Method of Characteristics for Linear Equations - Revision history

Levi: /* Example 1 */

2025-11-11T11:52:24Z

Example 1

← Older revision		Revision as of 11:52, 11 November 2025
Line 81:		Line 81:
	</math></center>		</math></center>

	Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = x</math> <math>u(x,t)</math> must be a constant.		Therefore along the curve <math>\frac{\mathrm{d} X}{\mathrm{d}t} = X</math> <math>u(x,t)</math> must be a constant.
	These are the curves <math>x = ce^t</math>		These are the curves <math>x = ce^t</math>
	This means that we have		This means that we have

2025-11-11T11:52:07Z

General Form

← Older revision		Revision as of 11:52, 11 November 2025
Line 51:		Line 51:
	<center>		<center>
	<math>		<math>
	\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - a(x,t) \right).		\frac{\mathrm{d} }{\mathrm{d} t} u(X(t),t) = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d} X}{\mathrm{d}t} - a(X,t) \right).
	</math>		</math>
	</center>		</center>
Line 57:		Line 57:
	<center>		<center>
	<math>		<math>
	\frac{\mathrm{d} X}{\mathrm{d} t} = a(x,t) .		\frac{\mathrm{d} X}{\mathrm{d} t} = a(X,t) .
	</math>		</math>
	</center>		</center>

2020-08-07T05:25:06Z

Part 1

2020-08-07T05:11:51Z

Non-homogeneous Example

← Older revision		Revision as of 05:11, 7 August 2020
Line 145:		Line 145:
	\frac{\mathrm{d}}{\mathrm{d}t}u(x,t) = xt = t^3/2 + c t		\frac{\mathrm{d}}{\mathrm{d}t}u(x,t) = xt = t^3/2 + c t
	</math>		</math>


	</center>		</center>
	Therefore		Therefore
Line 160:		Line 162:
	Therefore the solution is given <math>u(x,t) = t^4/8 + (x -t^2/2) t^2/2 + f(x-t^2/2)\,</math> or		Therefore the solution is given <math>u(x,t) = t^4/8 + (x -t^2/2) t^2/2 + f(x-t^2/2)\,</math> or
	<math>u(x,t) = -t^4/8 + x t^2/2 + f(x-t^2/2)\,</math>		<math>u(x,t) = -t^4/8 + x t^2/2 + f(x-t^2/2)\,</math>

			== Lecture Videos ==

			=== Part 1 ===

			{{#ev:youtube\|vRRVU4VrxV0}}

2017-11-03T00:47:38Z

Non-homogeneous Example

← Older revision		Revision as of 00:47, 3 November 2017
Line 126:		Line 126:
	<center>		<center>
	<math>		<math>
	\partial_t u + t \partial_x u = xt,\,\,-\infty<x<\infty,\,\,t>0,		\partial_t u + t \partial_x u = xt,\,\,-\infty < x < \infty,\,\,t>0,
	</math></center>		</math></center>
	subject to the initial conditions		subject to the initial conditions

2017-11-03T00:47:21Z

Example 2

← Older revision		Revision as of 00:47, 3 November 2017
Line 97:		Line 97:
	<center>		<center>
	<math>		<math>
	\partial_t u + t \partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,		\partial_t u + t \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
	</math></center>		</math></center>
	subject to the initial conditions		subject to the initial conditions

2017-11-03T00:47:01Z

Example 1

← Older revision		Revision as of 00:47, 3 November 2017
Line 67:		Line 67:
	<center>		<center>
	<math>		<math>
	\partial_t u + x \partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,		\partial_t u + x \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
	</math></center>		</math></center>
	subject to the initial conditions		subject to the initial conditions

2017-11-03T00:46:44Z

General Form

← Older revision		Revision as of 00:46, 3 November 2017
Line 45:		Line 45:
	<center>		<center>
	<math>		<math>
	\partial_t u + a(x,t)\partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,		\partial_t u + a(x,t)\partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
	</math></center>		</math></center>
	then we can apply the method of characteristics.		then we can apply the method of characteristics.

2017-11-03T00:46:23Z

Introduction

← Older revision		Revision as of 00:46, 3 November 2017
Line 17:		Line 17:
	<center>		<center>
	<math>		<math>
	\partial_t u + \partial_x u = 0,\,\,-\infty<x<\infty,\,\,t>0,		\partial_t u + \partial_x u = 0,\,\,-\infty < x < \infty,\,\,t>0,
	</math></center>		</math></center>
	subject to the initial conditions		subject to the initial conditions

2011-09-13T00:55:48Z

Non-homogeneous Example

← Older revision		Revision as of 00:55, 13 September 2011
Line 137:		Line 137:
	<center>		<center>
	<math>		<math>
	\frac{\mathrm{d} U}{\mathrm{d} t} = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d}X}{\mathrm{d}t} - t \right) + xt		\frac{\mathrm{d} u}{\mathrm{d} t} = \partial_t u + \frac{\mathrm{d} X}{\mathrm{d}t}\partial_x u = \partial_x u \left(\frac{\mathrm{d}X}{\mathrm{d}t} - t \right) + xt
	</math></center>		</math></center>