
#1
Feb909, 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 :( 



#2
Feb1309, 01:26 PM

Sci Advisor
P: 1,562

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 onevariable 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 + (1a)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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