Re: numerical solution to a nonlinear 2nd order PDE
- From: sbh <steve.sbh77@xxxxxxxxx>
- Date: Fri, 18 Jul 2008 13:19:35 -0700 (PDT)
On Jul 18, 6:35 am, Torsten Hennig <Torsten.Hen...@xxxxxxxxxxxxxx>
wrote:
Hello all,
I am trying to solve this nonlinear 2nd order, 1-D
PDE (independent
variables: time and space) using the method of lines,
numerically. I
have as the initial conditions all zeros at the
initial time. The
left boundary (at zero) has the Dirichlet boundary
condition of its
value being zero while its Neumann boundary condition
is a number that
results from a function of the independent variable
(time). This is
where I become a little confused. In school we were
always taught
that 2nd order needed two boundary conditions and
that is it..... I
know the real solution should approach a flat line in
the far-field
and therefore I know that in the far-field the first
and second
derivatives will be zero. Should I try to enforce
these far-field
boundary conditions?
Thanks much for any advice out there.
sbh
Could you be more specific and give us the equation
of your PDE ?
Best wishes
Torsten.
Sure. Thanks for your interest.
Here is the PDE: (z = z(x,t))
z_t = A*sqrt{1 + ( tan(phi(t)) - z_x )^2 }
+ B*z_xx*(1/( 1 + (tan(phi(t)) - z_x)^2 ))
+ (B/r)*(sin(phi(t)) + cos(phi(t))*[ z_x - tan(phi(t)) ] )
- (x+R)*phi_t
- (z_x - tan(phi(t))*(z - x*tan(phi(t)))*phi_t
+ x*phi_t*(1/(cos(phi(t))*cos(phi(t)))
where r = (R+x)*cos(phi(t)) - sin(phi(t))*(z - x*tan(phi(t)))
and phi_t is the time derivative of phi, z_x is the spatial first
derivative of x, etc. A and B are constants.
I also know phi(t) = arcsin((A*t - R)/R)
The initial conditions are zero for each ODE in the method of lines.
The left edge of the domain (x = 0) has the Dirichlet boundary
condition of zero and the Neumann boundary condition of tan(phi(t))+C
(where C is a constant)
I hope this is easier to read than it was to write!
Thanks again,
sbh
.
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