Burgers Equation - Revision history

Levi: /* Numerical Solution of Burgers equation */

2025-11-06T10:35:03Z

Numerical Solution of Burgers equation

← Older revision		Revision as of 10:35, 6 November 2025
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	\exp \left( -\nu k^{2}\Delta t\right)		\exp \left( -\nu k^{2}\Delta t\right)
	\end{matrix}</math></center>		\end{matrix}</math></center>
			Or, in convolution form:
			<center><math>u(x, t+\Delta t)=\left(u(x, t)-\dfrac{\Delta t}2\mathcal F^{-1}\left\{ik\right\}u^2(x, t)\right)\mathcal F^{-1}\left\{e^{-\nu k^2\Delta t}\right\}</math></center>
			And for our previously defined <math>x_m</math> and <math>k_n</math> it can be proven that <math>\mathcal F^{-1}\left\{ik\right\}_m=\dfrac\pi{2L}(-1)^m\left(\cot\dfrac{\pi m}N+i\right)</math> (<math>\cot 0</math> taken to be <math>0</math>).

	{\| class="wikitable"		{\| class="wikitable"

Levi: /* Exact Solution of Burgers equations */ more beautiful solution form

2025-11-04T08:06:38Z

Exact Solution of Burgers equations: more beautiful solution form

← Older revision		Revision as of 08:06, 4 November 2025
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	\phi \left( x,t\right) &=&\Phi \left( x\right) * \mathcal{F}^{-1}\left[		\phi \left( x,t\right) &=&\Phi \left( x\right) * \mathcal{F}^{-1}\left[
	e^{-k^{2}\nu t}\right] \\		e^{-k^{2}\nu t}\right] \\
	&=&\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)		&=&\Phi \left( x\right) * \dfrac1{\sqrt{4\pi\nu t}} e^{-x^2 / 4\nu t} \\
			&=&\frac{1}{\sqrt{4\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)
	\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] \mathrm{d}y		\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] \mathrm{d}y
	\end{matrix}</math></center>		\end{matrix}</math></center>
	Which can be expressed as		Which can be expressed as
	<center><math>		<center><math>
	\phi \left( x,t\right) =\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty		\phi \left( x,t\right) =\frac{1}{\sqrt{4\pi \nu t}}\int_{-\infty }^{\infty
	}\exp \left[ -\frac{f}{2\nu }\right] \mathrm{d}y		}\exp \left[ -\frac{f}{2\nu }\right] \mathrm{d}y
	</math></center>		</math></center>
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	u\left( x,t\right) &=&-2\nu \dfrac{\partial _{x}\phi \left( x,t\right) }{\phi		u\left( x,t\right) &=&-2\nu \dfrac{\partial _{x}\phi \left( x,t\right) }{\phi
	\left( x,t\right) } \\		\left( x,t\right) } \\
			&=&-2\nu \dfrac{\partial _{x}\left( \Phi \left( x\right) * e^{-x^2 / 4\nu t} \right) }{\Phi \left( x\right) * e^{-x^2 / 4\nu t} } \\
			&=&-2\nu \dfrac{\partial _{x} \Phi \left( x\right) * e^{-x^2 / 4\nu t} }{\Phi \left( x\right) * e^{-x^2 / 4\nu t} } = \dfrac{\Phi \left( x\right) * x e^{-x^2 / 4\nu t} }{\Phi \left( x\right) * t e^{-x^2 / 4\nu t} } \\
	&=&\dfrac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -		&=&\dfrac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -
	\dfrac{f}{2\nu }\right] \mathrm{d}y}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{		\dfrac{f}{2\nu }\right] \mathrm{d}y}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{
	2\nu }\right] \mathrm{d}y}		2\nu }\right] \mathrm{d}y}
	\end{matrix}</math></center>		\end{matrix}</math></center>
			as <math>\partial_x(f * y)=\partial_x f * y = f * \partial_x y</math>.

	== Lecture Videos ==		== Lecture Videos ==

Meylan: /* Exact Solution of Burgers equations */

2020-10-24T02:14:18Z

Exact Solution of Burgers equations

← Older revision		Revision as of 02:14, 24 October 2020
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	2\nu }\right] \mathrm{d}y}		2\nu }\right] \mathrm{d}y}
	\end{matrix}</math></center>		\end{matrix}</math></center>

			== Lecture Videos ==

			=== Part 1 ===

			{{#ev:youtube\|tVXQmxOG_6Y}}

			=== Part 2 ===

			{{#ev:youtube\|hzgpMM_wWts}}

			=== Part 3 ===

			{{#ev:youtube\|uH4B1XsGB-0}}

			=== Part 4 ===

			{{#ev:youtube\|h6aDmCtJygM}}

			=== Part 5 ===

			{{#ev:youtube\|CsnUKrLjtyQ}}

Meylan: /* Exact Solution of Burgers equations */

2011-10-14T01:09:42Z

Exact Solution of Burgers equations

← Older revision		Revision as of 01:09, 14 October 2011
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	<center><math>		<center><math>
	-2\nu \frac{\partial _{t}\phi }{\phi }=2\nu ^{2}\left( \frac{\partial		-2\nu \frac{\partial _{t}\phi }{\phi }=2\nu ^{2}\left( \frac{\partial
	_{x}\phi }{\phi }\right) ^{2}+\frac{1}{2}\left( 2\nu \frac{\partial _{x}\phi		_{x}\phi }{\phi }\right) ^{2}
			-2\nu^2 \frac{\partial_x^2\phi}{\phi}
			-\frac{1}{2}\left( 2\nu \frac{\partial _{x}\phi
	}{\phi }\right) ^{2}		}{\phi }\right) ^{2}
	</math></center>		</math></center>

Meylan: /* Numerical Solution of Burgers equation */

2011-10-12T22:54:03Z

Numerical Solution of Burgers equation

← Older revision		Revision as of 22:54, 12 October 2011
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	We solve this by solving in Fourier space to give		We solve this by solving in Fourier space to give
	<center><math>		<center><math>
	\partial _{t}\hat{u}=-\frac{1}{2}ik\left( u^{2}\right) -\nu k^{2}\hat{u}		\partial _{t}\hat{u}=-\frac{1}{2}ik \widehat{\left( u^{2}\right)} -\nu k^{2}\hat{u}
	~~</math></center>~~
	~~Then we solve each of the steps in turn to get~~
	~~<center><math>~~
	~~\partial _{t}u=\partial _{x}\left( u^{2}\right)~~
	</math></center>		</math></center>
			Then we solve each of the steps in turn
	for a small time interval to give		for a small time interval to give
	<center><math>\begin{matrix}		<center><math>\begin{matrix}

Meylan: /* Travelling Wave Solution */

2011-10-04T04:16:39Z

Travelling Wave Solution

← Older revision		Revision as of 04:16, 4 October 2011
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	can be integrated to give		can be integrated to give
	<center><math>		<center><math>
	-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u=c_{1}		-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u^{\prime} =c_{1}
	</math></center>		</math></center>
	which can be rearranged to give		which can be rearranged to give
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	</math></center>		</math></center>
	We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu		We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu
	u~~^{\prime }~~-2c_{1}=0</math> by <math>u_{1}</math> and <math>u_{2}</math>		u-2c_{1}=0</math> by <math>u_{1}</math> and <math>u_{2}</math>
	and we assume that <math>u_{2} < u_{1}</math>. Note that there is only a bounded		and we assume that <math>u_{2} < u_{1}</math>. Note that there is only a bounded
	solution if we have two real roots and for the bounded solution		solution if we have two real roots and for the bounded solution
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	is		is
	<center><math>		<center><math>
	v=\frac{1}{2}\left( u_{1}+u_{2}\right)		c=\frac{1}{2}\left( u_{1}+u_{2}\right)
	</math></center>		</math></center>
	The equation can therefore be written as		The equation can therefore be written as

Meylan: /* Travelling Wave Solution */

2011-10-04T04:13:43Z

Travelling Wave Solution

← Older revision		Revision as of 04:13, 4 October 2011
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	This leads to the equations		This leads to the equations
	<center><math>		<center><math>
	-vu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0		-cu^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0
	</math></center>		</math></center>
	We begin by looking at the phase plane for this system, writing <math>w=u^{\prime		We begin by looking at the phase plane for this system, writing <math>w=u^{\prime
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	can be integrated to give		can be integrated to give
	<center><math>		<center><math>
	-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u~~^{\prime }~~=c_{1}		-cu+\frac{1}{2}\left( u\right) ^{2}-\nu u=c_{1}
	</math></center>		</math></center>
	which can be rearranged to give		which can be rearranged to give
	<center><math>		<center><math>
	u^{\prime }=\frac{1}{2\nu }\left( \left( u\right) ^{2}-2cu~~^{\prime~~		u^{\prime }=\frac{1}{2\nu }\left( \left( u\right) ^{2}-2cu-2c_{1}\right)
	}-2c_{1}\right)
	</math></center>		</math></center>
	We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu		We define the two roots of the quadratic <math>\left( u\right) ^{2}-2\nu

Adi Kurniawan: /* Exact Solution of Burgers equations */

2010-11-06T10:23:40Z

Exact Solution of Burgers equations

← Older revision		Revision as of 10:23, 6 November 2010
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	To find <math>u</math> we recall that		To find <math>u</math> we recall that
	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	u\left( x,t\right) &=&-2\nu \~~frac~~{\partial _{x}\phi \left( x,t\right) }{\phi		u\left( x,t\right) &=&-2\nu \dfrac{\partial _{x}\phi \left( x,t\right) }{\phi
	\left( x,t\right) } \\		\left( x,t\right) } \\
	&=&\~~frac~~{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -		&=&\dfrac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -
	\~~frac~~{f}{2\nu }\right] \mathrm{d}y}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{		\dfrac{f}{2\nu }\right] \mathrm{d}y}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{
	2\nu }\right] \mathrm{d}y}		2\nu }\right] \mathrm{d}y}
	\end{matrix}</math></center>		\end{matrix}</math></center>

Adi Kurniawan: /* Exact Solution of Burgers equations */

2010-11-06T10:22:33Z

Exact Solution of Burgers equations

← Older revision		Revision as of 10:22, 6 November 2010
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	We can write this as		We can write this as
	<center><math>		<center><math>
	\frac{d}{dx}\left( \log \left( \phi \right) \right) =-\frac{1}{2\nu }F\left(		\frac{\mathrm{d}}{\mathrm{d}x}\left( \log \left( \phi \right) \right) =-\frac{1}{2\nu }F\left(
	x\right)		x\right)
	</math></center>		</math></center>
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	<center><math>		<center><math>
	\phi \left( x,0\right) =\Phi \left( x\right) =\exp \left( -\frac{1}{2\nu }		\phi \left( x,0\right) =\Phi \left( x\right) =\exp \left( -\frac{1}{2\nu }
	\int_{0}^{x}F\left( s\right) ds\right)		\int_{0}^{x}F\left( s\right) \mathrm{d}s\right)
	</math></center>		</math></center>
	We need to solve		We need to solve
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	e^{-k^{2}\nu t}\right] \\		e^{-k^{2}\nu t}\right] \\
	&=&\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)		&=&\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty }\Phi \left( y\right)
	\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] dy		\exp \left[ -\frac{\left( x-y\right) ^{2}}{4\nu t}\right] \mathrm{d}y
	\end{matrix}</math></center>		\end{matrix}</math></center>
	Which can be expressed as		Which can be expressed as
	<center><math>		<center><math>
	\phi \left( x,t\right) =\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty		\phi \left( x,t\right) =\frac{1}{2\sqrt{\pi \nu t}}\int_{-\infty }^{\infty
	}\exp \left[ -\frac{f}{2\nu }\right] dy		}\exp \left[ -\frac{f}{2\nu }\right] \mathrm{d}y
	</math></center>		</math></center>
	where		where
	<center><math>		<center><math>
	f\left( x,y,t\right) =\frac{1}{2\nu }\int_{0}^{y}F\left( s\right) ds+\frac{		f\left( x,y,t\right) =\frac{1}{2\nu }\int_{0}^{y}F\left( s\right) \mathrm{d}s+\frac{
	\left( x-y\right) ^{2}}{2t}		\left( x-y\right) ^{2}}{2t}
	</math></center>		</math></center>
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	\left( x,t\right) } \\		\left( x,t\right) } \\
	&=&\frac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -		&=&\frac{\int_{-\infty }^{\infty }\left( \frac{x-y}{t}\right) \exp \left[ -
	\frac{f}{2\nu }\right] dy}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{		\frac{f}{2\nu }\right] \mathrm{d}y}{\int_{-\infty }^{\infty }\exp \left[ -\frac{f}{
	2\nu }\right] dy}		2\nu }\right] \mathrm{d}y}
	\end{matrix}</math></center>		\end{matrix}</math></center>

Adi Kurniawan: /* Travelling Wave Solution */

2010-11-06T10:21:18Z

Travelling Wave Solution

← Older revision		Revision as of 10:21, 6 November 2010
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	}</math> so that		}</math> so that
	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	\~~frac~~{du}{d\xi } &=&w \\		\dfrac{\mathrm{d}u}{\mathrm{d}\xi } &=&w \\
	\~~frac~~{dw}{d\xi } &=&\frac{1}{\nu }\left( w\left( u-c\right) \right)		\dfrac{\mathrm{d}w}{\mathrm{d}\xi } &=&\frac{1}{\nu }\left( w\left( u-c\right) \right)
	\end{matrix}</math></center>		\end{matrix}</math></center>
	This is a degenerate system with the entire <math>u</math> axis being equilibria.		This is a degenerate system with the entire <math>u</math> axis being equilibria.