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Homework Help: 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


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    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


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    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]
    [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]
    [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.
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