Thread Subject:

HOw to solve these type of equations

Hi

I have a three deferential equations as

y1' = .......
y2'=.....
0=..... + X(t) in the third equation i dont have y3' term and more over I have the extra term which varies with time . I know how to solve these if there is y3' and no X(t) by using ODE 23 , but how to do this ?

Thank You

On Aug 4, 6:47am, "Dinesh " <revooridin...@gmail.com> wrote:
> Hi
>
> I have a three deferential equations as
>
> y1' = .......
> y2'=.....
> 0=..... + X(t) in the third equation i dont have y3' term and more over I have the extra term which varies with time . I know how to solve these if there is y3' and no X(t) by using ODE 23 , but how to do this ?
>
> Thank You

if no y3' then it not differential equation but algebra like in
school. just move the known to the left and unknown to the right to
solve and for the first 2 differentials use the matlab od54 functions
but need initial conditions

 
> if no y3' then it not differential equation but algebra like in
> school. just move the known to the left and unknown to the right to
> solve and for the first 2 differentials use the matlab od54 functions
> but need initial conditions

Thank you for the reply
the three equations are functions of y1,y2,y3 and i know the values before time 0-

my original equations are C1*y1' = F*C2*y2 - C3*y1
y2' = C4 -B* C5*y3 - C6*y1
y2-C7=x(t)*y3

X(t) is a known function
where F = 0 when y3<y1
                1 when y3>y1
B=0 when C7<y3
     1 when C7>y3

so what I am thinking is at time t1 i will get the values for F and B depending on previous values and X value will be known and I need the solution for the unknowns at time t1. I tried solving this by newton raphshon method but the solution is not converging.

Thanks a lot

"Dinesh " <revooridinesh@gmail.com> wrote in message <jvkiq9$ipf$1@newscl01ah.mathworks.com>...
>
> > if no y3' then it not differential equation but algebra like in
> > school. just move the known to the left and unknown to the right to
> > solve and for the first 2 differentials use the matlab od54 functions
> > but need initial conditions
>
> Thank you for the reply
> the three equations are functions of y1,y2,y3 and i know the values before time 0-
>
> my original equations are C1*y1' = F*C2*y2 - C3*y1
> y2' = C4 -B* C5*y3 - C6*y1
> y2-C7=x(t)*y3
>

From the last equation y3 can be expressed as y2. Your equations is an ODE, that can be solved with ode solvers.

Bruno

will there be any problem due to functions 'F' and 'B'.
which ODE function should I use for this ?
"Bruno Luong" <b.luong@fogale.findmycountry> wrote in message <jvl9up$5ir$1@newscl01ah.mathworks.com>...
> "Dinesh " <revooridinesh@gmail.com> wrote in message <jvkiq9$ipf$1@newscl01ah.mathworks.com>...
> >
> > > if no y3' then it not differential equation but algebra like in
> > > school. just move the known to the left and unknown to the right to
> > > solve and for the first 2 differentials use the matlab od54 functions
> > > but need initial conditions
> >
> > Thank you for the reply
> > the three equations are functions of y1,y2,y3 and i know the values before time 0-
> >
> > my original equations are C1*y1' = F*C2*y2 - C3*y1
> > y2' = C4 -B* C5*y3 - C6*y1
> > y2-C7=x(t)*y3
> >
>
> From the last equation y3 can be expressed as y2. Your equations is an ODE, that can be solved with ode solvers.
>
> Bruno

"Dinesh " <revooridinesh@gmail.com> wrote in message <jvljvf$803$1@newscl01ah.mathworks.com>...
> will there be any problem due to functions 'F' and 'B'.

Could be, ODEs with non-Lipschitz RHS are nasty.

> which ODE function should I use for this ?

I don't know. Make sure your equation makes sense mathematically and well-posed first.

Bruno

"Dinesh " <revooridinesh@gmail.com> wrote in message <jvkiq9$ipf$1@newscl01ah.mathworks.com>...
...
> my original equations are C1*y1' = F*C2*y2 - C3*y1
> y2' = C4 -B* C5*y3 - C6*y1
> y2-C7=x(t)*y3
>
> X(t) is a known function
> where F = 0 when y3<y1
> 1 when y3>y1
> B=0 when C7<y3
> 1 when C7>y3

Is C7 a constant? A known constant? Or are you trying to solve the equations with C1,...,C7 as parameters?
Here is what I think: maybe what I say is all wrong if I am misunderstanding your question.
You can solve for y3 in terms of y2, provided x(t) is not zero. If it is zero, then y2=C7, so let's cut 3-dimensional space (y1,y2,y3) into two pieces along y2=C7. Similarly cut along y3=C7 and along y1=y3. Is it possible to solve separately in each of the 8 pieces of 3-space thus obtained?
For specific values of C1,...,C7, it may be possible to cleanly specify initial values along certain parts of the boundary of a piece and obtain solutions. Where two pieces intersect, the values of the solutions must coincide.