Solve differential equation with anonymous functions

Hello, everyone. I am already quite familiar when it comes to the resolution of differential equations(DE’S) in MATLAB with “ode45” function. I have already solved this problem by making a function dFdV that contained the DE’S which has as inputs (V,F), being V the independent and F the dependent variable.

function dFdV=funcion(V,F)

CTo=0.286;

k=0.4;

FT=F(1)+F(2)+F(3);    % FT=FA+FB+FC;

CA=(CTo*F(1))/FT;

rA=-k*CA^2;

rB=-rA;

rC=-0.5*rA;

dFdV=zeros(3,1);

dFdV(1)=rA;  

dFdV(2)=rB;  

dFdV(3)=rC; 

end

What I now want is to solve the same problem creating an anonymous function that contains the DE’S and using again “ode45” to solve them.

CTo=0.286;

k=0.4;

FT=@(F)(F(1)+F(2)+F(3));    % FT=FA+FB+FC;

CA=@(F)((CTo*F(1))/FT(F));

rA=-k*(CA(F))^2;

rB=-rA;

rC=-0.5*rA;

dFdV=@(V,F)[rA;rB;rC];

The problem is that when executing It pops up this message (“Undefined function or variable 'F'), which is obvious because now my function dFdV depends on V and F, but F has not been defined (as an input) as the former case.

What should I do?. Thanks

Answer 1

[Walter Roberson]

CTo=0.286;

k=0.4;

FT=@(F)(F(1)+F(2)+F(3));    % FT=FA+FB+FC;

CA=@(F)((CTo*F(1))/FT(F));

rA=@(F) -k*(CA(F))^2;

rB=@(F)-rA(F);

rC=@(F)-0.5*rA(F);

dFdV=@(V,F)[rA(F);rB(F);rC(F)];

[Walter Sanchez]

Thanks, Walter. It worked right away.