ODE15s: segmentation violation and imaginary parts

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Marco Sammito
Marco Sammito 2016-11-21
编辑： Jan 2016-11-23
Hi. I am solving
If I set P0=2100 I get a result, while if I set P0=2000: the first time I run the code, MATLAB returns "Segmentation violation"; if I try a second time running the code, MATLAB returns "Warning: Imaginary parts of complex X and/or Y arguments ignored." Can anyone help me out? Thank you.
function out = convezione_20(t,y,R0)
S = 0.073;
rho = 998;
mi = 1.005e-3;
P0 = 2000; %<-------------------------------------------
L = 2454000;
Tinf = 293;
rho_g = 1/57.79;
lambda = 0.5982;
cp = 4184;
Pv_Tinf = 2329.6;
%
out=[y(2)
-1.5 * y(2)^2 / y(1) + 1 / rho / y(1) * (Pv_Tinf - P0 +...
(L * rho_g)^2 / Tinf * y(2) / y(3) -...
4 * mi * y(2) / y(1) - 2 * S / y(1))
-y(3) + lambda * 0.5 / y(1) * (2 + (-6 * rho * cp * y(2) / pi / y(3))^(1/3)...
- 12 * rho * cp * y(2) / pi / y(3)) ];
run file
R0 = 510e-6;
y0 = [R0; 0; 1172.94];
M = [1 0 0; 0 1 0; 0 0 0];
options = odeset('Mass',M);%,'RelTol',1e-10,'AbsTol',[1e-10 1e-10 1e-10]);
[t,y] = ode15s(@(t,y) convezione_20(t,y,R0),[0 40e-6],y0,options);
t = t * 1000000;
y = y * 1000000;
[t,y(:,1)];
plot(t,y(:,1))
Edit: this is what MATLAB says:
(...)
Register State:
EAX = 00000008 EBX = 00000000
ECX = 02723d90 EDX = 00000000
ESI = 00cdb99c EDI = 00cdb988
EBP = 00000000 ESP = 00cdb750
EIP = 00000000 FLG = 00210246
(...)
Error in ==> convezione_20 at 13
out=[y(2)
Error in ==> run_convezione_20>@(t,y) convezione_20(t,y,R0) at 5
[t,y] = ode15s(@(t,y) convezione_20(t,y,R0),[0 40e-6],y0,options);
Error in ==> funfun\private\odenumjac at 131
Fdel(:,j) = feval(F,Fargs{1:diffvar-1},ydel(:,j),Fargs{diffvar+1:end});
Error in ==> funfun\private\daeic12 at 38
[DfDy,Joptions.fac,nF] = odenumjac(fun, {t0,y,args{:}}, f, Joptions);
Error in ==> ode15s at 374
[y,yp,f0,dfdy,nFE,nPD,Jfac] = daeic12(odeFcn,odeArgs,t,ICtype,Mt,y,yp0,f0,...
Error in ==> run_convezione_20 at 5
[t,y] = ode15s(@(t,y) convezione_20(t,y,R0),[0 40e-6],y0,options);
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Torsten
Torsten 2016-11-23
Does your initial value for h (h0) together with the initial values for R and Rdot (R0 and Rdot0) satisfy the algebraic equation, i.e.
-h0 + lambda * 0.5 / R0 * (2 + (-6 * rho * cp * Rdot0 / pi / h0)^(1/3) - 12 * rho * cp * Rdot0 / pi / h0) = 0
?
If not, first try to adjust h0 such that this condition holds.
Best wishes
Torsten.
Jan
Jan 2016-11-23
@Marco: Here are 2 problems:
  1. The segementation violation
  2. The mathematical problem
It is not worth to examine the 2nd one before the 1st is solved, because a segmentation violation means, that either the RAM, the processor, Matlab itself or a MEX function works incorrectly. You cannot solve the mathematical problem, when the computing system is not working reliably.
See [EDITED in my answer]

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回答（1 个）

Jan
Jan 2016-11-22
编辑：Jan 2016-11-23
A segmentation violation is severe. You cannot trust any computations afterwards. I cannot imagine that the posted code produces such an error. Do you use any user-defined Mex-functions?
[EDITED] The segmentation violation must be solved at first:
  • Check again in any user defined C-Mex code is running:
[mFun, mexFun] = inmem
after your code ran.
  • Run exhaustive unit-tests, if you have some.
  • Run examples from the documentation with ode15s.
  • Check the hardware: CPU, memory, ...
  • Re-install Matlab
When the system is running reliably, caring about the mathematical problem is the next step. The input parameters of the mathematical problem might trigger the segmentation fault, but they are not the reason.
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Jan
Jan 2016-11-22
编辑：Jan 2016-11-22
A segmentation vialoation means, that some code has written to a wrong memory address. This could cause any arbitrary behavior. Restarting Matlab is obligatory (and as far as I know this happens automatically). But if the code runs the next time, this might be luck only: Code, which creates memory violations is not trustworthy.
Do you get any further information? "MATLAB returns 'Segmentation violation'" is a lean description of a serios error.

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