How to express a function both time and spatial dimension dependent?

Hi there,
It seems an easy one but I guess it is not that easy. I have a fluid flowing through a vessel. I want to measure its velocity u1 and u2. the vessel coordinetes from the lateral position x=0 to x=1 in time from t=0 to t=3. The code is :
ii=101;
dt=0.1;
t = 0:dt:10;
dx=0.01; %step size
fik=0.4;
for i=1:ii;
x=(i-1)*dx;
H1D=0.1;
H1=0;
A=0;
AD=0.1;
fikness=fik*sin(pi*x);
ub1(i)=(c1b-H1D*(x-0.5)+AD/2*(x-0.5)^2)/(H1-0.5*fikness-A*(x-0.5));
ub2(i)=(c2b+H1D*(x-0.5)-AD/2*(x-0.5)^2)/(1-H1+0.5*fikness+A*(x-0.5));
end
H1D=H1D+dt*H1DDOT %------lets say these H1D, H1DDOT, H1 are time dependent functions.
H1=H1+dt*H1
AD=AD+dt*ADDOT;
A=A+dt*AD
Matlab runs the program above but it is not showing time. For example if I want to see in time t=2 what are the velocities u1 and u2 over the whole vessel (from x=0 to x=1)? How can I plot and see the results Thanks in advance

Answers (1)

Well, your time is stored in vector t. So lets say you want to plot ub1 versus time, it will be:
plot(t,ub1);

3 Comments

I think there is something wrong in this code too. You are not looping through time actually. I think this code is done to get velocities at t = 0.1 sec. You did not go further.
You have to loop through time as well.
Just a piece of advice: Matlab have good numerical integrators like ode45 which is more stable that simple euler forward integration that you're doing here.
plot (t,ub1) is not what I want since x axes must be x(lateral position) y axes must be ub1 and ub2 at a special time say t=1 or t=2. I mean we want to see the velocities versus x (lateral position) at time 1st second or 2nd second like that. Thanks a lot.
Like I said, you have done your calculation for the first time step only. You need to go through more time loops.

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Asked:

on 27 Jan 2014

Commented:

on 27 Jan 2014

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