2 \beta
	+ k a_0^2 \alpha\beta		+ k a_0^2 \alpha\beta
	</math>		</math>

Meylan: /* The discrete Fourier transform */

2024-10-24T23:15:35Z

The discrete Fourier transform

← Older revision		Revision as of 23:15, 24 October 2024
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	The '''inverse discrete Fourier transform (IDFT)''' is given by		The '''inverse discrete Fourier transform (IDFT)''' is given by
	<center>		<center>
	<math>c_n = \frac{1}{2N} \sum_{m=0}^{N-1} \hat{c}_m e^{2\pi \mathrm{i}mn/N} \quad \quad n = 0,\dots,N-1.</math>		<math>c_n = \frac{1}{N} \sum_{m=0}^{N-1} \hat{c}_m e^{2\pi \mathrm{i}mn/N} \quad \quad n = 0,\dots,N-1.</math>
	</center>		</center>

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	</math>		</math>
	</center>		</center>
			Note that the choice of where to put the <math>1/N</math> is arbitrary in the definition of FFT and IFFT and
			does not exactly match here. Of course since it appears once the formula above is correct regardless.
	The only difficulty is that we need to define carefully the values of		The only difficulty is that we need to define carefully the values of
	<math>k_n</math>		<math>k_n</math>

Meylan at 05:06, 8 October 2021

2021-10-08T05:06:59Z

← Older revision		Revision as of 05:06, 8 October 2021
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	=== Part 2 ===		=== Part 2 ===

	{{#ev:youtube\|~~MR7xSsJ_zLs~~}}		{{#ev:youtube\|D0NwYlM-uOg}}

	=== Part 3 ===		=== Part 3 ===

	{{#ev:youtube\|~~l9tXlpnlqn8~~}}		{{#ev:youtube\|5IEZJtJaDHk}}

	=== Part 4 ===		=== Part 4 ===

	{{#ev:youtube\|~~Ip353jAdBkk~~}}		{{#ev:youtube\|t_OjTSwVgdo}}

	=== Part 5 ===		=== Part 5 ===
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	=== Part 8 ===		=== Part 8 ===

	{{#ev:youtube\|~~t3TiemLMxDg~~}}		{{#ev:youtube\|yQ-O2KIqu44}}



	[[Category:Simple Nonlinear Waves]]		[[Category:Simple Nonlinear Waves]]

Meylan: /* Lecture Videos */

2020-10-21T01:28:05Z

Lecture Videos

← Older revision		Revision as of 01:28, 21 October 2020
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	=== Part 1 ===		=== Part 1 ===

	{{#ev:youtube\|~~Yi8fytUszrQ~~}}		{{#ev:youtube\|rF4X42jP0v8}}

	=== Part 2 ===		=== Part 2 ===

	{{#ev:youtube\|~~_668xmJ9DAQ~~}}		{{#ev:youtube\|MR7xSsJ_zLs}}

	=== Part 3 ===		=== Part 3 ===

	{{#ev:youtube\|~~yTat11ERJMg~~}}		{{#ev:youtube\|l9tXlpnlqn8}}

	=== Part 4 ===		=== Part 4 ===

	{{#ev:youtube\|~~qG8bPJwX96k~~}}		{{#ev:youtube\|Ip353jAdBkk}}

	=== Part 5 ===		=== Part 5 ===

	{{#ev:youtube\|~~38ZWW2dM1Qs~~}}		{{#ev:youtube\|kTHVZaYezLk}}

	=== Part 6 ===		=== Part 6 ===

	{{#ev:youtube\|~~AZHlhA5pQBg~~}}		{{#ev:youtube\|MVuSg5_sfYI}}

	=== Part 7 ===		=== Part 7 ===

	{{#ev:youtube\|~~VhlY5lQcTlM~~}}		{{#ev:youtube\|hxKMOHyy6Bw}}

	=== Part 8 ===		=== Part 8 ===

	{{#ev:youtube\|~~VhlY5lQcTlM~~}}		{{#ev:youtube\|t3TiemLMxDg}}

	[[Category:Simple Nonlinear Waves]]		[[Category:Simple Nonlinear Waves]]

Meylan: /* Travelling Waves solution */

2020-10-21T01:08:01Z

Travelling Waves solution

@@ Line 405: / Line 405: @@
 so that <math>k^2 - vk +1 < 0</math>, which is possible provided <math>v\geq 2</math>, for example
 <math>k = \dfrac{v}{2}</math>.
 [[Category:Simple Nonlinear Waves]]

Meylan: /* Travelling Waves solution */

2010-11-10T00:34:26Z

Travelling Waves solution

← Older revision		Revision as of 00:34, 10 November 2010
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	so that <math>\alpha + \beta = c_0 + c_1 e^{-vy}</math>. Boundary conditions		so that <math>\alpha + \beta = c_0 + c_1 e^{-vy}</math>. Boundary conditions
	are that as <math>y\to\infty </math> <math>\alpha = 1</math> and		are that as <math>y\to\infty </math> <math>\alpha = 1</math> and
	<math>\beta = 0</math>~~. We also require the solution to be bounded~~		<math>\beta = 0</math> and if <math>y\to-\infty</math> then <math>\alpha = 0</math> and
	as <math>y\to-\infty</math> ~~so that~~ <math>\alpha + \beta = 1</math>.		<math>\beta = 1</math>. Therefore <math>\alpha + \beta = 1</math>.
	This means that, since <math>\alpha \geq 0</math>, we must have		This means that, since <math>\alpha \geq 0</math>, we must have
	<math>0\leq \beta \leq 1</math>.		<math>0\leq \beta \leq 1</math>.

Adi Kurniawan: /* Diffusion */

2010-11-06T10:19:55Z

Diffusion

← Older revision		Revision as of 10:19, 6 November 2010
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	<center>		<center>
	<math>		<math>
	c(x,t) = \frac{1}{\sqrt{4\pi D t}} \int_{-\infty}^{\infty} c_0(x) e^{(x-s)^2/4Dt}ds		c(x,t) = \frac{1}{\sqrt{4\pi D t}} \int_{-\infty}^{\infty} c_0(x) e^{(x-s)^2/4Dt}\mathrm{d}s
	</math>		</math>
	</center>		</center>
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	<center>		<center>
	<math>		<math>
	\hat{c}_n(0) = \frac{1}{2L} \int_{-L}^{L} e^{-\mathrm{i} k_n x} c_0(x) dx		\hat{c}_n(0) = \frac{1}{2L} \int_{-L}^{L} e^{-\mathrm{i} k_n x} c_0(x) \mathrm{d}x
	</math>		</math>
	</center>		</center>