1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Use separation of variables to find particular solutions

  1. Oct 15, 2006 #1
    "Use separation of variables to find particular solutions of

    [tex]u_t-u_{xx}-2u_x=0, 1<x<2, 0<t, u(1,t)=u(2,t)=0[/tex]

    hint: change coordinates"

    I can't find the solution. The equation seems already separated, so all I need to do is to find a change of variables, I think. But I can't find one that works.
     
  2. jcsd
  3. Oct 15, 2006 #2

    Fermat

    User Avatar
    Homework Helper

    You're looking at it the wrong way.
    It's not a change of variables that's wanted, but separation of variables.

    Remember this ?

    [tex]\frac{dy}{dx} = \frac{y}{x}[/tex]

    separate the variables

    [tex]\frac{dy}{y} = \frac{dx}{x}[/tex]

    now integrate
    [tex]\int\frac{dy}{y} = \int\frac{dx}{x}[/tex]
    [tex]ln(y) = ln(x) + c[/tex]

    You are expected to do something similar.

    First of all, you assume a solution of the form u(x,t) = f(x)g(t)
    Now substitute that into your original pde and apply separation of variables.
    You should end up with functions of x on one side of the equals sign and functions of t on the other side.
    Since the lhs is a function of x and the rhs is a function of t, then the only way they (the two functions) can be equal to each other is to set them both equal to the same constant value. i.e. lhs = C and rhs = C
    You now have two independent pdes, (in fact, now they are ode's) one in x and one in t, that you can solve separately.
     
    Last edited: Oct 15, 2006
  4. Oct 16, 2006 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Then you need to review what "separation of variables" means for partial differential equations.

    Write u(x,t)= X(x)T(t). That is, X is a function of x only, T a function of t only. The the differential equation becomes
    XT'- TX"- 2TX'= 0. Divide both sides by XT to get
    [tex]\frac{T'}{T}- \frac{X"}{X}- \frac{2X'}{X}= 0[/tex]
    or
    [tex]\frac{T'}{T}= \frac{X"- 2X'}{X}[/tex]
    Since the left side depends on t only and the right side depends on x only the only way they can be equal (for all x and t) is if each side is a constant:
    [tex]\frac{T'}{T}= \alpha[/tex]
    so that [itex]T'= \alpha T[/itex]
    and
    [tex]\frac{X"- 2X'}= \alpha[/tex]
    so that [itex]X"- 2X'= \alpha X[/itex]
    where [itex]\alpha[/itex] is some unknown constant. The general solution will typically be a sum of such things.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Use separation of variables to find particular solutions
Loading...