Time dependent domain

[kleinwolf]

How to compute the solution of a partial differential equation problem of the type :

 

 

 

a) [math] \frac{\partial^2 f}{\partial x^2}(x,t)=D\frac{\partial f}{\partial t}(x,t)[/math]

 

D is a non vanishing constant and

 

b) an initial condition, for example [math] f(x,t=0)=\delta(x)[/math]

 

c) a boundary condition [math]f(at,t)=f(-at,t)[/math] with 'a' nonzero

 

it can be seen that

 

* separation of variables does not work because of the time-dependence

of the boundary

* pseudo spectral method fails due to the non-orthogonality of the eigenvectors of the space-operator and their time derivatives.

 

Is there another way to solve that kind of differential problem ?

 

thx

[Bignose]

These kind of problems are made for separation of variables:

 

Let [math]f(x,t) = X(x)T(t)[/math]

 

Note that we are letting the unknown function, f, be a product of two unknown functions, X & T, and that X(x) is only a function of x, and T(t) is only a function of t.

 

Now, put this into the eqn:

 

(I'll do the RHS, you can do the LHS)

 

[math]D\frac{\partial f(x,t)}{\partial t} = D\frac{\partial X(x)T(t)}{\partial t} = D X(x) \frac{\partial T(t)}{\partial t}[/math]

 

The X(x) part comes out, just like the T(t) part will come out of the left hand side.

 

Now, divide the equation by X(x) and T(t) and you should get

 

LHS that has only derivatives and functions only of x = [math]\frac{D}{T(t)} \frac{\partial T(t)}{\partial t}[/math]

 

Now, since the LHS is only a function of x and the RHS is only a function of t, the only way that is possible is if both sides are equal to a constant, call it C.

 

LHS = C = RHS

 

This actually is now two ODEs (LHS=C & C=RHS) which hopefully are easier to solve. The constant is determined by boundary conditions and initial conditions.

[kleinwolf]

This is right, a often made supposition on the solution. Let see the next steps :

 

 

Now take the space-dependent part : [math] X''(x)=C*X(x) [/math]

 

with C the given constant. (indep. of x and t)

 

But the boundary condition is "time-dependent"

 

and X(x) should be a function of time too.

 

hence a contradiction, and we deduce the separation Ansatz is not the right hypothesis on the solution.

 

Note :

 

for example that the brownian motion, which is a solution of parabolic problem, is not separable (gaussian of the type exp(x²/t)<>X(x)*T(t))

 

The general solution Ansatz, which is just a change of basis (Fourier basis, aso.) is just using the linearity of the equation :

 

[math] f(x,t)=\sum_n X_n(x)T_n(t) [/math] with [math] X_n [/math] basis functions, often the eigenvectors of the space-operator. The time-dependent part is found out of projection on [math] X_m(x)[/math].

 

However here the BC (boundary condition) makes that [math] X_n=X_n(x,t)[/math].

Edited December 1, 2008 by kleinwolf