Prev: Reading and modifying XML file using MATLAB
Next: Change size of Principal Component Visualization Tool (mapcaplot)
From: Kamel on 19 Apr 2010 06:41
Thank you Tortsen for your help.
By the way, I found also an other way to resolve this problem using one variable, u.
I derived manually the term dsig/dY to obtain an expression depending only on DuDz to form the expression of 1/H.
And for the left boundary condition, I wrote Y=Y-1^(sig0) as 1-Y-1^(sig0)+DuDz=0, Y-1^(sig0) being resolved beforehand, and it works.

c = 1/H;
f = DuDx;
s = 0;

BC:
pl = ul;
ql = 0;
pr = 1-Y-1^(sig0);
qr = 1;



Regards,

Kamel



Torsten Hennig <Torsten.Hennig(a)umsicht.fhg.de> wrote in message <1191033208.35186.1271658944531.JavaMail.root(a)gallium.mathforum.org>...
> > Dear Torsten,
> >
> > Resolution of the nonlinear equation is now easy to
> > do.
> > I have just a small trouble with one of the boundary
> > cdt using pdepe.
> >
> > I wrote the equations like this for the resolution
> > using pdepe:
> > (u1=u, u2=Y)
> >
> > (1/H, 0)*d/dt(u1, u2)=d/dz*(du1/dz, u1)+(0, 1-u2)
> > and the pdepe coefficients are normally:
> > c = [1/H; 0];
> > f = [DuDz; u] ;
> > s = [0; 1-u2];
> >
> > What I am suggesting for the boundary conditions is
> > left:
> > (u1, u2-1) + (0, -du1/dz*(du2/dz)^-1)*(du1/dz,du2/dz)
> > = (0, 0)
> > right:
> > (u2-u1-1, u2-u2^-1(sig0)) + (0, 0)*(du1/dz,du2/dz) =
> > (0, 0)
> >
> > which should lead to pdex1bc coefficients in the
> > following form:
> > pl = [ul(1); ul(2)-1];
> > ql = [0; ??];
> > pr = [ur(2)-ur(1)-1; ur(2)-ur2^-1(sig0)];
> > qr = [0; 0];
> >
> > My question is: how can I write the second ql
> > condition (in ??) corresponding to the term
> > -du1/dz*(du2/dz)^-1 ?
> >
> > Thank you in advance for your precious help.
> >
> > Best regards,
> >
> > Kamel
> >
> >
>
> I thought to take
>
> c = [1/H; 0];
> f = [DuDz(1); 0] ;
> s = [0; Dudz(1)+1-u2];
>
> pl = [1-ul(2);ul(1)];
> ql = [1 ; 0];
> pr = [1-ur(2);ur(2)-u(2)^(-1)(sig0)];
> qr = [1 ; 0];
>
> A problem may arise at the right boundary because
> at t=0, du/dz=0 at z=h, but Y=Y^(-1)(sig0) <> 1 in
> general.
> Maybe you can adjust the initial value for u
> at z=h such that du/dz|z=h = Y^(-1)(sig0) - 1.
>
> Best wishes
> Torsten.
First  |  Prev  | 
Pages: 1 2 3
Prev: Reading and modifying XML file using MATLAB
Next: Change size of Principal Component Visualization Tool (mapcaplot)