solve() is unable to find a solution
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
SOURAV KUMAR
am 26 Mär. 2021
Kommentiert: Alan Stevens
am 14 Apr. 2021
Hello everyone,
I was trying to solve a transcendental eqation in MATLAB
I performed two programs one graphical approach and another by solve()
In the graphical approach, I find one solution exists in the range
∈ (
) for
, as shown in the figure below




i.e., the intersection of blue and red curve (
)

Now, when i performed another program (T2.m) using solve() :
Command Prompt is giving empty sym with the following message:
Warning: 3 equations in 1 variables.
> In C:\Program Files\MATLAB\R2013a\toolbox\symbolic\symbolic\symengine.p>symengine at 56
In mupadengine.mupadengine>mupadengine.evalin at 97
In mupadengine.mupadengine>mupadengine.feval at 150
In solve at 170
In T2 at 16
Warning: Explicit solution could not be found.
> In solve at 179
In T2 at 16
k1_sol = [ empty sym ]
My T2.m is as follows:
clc
clear all
close all
one_eV= 1.6/(10^19);
Vo=0.3 * one_eV;
b=5/(10^9);
S=Vo*b;
a=10/(10^9);
m=0.067 * 9.1/(10^31);
h=6.626/(10^34);
hbar= h/(2*pi);
k=pi/(4*a);
lhs= cos(k*a);
syms k1
rhs=cos(k1*a) + (m*S)*sin(k1*a)/(hbar^2 * k1);
k1_sol= solve(lhs== rhs,k1 >= 0.27*(10^9), k1 <= 0.32*(10^9))
so, how to rectify this?
0 Kommentare
Akzeptierte Antwort
Alan Stevens
am 26 Mär. 2021
Just solve it numerically using fzero. Take the initial guess close to the average value of the bounds you specify:
k1_0 = 0.29E9; % initial guess
k1 = fzero(@k1fn,k1_0);
disp(k1)
disp(cos(pi/4)) % = cos(k*a)
function Z = k1fn(k1)
one_eV= 1.6/(10^19);
Vo=0.3 * one_eV;
b=5/(10^9);
S=Vo*b;
a=10/(10^9);
m=0.067 * 9.1/(10^31);
h=6.626/(10^34);
hbar= h/(2*pi);
k=pi/(4*a);
Z = cos(k1*a) + (m*S)*sin(k1*a)/(hbar^2 * k1) - cos(k*a);
end
this reults in
2.7859e+08 for k1 and 0.7071 for left and right hand sides of your equation.
2 Kommentare
Weitere Antworten (0)
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!