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Method of characteristics. pde. limits of integral question

  • Thread starter binbagsss
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  • #1
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I'm using the method of characteristics to solve a pde of the from ## au_{x}+bu_{y}=c##

where ## a=\frac{dx}{d \tau} , b= a=\frac{dy}{d \tau}, c=a=\frac{du}{d \tau}##

where initial data is parameterised by ##s## and initial curve given by ##x( \tau)=x_{0}(s)##, ##y( \tau)=y_{0}(s)## and ##u( \tau)=u_{0}(s)## at ## \tau = 0## .


Question:

Questions been given so far on my course have ## c ## given explicitly. I'm trying to solve:
##u_{x} + u_{y} =f(x,y)##. subject to initial data ##u=g(y)## on ##x=0##, functions ##f## and ##g## given.

Attempt:

Solving ## \frac{du}{d\tau} = f(x,y)##
i get ##u= \int f(x(\tau),y(\tau))d\tau+ u_{0}.##

QUESTION: In the solution the limits of this integral run from ##\tau## to 0. I'm not used to doing these with limits, and if I'm honest, I'm totally clueless where this comes from... can someone please explain the integral limits to me? usually when it's just given explicitly I just integrate.

So the solution has : ##u= \int^{\tau} _{0} f(x(\tau '),y(\tau ')) d\tau '+ u_{0}##

Many thanks in advance.
 
Last edited:

Answers and Replies

  • #2
RUber
Homework Helper
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Your solution is in the indefinite form. If you wanted to find ##u(t)## you would need limits on your integral. The solution you posted is for ##u(\tau)##.
 
  • #3
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Let du=uxdx+uydy. From this, it follows that
[tex]\frac{du}{dt}=u_x\frac{dx}{dt}+u_y\frac{dy}{dt}=V_xu_x+V_yu_y[/tex]
Think of Vx and Vy as the x and y components of velocity, and think of u as traveling along with this velocity. So du/dt is the rate of change of u moving along with velocity components Vx and Vy.

So, in your problem,

##\frac{dx}{dt}=a##

##\frac{dy}{dt}=b##

and, from your differential equation ##\frac{du}{dt}=c##

Hope this helps.

Chet
 

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