fsolve no solution found.

Hi everybody, I'm trying to resolve a system equation with the fsolve() command. The problem is that if I choose a different value for the addictional parameter the solution converge until a certain value. For a value below 2.3e3 the solution converge,instead for a value above: "fsolve stopped because the problem appears regular as measured by the gradient, but the vector of function values is not near zero as measured by the default value of the function tolerance."

Ptot = 2.3e4;
x0 = [10 10];
P=Ptot;
f = @(X)myfunFC(X,P);
options = optimset('MaxFunEvals',10000,'MaxIter',10000);
[x,fval,exitflag] = fsolve(f,x0,options);

The function myfunFC is:

function F = myfunFC(X,P)

G0 = 237*10^3;  
F = 96485.3415; 
E0 = G0/(2*F);  
T = 25+273.15;  
R = 8.314;      
pH2 = 5;        
pO2 = 5;        

A=0.03;
r=0.000245;
m=2.11E-5;
n=0.008;
i0 = 0.01;
i= 0.01:0.1:1500;

E = E0 + ((R*T)/(2*F))*log(pH2*pO2^0.5);

Eoc = E + A*log(i0);

C = 0.06;
PO1=1;  
PO2=5;  
DeltaVgain = C*log(PO2/PO1);

T1 = 25+273.15;     
eta_m = 0.9;        
eta_c= 0.7;         
lambda = 2;         
Pratio=PO2/PO1;     

Vloss = 3.58*10^(-4)*(15/(eta_m*eta_c))*(Pratio^0.286 - 1)*lambda;

i = X(1);
Vtot = X(2);

F(1)= Vtot - ((Eoc - r*i - A*log(i) - m*exp(n*i))+DeltaVgain -Vloss);

F(2)= P-24*(Vtot*i)*10^(-3)*232;

I tried to change the initial value but the problem still the same.

[Sean de Wolski]

So you know, the first assignment of 'i' and 'T1' are never used. Are they supposed to be?

[Mario]

Thanks, I forgot to insert the T1 value in the expression:

Vloss = 3.58*10^(-4)*(T1/(eta_m*eta_c))*(Pratio^0.286 - 1)*lambda;

'i' is one of the two variables, and the problem still the same

[Walter Roberson]

Please avoid using a variable named "i", as "i" can also mean the square root of negative one.

[Mario]

You are right, I changed with "I" but the problem still not resolve it.

Answer 1

[John D'Errico]

Consider this simple function of one variable.

fun = @(x) (x-1).*exp(-x) + 0.5;
fplot(fun,[0,10])
yline(0);
grid on

As you can see, there is a root roughly at x = 0.3. What happens if we start fsolve near there?

[xsol,fsol,exitfflag] = fsolve(fun,1)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

xsol = 0.3149

fsol = -2.5123e-07

exitfflag = 1

So fsolve was happy, finding the expected solution. Any starting value less than 2 would seem to result in the desired solution.

But what happens if we start fsolve in a bad place?

[xsol,fsol,exitfflag] = fsolve(fun,3)

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,
but the vector of function values is not near zero as measured by the
value of the function tolerance.

xsol = 16.5000

fsol = 0.5000

exitfflag = -2

So fsolve knows it has not found a solution. It keeps trying to reduce the function by increasing x. But clearly, you can see this will never work. Eventually, fsolve gives up, around x==16.5 as you can see. There the function is quite close to a constant.

format long g
fun(16.5 + [0 .1 .2])

ans = 1×3

0.500001057968523 0.500000963465568 0.500000877368029

As you can see, fun is asymptotically approaching a constant.

No matter how large x is made, fun(x) will continuously decrease, but it will never be less that 0.5. fsolve does not understand what the function does, or that it needed to look below x==2.

This is the same behaviour you saw, although in a simple function we can easily visualize.