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I can't find anywhere how to solute PDE's.

by lukaszh
Tags: solute
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lukaszh
#1
Feb9-09, 09:21 AM
P: 32
I can't find anywhere how to solute PDE's. For exaple ODE:
[tex]\frac{du}{dt}=u\Rightarrow\ln u=t+\ln C\Rightarrow u=Ce^t[/tex]
But this?
[tex]\frac{du}{dt}+\frac{du}{dx}=u[/tex]

[tex]dudx+dudt=udtdx\Rightarrow udx+udt=udtdx[/tex]

I haven't had pde's yet, but i'm interested in solving these equations :-(
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Ben Niehoff
#2
Feb13-09, 01:26 PM
Sci Advisor
PF Gold
P: 1,594
For many equations, there is a technique that works called "separation of variables". To separate the equation, we assume u(x,t) can be factored into two one-variable functions:

[tex]u(x,t) = X(x)T(t)[/tex]

And then

[tex]\frac{\partial u}{\partial x} = T(t)\frac{dX(x)}{dx}[/tex]

[tex]\frac{\partial u}{\partial t} = X(x)\frac{dT(t)}{dt}[/tex]

Then, for your equation, we can write:

[tex]X(x)\frac{dT(t)}{dt} + T(t)\frac{dX(x)}{dx} = X(x)T(t)[/tex]

Next, divide by X(x)T(t):

[tex]\frac{1}{T(t)}\frac{dT(t)}{dt} + \frac{1}{X(x)}\frac{dX(x)}{dx} = 1[/tex]

Now, notice that we've separated the variables into the form

[tex]P(t) + Q(x) = 1[/tex]

Since this must be true for all x and t, the functions P and Q must individually be constant. This gives

[tex]a + b = 1[/tex]

[tex]\frac{1}{T(t)}\frac{dT(t)}{dt} = a[/tex]

[tex]\frac{1}{X(x)}\frac{dX(x)}{dx} = b[/tex]

These have the solution

[tex]X(x) = e^{bx}[/tex]

[tex]T(t) = e^{at}[/tex]

and so

[tex]u(x,t) = Ce^{at}e^{bx} = Ce^{at + (1-a)x}[/tex]

for arbitrary constants C and a. Also note that any linear combination of u's (with different values for a) are also solutions, because the equation is linear. This can be used to construct a more general solution, given some boundary conditions.


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