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From: Adam Jacob on 11 Apr 2008 17:54
Hello.. wondering if anyone can help out.
I'm trying to solve an ode final value problem. So for e.g.
solving dydt=f(y,t) for t0 (initial time) to tfinal (final
time) with y(tfinal)=yf.
I tried solving backwards in time using

tspan = LINSPACE(tf1,t01);
ode45(@(t,Y) odefun(t,Y,vars),tspan,yf);

So my timespan is backwards. I got a very bad solution
(very large) but when I solved it as an initial value
problem

tspan = LINSPACE(t01,tf1);
ode45(@(t,Y) odefun(t,Y,vars),tspan,yf);

The solution seemed reasonable.
Just wondering if I "flipped" the result of the initial
value problem, does that make it a valid final value
problem? Wondering how ode45 solves backwards in time (when
tspan has decreasing values).

Any help at all would be appreciated..
Adam
From: Yi Cao on 12 Apr 2008 10:28
"Adam Jacob" <ajacob73(a)yahoo.com> wrote in message
<ftomlp$d4q$1(a)fred.mathworks.com>...
> Hello.. wondering if anyone can help out.
> I'm trying to solve an ode final value problem. So for
e.g.
> solving dydt=f(y,t) for t0 (initial time) to tfinal
(final
> time) with y(tfinal)=yf.
> I tried solving backwards in time using
>
> tspan = LINSPACE(tf1,t01);
> ode45(@(t,Y) odefun(t,Y,vars),tspan,yf);
>
> So my timespan is backwards. I got a very bad solution
> (very large) but when I solved it as an initial value
> problem
>
> tspan = LINSPACE(t01,tf1);
> ode45(@(t,Y) odefun(t,Y,vars),tspan,yf);
>
> The solution seemed reasonable.
> Just wondering if I "flipped" the result of the initial
> value problem, does that make it a valid final value
> problem? Wondering how ode45 solves backwards in time
(when
> tspan has decreasing values).
>
> Any help at all would be appreciated..
> Adam

It may be because of your odefun. Could you post your
odefun here?

Yi Cao
From: Adam Jacob on 13 Apr 2008 12:59
Thanks Yi... I'll put in the whole m-script so that if you
run it you can see what I mean... if I switch tf and t0 in
tspan (so ODE45 is solving an initial value problem
forwards in time) than it gets a somewhat reasonable
solution than backwards in time. Not sure since if I do
that with other functions I get the same result (backwards
in time ODE45 solution is the same as forwards in time
ODE45 solution but reversed in time)...
Thanks,
Adam

%-----MAIN M-FILE--------------
clear all
%-----VARIABLE SET-UP--------------
brakestep=1;
brakestart=1;
brakerelease=11;
exptime=30;
c=0.1;
g=9.81;
A = [0 1 0 0 0; 0 0 -g 0 1; 0 0 0 1 0; 0 0 0 -1/c 0;0 0 0
0 -20];
B = [0; 0; 0; 1/c; 0];
R = 10;
Q = [1 0 0 0 0; 0 0 0 0 0; 0 0 0 0 0; 0 0 0 0 0;0 0 0 0 0];
S = [0; 0; 0; 0; 0];
E = eye(5);
dps=70;
stepnum=dps*exptime
I5=eye(5)
%-----TO RUN and PLOT ODE SOLUTION--------------
t01= 0;
tf1=exptime;
tspan1 = LINSPACE(tf1,t01,dps*(tf1-t01));
X01=zeros(5);
[Time1,X1] = ode45(@(t,X) odefuncare2
(t,X,A,B,Q,R,S,E),tspan1,X01);
plot(Time1,X1)
%-------ODEFUNCTION------------------
function dxdt = odefuncare(t,X,A,B,Q,R,S,E)
X = reshape(X,5,5); %converting X from a column vector
generated by ode45 into a 5 x 5 Matrix
dxdt=((A')*X*E)+((E')*X*A)-(((E')*(X)*(B)+S)*(inv(R))*((B')*
(X)*(E)+(S')))+Q;
dxdt = dxdt(:); %converting dxdt into a column vector
as expected by ode45
%-------END OF SCRIPT FILES--------




"Yi Cao" <y.cao(a)cranfield.ac.uk> wrote in message
<ftqgth$4t6$1(a)fred.mathworks.com>...
> "Adam Jacob" <ajacob73(a)yahoo.com> wrote in message
> <ftomlp$d4q$1(a)fred.mathworks.com>...
> > Hello.. wondering if anyone can help out.
> > I'm trying to solve an ode final value problem. So for
> e.g.
> > solving dydt=f(y,t) for t0 (initial time) to tfinal
> (final
> > time) with y(tfinal)=yf.
> > I tried solving backwards in time using
> >
> > tspan = LINSPACE(tf1,t01);
> > ode45(@(t,Y) odefun(t,Y,vars),tspan,yf);
> >
> > So my timespan is backwards. I got a very bad solution
> > (very large) but when I solved it as an initial value
> > problem
> >
> > tspan = LINSPACE(t01,tf1);
> > ode45(@(t,Y) odefun(t,Y,vars),tspan,yf);
> >
> > The solution seemed reasonable.
> > Just wondering if I "flipped" the result of the initial
> > value problem, does that make it a valid final value
> > problem? Wondering how ode45 solves backwards in time
> (when
> > tspan has decreasing values).
> >
> > Any help at all would be appreciated..
> > Adam
>
> It may be because of your odefun. Could you post your
> odefun here?
>
> Yi Cao

From: Yi Cao on 14 Apr 2008 10:56
"Adam Jacob" <ajacob73(a)yahoo.com> wrote in message
<ftte4m$935$1(a)fred.mathworks.com>...
> Thanks Yi... I'll put in the whole m-script so that if
you
> run it you can see what I mean... if I switch tf and t0
in
> tspan (so ODE45 is solving an initial value problem
> forwards in time) than it gets a somewhat reasonable
> solution than backwards in time. Not sure since if I do
> that with other functions I get the same result
(backwards
> in time ODE45 solution is the same as forwards in time
> ODE45 solution but reversed in time)...
> Thanks,
> Adam
>
> %-----MAIN M-FILE--------------
> clear all
> %-----VARIABLE SET-UP--------------
> brakestep=1;
> brakestart=1;
> brakerelease=11;
> exptime=30;
> c=0.1;
> g=9.81;
> A = [0 1 0 0 0; 0 0 -g 0 1; 0 0 0 1 0; 0 0 0 -1/c 0;0 0 0
> 0 -20];
> B = [0; 0; 0; 1/c; 0];
> R = 10;
> Q = [1 0 0 0 0; 0 0 0 0 0; 0 0 0 0 0; 0 0 0 0 0;0 0 0 0
0];
> S = [0; 0; 0; 0; 0];
> E = eye(5);
> dps=70;
> stepnum=dps*exptime
> I5=eye(5)
> %-----TO RUN and PLOT ODE SOLUTION--------------
> t01= 0;
> tf1=exptime;
> tspan1 = LINSPACE(tf1,t01,dps*(tf1-t01));
> X01=zeros(5);
> [Time1,X1] = ode45(@(t,X) odefuncare2
> (t,X,A,B,Q,R,S,E),tspan1,X01);
> plot(Time1,X1)
> %-------ODEFUNCTION------------------
> function dxdt = odefuncare(t,X,A,B,Q,R,S,E)
> X = reshape(X,5,5); %converting X from a column vector
> generated by ode45 into a 5 x 5 Matrix
> dxdt=((A')*X*E)+((E')*X*A)-(((E')*(X)*(B)+S)*(inv(R))*
((B')*
> (X)*(E)+(S')))+Q;
> dxdt = dxdt(:); %converting dxdt into a column
vector
> as expected by ode45
> %-------END OF SCRIPT FILES--------
>
>
>
>
> "Yi Cao" <y.cao(a)cranfield.ac.uk> wrote in message
> <ftqgth$4t6$1(a)fred.mathworks.com>...
> > "Adam Jacob" <ajacob73(a)yahoo.com> wrote in message
> > <ftomlp$d4q$1(a)fred.mathworks.com>...
> > > Hello.. wondering if anyone can help out.
> > > I'm trying to solve an ode final value problem. So
for
> > e.g.
> > > solving dydt=f(y,t) for t0 (initial time) to tfinal
> > (final
> > > time) with y(tfinal)=yf.
> > > I tried solving backwards in time using
> > >
> > > tspan = LINSPACE(tf1,t01);
> > > ode45(@(t,Y) odefun(t,Y,vars),tspan,yf);
> > >
> > > So my timespan is backwards. I got a very bad
solution
> > > (very large) but when I solved it as an initial value
> > > problem
> > >
> > > tspan = LINSPACE(t01,tf1);
> > > ode45(@(t,Y) odefun(t,Y,vars),tspan,yf);
> > >
> > > The solution seemed reasonable.
> > > Just wondering if I "flipped" the result of the
initial
> > > value problem, does that make it a valid final value
> > > problem? Wondering how ode45 solves backwards in time
> > (when
> > > tspan has decreasing values).
> > >
> > > Any help at all would be appreciated..
> > > Adam
> >
> > It may be because of your odefun. Could you post your
> > odefun here?
> >
> > Yi Cao
>

Dear Adma

Now I know what is your problem. You want to integrate the
differential Riccati equation in reverse time. The Riccati
equation in reverse time is unstable. Hence, a small
numerical error will eventually becomes infinite large.
That is why you feel you result is strange. If you look the
results in a short period, you will see this instability.

Because your equation is not time varying, therefore, as
you mentioned before, it is better to integrate in forward
time, i.e. from [-30 0], then explain the results
reversely.

hth.
Yi Cao
From: Adam Jacob on 17 Apr 2008 20:50
Thanks very much for the help Yi..
Adam


"Yi Cao" <y.cao(a)cranfield.ac.uk> wrote in message <ftvra1
$8b3$1(a)fred.mathworks.com>...
> "Adam Jacob" <ajacob73(a)yahoo.com> wrote in message
> <ftte4m$935$1(a)fred.mathworks.com>...
> > Thanks Yi... I'll put in the whole m-script so that if
> you
> > run it you can see what I mean... if I switch tf and t0
> in
> > tspan (so ODE45 is solving an initial value problem
> > forwards in time) than it gets a somewhat reasonable
> > solution than backwards in time. Not sure since if I do
> > that with other functions I get the same result
> (backwards
> > in time ODE45 solution is the same as forwards in time
> > ODE45 solution but reversed in time)...
> > Thanks,
> > Adam
> >
> > %-----MAIN M-FILE--------------
> > clear all
> > %-----VARIABLE SET-UP--------------
> > brakestep=1;
> > brakestart=1;
> > brakerelease=11;
> > exptime=30;
> > c=0.1;
> > g=9.81;
> > A = [0 1 0 0 0; 0 0 -g 0 1; 0 0 0 1 0; 0 0 0 -1/c 0;0 0
0
> > 0 -20];
> > B = [0; 0; 0; 1/c; 0];
> > R = 10;
> > Q = [1 0 0 0 0; 0 0 0 0 0; 0 0 0 0 0; 0 0 0 0 0;0 0 0 0
> 0];
> > S = [0; 0; 0; 0; 0];
> > E = eye(5);
> > dps=70;
> > stepnum=dps*exptime
> > I5=eye(5)
> > %-----TO RUN and PLOT ODE SOLUTION--------------
> > t01= 0;
> > tf1=exptime;
> > tspan1 = LINSPACE(tf1,t01,dps*(tf1-t01));
> > X01=zeros(5);
> > [Time1,X1] = ode45(@(t,X) odefuncare2
> > (t,X,A,B,Q,R,S,E),tspan1,X01);
> > plot(Time1,X1)
> > %-------ODEFUNCTION------------------
> > function dxdt = odefuncare(t,X,A,B,Q,R,S,E)
> > X = reshape(X,5,5); %converting X from a column vector
> > generated by ode45 into a 5 x 5 Matrix
> > dxdt=((A')*X*E)+((E')*X*A)-(((E')*(X)*(B)+S)*(inv(R))*
> ((B')*
> > (X)*(E)+(S')))+Q;
> > dxdt = dxdt(:); %converting dxdt into a column
> vector
> > as expected by ode45
> > %-------END OF SCRIPT FILES--------
> >
> >
> >
> >
> > "Yi Cao" <y.cao(a)cranfield.ac.uk> wrote in message
> > <ftqgth$4t6$1(a)fred.mathworks.com>...
> > > "Adam Jacob" <ajacob73(a)yahoo.com> wrote in message
> > > <ftomlp$d4q$1(a)fred.mathworks.com>...
> > > > Hello.. wondering if anyone can help out.
> > > > I'm trying to solve an ode final value problem. So
> for
> > > e.g.
> > > > solving dydt=f(y,t) for t0 (initial time) to tfinal
> > > (final
> > > > time) with y(tfinal)=yf.
> > > > I tried solving backwards in time using
> > > >
> > > > tspan = LINSPACE(tf1,t01);
> > > > ode45(@(t,Y) odefun(t,Y,vars),tspan,yf);
> > > >
> > > > So my timespan is backwards. I got a very bad
> solution
> > > > (very large) but when I solved it as an initial
value
> > > > problem
> > > >
> > > > tspan = LINSPACE(t01,tf1);
> > > > ode45(@(t,Y) odefun(t,Y,vars),tspan,yf);
> > > >
> > > > The solution seemed reasonable.
> > > > Just wondering if I "flipped" the result of the
> initial
> > > > value problem, does that make it a valid final
value
> > > > problem? Wondering how ode45 solves backwards in
time
> > > (when
> > > > tspan has decreasing values).
> > > >
> > > > Any help at all would be appreciated..
> > > > Adam
> > >
> > > It may be because of your odefun. Could you post your
> > > odefun here?
> > >
> > > Yi Cao
> >
>
> Dear Adma
>
> Now I know what is your problem. You want to integrate
the
> differential Riccati equation in reverse time. The
Riccati
> equation in reverse time is unstable. Hence, a small
> numerical error will eventually becomes infinite large.
> That is why you feel you result is strange. If you look
the
> results in a short period, you will see this instability.
>
> Because your equation is not time varying, therefore, as
> you mentioned before, it is better to integrate in
forward
> time, i.e. from [-30 0], then explain the results
> reversely.
>
> hth.
> Yi Cao

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