Why do I get empty result?

Hello all,

I have the following equation that needs to be solved by MATLAB but it gives me empty sym. I don't know why

syms T a b
eqn= (a*b*exp(b/T)*((1 - exp(a*b*ei(b/T) - T*a*exp(b/T))*exp(294*a*exp(b/294) - a*b*ei(b/294))) - 1))/T^2 == 0;
solT = solve(eqn,T)

Even when I give values for a and b, MATLAb gives me the same message:

syms T 
a=(exp(183.8)/1.5);
b=-6.2662e+04;
eqn= (a*b*exp(b/T)*((1 - exp(a*b*ei(b/T) - T*a*exp(b/T))*exp(294*a*exp(b/294) - a*b*ei(b/294))) - 1))/T^2 == 0;
solT = solve(eqn,T)

I would be thankful if someone help me with this.

[darova]

It's too complicated for symbolic calculations. Try numerical - fsolve

[Faezeh Manesh]

I changed my code as follows:

clc
clear all
%Substitute n(T) from the abobe solution into d2n/dT2

syms T 
a=(exp(183.8)/1.5);
b=-6.2662e+04;
odefun = @func;
x0 = [300];
x = fsolve(odefun,x0);
% eqn= (a*b*exp(b/T)*((1 - exp(a*b*ei(b/T) - T*a*exp(b/T))*exp(294*a*exp(b/294) - a*b*ei(b/294))) - 1))/T^2 == 0;
% 
% solT = solve(eqn,T)

function F = func(x)
a=(exp(183.8)/1.5);
b=-6.2662e+04;
F(1) = (a*b*exp(b/x(1))*((1 - exp(a*b*ei(b/x(1)) - x(1)*a*exp(b/x(1)))*exp(294*a*exp(b/294) - a*b*ei(b/294))) - 1))/x(1)^2 == 0;
end

But it gives the message that "FSOLVE requires all values returned by functions to be of data type double". I don't know why

[darova]

Try this

[Faezeh Manesh]

Now it works. But it gives the initial point (x=300) as the final answer which is incorrect.

[darova]

I draw you function

a = (exp(183.8)/1.5);
b = -6.2662e+04;
F = @(x) (a*b*exp(b/x)*((1 - exp(a*b*ei(b/x) - x*a*exp(b/x))*exp(294*a*exp(b/294) - a*b*ei(b/294))) - 1))/x^2;
x = linspace(100,400);
y = arrayfun(F,x);
plot(x,y)
% x0 = [300];
% x = fsolve(odefun,x0);
ylim([-1 1]*0.1)

Result

[Faezeh Manesh]

Thanks. Then how can I find the peak? My ultimate goal is to find the peak in terms of a and b

Answer 1

[Steven Lord]

When I display the expression in the Live Editor, I see that the numerator of the expression is the product of a (very large) constant positive value and two exponentials. That numerator, divided by T^2, is supposed to be equal to 0.

The constant is positive. The values of the two exponentials are only 0 in the limit; there's no finite value z for which exp(z) is 0.

So MATLAB correctly gave the answer that there is no solution to your equation as written.

Your expression does look a little unusual, though. Did you intend to have the Ei and one of the exponentials inside the exponent of another exponential call? Maybe there's a parenthesis in the wrong place?

[Faezeh Manesh]

Actually what I am trying to do is as follows:

I have a diferential equation which has the following form:

when I solved this (with a and b known) and thenfind dy/dT in terms of T, it becomes a figure like this:

I am trying to find the peak in terms of a and b. So, firstly I solved my differential equation symbollicaly in MATLAB to find y(T), then I differentiate dy/dT once more to find dy2/dT2 in terms of y and T. Then I substituted y(T) which had been obtained previously in it. So, my equation would only be in terms of T and then I want to put it equal to zero to find the peak in terms of a and b. but here I substituted my experimental value for a and b o check my answer. So, all I found is from MATLAB