# Boundary condition in Newton method
**URL:** https://community.freefem.org/t/boundary-condition-in-newton-method/1284
**Category:** General Discussion
**Created:** [October 25, 2021, 9:48pm UTC](https://community.freefem.org/t/boundary-condition-in-newton-method/1284 "2021-10-25T21:48:01Z")
**Posts on this page:** 4
**Page:** 1
### Author:  [@tess35](https://community.freefem.org/u/tess35)
#### Post date: [October 25, 2021, 9:48pm UTC](https://community.freefem.org/t/boundary-condition-in-newton-method/1284/1 "2021-10-25T21:48:01Z")
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Hi everybody,
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I search how to put the boundary condition in the Newton method , for Dirichlet , Neumann or Robin condition.
For example, solving the toy equation d2 u(x) /dx2 + u(x) -1 =0, u(0)=1, u(1)=10. It’s just to try the program. Here limits are added to each loop, so it’s an error.
fespace Vh(Th,P1);
Vh u, v, du, dv;
varf right(u, v) = int2d(Th)( -dx(u) \* dx(v) +v \* u ) -int2d(Th)( v\*1.0 )
- on(2,u=10 ) + on(4,u=1);
varf left(du , dv ) = int2d(Th)( -dx(du) \*dx(dv) +dv \*du )
- on(2,du=0 ) + on(4,du=0) ;
for (int n = 0; n \< 6; n++){
real[int] b= right(0, Vh) ;
matrix a=left(Vh, Vh );
du[]= a^-1 \* b ;
u[] += du[];
}
```
Thank you
```
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### Author:  [@frederichecht](https://community.freefem.org/u/frederichecht)
#### Post date: [October 26, 2021, 3:18pm UTC](https://community.freefem.org/t/boundary-condition-in-newton-method/1284/2 "2021-10-26T15:18:28Z")
Some remark on Newton method and Boundary condition of every kind
You solve F(u) = 0
let call DF (u)v = the derivative in direction v
The Newton method is
let call w^n solution of $ DF(u^n) w = F(u^n)$ ;
when $ u^{n=1} = u^n - w^n $
and the Boundary condition on w^n must be homogenous
now remember $DF(u^n) (w)$ is linear in so we can commute directly $u^{n+1} u^n - w^n $
so
DF(u^n) u^{n+1} = DF(u^n) u^{n} - F(u^n)
en the B.C on $u^{n+1}$ are the given B.C and the unitialisation is more simple (without B.C) in
general:
exemple
mesh Th=square(10,10); // mesh definition of $\Omega$
fespace Vh(Th,P1); // finite element space
macro grad(u) [dx(u),dy(u)]//
macro F(u,v) (int2d(Th)( grad(u)'_grad(v) - u^3_v) + int2d(Th)((x+y)_v))// + on(1,2,3,4,u=1);
// the probleme is Find u : F(u,v) = 0 + on(1,2,3,4,u=1);
macro DF(u,v,w) int2d(Th)( grad(w)'grad(v) - 3u^2_w\*v) // + on(1,2,3,4,u=1);
Vh u=0, up,v ; // the current value of the solution
for (int iter=1; iter \< 100; ++iter)
{
up[]=u[];
solve PbNewton(u,v) = DF(up,v,u) - DF(up,v,up) + F(up,v) + on(1,2,3,4,u=1);
up[] -=u[]; // to compute the err
real err = up[].linfty ;
cout \<\< iter \<\< " " \<\< err \<\< endl;
plot(u);
if(err\<1e-6) break;
}
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### Author:  [@tess35](https://community.freefem.org/u/tess35)
#### Post date: [October 28, 2021, 3:27pm UTC](https://community.freefem.org/t/boundary-condition-in-newton-method/1284/3 "2021-10-28T15:27:32Z")
```
Hi,
```
Thanks you very much Professor Hecht.
If I understand correctly, we transform the equation into a sequence , u^n → u^(n+1) , w^n is no more used.
I would like to share an another toy equation in 1d :
y’(x) -y(x)^2 -1 =0 , which have a analytical solution, y(x) = tan(x + constant)
PbNewton(u,v) = int2d(Th)( dx(u)_v - 2_u_up_v ) -int2d(Th)( dx(up)_v - 2_up_up_v )
+int2d(Th)( dx(up)_v - up^2_v -v )
+on(4, u= 0.2 ) ; // the B.C
But I have a difficulty for Neumann condition.
If I take y(x)=tan(x), cst=0 for simplicity, y’(x)=1+tan² (x) , y’(0)=1
and change the last line by :
-int1d(Th,4) ( 1.0_v ) + int2d(Th)( 1E-6_u\*v)
It don’t work, with or without the penalty term. Someone could remind me how to put cleanly the B.C ?
```
Sincerely
```
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### Author:  [@frederichecht](https://community.freefem.org/u/frederichecht)
#### Post date: [October 29, 2021, 4:28pm UTC](https://community.freefem.org/t/boundary-condition-in-newton-method/1284/4 "2021-10-29T16:28:42Z")
You want to solve u’(x) -u(x)^2 -1 =0 on ]0,1[, u’(0)=1 but you have not good variational form
The problem not a pb of Newton but a Pb of formulation in FreeFem++``
Sorry no idea your problem is not a good problem for variational formulation.