Numerically solve equations with variables inside numerical integration

I am trying to solve equations in a form of

where C1, C2, C3 are known and , M, F are variables to be solved. I have tried fsolve() like (below is definitely a simplified version for better illustration)

x = fsolve(@tobesolve,[0,0,0])

function rei = tobesolve(mud,M,F)
    rei(1) = integral2(@(p,theta)intfun1(p,theta,mud,M,F),0,631,0,pi)-600;
    rei(2) = integral2(@(p,theta)intfun2(p,theta,mud,M,F),0,631,0,pi)-400;
    rei(3) = integral2(@(p,theta)intfun3(p,theta,mud,M,F),0,631,0,pi)-200;
end

function i1 = int1(p,theta,mud,M,F)
i1 = p+theta+mud+M+F;
end

function i2 = int2(p,theta,mud,M,F)
i2 = p-theta-mud-M-F;
end

function i3 = int3(p,theta,mud,M,F)
i3 = p+theta-mud+M-F;
end

What is unexpected is that the error read

>> main
函数或变量 'M' 无法识别。

出错 main>@(p,theta)int1(p,theta,mud,M,F) (第 24 行)
    rei(1) = integral2(@(p,theta)int1(p,theta,mud,M,F),0,631,0,pi)-600;

出错 integral2Calc>integral2t/tensor (第 228 行)
        Z = FUN(X,Y);  NFE = NFE + 1;

出错 integral2Calc>integral2t (第 55 行)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);

出错 integral2Calc (第 9 行)
    [q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);

出错 integral2 (第 105 行)
    Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);

出错 main>tobesolve (第 24 行)
    rei(1) = integral2(@(p,theta)int1(p,theta,mud,M,F),0,631,0,pi)-600;

出错 fsolve (第 264 行)
            fuser = feval(funfcn{3},x,varargin{:});

出错 main (第 21 行)
x = fsolve(@tobesolve,[0,0,0])

原因:
    Failure in initial objective function evaluation. FSOLVE cannot continue.

I wonder why M is undefined, although I have provided guess values and the same error does not affect mud.

And how to solve such equations with numeric integration in MATLAB.

Answer 1

[Torsten]

x = fsolve(@(y)tobesolve(y(1),y(2),y(3)),[0,0,0])

instead of

x = fsolve(@tobesolve,[0,0,0])

And the function names must be intfun1, intfun2 and intfun3, not int1, int2 and int3.

[泽城 邹]

Thanks very much! This actually fix the problem. (although I'm still wondering why)

Yes, indeed the functions are intfun1, 2, 3. These are typos when I tried to simplified the actual expression to the shortened version presented here.

[Torsten]

x = fsolve(@tobesolve,[0,0,0])

expects the input to tobesolve to be the standard input, i.e. a vector of length 3.

But your input are 3 scalars of length 1.

How should the solver know this if you don't tell him in advance by modifying the input list as

@(y)tobesolve(y(1),y(2),y(3))

?

[泽城 邹]

Yes, I fully understand this now. Thank you again!