Solving Non-Linear Equation in a loop

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Alakesh Upadhyaya
Alakesh Upadhyaya 2020년 11월 17일
댓글: Alakesh Upadhyaya 2020년 12월 18일
I want to solve this equation for different values of E0, I want to vary the value of E0 from 1 to 100, for which I will get 99x3 equations+1 as E0 is present in 3 out of 4 equations given below. I can solve the equation for a fix value of E0 using VPA solve but I can't write a program to iterate the value of E0 for which I should get 99 different values of E,P,F and S from the equation for 99 different values of E0.
clc;clear;close;
syms E F S P
E00=1; %Total concetration of E0
F0=100; %TotalCOncentration of F0
k1=4; %rate constants
k2=6;
k3=4;
k4=1;
k5=6;
k6=4;
k7=5;
k8=8;
for i=1:99
E0(i)=E00+i;
eqn1=(E0-E)*(k1+k2)-k1*E*S-k4*E*P==0; % First equation
eqn2=(E0-E)*k3+(F0-F)*k6-E*S*k1-k8*S*F==0; %Second equation
eqn3=(F0-F)*(k7+k6)-F*P*k5-k8*S*F==0; %Third equation
eqn4=k2*(E0-E)+(F0-F)*k7-k5*F*P-k4*E*P==0; %Fourth Equation
sol(i)=solve([eqn1,eqn2,eqn3,eqn4,E>0,S>0,P>0,F>0,E<100,F<100,S<100,P<100],[E,P,S,F]);
end
heres my second code for a single value of E0
clc;clear;close;
%% Initial conditons and values of rate constants
syms E S F P
E0=1; %Total concetration of E0
F0=100; %TotalCOncentration of F0
k1=4; %rate constants
k2=6;
k3=4;
k4=1;
k5=6;
k6=4;
k7=5;
k8=8;
%% The equations
eqn1=(E0-E)*(k1+k2)-k1*E*S-k4*E*P==0; % First equation
eqn2=(E0-E)*k3+(F0-F)*k6-E*S*k1-k8*S*F==0; %Second equation
eqn3=(F0-F)*(k7+k6)-F*P*k5-k8*S*F==0; %Third equation
eqn4=k2*(E0-E)+(F0-F)*k7-k5*F*P-k4*E*P==0; %Fourth Equation
equations=[eqn1 eqn2 eqn3 eqn4]; %Total equations
vars=[ E S P F];
range=[0 E0-0.1;0.1 E0-0.1;0.1 E0-0.1;0.1 F0-0.1]; %Range of variables to omit unwanted values
% Range has been taken from 0 to E0-0.1 so that the value of E does not
% exceed the value of E0, same is done for S P and F
%% Solver
sol=vpasolve(equations,vars,range);
[sol.E sol.S sol.F sol.P];
%% Values
E=sol.E %Numerical value of E
P=sol.P %Numerical value of P
S=sol.S %Numerical value of S
F=sol.F %Mumerical value of F
digits(4);
x=P/(P+S)
b=E0/F0; %E0/F0
I just can't write a code for E0=1 to 99, I tried to put vpa into loop, but it doesn't work. Please help me out

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Mahesh Taparia
Mahesh Taparia 2020년 12월 17일
Hi
The function vpasolve returns the structure which can be stored in a cell array. Also there were some missing initialization. The below code will work for your problem:
clc;clear;close;
syms E F S P
E00=1; %Total concetration of E0
F0=100; %TotalCOncentration of F0
k1=4; %rate constants
k2=6;
k3=4;
k4=1;
k5=6;
k6=4;
k7=5;
k8=8;
E0=zeros(100,1);
sol=cell(100,1);
for i=1:99
E0(i)=E00+i;
eqn1=(E0(i)-E)*(k1+k2)-k1*E*S-k4*E*P==0; % First equation
eqn2=(E0(i)-E)*k3+(F0-F)*k6-E*S*k1-k8*S*F==0; %Second equation
eqn3=(F0-F)*(k7+k6)-F*P*k5-k8*S*F==0; %Third equation
eqn4=k2*(E0(i)-E)+(F0-F)*k7-k5*F*P-k4*E*P==0; %Fourth Equation
equations=[eqn1 eqn2 eqn3 eqn4]; %Total equations
vars=[ E S P F];
range=[0 E0(i)-0.1;0.1 E0(i)-0.1;0.1 E0(i)-0.1;0.1 F0-0.1]; %Range of variables to omit unwanted values
sol{i}=vpasolve(equations,vars,range);
end
Hope it will help!

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