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Apr25-06, 05:40 PM
P: 837
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.


[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]


[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.
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