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Aide de Scilab >> Intégration - dérivation > ode_roots

# ode_roots

ordinary differential equation solver with roots finding

### Syntax

`[y, rd, w, iw] = ode("roots", y0, t0, t [,rtol [,atol]], f [,jac], ng, g [,w,iw])`

### Arguments

y0

a real vector or matrix (initial conditions).

t0

a real scalar (initial time).

t

a real vector (times at which the solution is computed).

f

an external i.e. function or character string or list.

rtol, atol

a real constants or real vectors of the same size as `y`.

jac

an external i.e. function or character string or list.

ng

an integer.

g

an external i.e. function or character string or list.

y

a real vector or matrix. The solution.

rd

a real vector.

w, iw

vectors of real numbers. See ode() optional output

### Description

With this syntax (first argument equal to `"roots"`) `ode` computes the solution of the differential equation `dy/dt=f(t,y)` until the state `y(t)` crosses the surface `g(t,y)=0`.

`g` should give the equation of the surface. It is an external i.e. a function with specified syntax, or the name of a Fortran subroutine or a C function (character string) with specified syntax or a list.

If `g` is a function the syntax should be as follows:

`z = g(t,y)`

where `t` is a real scalar (time) and `y` a real vector (state). It returns a vector of size `ng` which corresponds to the `ng` constraints. If `g` is a character string it refers to the name of a Fortran subroutine or a C function, with the following calling sequence: `g(n,t,y,ng,gout)` where `ng` is the number of constraints and `gout` is the value of `g` (output of the program). If `g` is a list the same conventions as for `f` apply (see ode help).

Output `rd` is a `1 x k` vector. The first entry contains the stopping time. Other entries indicate which components of `g` have changed sign. `k` larger than 2 indicates that more than one surface (`(k-1)` surfaces) have been simultaneously traversed.

Other arguments and other options are the same as for `ode`, see the ode help.

### Examples

```// Integration of the differential equation
// dy/dt=y , y(0)=1, and finds the minimum time t such that y(t)=2
deff("[ydot]=f(t,y)","ydot=y")
deff("[z]=g(t,y)","z=y-2")
y0=1;ng=1;
[y,rd]=ode("roots",y0,0,2,f,ng,g)

deff("[z]=g(t,y)","z=y-[2;2;33]")
[y,rd]=ode("roots",1,0,2,f,3,g)```