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

[lukaszh]

I can't find anywhere how to solute PDE's. For exaple ODE:
\frac{du}{dt}=u\Rightarrow\ln u=t+\ln C\Rightarrow u=Ce^t
But this?
\frac{du}{dt}+\frac{du}{dx}=u

dudx+dudt=udtdx\Rightarrow udx+udt=udtdx

I haven't had pde's yet, but I'm interested in solving these equations :-(

 

[Ben Niehoff]

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:

u(x,t) = X(x)T(t)

And then

\frac{\partial u}{\partial x} = T(t)\frac{dX(x)}{dx}

\frac{\partial u}{\partial t} = X(x)\frac{dT(t)}{dt}

Then, for your equation, we can write:

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

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

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

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

P(t) + Q(x) = 1

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

a + b = 1

\frac{1}{T(t)}\frac{dT(t)}{dt} = a

\frac{1}{X(x)}\frac{dX(x)}{dx} = b

These have the solution

X(x) = e^{bx}

T(t) = e^{at}

and so

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

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.