How can I calculate unknown parameters in a integration with given boundary conditions

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john zhang
john zhang 2020-5-11
评论： john zhang 2020-5-13
Hi I am trying to calculate unknown parameters in a integration with given boundary conditions. The integration and boundary conditions are shown below:
a11-a23 are the six unknown parameters to be calculated. z1 is a known vector when t=1 (I can give z2, z3... if needed).
The x could be discretized and thus the z1 can be a known vector constant.
I am trying to use integration (int in matlab) to represent z(x,t) for a given x (e.g. x=-1.5:0.1:1.5) and then use fmincon to find a11-a23 to make modulus of (Z(x,t)-z1) the minimum. However, the calculation of the integration is super time-consuming.
c=0.011;
syms x t a1 a2 a3 a4 a5 a6
a=[a1 a2 a3 a4 a5 a6];
x=-1.5:0.1:1.5;
Er=@(t)-(a(1)+a(2)*exp(-t./a(3))).*(1-x.^2./(a(4)+a(5).*t-a(6).*t.^2).^2).^c;
intE=int(@(t)Er(t),t,0,1);
tt=1;% integration time
FEr=Er(x,t,a)
fun=@(t,a)(norm(int(Er,t,0,tt)-z1));
a0=zeros(6,1);
A=[];
b=[];
Aeq=[];
beq=[];
lb=[0,0,0,0,0,0];
ub=[1,1,10,1,1,1];
aa=fmincon(@(a)fun,a0,A,b,Aeq,beq,lb,ub);
Any ideas appreciated. Thank you in advance!
  2 个评论
Walter Roberson
Walter Roberson 2020-5-11
x is continuous, which would imply that Z(x,1) takes on an infinite number of values, not a finite vector of values. Unless, that is, z1 is a scalar constant ? But for that to be the case, the first term of the integral, independant of x, would have to work out as 0 which would require a11 = 0, a12 = 0, and a13 not negative infinity.
john zhang
john zhang 2020-5-11
Hi Walter, many thanks for your prompt comments. In my problem, x can be discretized so that z1 can be a vector constant, i.e. for each discreted x along -1.5 to 1.5, I will have a know z for this x when t=1.

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回答（1 个）

Walter Roberson
Walter Roberson 2020-5-11
Probably the easiest way is to use the CurveFitting Toolbox with 'fittype' being an anonymous function with 6 model parameters. See https://www.mathworks.com/help/curvefit/fittype.html#btpaend-8

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