# Old example of Newton3D Using Boussinesq hypothesis gives seg-fault

Hello,

The example code from: FreeFem++ users | Main >> Newton3D is giving a seg-fault in the while loop, specifically the int3d(Th) functions are not working properly.

I would like to solve this, any pointers would be appreciated.

Best,
Isha

I have try the example, the converge is bad and test test explose but the code works.

– FreeFem++ v4.14 (Ven 7 jui 2024 11:35:37 CEST - git v4.14-60-g660dbd36)
file : boussinesq3d.edp
1 : ////
2 : //// Adapted in April 2014 from a code by F. Hecht
3 : ////
4 : /**
5 : Incompressible steady Navier Stokes + boussinesq coupled system
6 : with Taylor-Hood Finite element
7 : Non-linearity : Newton method
8 : continuation on Rayleigh Number
9 : Solution computed on a cube [0,1]^3
10 :
11 : Problem: Navier-Sokes and Temperature on vertical direccion (Boussinesq hyp.)
12 :
13 : Equations:
14 :
16 : (u.grad)T-kLap(T) = 0 in the cube
17 :
18 : Boundary Conditions:
19 :
20 : u=0 on boundary of cube
21 : T=0 on opposite sidewalls 2 and 4
22 : (d/dn)T=0 on the rest of walls
23 :
24 : /
25 :
26 : ////
27 : //// 2D mesh
28 : ////
29 : int nn=10;
30 : mesh th2d=square(nn,nn);
31 : ////
32 : //// labels for the cube 3d mesh
33 : ////
34 : int[int] rup=[0,1]; ////top of the cube labelled by cero
35 : int[int] rdown=[0,1];////bottom of the cube labelled by cero
36 : int[int] rmid=[1,1,2,2,3,3,4,4];//// sidewalls labelled as the 2d edges they stand on
37 : ////
38 : //// 3D mesh
39 : ////
41 : mesh3 Th=buildlayers(square(nn,nn),nn, zbound=[0.,1.],
42 : labelmid=rmid, labelup = rup, labeldown = rdown);
43 : real errm=1e-2;
44 : ////
45 : //// MACROS…they mean
46 : ////
48 : ////
49 : macro grad(u) [dx(u),dy(u),dz(u)] ) ////
50 : ////
51 : //// For a vector u of componentes u#1,u#2,u#3
52 : //// here # acts as wildcar character
54 : ////
56 : ////
57 : //// For a vector u of componentes u#1,u#2,u#3
58 : //// div(u) compute the divergence
59 : ////
60 : macro div(u) (dx(u#1)+dy(u#2)+dz(u#3)) ) ////
61 : ////
62 : //// for a vector u of componentes u#1,u#2,u#3 a scalar v
64 : ////
66 : ////
67 : //// for a vector u of componentes u#1,u#2,u#3
68 : //// for a vector v of componentes v#1,v#2,v#3
69 : //// the temperature T
70 : //// UgradV(u,v,T) gives a vector of componente
72 : ////
73 : //// This is a non linear term when u=v
74 : ////
76 : ////
77 : //// Finite element spaces for computations
78 : ////
79 : //// Taylor-Hood P2-P1 pair for better precision. Migth be a bit expensive and
80 : //// could be better to use P1b-P1
81 : ////
82 : fespace XXXMh(Th,[P2,P2,P2,P1,P1]);////vectorial FE space for velocity and pressure
83 :
84 : XXXMh [u1,u2,u3,p,T]; ////
85 : XXXMh [uw1,uw2,uw3,pw,Tw]; ////increments during the Newton iteration
86 : XXXMh [v1,v2,v3,q,TT]; //// test functions
87 : XXXMh [uc1,uc2,uc3,pc,Tc]; //// Current computed values out of the Newton iterate
88 : ////
89 : //// Physical constants
90 : ////
91 : real nu= 1./100; //// starting viscosity
92 : real nufinal=1/1000.; //// wanted final viscosity
93 : ////
94 : //// cnu reducing factor for viscosity
95 : //// We reduce viscosity by dividing by cnu
96 : ////
97 : real cnu=2;
98 : ////
99 : //// Number of reductions on viscosity
100 : ////
101 : int countermax=10;
102 : int counter=0;
103 : ////
104 : real Pr = 56.2; //// Prandtl number
105 : real k = 1. / Pr ; //// Heat diffusion, inverse of Prandtl
106 : ////
107 : ////
108 : verbosity=0;
109 : real err=0;
110 : //// stop test for Newton
111 : real eps=1e-6;
112 :
113 : real nuc=nu;//// current viscosity
114 : cout << " ---------------------------- " << endl;
115 : cout << " — START COMPUTATIONS — " << endl;
116 : cout << " ---------------------------- " << endl;
117 : ////
120 : ////
121 : //// initial data for Newton process
122 : ////
123 : [u1,u2,u3,p,T]=[0,0,0,0,x]; //// cero velocity and heating from above
124 : plot(T,wait=0,cmm=“Initial temperature T=x”, coef=0.3,fill=1,value=1,nbiso=20);
125 : ////
126 : //// Continuation on viscosity number
127 : //// Compute first for a given viscosity and then after convergence of
128 : //// Newton method reduce the viscosity number until nufinal
129 : ////
130 : while((nuc>nufinal) && (counter<countermax)&&(cnu>1.05))
131 : { //// start loop on viscosity
132 : counter =counter+1;
133 : real rey=1./nuc;
134 : real c= - rey/Pr; //// ratio Reynolds/Prandtl
135 :
136 : int n; //// counter for Newton iteration
137 : for (n=0;n<=20;n++)//// nonlinear Newton iteration
138 : {
139 : //// initial data very first iteration is
140 : ////[u1,u2,u3,p,T]=[0,0,0,0,1-x];
141 : ////
142 : //// variables uw1,uw2,uw3,pw,Tw contain the increments computed by
143 : //// using the Gateaux o Frechet derivative
144 : ////
145 : //// v represents—> [v1,v2,v3] (test functions)
146 : //// TT test function for temperature
147 : ////
148 : solve BoussinesqNL([uw1,uw2,uw3,pw,Tw],[v1,v2,v3,q,TT])
149 : =
150 : //// start now with the term
151 : //// <dF(u,T),(uw,Tw)>
152 : ////
153 : int3d(Th) (
154 : ////
155 : ////result of the nonlinear parts on the differential mappping
156 : ////nonlinear parts (u
157 : ////
158 : ////
159 : //// u represents—> [u1,u2,u3]
160 : //// uw represents—> [uw1,uw2,uw3]
161 : ////
162 : //// Solving <dF(u,T),(uw,Tw)>=-F(u,T) using variational formulation
163 : //// and then
164 : //// u:=u+uw
165 : //// T:=T+Tw
166 : ////
169 : ////
170 : //// nu
Lap(uw)v (here uw represents—> [uw1,uw2,uw3])
171 : ////
173 : ////
174 : //// effect of Boussinesq hypothesis body force is -c
T
(0,0,1)’
175 : ////
176 : + c
Tw
v3
177 : ////
178 : ////-k
Lap(Tw)
180 : ////
181 : //// incompressibility for uw
182 : ////
183 : - div(uw) (dx(uw1)+dy(uw2)+dz(uw3)) q
184 : ////
186 : ////
187 : -div(v) (dx(v1)+dy(v2)+dz(v3)) pw
188 : ////
189 : //// stabilization term
190 : ////
191 : + 1e-8
pw
q
192 : )
193 : ////
194 : //// Now the -F(u,T) term goes +F(u,T) on left hand side
195 : + int3d(Th)(
196 : ////
197 : //// non linear term on data (u,T)
198 : //// (u
199 : ////
201 : ////
202 : ////the -nu
Lap(u)
203 : ////
205 : ////
206 : //// the body force c
T
(0,0,1)’
207 : ////
208 : + c
T
v3
209 : ////
210 : //// the -k
Lap(T)
211 : ////
213 : ////
214 : //// the incompressibility condition
215 : ////
216 : - div(u) (dx(u1)+dy(u2)+dz(u3)) q
217 : ////
219 : ////
220 : -div(v) (dx(v1)+dy(v2)+dz(v3)) p
221 : //// the stabilization term
222 : + 1e-8
p
q
223 : )
224 : //// Boundary Conditions for velocity field
225 : + on(1,2,3,4, uw1=0,uw2=0,uw3=0)//// velocity cero top, bottom and sidewalls
226 : //// Boundary Conditions for Temperature field
227 : + on(2,4,Tw=0);//// Temperature cero opposite sidewalls 2 and 4
228 : //// on the rest…normal derivative of T is cero.
229 : ////
230 : //// update all the fields…velocities and temperature whithin Newton iteration
231 : ////
232 : ////
233 : //// update values to compute again
234 : ////
235 : u1[] += uw1[]; ////u1[]=u1[]+uw1[] an update the rest in the same way
236 :
237 : //// u2[] += uw2[]; //u2[]=u2[]+uw2[]
238 : //// u3[] += uw3[]; //u3[]=u3[]+uw3[]
239 : //// p[] += pw[]; //p[]=p[]+pw[]
240 : //// T[] += Tw[]; //T[]=T[]+Tw[]
241 :
242 :
243 : ////
244 : //// computation of norms for the increments, takes five components at the same time
245 : ////
246 : real solutionl2=u1[].l2; //// l2 norm of all components [u1,u2,u3,p,T]
247 : real incrementl2=uw1[].l2; //// l2 norm of all components [uw1,uw2,uw3,pw,Tw]
248 : ////
249 : //// normalization of the norms
250 : ////
251 : err= incrementl2/solutionl2;
252 :
253 : cout << " iter = "<< n << " norm l2 of the solution is = " << solutionl2
254 : << " nu = " << nuc << endl;
255 : cout << " iter = "<< n << " norm l2 of the increment is = " << incrementl2
256 : << " nu = " << nuc << endl;
257 :
258 : cout << " iter = "<< n << " err " <<err
259 : << " nu = " << nuc << endl;
260 : ////
261 : //// plot iterate
262 : ////
263 : plot(T,wait=0,cmm=“Temperature using Newton Method for viscosity = " + nuc +”, it = "+ n+ ", err = “+err, coef=0.3,fill=1,value=1,nbiso=20);
264 : ////
265 : ////–> In case of no convergence…
266 : //// (n>3 AND err>2) is blowup
267 : //// (n>12 AND err>eps) is lack of convergence or very slow
268 : ////
269 : if( (n>3 && err > 2.) || (n>12 && err > 1) ) //// (n>3 AND err>2) or (n>12 AND err>eps)
270 : {//// start —> if
271 : cout<<” NO convergence for nu = "+nuc+ ", it = "+ n+ ", err = "+err <<endl;
272 : nuc=nuc
cnu; //// get back to previous nu
273 : cout << " LAST viscosity TO CONVERGE = " << nuc<< endl;
274 : ////
275 : //// take computed values for previous nuc to start again
276 : ////
277 : u1=uc1;
278 : ////
279 : //// New reduction for cnu
280 : ////
281 : cnu= cnu^(0.6); //// no conv. => change lower (cnu^(0.6) makes cnu smaller)
282 : nuc = nuc/cnu; //// new vicosity
283 : cout << " RESTART with NEW viscosity = " << nuc << endl;
284 : cout << " Using as Inital Data the solution for PREVIOUS v
… : iscosity " << endl;
285 : break; //// Blowup ??? —>out of for loop go to while loop
286 : }//// end —> if( n>3 && err > 2.)
287 : ////
288 : ////–>In case of Convergence then…
289 : ////
290 : if(err < eps)
291 : {//// start —> if(err < eps)
292 : cout << " CONVERGENCE for viscosity = " << nuc << " on iters = “<< n <<”, err = “+err << endl;
293 : ////uc1=u1; uc2=u2; uc3=u3; pc=p; Tc=T; // save computed values of this convergent solution
294 : uc1=u1;
295 : plot(T,wait=0,cmm=“Solution for viscosity = " + nuc,coef=0.3,fill=1,value=1,nbiso=20);
296 :
297 : if( n < 4) cnu=cnu^1.5; //// fast converge => change faster ( cnu^ 1.5 makes bigger cnu)
298 : nuc = max(nufinal, nuc/cnu); //// reduce to a new vicosity
299 : cout << " Continuation to new viscosity = " << nuc <<” (dividing factor “<<cnu<<”)”<< endl;
300 : cout << " Initial data IS the FINAL computation for previ
… : ous viscosity " << endl;
301 : break; //// converge —> out of for loop. Start again with u1,u2,p and new nuc
302 : }//// end —> if(err < eps)
303 : }////end —> for loop
304 : } //// end while loop on viscosity
305 : plot(T,cmm=“Temperature Last converge of Newton Method for nu
… : = " + nuc,coef=0.3,wait=0,fill=1,value=1,nbiso=10);
306 : cout << " LAST viscosity TO CONVERGE nu = " << nuc << endl;
307 : plot(coef=0.2,cmm=” Temperature for viscosity "+nuc,T,fill=1,value=1,nbiso=20); sizestack + 1024 =8456 ( 7432 )

## — START COMPUTATIONS —

FALLBACK (log once): Fallback to SW vertex for line stipple
FALLBACK (log once): Fallback to SW vertex processing, m_disable_code: 2000
FALLBACK (log once): Fallback to SW vertex processing in drawCore, m_disable_code: 2000
iter = 0 norm l2 of the solution is = 54.5936 nu = 0.01
iter = 0 norm l2 of the increment is = 50.1886 nu = 0.01
iter = 0 err 0.919313 nu = 0.01
iter = 1 norm l2 of the solution is = 36.0245 nu = 0.01
iter = 1 norm l2 of the increment is = 24.0584 nu = 0.01
iter = 1 err 0.667836 nu = 0.01
iter = 2 norm l2 of the solution is = 30.3662 nu = 0.01
iter = 2 norm l2 of the increment is = 9.25408 nu = 0.01
iter = 2 err 0.30475 nu = 0.01
iter = 3 norm l2 of the solution is = 28.8534 nu = 0.01
iter = 3 norm l2 of the increment is = 2.88402 nu = 0.01
iter = 3 err 0.0999543 nu = 0.01
iter = 4 norm l2 of the solution is = 28.775 nu = 0.01
iter = 4 norm l2 of the increment is = 0.252122 nu = 0.01
iter = 4 err 0.00876183 nu = 0.01
iter = 5 norm l2 of the solution is = 28.7755 nu = 0.01
iter = 5 norm l2 of the increment is = 0.00210796 nu = 0.01
iter = 5 err 7.32553e-05 nu = 0.01
iter = 6 norm l2 of the solution is = 28.7755 nu = 0.01
iter = 6 norm l2 of the increment is = 8.17775e-08 nu = 0.01
iter = 6 err 2.84191e-09 nu = 0.01
CONVERGENCE for viscosity = 0.01 on iters = 6, err = 2.84191e-09
Continuation to new viscosity = 0.005 (dividing factor 2)
Initial data IS the FINAL computation for previous viscosity
iter = 0 norm l2 of the solution is = 41.6728 nu = 0.005
iter = 0 norm l2 of the increment is = 18.7846 nu = 0.005
iter = 0 err 0.450764 nu = 0.005
iter = 1 norm l2 of the solution is = 41.504 nu = 0.005
iter = 1 norm l2 of the increment is = 4.46038 nu = 0.005
iter = 1 err 0.107469 nu = 0.005
iter = 2 norm l2 of the solution is = 41.7005 nu = 0.005
iter = 2 norm l2 of the increment is = 0.755348 nu = 0.005
iter = 2 err 0.0181136 nu = 0.005
iter = 3 norm l2 of the solution is = 41.7012 nu = 0.005
iter = 3 norm l2 of the increment is = 0.0127765 nu = 0.005
iter = 3 err 0.000306383 nu = 0.005
iter = 4 norm l2 of the solution is = 41.7012 nu = 0.005
iter = 4 norm l2 of the increment is = 6.72466e-06 nu = 0.005
iter = 4 err 1.61258e-07 nu = 0.005
CONVERGENCE for viscosity = 0.005 on iters = 4, err = 1.61258e-07
Continuation to new viscosity = 0.0025 (dividing factor 2)
Initial data IS the FINAL computation for previous viscosity
iter = 0 norm l2 of the solution is = 66.0549 nu = 0.0025
iter = 0 norm l2 of the increment is = 31.8274 nu = 0.0025
iter = 0 err 0.481832 nu = 0.0025
iter = 1 norm l2 of the solution is = 65.3784 nu = 0.0025
iter = 1 norm l2 of the increment is = 7.43208 nu = 0.0025
iter = 1 err 0.113678 nu = 0.0025
iter = 2 norm l2 of the solution is = 65.4833 nu = 0.0025
iter = 2 norm l2 of the increment is = 1.6268 nu = 0.0025
iter = 2 err 0.0248429 nu = 0.0025
iter = 3 norm l2 of the solution is = 65.4821 nu = 0.0025
iter = 3 norm l2 of the increment is = 0.147724 nu = 0.0025
iter = 3 err 0.00225594 nu = 0.0025
iter = 4 norm l2 of the solution is = 65.4822 nu = 0.0025
iter = 4 norm l2 of the increment is = 0.00298921 nu = 0.0025
iter = 4 err 4.56493e-05 nu = 0.0025
iter = 5 norm l2 of the solution is = 65.4822 nu = 0.0025
iter = 5 norm l2 of the increment is = 1.44427e-06 nu = 0.0025
iter = 5 err 2.2056e-08 nu = 0.0025
CONVERGENCE for viscosity = 0.0025 on iters = 5, err = 2.2056e-08
Continuation to new viscosity = 0.00125 (dividing factor 2)
Initial data IS the FINAL computation for previous viscosity
iter = 0 norm l2 of the solution is = 113.327 nu = 0.00125
iter = 0 norm l2 of the increment is = 61.1883 nu = 0.00125
iter = 0 err 0.539928 nu = 0.00125
iter = 1 norm l2 of the solution is = 138.326 nu = 0.00125
iter = 1 norm l2 of the increment is = 83.6301 nu = 0.00125
iter = 1 err 0.604589 nu = 0.00125
iter = 2 norm l2 of the solution is = 327.972 nu = 0.00125
iter = 2 norm l2 of the increment is = 309.394 nu = 0.00125
iter = 2 err 0.943356 nu = 0.00125
iter = 3 norm l2 of the solution is = 5173.89 nu = 0.00125
iter = 3 norm l2 of the increment is = 5173 nu = 0.00125
iter = 3 err 0.999828 nu = 0.00125
iter = 4 norm l2 of the solution is = 43883.1 nu = 0.00125
iter = 4 norm l2 of the increment is = 43792.8 nu = 0.00125
iter = 4 err 0.997941 nu = 0.00125
iter = 5 norm l2 of the solution is = 51760.5 nu = 0.00125
iter = 5 norm l2 of the increment is = 64390.4 nu = 0.00125
iter = 5 err 1.24401 nu = 0.00125
iter = 6 norm l2 of the solution is = 652427 nu = 0.00125
iter = 6 norm l2 of the increment is = 657325 nu = 0.00125
iter = 6 err 1.00751 nu = 0.00125
iter = 7 norm l2 of the solution is = 6.52001e+06 nu = 0.00125
iter = 7 norm l2 of the increment is = 6.57713e+06 nu = 0.00125
iter = 7 err 1.00876 nu = 0.00125
iter = 8 norm l2 of the solution is = 1.40046e+07 nu = 0.00125
iter = 8 norm l2 of the increment is = 1.52788e+07 nu = 0.00125
iter = 8 err 1.09099 nu = 0.00125
iter = 9 norm l2 of the solution is = 4.62e+08 nu = 0.00125
iter = 9 norm l2 of the increment is = 4.61924e+08 nu = 0.00125
iter = 9 err 0.999836 nu = 0.00125
iter = 10 norm l2 of the solution is = 1.45225e+09 nu = 0.00125
iter = 10 norm l2 of the increment is = 1.47547e+09 nu = 0.00125
iter = 10 err 1.01599 nu = 0.00125
iter = 11 norm l2 of the solution is = 3.73892e+10 nu = 0.00125
iter = 11 norm l2 of the increment is = 3.71997e+10 nu = 0.00125
iter = 11 err 0.994932 nu = 0.00125
iter = 12 norm l2 of the solution is = 1.17798e+11 nu = 0.00125
iter = 12 norm l2 of the increment is = 1.2022e+11 nu = 0.00125
iter = 12 err 1.02056 nu = 0.00125
iter = 13 norm l2 of the solution is = 3.50408e+11 nu = 0.00125
iter = 13 norm l2 of the increment is = 3.74938e+11 nu = 0.00125
iter = 13 err 1.07 nu = 0.00125
NO convergence for nu = 0.00125, it = 13, err = 1.07
LAST viscosity TO CONVERGE = 0.0025
Using as Inital Data the solution for PREVIOUS viscosity
iter = 0 norm l2 of the solution is = 88.8106 nu = 0.00164938
iter = 0 norm l2 of the increment is = 30.0889 nu = 0.00164938
iter = 0 err 0.338799 nu = 0.00164938
iter = 1 norm l2 of the solution is = 88.0441 nu = 0.00164938
iter = 1 norm l2 of the increment is = 6.9484 nu = 0.00164938
iter = 1 err 0.0789195 nu = 0.00164938
iter = 2 norm l2 of the solution is = 88.1804 nu = 0.00164938
iter = 2 norm l2 of the increment is = 5.63273 nu = 0.00164938
iter = 2 err 0.0638774 nu = 0.00164938
iter = 3 norm l2 of the solution is = 88.0055 nu = 0.00164938
iter = 3 norm l2 of the increment is = 1.63483 nu = 0.00164938
iter = 3 err 0.0185764 nu = 0.00164938
iter = 4 norm l2 of the solution is = 88.0043 nu = 0.00164938
iter = 4 norm l2 of the increment is = 0.14972 nu = 0.00164938
iter = 4 err 0.00170128 nu = 0.00164938
iter = 5 norm l2 of the solution is = 88.0042 nu = 0.00164938
iter = 5 norm l2 of the increment is = 0.00183785 nu = 0.00164938
iter = 5 err 2.08837e-05 nu = 0.00164938
iter = 6 norm l2 of the solution is = 88.0042 nu = 0.00164938
iter = 6 norm l2 of the increment is = 5.67813e-07 nu = 0.00164938
iter = 6 err 6.45211e-09 nu = 0.00164938
CONVERGENCE for viscosity = 0.00164938 on iters = 6, err = 6.45211e-09
Continuation to new viscosity = 0.00108819 (dividing factor 1.51572)
Initial data IS the FINAL computation for previous viscosity
iter = 0 norm l2 of the solution is = 133.75 nu = 0.00108819
iter = 0 norm l2 of the increment is = 68.0884 nu = 0.00108819
iter = 0 err 0.509074 nu = 0.00108819
iter = 1 norm l2 of the solution is = 163.842 nu = 0.00108819
iter = 1 norm l2 of the increment is = 108.07 nu = 0.00108819
iter = 1 err 0.6596 nu = 0.00108819
iter = 2 norm l2 of the solution is = 772.998 nu = 0.00108819
iter = 2 norm l2 of the increment is = 759.157 nu = 0.00108819
iter = 2 err 0.982095 nu = 0.00108819
iter = 3 norm l2 of the solution is = 1561.14 nu = 0.00108819
iter = 3 norm l2 of the increment is = 1555.72 nu = 0.00108819
iter = 3 err 0.996531 nu = 0.00108819
iter = 4 norm l2 of the solution is = 10970.7 nu = 0.00108819
iter = 4 norm l2 of the increment is = 11042.6 nu = 0.00108819
iter = 4 err 1.00656 nu = 0.00108819
iter = 5 norm l2 of the solution is = 1.81488e+06 nu = 0.00108819
iter = 5 norm l2 of the increment is = 1.81507e+06 nu = 0.00108819
iter = 5 err 1.00011 nu = 0.00108819
iter = 6 norm l2 of the solution is = 6.91653e+06 nu = 0.00108819
iter = 6 norm l2 of the increment is = 7.1665e+06 nu = 0.00108819
iter = 6 err 1.03614 nu = 0.00108819
iter = 7 norm l2 of the solution is = 8.98585e+06 nu = 0.00108819
iter = 7 norm l2 of the increment is = 1.06925e+07 nu = 0.00108819
iter = 7 err 1.18993 nu = 0.00108819
iter = 8 norm l2 of the solution is = 3.07881e+07 nu = 0.00108819
iter = 8 norm l2 of the increment is = 3.19718e+07 nu = 0.00108819
iter = 8 err 1.03845 nu = 0.00108819
iter = 9 norm l2 of the solution is = 6.88865e+07 nu = 0.00108819
iter = 9 norm l2 of the increment is = 7.16418e+07 nu = 0.00108819
iter = 9 err 1.04 nu = 0.00108819
iter = 10 norm l2 of the solution is = 6.74559e+08 nu = 0.00108819
iter = 10 norm l2 of the increment is = 6.71772e+08 nu = 0.00108819
iter = 10 err 0.995868 nu = 0.00108819
iter = 11 norm l2 of the solution is = 4.34913e+09 nu = 0.00108819
iter = 11 norm l2 of the increment is = 4.40264e+09 nu = 0.00108819
iter = 11 err 1.0123 nu = 0.00108819
iter = 12 norm l2 of the solution is = 1.39342e+10 nu = 0.00108819
iter = 12 norm l2 of the increment is = 1.44947e+10 nu = 0.00108819
iter = 12 err 1.04023 nu = 0.00108819
iter = 13 norm l2 of the solution is = 8.43101e+11 nu = 0.00108819
iter = 13 norm l2 of the increment is = 8.43635e+11 nu = 0.00108819
iter = 13 err 1.00063 nu = 0.00108819
NO convergence for nu = 0.00108819, it = 13, err = 1.00063
LAST viscosity TO CONVERGE = 0.00164938
Using as Inital Data the solution for PREVIOUS viscosity
iter = 0 norm l2 of the solution is = 107.797 nu = 0.00128514
iter = 0 norm l2 of the increment is = 27.2448 nu = 0.00128514
iter = 0 err 0.252743 nu = 0.00128514
iter = 1 norm l2 of the solution is = 108.026 nu = 0.00128514
iter = 1 norm l2 of the increment is = 20.6578 nu = 0.00128514
iter = 1 err 0.191231 nu = 0.00128514
iter = 2 norm l2 of the solution is = 107.749 nu = 0.00128514
iter = 2 norm l2 of the increment is = 12.5265 nu = 0.00128514
iter = 2 err 0.116256 nu = 0.00128514
iter = 3 norm l2 of the solution is = 112.77 nu = 0.00128514
iter = 3 norm l2 of the increment is = 33.9331 nu = 0.00128514
iter = 3 err 0.300905 nu = 0.00128514
iter = 4 norm l2 of the solution is = 109.131 nu = 0.00128514
iter = 4 norm l2 of the increment is = 26.5275 nu = 0.00128514
iter = 4 err 0.243078 nu = 0.00128514
iter = 5 norm l2 of the solution is = 1384.5 nu = 0.00128514
iter = 5 norm l2 of the increment is = 1382.35 nu = 0.00128514
iter = 5 err 0.998452 nu = 0.00128514
iter = 6 norm l2 of the solution is = 1188 nu = 0.00128514
iter = 6 norm l2 of the increment is = 1178.22 nu = 0.00128514
iter = 6 err 0.99177 nu = 0.00128514
iter = 7 norm l2 of the solution is = 15670.2 nu = 0.00128514
iter = 7 norm l2 of the increment is = 15673.3 nu = 0.00128514
iter = 7 err 1.0002 nu = 0.00128514
iter = 8 norm l2 of the solution is = 32865 nu = 0.00128514
iter = 8 norm l2 of the increment is = 34103.6 nu = 0.00128514
iter = 8 err 1.03769 nu = 0.00128514
iter = 9 norm l2 of the solution is = 229752 nu = 0.00128514
iter = 9 norm l2 of the increment is = 231107 nu = 0.00128514
iter = 9 err 1.0059 nu = 0.00128514
iter = 10 norm l2 of the solution is = 1.7383e+06 nu = 0.00128514
iter = 10 norm l2 of the increment is = 1.78141e+06 nu = 0.00128514
iter = 10 err 1.0248 nu = 0.00128514
iter = 11 norm l2 of the solution is = 4.5312e+06 nu = 0.00128514
iter = 11 norm l2 of the increment is = 5.00278e+06 nu = 0.00128514
iter = 11 err 1.10407 nu = 0.00128514
iter = 12 norm l2 of the solution is = 7.86183e+07 nu = 0.00128514
iter = 12 norm l2 of the increment is = 7.84382e+07 nu = 0.00128514
iter = 12 err 0.997709 nu = 0.00128514
iter = 13 norm l2 of the solution is = 1.34233e+08 nu = 0.00128514
iter = 13 norm l2 of the increment is = 1.39051e+08 nu = 0.00128514
iter = 13 err 1.03589 nu = 0.00128514
NO convergence for nu = 0.00128514, it = 13, err = 1.03589
LAST viscosity TO CONVERGE = 0.00164938
Using as Inital Data the solution for PREVIOUS viscosity
iter = 0 norm l2 of the solution is = 99.0145 nu = 0.00142004
iter = 0 norm l2 of the increment is = 14.6384 nu = 0.00142004
iter = 0 err 0.147841 nu = 0.00142004
iter = 1 norm l2 of the solution is = 99.0791 nu = 0.00142004
iter = 1 norm l2 of the increment is = 5.42687 nu = 0.00142004
iter = 1 err 0.0547731 nu = 0.00142004
iter = 2 norm l2 of the solution is = 98.8848 nu = 0.00142004
iter = 2 norm l2 of the increment is = 1.85827 nu = 0.00142004
iter = 2 err 0.0187923 nu = 0.00142004
iter = 3 norm l2 of the solution is = 98.8679 nu = 0.00142004
iter = 3 norm l2 of the increment is = 0.50027 nu = 0.00142004
iter = 3 err 0.00505999 nu = 0.00142004
iter = 4 norm l2 of the solution is = 98.8675 nu = 0.00142004
iter = 4 norm l2 of the increment is = 0.0185601 nu = 0.00142004
iter = 4 err 0.000187727 nu = 0.00142004
iter = 5 norm l2 of the solution is = 98.8675 nu = 0.00142004
iter = 5 norm l2 of the increment is = 3.41836e-05 nu = 0.00142004
iter = 5 err 3.45751e-07 nu = 0.00142004
CONVERGENCE for viscosity = 0.00142004 on iters = 5, err = 3.45751e-07
Continuation to new viscosity = 0.00122258 (dividing factor 1.16151)
Initial data IS the FINAL computation for previous viscosity
iter = 0 norm l2 of the solution is = 112.495 nu = 0.00122258
iter = 0 norm l2 of the increment is = 21.4033 nu = 0.00122258
iter = 0 err 0.19026 nu = 0.00122258
iter = 1 norm l2 of the solution is = 115.015 nu = 0.00122258
iter = 1 norm l2 of the increment is = 30.1166 nu = 0.00122258
iter = 1 err 0.261848 nu = 0.00122258
iter = 2 norm l2 of the solution is = 1952.41 nu = 0.00122258
iter = 2 norm l2 of the increment is = 1958 nu = 0.00122258
iter = 2 err 1.00287 nu = 0.00122258
iter = 3 norm l2 of the solution is = 979.026 nu = 0.00122258
iter = 3 norm l2 of the increment is = 981.707 nu = 0.00122258
iter = 3 err 1.00274 nu = 0.00122258
iter = 4 norm l2 of the solution is = 1362.15 nu = 0.00122258
iter = 4 norm l2 of the increment is = 1363.92 nu = 0.00122258
iter = 4 err 1.0013 nu = 0.00122258
iter = 5 norm l2 of the solution is = 15041.9 nu = 0.00122258
iter = 5 norm l2 of the increment is = 15045.3 nu = 0.00122258
iter = 5 err 1.00022 nu = 0.00122258
iter = 6 norm l2 of the solution is = 31833.9 nu = 0.00122258
iter = 6 norm l2 of the increment is = 32619.8 nu = 0.00122258
iter = 6 err 1.02469 nu = 0.00122258
iter = 7 norm l2 of the solution is = 122936 nu = 0.00122258
iter = 7 norm l2 of the increment is = 128315 nu = 0.00122258
iter = 7 err 1.04376 nu = 0.00122258
iter = 8 norm l2 of the solution is = 1.69228e+06 nu = 0.00122258
iter = 8 norm l2 of the increment is = 1.69446e+06 nu = 0.00122258
iter = 8 err 1.00129 nu = 0.00122258
iter = 9 norm l2 of the solution is = 5.27616e+06 nu = 0.00122258
iter = 9 norm l2 of the increment is = 5.33833e+06 nu = 0.00122258
iter = 9 err 1.01178 nu = 0.00122258
iter = 10 norm l2 of the solution is = 5.51745e+07 nu = 0.00122258
iter = 10 norm l2 of the increment is = 5.5605e+07 nu = 0.00122258
iter = 10 err 1.0078 nu = 0.00122258

Thank you for trying out the example, I am not using the cocoa executable. I installed it on a RHEL(Oracle linux version 9) and with that binary I get a seg-fault.
More specifically I am getting an exception of “Broken pipe” at the following line:
‘fwrite(ptr,size,nmemb,stream);’ in ffapi.cpp line 198

The call stack looks like:

You should deactivate plotting, probably by using the additional command line parameter `-nw`, then the code will run fine.

Maybe these are two different problems indeed but similar outcome. Without plotting, I am getting a seg fault exception somewhere in the UMF_kernel, GEMV function. Probably one of the column vectors is unfound, one symbol is optimised out so can’t say for sure.

There is something wrong with your installation. How did you compile FreeFEM? What version are you using?

I compiled the develop branch of FreeFem-sources, because there were issues related to 3rd-party downloads on the master branch. Some checks are indeed failing when I run make check:

test-suite.log (173.2 KB)

The only little hack I made to get it all compiled was commenting out a couple of lines in plugin mpi: