i have a nonlinear equation 2

Question

beso ss 2018-5-4

1
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/399149-i-have-a-nonlinear-equation-2

编辑： Walter Roberson 2018-5-14

在 MATLAB Online 中打开

i have three equations which are nonlinear equations , i want to use only Biesction method to solve it in matlab ,,

E1= E0 – (alfa* T^2)/(theta+T)

E2= E0 – (alfa* T^2) / (theta+T) – x * KB * T

(x * e^x)= [(sigma /(KB * T))^2 – X ] * (τaur/τautr ) * e^(((E0 – Ea)) ⁄ KB * T)

where E1 and E2 are the energy , and they are related to each other the diffrent is in E2 we have the part of x incloud ,, and x change each time when T change , it is a small change but it apear in the figer

,,T is the temperature and it is change from 0 to 300 ,, x is the temperature-dependent coefficient and i have a condition on x which is : 0<X < (sigma/(KB.*T(i)))^2

that x is larger than zereo and less than (sigma/(KB*T(i)))^2

sigma = 33*10^-3; %eV
deltaE = -0.80; %eV   , deltaE means E0 – Ea
E0 = 3.184; %eV
alfa=0.011; % ev/k
tautr/taur =0.08;
theta=630; %K
KB= 8.617*10^-5; %eV/K
T= 0:300;

......................

I write the equation for x in biesction method but i dont know how to write the condition on x inside it ,, also ,, the plot give me a result for x which are minuse and thats wrong

syms  sigma deltaE E0 tautr taur KB E2 alfa theta
sigma = 33*10^-3; %eV
deltaE = -0.080; %eV
E0 = 3.184; %eV
tautr =0.08;
taur=1.0;
theta= 630;% K
alfa= 0.0011 ; % ev/k
KB= 8.617*10^-5; %eV/K
T=0:300;
X=zeros(1,numel(T));
E2 = zeros(size(T));
for i=1:numel(T)
syms x
min=0;
max=(sigma/(KB*T(i)))^2;
f=@(x) ((sigma/(KB*T(i)))^2-x)*(tautr/taur)*exp(deltaE/(KB*T(i)))-x*exp(x);
X(i)=bisection(f,f(max),f(min))
E2(i) = E0  -( alfa * T(i)^2 )/(theta + T(i)) - X(i) * KB * T(i);
clear x
end
plot(T,X)
plot(T,E2)

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beso ss 2018-5-5

anyone??

:(

Yi-Lin Tsai 2018-5-14

编辑：Walter Roberson 2018-5-14

在 MATLAB Online 中打开

Dear beso ss do you know how to get this value

sigma = 33*10^-3; %eV
deltaE = -0.080; %eV
E0 = 3.184; %eV
tautr =0.08;
taur=1.0;
theta= 630;% K
alfa= 0.0011 ; % ev/k

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Answer 1

Walter Roberson 2018-5-6

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/399149-i-have-a-nonlinear-equation-2#answer_318873

Because of the division by T, there is no solution for T = 0. As T approaches 0, it becomes difficult to track down what the solutions, x, are.

For example as T approaches 1/100, I am seeing some suggestion that there is a solution between 10^-40311.72 and 10^-40311.71 but it is tricky to nail down, as the values switch between roughly -2.4018*10^(-40314) and 2.0366*10^(-40314)

For T = 1, there is a solution, but it is too small to represent.

The solutions from T = 1 to 300 are approximately

[0.74397e-399, 0.73862e-198, 0.51985e-131, 0.11637e-97, 0.10762e-77, 0.20581e-64, 0.60189e-55, 0.73029e-48, 0.22970e-42, 0.56185e-38, 0.21492e-34, 0.20475e-31, 0.67035e-29, 0.94908e-27, 0.68765e-25, 0.28927e-23, 0.77801e-22, 0.14420e-20, 0.19542e-19, 0.20298e-18, 0.16791e-17, 0.11413e-16, 0.65405e-16, 0.32290e-15, 0.13983e-14, 0.53932e-14, 0.18768e-13, 0.59589e-13, 0.17428e-12, 0.47341e-12, 0.12031e-11, 0.28785e-11, 0.65202e-11, 0.14050e-10, 0.28929e-10, 0.57130e-10, 0.10858e-9, 0.19924e-9, 0.35389e-9, 0.61001e-9, 0.10227e-8, 0.16709e-8, 0.26655e-8, 0.41583e-8, 0.63537e-8, 0.95218e-8, 0.14013e-7, 0.20275e-7, 0.28873e-7, 0.40505e-7, 0.56031e-7, 0.76488e-7, 0.10312e-6, 0.13740e-6, 0.18105e-6, 0.23608e-6, 0.30480e-6, 0.38982e-6, 0.49413e-6, 0.62107e-6, 0.77436e-6, 0.95814e-6, 0.11770e-5, 0.14357e-5, 0.17399e-5, 0.20954e-5, 0.25083e-5, 0.29854e-5, 0.35339e-5, 0.41614e-5, 0.48758e-5, 0.56855e-5, 0.65993e-5, 0.76264e-5, 0.87762e-5, 0.10059e-4, 0.11484e-4, 0.13062e-4, 0.14804e-4, 0.16720e-4, 0.18822e-4, 0.21121e-4, 0.23628e-4, 0.26355e-4, 0.29312e-4, 0.32512e-4, 0.35966e-4, 0.39685e-4, 0.43681e-4, 0.47965e-4, 0.52548e-4, 0.57441e-4, 0.62654e-4, 0.68199e-4, 0.74086e-4, 0.80325e-4, 0.86925e-4, 0.93896e-4, 0.10125e-3, 0.10899e-3, 0.11713e-3, 0.12567e-3, 0.13463e-3, 0.14400e-3, 0.15381e-3, 0.16405e-3, 0.17472e-3, 0.18585e-3, 0.19742e-3, 0.20945e-3, 0.22194e-3, 0.23490e-3, 0.24832e-3, 0.26221e-3, 0.27657e-3, 0.29141e-3, 0.30673e-3, 0.32252e-3, 0.33879e-3, 0.35554e-3, 0.37277e-3, 0.39048e-3, 0.40867e-3, 0.42733e-3, 0.44647e-3, 0.46608e-3, 0.48616e-3, 0.50671e-3, 0.52772e-3, 0.54920e-3, 0.57114e-3, 0.59353e-3, 0.61637e-3, 0.63965e-3, 0.66338e-3, 0.68755e-3, 0.71214e-3, 0.73716e-3, 0.76260e-3, 0.78845e-3, 0.81472e-3, 0.84138e-3, 0.86844e-3, 0.89588e-3, 0.92371e-3, 0.95191e-3, 0.98048e-3, 0.10094e-2, 0.10387e-2, 0.10683e-2, 0.10983e-2, 0.11286e-2, 0.11592e-2, 0.11901e-2, 0.12214e-2, 0.12529e-2, 0.12847e-2, 0.13169e-2, 0.13492e-2, 0.13819e-2, 0.14148e-2, 0.14480e-2, 0.14814e-2, 0.15150e-2, 0.15489e-2, 0.15830e-2, 0.16173e-2, 0.16518e-2, 0.16864e-2, 0.17213e-2, 0.17564e-2, 0.17916e-2, 0.18270e-2, 0.18625e-2, 0.18982e-2, 0.19340e-2, 0.19700e-2, 0.20061e-2, 0.20423e-2, 0.20786e-2, 0.21150e-2, 0.21515e-2, 0.21881e-2, 0.22248e-2, 0.22615e-2, 0.22983e-2, 0.23352e-2, 0.23721e-2, 0.24091e-2, 0.24461e-2, 0.24832e-2, 0.25203e-2, 0.25574e-2, 0.25945e-2, 0.26316e-2, 0.26687e-2, 0.27058e-2, 0.27429e-2, 0.27800e-2, 0.28171e-2, 0.28541e-2, 0.28911e-2, 0.29281e-2, 0.29651e-2, 0.30019e-2, 0.30388e-2, 0.30756e-2, 0.31123e-2, 0.31489e-2, 0.31855e-2, 0.32220e-2, 0.32584e-2, 0.32948e-2, 0.33310e-2, 0.33672e-2, 0.34032e-2, 0.34392e-2, 0.34751e-2, 0.35108e-2, 0.35465e-2, 0.35820e-2, 0.36174e-2, 0.36527e-2, 0.36878e-2, 0.37228e-2, 0.37577e-2, 0.37925e-2, 0.38271e-2, 0.38616e-2, 0.38959e-2, 0.39301e-2, 0.39642e-2, 0.39981e-2, 0.40318e-2, 0.40654e-2, 0.40988e-2, 0.41321e-2, 0.41652e-2, 0.41981e-2, 0.42309e-2, 0.42635e-2, 0.42959e-2, 0.43282e-2, 0.43603e-2, 0.43922e-2, 0.44239e-2, 0.44554e-2, 0.44868e-2, 0.45180e-2, 0.45490e-2, 0.45798e-2, 0.46104e-2, 0.46409e-2, 0.46711e-2, 0.47012e-2, 0.47311e-2, 0.47608e-2, 0.47903e-2, 0.48195e-2, 0.48486e-2, 0.48776e-2, 0.49063e-2, 0.49348e-2, 0.49631e-2, 0.49912e-2, 0.50191e-2, 0.50468e-2, 0.50744e-2, 0.51017e-2, 0.51288e-2, 0.51557e-2, 0.51825e-2, 0.52090e-2, 0.52353e-2, 0.52614e-2, 0.52873e-2, 0.53130e-2, 0.53385e-2, 0.53638e-2, 0.53889e-2, 0.54139e-2, 0.54386e-2, 0.54631e-2, 0.54873e-2, 0.55114e-2, 0.55353e-2, 0.55590e-2, 0.55825e-2, 0.56058e-2, 0.56289e-2, 0.56518e-2, 0.56745e-2, 0.56970e-2, 0.57193e-2, 0.57413e-2, 0.57632e-2, 0.57849e-2, 0.58064e-2, 0.58277e-2, 0.58488e-2]

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beso ss 2018-5-6

but even when i but T= 10 : 300 start with 10 not 0 ,,, the same thing .. there are som muins values of x

how can i write the conditons of x that

0<X < (sigma/(KB.*T(i)))^2

in this code .. to privent the minus values ???

Walter Roberson 2018-5-6

在 MATLAB Online 中打开

We do not know. You are using a routine, "bisection", that we do not have the source to. Perhaps the code for it is wrong. Perhaps you are calling it incorrectly. Currently you are calling

X(i)=bisection(f,f(max),f(min))

but it would seem more likely that you should be calling

X(i) = bisection(f, min, max)

I usually find it a waste of time to try to debug a program that uses "min" or "max" or "sum" as a variable name.

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