Why do I get empty result?

Question

Faezeh Manesh 2020-5-4

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评论： Faezeh Manesh 2020-5-5

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.

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darova 2020-5-5

在 MATLAB Online 中打开

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 2020-5-5

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

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Answer 1

Steven Lord 2020-5-4

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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?

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Faezeh Manesh 2020-5-5

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

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