|April 28, 2004, 13:31||
Question on Hilbert Transform
Have you ever tried to use Hilbert Transform to solve PDE/ODE?
For example for an ocean wave bottom boundary layer problem we got equations like this:
d(u-U)/dt = nu d^2(u-U)/dz^2
with BC: u=0 at z=0, u=U at z--> infinity
where u=u(x,z,t), U=U(x,t), and x is the direction of wave propagation. And U(x,t) is velocity outside boundary layer which can be obtained from wave models, nu is kinematic viscosity=const for laminar problem.
Let's say U=Um*cos(wt-kx), and w being angular frequency, k being wave number, it's a wave propagating in +x direction.
To solve the upper problem we can't use separation of variables easily. But when we use complex variables, we can!!
Let v=u-U to be defect velocity, and let v= Re(Vc) which indroduces a complex velocity Vc. And let U=Re(Uc)=Re[Um*cos(wt-kx)+ i Um*sin(wt-kx)]=Re[Um*exp(i *theta)]with theta =wt-kx the phase angle.
Now solve the equation for Vc:
with BC: Vc=0 at z-->infinity, Vc=-Uc at z=0.
Separation of variables Vc(z,t)= V1(z)exp(i*theta) we find nicely that
where beta=sqrt(w/2/nu) , and further find results:
From the upper result of v, we find that there is no way we can seperate variable z and t(in theta) in real space. But we did it nicely in complex space.
Now, my real question being, what if the velocity outside boundary U(x,t) is arbitrary and I don't want to use Fourier transform to get U in terms of sum of sin and cos functions? Do i still have the trick of usign complex velocity? For U=Um*cos(theta) we find its complex counterpart easily, what does it look like for arbitrary U? Then I thought about Hilbert Transform, but I have never seen Hilbert Transform in solving ODE/PDE,
can you guys point me any information?
|April 29, 2004, 05:58||
Re: Question on Hilbert Transform
Since your domain of definition is z>0 (semi-infinite) you can't use a Hilbert transform since you then need the problem defined for all z<>0.
The obvious way to solve this is to take the Fourier cosine transform sovle the resulting ode and then invert the transform (you also need to specify your initial condition at t=0). The result you've written down can be obtained in this way after letting t->infinity to remove the transient contribution ( Your solution is simply that of the classical Stokes layer which can be found in many books).
If w(k,t) is the Fourier transform of v(z,t) and Q(k,t) is the FT of U(x,t) then the solution with v=0 at t=0 is
w(k,t) = -Q(k,t)( 1 - exp(-nu.k^2.t) )
and inverting the transform gives
v(x,t) = 2/pi int_0^\infty w(k,t)cos(kz).dk
Note I haven't used complex numbers!
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