# How to find the zero of a function defined by a non linear differential equation

I need to find the zero of a mysterious function sigma defined by a differential equation of the following form
u∗σ′(u)+(1−χ0)∗σ(u)+D∑k=1σ(u−xk)(χk−1−χk)=0,u*\sigma'(u)+(1-\chi_0)*\sigma(u)+\sum_{k=1}^{D} \sigma(u-x_k)(\chi_{k-1}-\chi_{k})=0,
with the initial condition σ(u)=uχ(0)−1\sigma(u)=u^{\chi(0)-1} if u<=x1u<=x_1 and σ(u)=0\sigma(u)=0 if u<0u<0 in these two cases: (D=10D=10 for both cases and x0=0x_0=0) 1) xk=1/(3∗(D−k+1))x_k=1/(3*(D-k+1)) and χk=8∗(cos((2∗(D−k)+3)Ï€/(3∗(D−k)+4)))3−4∗cos((2∗(D−k)+3)Ï€/(3∗(D−k)+4)).\chi_k=8* (\cos((2*(D-k)+3)Ï€/(3*(D-k)+4)))^3-4* \cos((2*(D-k)+3)Ï€/(3*(D-k)+4)). 2) x_k= 1/(2*(D-k+1))x_k= 1/(2*(D-k+1)) and \chi_k=16* \cos(((D-k)+2)Ï€/(2*(D-k)+5))^4 -12*\cos(((D-k)+2)Ï€/(2*(D-k)+5)))^2 +1.\chi_k=16* \cos(((D-k)+2)Ï€/(2*(D-k)+5))^4 -12*\cos(((D-k)+2)Ï€/(2*(D-k)+5)))^2 +1. I want just to mention that the sequence (x_k)(x_k) is increasing and we have x_0=0 {-2 10^-8, 2 10^-8}, AxesLabel -> {u, d}]

FindRoot[s[u], {u, 2.2}]
{* {u -> 2.25542} *}

The second case is similar

x[k_] := If[k == 0, 0, 1/(2*(n – k + 1))]
chi[k_] := 16*(Cos[(n – k + 2) Ï€/(2*(n – k) + 5)])^4 –
12*(Cos[(n – k + 2) Ï€/(2*(n – k) + 5)])^2 + 1

with a zero at 3.20177.

@Khadija, why don’t you try running it yourself? Also, I removed a comment asking to be e-mailed the results; this is considered disrespectful.
– J. M.♦
Jun 14 ’15 at 20:51

I am sorry if I caused any disturbance. I tried with my Mathemtica version but it didn’t work that’s why I asked to compile this one by someone who has a more advanced version than the mine.