Vector input for ODE45

Sirs, I have something like the below formula:

T=(A+B+C)+(D*T) dt where A, B and C are vectors with let me say (1,6000) length How Can I integrate this formula using ODE45, please help, and thanks in advance

Answer 1

[Walter Roberson]

   x0 = ... something same length as A;

   [T, X] = ode45( @(t, x) f(t, X, A, B, C, D), tspan, x0);

function dy = f(t, y, A, B, C, D)
  dy = (A + B + C) + (D*t);

... except of course if that were really your formula you would construct

    x0 = ... something same length as A;
    ApBpC = A + B + C;

   [T, X] = ode45( @(t, x) f(t, X, ApBpC, D), tspan, x0);

function dy = f(t, y, ApBpC, D)
  dy = ApBpC + (D*t);

[Oday Shahadh]

Thanks Mr. Walter..

I will re-arrange the equation just in case this make a difference

Temp/dt=(A+B+C)+(D*Temp) I estimated the initial value of (Temp) D: is a constant dt: time interval

[Walter Roberson]

Could you confirm that the form of the equation is

T(t) = (A+B+C) + (D*T(t))*diff(T(t),t)

If it is, then provided that you meet the theoretical boundary conditions, this has two solutions:

1)

T(t) = A + B + C      %that is, constant in time

or 2)

T(t) =  (lambertw(0, exp(-(A + B + C - t/D - (log(-exp(-(A + B + C - T0)/(A + B + C))*(A + B + C - T0)) + 1)*(A + B + C))/(A + B + C))/(A + B + C)) + 1)*(A + B + C)

where T0 is the boundary value at T(0)

[Oday Shahadh]

 T(t)=(A/dt+B/dt+C/dt)+D(T/dt);

A,B and C has been calculated by another algorithm in the script for(6000,1) length, and just(D= constant number)

[Walter Roberson]

So to confirm,

T(t) == (A / diff(T(t),t) + B / diff(T(t),t) + C / diff(T(t),t)) + D * T(t) ./ diff(T(t),t)

? Which would simplify in part to (A+B+C)/diff(T(t),t) and then A+B+C, all being constants, could be replaced with the sum of the three?

When I see A/dt I tend to think you intend that A is a function of t and that you are expressing its derivative there, an operation that would more typically be expressed as dA/dt . But if that were the case then it is not clear why you would have a vector of constants for A.

[Oday Shahadh]

sorry for making you miss understanding me, Dear Sir A,B and C varying with time, that have a value at each time step (t), and they are estimated previously, so for run time of(16000 second) I have the values of A,B and C for each time step, what you wrote is right (dA/dt)

T(t)=(dA/dt+dB/dt+dC/dt)+D(T/dt);

[Oday Shahadh]

size(dA/dt) =(16000,1) and also dB/dt and dC/dt

[Walter Roberson]

Are A, B, C intended to be tables to be linearly interpolated at the real times, along the lines of

this_A = interp1(times_A_was_sampled_at, A, t);   %estimate at current time

? If so then there would be concern about whether the prior estimation and the interpolation would work together to produce values with continuous first derivatives; if not then you need to use a different approach because ode45() cannot support "jumps" at all.

[Oday Shahadh]

Dear Walter, it is really difficult to me , but I tried this, insert a loop inside the ODE45, is it correct, also it going to be very slow

global : A B C

for i=1:10:t

Tdot=((A(i)+B(i)+C(i))-(T);
end

F1T=[Tdot];

end

[Walter Roberson]

In your "for i" loop, you overwrite all of Tdot. The effect is going to be the same as if you had only done the last of the iterations.

You should avoid using global: global are the slowest form of variables. https://www.mathworks.com/help/matlab/math/parameterizing-functions.html

[Oday Shahadh]

how to develope the loop? to avoid overwriting a previues values?

[Walter Roberson]

i_vals = 1 : 10 : t;
num_i = length(i_vals);
Tdot = zeros(num_i, 1);
for i_idx = 1 : num_i
  i = i_vals(i_idx);
  Tdot(i_idx) = A(i) + B(i) + C(i) - T;
end

result = TDot;

Note that the resulting value changes length as t increases. If t represents the time parameter to the ode, then this would mean that you are trying to return a different number of derivatives as time goes on.