Solve Two Coupled ODEs with Picard's Iteration

[Nusc]

How do you use Picard's iteration to solve the solution for two coupled ODE's, given initial conditions?

 

[HallsofIvy]

Treat them as two separate equations, using both values in the calculations.

For a "one equation" Picard method, if you are solving the initial value problem dx/dt= f(x,t), with x(t₀)= x₀, you start by replacing x in f with x_0 and get x_1(t)= x_0+ \int_{t_0}^t f(x_0,\tau)d\tau, then x_2(t)= x_0+ \int_{t_0}^t f(x_1(\tau),\tau)d\tau, etc.

With two equations, say dx/dt= f(x,y,t) and dy/dt= g(x,y,t), with x(t₀)= x₀, y(t<sub>0)= y0, start by letting x and y in those function be x0, y0 and integrate to get x_1(t)= \int_{t_0}^t f(x_0, y_0, \tau) d\tau and y_1(t)= \int_{t_0}^t g(x_0, y_0, \tau)d\tau, then x_2(t)= x_0+ \int_{t_0}^t f(x_1, y_1, \tau)d\tau, y_2(t)= \int_{t_0}^t g(x_1,y_1,\tau)d\tau, etc.

(Note you can also do: x_1(t)= \int_{t_0}^t f(x_0, y_0, \tau) d\tau and y_1(t)= \int_{t_0}^t g(x_1, y_0, \tau)d\tau, x_2(t)= x_0+ \int_{t_0}^t f(x_1(\tau),y_1(\tau),\tau)d\tau, y_2(t)= y_0+ \int_{t_0}^t f(x_2(\tau),y_1(\tau),\tau)d\tau, etc., using each new value as soon as we have it. That will give a sightly different answer but still a valid approximation to the true solution.)</sub>

 

[Nusc]

Is there a name for the later? How come the later is true?

Is there a reference for these methods? Most elementary textbooks on ODE's that I know of don't cover Picard's method.

 

[Nusc]

, then x_2(t)= x_0+ \int_{t_0}^t f(x_1, y_1, \tau)d\tau, y_2(t)= \int_{t_0}^t g(x_1,y_1,\tau)d\tau, etc.

You meant:

, then x_2(t)= x_0+ \int_{t_0}^t f(x_1, y_1, \tau)d\tau, y_2(t)= y_0 + \int_{t_0}^t g(x_1,y_1,\tau)d\tau, etc.

Right?

 

[HallsofIvy]

Oh, yes! Forgot all about the y_0! Thank you.

 

[HallsofIvy]

Is there a name for the later? How come the later is true?

Is there a reference for these methods? Most elementary textbooks on ODE's that I know of don't cover Picard's method.

Actuall most elementary textbooks mention Picard's method in reference to the "Existence and Uniqueness Theorem" for initial value problems. They don't "cover" it because it has a very slow convergence rate. There are much better methods for approximate solution to differential equations.

 

[HallsofIvy]

Is there a name for the later? How come the later is true?

Is there a reference for these methods? Most elementary textbooks on ODE's that I know of don't cover Picard's method.

Actuall most elementary textbooks mention Picard's method (perhaps not by that name) in reference to the "Existence and Uniqueness Theorem" for initial value problems. It is Picard's method that gives the fixed point formula needed for the proof. Actually, I believe Picard himself developed it for that purposeThey don't "cover" it because it has a very slow convergence rate. There are much better methods for solving differential equations.

For example, to solve the problem x'(t)= x, with x(0)= 1, you take as your "first approximation" x_0(t)= 1 so that x'(t)= dx/dt= 1 and, integrating both sides, x_0= y_0+ \int_0^t 1 dt= 1+ t.

Now, take x_1(t)= 1+ t so that x'= dx/dt= 1+ t and, integrating both sides, x_1= y_0+ \int_0^t (1+ t)dt= 1+ t+ (1/2)t^2.

Now, take x_2= 1+ t+ (1/2)t^2 so that x&#039;= dx/dt= 1+ t+ (1/2)t^2 and, integrating both sides
x_3= 1+ \int_0^t (1+ t+ (1/2)t^2)dt= 1+ t+ (1/2)t^2+ (1/6)t^3.

At this point you should be able to see that if you continued this forever, you would get the MacLaurin series expansion of e^x which is,in fact, the solution to this problem- but you are going to take an unGodly long time getting their!

 

[Nusc]

So if you wanted to find a solution can be obtained for a value for, say, |x| <0.5 how would you do this?