# Solve a nonlinear equation

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cm on 18 Nov 2013
Edited: Walter Roberson on 23 Apr 2017
1.082 [ (q-1) log((3/4 -3q/4)/(1-3q/4))+ q*log((3q/4) *(3/4 -3q/4)] + 1.4427[-3log(1-q/2) *((q/2) -1)+(q/2+1/2)^2 *log(q/2+1/2)- (3q/4 -1)*log(1-3q/4)- (3q/4+1/4)*log(3q/4+1/4)] + 0.7213[q*log(q/2)* ( -q-3/2 +1/q)+ (-q-1)*log(q/2 +1/2) + q*(-q/2+1/2)*log(1/2-q/2)-3*log(1-q/2)-(q-1)*log(1/2 - q/2)] -0.36 *[(q-1)^2*log(1/2 - q/2)]=0
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cm on 18 Nov 2013
Edited: Walter Roberson on 23 Apr 2017
Natural Logarithm
Let me re write it as it is shown in MATLAB command window:
1.082 *( (q-1) *log((3/4 -3*q/4)/(1-3*q/4))+ q*log((3*q/4)* (3/4 -3*q/4)) )+ 1.4427*(-3*log(1-q/2) *((q/2) -1)+((q/2+1/2)^2) *log(q/2+1/2)- (3*q/4 -1)*log(1-3*q/4)- (3*q/4+1/4)*log(3*q/4+1/4)) + 0.7213*(q*log(q/2)*( -q-3/2 +1/q)+ (-q-1)*log(q/2 +1/2) + q*(-q/2+1/2)*log(1/2-q/2)-3*log(1-q/2)-(q-1)*log(1/2 - q/2)) -0.36 *((q-1)^2*log(1/2 - q/2))
thanks,

Walter Roberson on 18 Nov 2013
If you assume natural logs, then there are no non-complex solutions.
The value of the expression (assuming natural logs) is negative between 0 and 1 (going to negative infinity at those two bounds.) Outside of that range, the expression is real-valued at exactly one point near -3.000138648 and at exactly one point near 1.851008154. At those two exceptional locations, the expression has positive non-zero values. The fact that there are negative values and positive values does not, however, imply that there are points at which the function is real-valued and 0, as the function is discontinuous.
The results have the same shape if log10 is used instead of ln.
cm on 18 Nov 2013
yes I was looking for a value between 0 and 1 but it seems that the equation can not satisfy this constraint

Matt J on 18 Nov 2013
solve, fzero, fsolve?
cm on 18 Nov 2013
I used "solve" but the process took so long without any appropriate result! i will be so pleased if you introduce some thing that works for this equation, some stronger function

Andrei Bobrov on 18 Nov 2013
syms q
>> ex = 1.082*( (q-1)*log((3/4 -3*q/4)/(1-3*q/4))+ q*log((3*q/4) *(3/4 -3*q/4)) + 1.4427*(-3*log(1-q/2) *((q/2) -1)+(q/2+1/2)^2 *log(q/2+1/2)- (3*q/4 -1)*log(1-3*q/4)- (3*q/4+1/4)*log(3*q/4+1/4)) + 0.7213*(q*log(q/2)* ( -q-3/2 +1/q)+ (-q-1)*log(q/2 +1/2) + q*(-q/2+1/2)*log(1/2-q/2)-3*log(1-q/2)-(q-1)*log(1/2 - q/2)) -0.36 *((q-1)^2*log(1/2 - q/2)));
>>solve(ex,q)
ans =
- 3.7601288392107154402269169733539 + 0.46896675884329957486140629064384*i

cagatay yilmaz on 23 Apr 2017
Hello,
Could you guys help me the solve following problem. This is the dispersion equation for a symmetric lamb wave. I am looking for the z values for different f (frequencies)? I have to find something similar to Fig 1 a.
close all
clear all
clc
%cl longitudunal wave speed
%ct transversal wave speed
%cp phase velocity of wave
%cg group velocity of wave
% z=ct/cp;
pi=3.14;
nu=0.33;% poisson ratio
ro=2700;%density kg/m3
E=70e9;% elastic modulus Pa
mu=E/(2*(1+nu)); %shear modulus
cl=((E*(1-nu))/(ro*(1+nu)*(1-2*nu)))^(0.5);
ct=(mu/ro).^(0.5);
k=ct/cl;
f=10:10:3e6;
w=2*pi*f;
d=(w*0.8e-3)/ct;
fzero (@(z) (2*z.^2-1).^2*(sin(sqrt((1-z.^2))*d))*cos(sqrt((k.^2-z.^2.*d)))-(sin(sqrt (k.^2-z.^2.*d)))*cos(sqrt((1-z.^2))*d)*(4*z.^2)*sqrt(1-z.^2)*sqrt(k.^2-z.^2),1)
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cagatay yilmaz on 23 Apr 2017
I have started a new question