Picard iteration on systems of DEs

[TaliskerBA]

Homework Statement

For the system
u' = v, v' = −u
with initial conditions u(0) = 1 and v(0) = 0, find an approximate solution by
performing 4 steps of Picard iteration. Compare the results with the actual solution.

Homework Equations

In general:

y'= f(x,y), y(x₀)= y₀

y(x) = y(x₀) + $$\int$$ f(t,y(t)) .dt with x and x₀ the upper and lower points of the integral (couldn't work out how to format this in)

The Attempt at a Solution

I know how to do picard iteration for a single first-order equation but don't know how to extend it to systems. To be honest my attempt is probably so far off it's not worth writing but here it is anyway.

u(t) = (x₀) + $$\int$$ v(s)ds with t and t₀ the upper and lower points of the integral (I won't write this below but these will always be the lower and upper points). As v₀ = 0 I proceed as follows:

u⁰ = 1 + $$\int$$ 0ds = 1
u¹ = 1 + $$\int$$ 1.ds = 1 + t
u² = 1 + $$\int$$(1 + s)ds = 1 + t + t²/2

etc.

I'm way off the answer given in the book. Please someone offer a rough explanation!

Thanks

 

[TaliskerBA]

bump. Help!

 

[HallsofIvy]

Bumping is a good way to get yourself banned.

What you have done is wrong because you have neglected the "v" after the first step.
$u_n= u_0+ \int v_{n-1}dx$ and $v_n= v(0)- \int u_{n-1}dx$
u(0)= 1 and v(0)= 0 so the first step gives
$u_1(x)= 1+ \int 0dx= 1$, $v_(x)= 0- \int 1dx= -x$.

Now, the second step:
$u_2(x)= 1+ \int -xdx= 1- (1/2)x^2$, $v_2(x)= 0- \int 1 dx= -x$

Third step:
$u_3(x)= 1+ \int -x dx= 1- (1/2)x^2$, $v_3= 0- \int 1-(1/2)x^2 dx= -x+ (1/6)x^3$

 

[TaliskerBA]

Thanks very much for your help, I get it now. I'm sorry about bumping I didn't know it was against the rules. I won't do it again.