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) (24 )
rei(1) = integral2(@(p,theta)int1(p,theta,mud,M,F),0,631,0,pi)-600;
出错 integral2Calc>integral2t/tensor (228 )
Z = FUN(X,Y); NFE = NFE + 1;
出错 integral2Calc>integral2t (55 )
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);
出错 integral2Calc (9 )
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);
出错 integral2 (105 )
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
出错 main>tobesolve (24 )
rei(1) = integral2(@(p,theta)int1(p,theta,mud,M,F),0,631,0,pi)-600;
出错 fsolve (264 )
fuser = feval(funfcn{3},x,varargin{:});
出错 main (21 )
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.

 采纳的回答

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.

3 个评论

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

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