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Homework Help: Seperation of variables

  1. Apr 25, 2006 #1
    This is the the first time I've encountered seperation with partial differential equations. There are no worked examples, so I need some help to work through this problem. The question seems to be somewhat hand holding, since it seems to be THE introduction.

    Q: Apply seperation of variables [itex] u_t = u_x [/itex] by substituting [itex] u=A(x)B(t) [/itex] and then dividing by AB. If one side depends only on [itex] t [/itex] and the other only on [itex] x [/itex], they must equal a constant [itex] k [/itex]; what are [itex] A [/itex] and [itex] B [/itex]?

    [tex] \frac{\partial u}{\partial t}-\frac{\partial u}{\partial x} = 0 [/tex]

    [tex] u = A(x)B(t) [/tex]

    [tex] \frac{\partial}{\partial t} \left[ A(x)B(t) \right] - \frac{\partial}{\partial x} \left[ A(x)B(t) \right] = 0 [/tex]

    [tex] A(x)B'(t)-A'(x)B(t)=0[/tex]

    [tex] \frac{A(x)B'(t)-A'(x)B(t)}{A(x)B(t)} [/tex]

    [tex] \frac{B'(t)}{B(t)}-\frac{A'(x)}{A(x)}=0[/tex]

    Now I was reading on various websites, that I can set each independent term equal to seperation constants to make two coupled (is this the proper word to use?) differential equations. I don't understand where this step comes from.

    but...

    [tex] \frac{B'(t)}{B(t)}=k[/tex]

    [tex] \frac{A'(x)}{A(x)}=k[/tex]


    Now solving for [itex] A(x) [/itex] and [itex] B(t) [/itex]. I'm a little rusty here, so I don't know if this part is correct.

    Rewriting the two equations above in Leibniz notation

    [tex] \frac{dB(t)}{dt} \cdot \frac{1}{B(t)} = k [/tex]

    Seperating:

    [tex] \frac{dB(t)}{B(t)} = k dt [/tex]

    [tex] \int \frac{dB(t)}{B(t)} = \int k\,\,dt [/tex]

    [tex] \ln B(t) = kt +c [/tex]

    [tex] B(t) = e^{kt+c} [/tex]

    And subsequently:

    [tex] A(x) = e^{kx+c} [/tex]

    Does this make sense? :)
    Thanks in advance.
     
    Last edited: Apr 25, 2006
  2. jcsd
  3. Apr 25, 2006 #2
    The part where I said I wasn't unsure. Is this line of reasoning correct?

    [tex]\frac{B'(t)}{B(t)}-\frac{A'(x)}{A(x)}=0[/tex]

    Since the "A term" is equal to the "B term" for some value of x,t we can seperate them as a system of equations. Thus, allowing the ODE to be solved with seperation of variables.
     
    Last edited: Apr 25, 2006
  4. Apr 26, 2006 #3

    HallsofIvy

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    Science Advisor

    More to the point, "A term" is equal to the "B term" for all values of x,t, so we can separate them.

    Write this as
    [tex]\frac{B'(t)}{B(t)}=\frac{A'(x)}{A(x)}[/tex]
    Fix x and change t. The left side of the equation does not change because x does not change. But the equation is valid for all x and t- therefore the right side must not have changed even though t changes:
    we must have
    [tex]\frac{B'(t)}{B(t)}= k[/tex]
    for some constant k. Now let x change while t is fixed. We get
    [tex]\frac{A'(x)}{A(x)}= k[/tex].
    Of course, since they are equal, they must be equal to the same constant.

    You have the equations
    [tex]\frac{dB}{B}= kdt[/tex]
    and
    [tex]\frac{dA}{A}= kdt[/tex] so
    as, you had before,
    [tex]B(t)= e^{kt}+ c[/tex]
    which you can rewrite as
    [tex]B(t)= Ce^{kt}[/tex]
    and
    [tex]A(x)= De^{kx}[/tex]
    (be careful to use different symbols for the "undetermined constant" in each- they are not necessarily the same)

    Finally, since you started by assuming that u(x,y)= A(x)B(t),
    a solution to the differential equation is u(x,y)= Cekxekt= Cek(x+t). (The C here is the product of the constants C and D.) Of course, nothing has been said about what k might be. That would be determined by the additional requirements.
     
  5. Apr 27, 2006 #4
    Could you be more awesome?
    no... no you can't. Thank you!

    It's always fun when something makes sense.
     
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