Difference between revisions of "Method of Characteristics for Linear Equations"

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<center>
 
<center>
 
<math>
 
<math>
\frac{d U}{d t} = \partial_t u + \frac{d X}{dt}\partial_x u = \partial_x u \left(\frac{d X}{dt} + 1 \right)
+
\frac{d U}{d t} = \partial_t u + \frac{d X}{dt}\partial_x u = \partial_x u \left(\frac{d X}{dt} - 1 \right)
 
</math></center>
 
</math></center>
 +
 +
Therefore along the curve <math>\frac{d X}{dt} = 1</math> <math>u(x,t)</math> must be a constant.
 +
These are nothing but the straight lines <math>x = t+c</math>
 +
This means that we have
 +
<center>
 +
<math>
 +
u(x,t) = u(-t+c,t) = u(c,0) = f(c) = f(x+t)
 +
</math></center>
 +
Therefore the solution is given

Revision as of 04:30, 23 July 2010

Nonlinear PDE's Course
Current Topic Method of Characteristics for Linear Equations
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We present here a brief account of the method of characteristic for linear waves.

Introduction

The method of characteristics is an important method for hyperbolic PDE's which applies to both linear and nonlinear equations.

We begin with the simplest wave equation

[math]\displaystyle{ \partial_t u + \partial_x u = 0,\,\,-\infty\lt x\lt \infty,\,\,t\gt 0, }[/math]

subject to the initial conditions

[math]\displaystyle{ \left. u \right|_{t=0} = f(x) }[/math]

We consider the solution along the curve [math]\displaystyle{ (x,t) = (X(t),t) }[/math]. We then have

[math]\displaystyle{ \frac{d U}{d t} = \partial_t u + \frac{d X}{dt}\partial_x u = \partial_x u \left(\frac{d X}{dt} - 1 \right) }[/math]

Therefore along the curve [math]\displaystyle{ \frac{d X}{dt} = 1 }[/math] [math]\displaystyle{ u(x,t) }[/math] must be a constant. These are nothing but the straight lines [math]\displaystyle{ x = t+c }[/math] This means that we have

[math]\displaystyle{ u(x,t) = u(-t+c,t) = u(c,0) = f(c) = f(x+t) }[/math]

Therefore the solution is given