Thread Subject:

ode with definite integral

Hello,
I wonder how to solve such a task:
Suppose we want to solve a differential equation in the form:
dy/dt = f(t) = exp(-te)
where:
te = \int from 0 to t (2^(y/10)) dt
Is it possible to do it with Matlab ode solvers?

Best regards
Grzegorz

On 6/29/2011 11:56 PM, Grzegorz Knor wrote:
> Hello,
> I wonder how to solve such a task:
> Suppose we want to solve a differential equation in the form:
> dy/dt = f(t) = exp(-te)
> where:
> te = \int from 0 to t (2^(y/10)) dt
> Is it possible to do it with Matlab ode solvers?
>
> Best regards
> Grzegorz

You integrand is 2^(y/10) but you are integrating w.r.t. ?

So teh integrand is constant? So the integral is just

      2^(y/10) * Integral[ 1 , {0,t} ]

which is

      2^(y/10) * t

?

Can you clarify first, then will give you code.

--Nasser

"Nasser M. Abbasi" <nma@12000.org> wrote in message <iuh7tn$81c$1@speranza.aioe.org>...
> On 6/29/2011 11:56 PM, Grzegorz Knor wrote:
> > Hello,
> > I wonder how to solve such a task:
> > Suppose we want to solve a differential equation in the form:
> > dy/dt = f(t) = exp(-te)
> > where:
> > te = \int from 0 to t (2^(y/10)) dt
> > Is it possible to do it with Matlab ode solvers?
> >
> > Best regards
> > Grzegorz
>
> You integrand is 2^(y/10) but you are integrating w.r.t. ?
>
> So teh integrand is constant? So the integral is just
>
> 2^(y/10) * Integral[ 1 , {0,t} ]
>
> which is
>
> 2^(y/10) * t
>
> ?
>
> Can you clarify first, then will give you code.
>
> --Nasser

y depends on the time, so y = y(t).
Here is a tex code for this equation:
\frac{dy(t)}{dt} = e^{- \int_0^t 2^{y(z))/10}dz}
or link:
http://latex.codecogs.com/gif.latex?\huge&space;\frac{dy(t)}{dt}&space;=&space;e^{-&space;\int_0^t&space;2^{y(z))/10}dz}


Grzegorz

"Grzegorz Knor" wrote in message <iuh6ma$ggi$1@newscl01ah.mathworks.com>...
> Hello,
> I wonder how to solve such a task:
> Suppose we want to solve a differential equation in the form:
> dy/dt = f(t) = exp(-te)
> where:
> te = \int from 0 to t (2^(y/10)) dt
> Is it possible to do it with Matlab ode solvers?
>
> Best regards
> Grzegorz
- - - - - - - - -
  Yes, that can easily be done with an ode solver. Define

 te(t) = int('2^(y(s)/10)','s',0,t)

Then we have

 dte(t)/dt = 2^(y(t)/10)

and

 dy(t)/dt = exp(-te(t))

so this is in the proper form to be solved with an ode solver. You must of course arrange that te(0) = 0. The initial value, y(0), is up to you.

Roger Stafford

"Roger Stafford" wrote in message <iuh9ga$mua$1@newscl01ah.mathworks.com>...
> "Grzegorz Knor" wrote in message <iuh6ma$ggi$1@newscl01ah.mathworks.com>...
> > Hello,
> > I wonder how to solve such a task:
> > Suppose we want to solve a differential equation in the form:
> > dy/dt = f(t) = exp(-te)
> > where:
> > te = \int from 0 to t (2^(y/10)) dt
> > Is it possible to do it with Matlab ode solvers?
> >
> > Best regards
> > Grzegorz
> - - - - - - - - -
> Yes, that can easily be done with an ode solver. Define
>
> te(t) = int('2^(y(s)/10)','s',0,t)
>
> Then we have
>
> dte(t)/dt = 2^(y(t)/10)
>
> and
>
> dy(t)/dt = exp(-te(t))
>
> so this is in the proper form to be solved with an ode solver. You must of course arrange that te(0) = 0. The initial value, y(0), is up to you.
>
> Roger Stafford

Thank you.

Grzegorz

On 6/30/2011 12:44 AM, Roger Stafford wrote:
> "Grzegorz Knor" wrote in message<iuh6ma$ggi$1@newscl01ah.mathworks.com>...
>> Hello,
>> I wonder how to solve such a task:
>> Suppose we want to solve a differential equation in the form:
>> dy/dt = f(t) = exp(-te)
>> where:
>> te = \int from 0 to t (2^(y/10)) dt
>> Is it possible to do it with Matlab ode solvers?
>>
>> Best regards
>> Grzegorz
> - - - - - - - - -
> Yes, that can easily be done with an ode solver. Define
>
> te(t) = int('2^(y(s)/10)','s',0,t)
>
> Then we have
>
> dte(t)/dt = 2^(y(t)/10)
>
> and
>
> dy(t)/dt = exp(-te(t))
>
> so this is in the proper form to be solved with an ode solver.
> You must of course arrange that te(0) = 0. The initial value, y(0), is up to you.
>
> Roger Stafford

Nice Roger, that is a nice way to do it. Below I show your method,
implemented in ode45, using initial value for te(0)=0, and then
show another method I just hacked using the original formula as
is by using quad directly, and interp1 to do the integration
at each step. It is a hack, but it gives the same answer.

Had to protect against 2 special cases: when ode45 uses
the same time step twice, else intepr1 will not be happy,
and for initial case at t=0.

Here we go:

Roger method: (the right way to do it)

------------------
functionfoo1

figure;
y0=1;
integralAtZero=0;
maxTime = 10;
    
[t,y]=ode45(@RHS,[0 maxTime],[y0 integralAtZero]);
plot(t,y(:,1));
title('solution y(t)');

     function ydot = RHS(t,y)
         ydot = [exp(-y(2)) ; 2^(y(1)/10) ];
     end
     
end
---------------------------

my method, just for fun :)

------------------------------
function foo

y0=1;
maxTime = 10;
globalY = zeros(1000,1); % change as needed
globalt = zeros(1000,1); % to prevent dynamic allocation
globalydot = zeros(1000,1); %just a hack
indx=0;
    
[t,y]=ode45(@RHS,[0 maxTime],y0);
plot(t,y);


     function ydot = RHS(currentTime,y)
         if currentTime(1)>0 && currentTime(1) == globalt(indx)
                ydot= globalydot(indx);
         else
           indx=indx+1;
           globalY(indx)=y(1);
           globalt(indx)=currentTime(1);
           ydot = exp(-quad(@f,0,currentTime));
           globalydot(indx)=ydot;
         end
     end

     function v = f(t) % integrate 2^(y(t)/10)
         if all(t==0)
            v=t;
         else
           yi = interp1(globalt(1:indx), globalY(1:indx),t) ;
           v = 2.^(yi/10);
         end
     end
end
-------------------------------------


--Nasser