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wtisunday木蟲 (正式寫手)
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[求助]
MATLAB 非線性擬合 fminsearch中包含非線性方程的求解問題
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format short global a1 a2 a3 a4 a5 a6 a7 a8 b1 b2 b3 b4 b5 b6 b7 b8 k0=[[3 3 3 2 2 2 1 1];[1 1 1 1 1 1 1 1]]'; %參數(shù)的初始值 x0=[1,0]'; %拉伸應(yīng)力初始值 t=[0.00423,0.01621,0.06216,0.23841,0.91437,3.50692,14.6856,101.343]'; G=[336370,64246.4,44849.7,10649.7,5151.99,1224.43,63.4524,2.67362]'; %relaxation time,relaxation modulus實(shí)驗(yàn)數(shù)據(jù)[t,G] xexp=[0.03162 0.1 0.31623 1 3.16228 10.5635 20.395]'; yexp=[84834.3 84834.3 79215.4 62322.2 52509.5 32501.6 25570.4]'; %xexp,yexp實(shí)驗(yàn)數(shù)據(jù)[頻率x、拉伸粘度y] [k,fval]=fminsearch(@ObjFunc,k0,x0,xexp,yexp);%非線性擬合 function f=ObjFunc(x0,xexp,yexp) G=[336370,64246.4,44849.7,10649.7,5151.99,1224.43,63.4524,2.67362]'; t=[0.00423,0.01621,0.06216,0.23841,0.91437,3.50692,14.6856,101.343]'; [x]=fsolve(@kineticseqs103,x0); ff=sum(x(:,2)-x(:,1))./xexp; f=(sum(ff'-yexp))^2; function pp=rheolomodel(k,x,g,xexp) global a1 a2 a3 a4 a5 a6 a7 a8 b1 b2 b3 b4 b5 b6 b7 b8 k=[[a1,a2,a3,a4,a5,a6,a7,a8];[b1,b2,b3,b4,b5,b6,b7,b8]]'; q=1+(x(:,1)+2*x(:,2))/3./G.^0.5; u=2.*k(:,1)*exp(2/k(:,2)-1)*(1-1/q.^2)+1/q.^2*(1-0.3/k(:,2).*... ((x(:,1)).^2+2*(x(:,2)).^2)/3/g.^2); pp1=u.*x(:,1)-2*t.*xexp.*(x(:,1)+G)+G.*(u-1)+... 0.3./k(:,2)./G.*(x(:,1))^2; pp2=u.*x(:,2)+t.*xexp.*x(:,2)+G.*(u-1)+... 0.3./k(:,2)./G.*(x(:,2))^2+t.*G.*xexp; pp=[pp1,pp2]'; 以上是我依照前人經(jīng)驗(yàn)編的程序,由于是新手,在程序出錯(cuò)和修改上解決不了,還望各位蟲友幫幫忙。 |
鐵桿木蟲 (職業(yè)作家)
木蟲 (正式寫手)
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我的問題是多模態(tài)問題 就是已知t的i個(gè)分量和G的i個(gè)分量,通過兩個(gè)方程求出x的i個(gè)分量和y的i個(gè)分量,再通過x和y分量和建立優(yōu)化函數(shù),通過優(yōu)化實(shí)現(xiàn)a的2i個(gè)分量的確定。 比如 i=3 u(i)=a(i)*exp(a(i+4)-t(i))+1/(x(i)+y(i))/G(i).....中間變量表達(dá)式 f1=u(i)*x(i)-t(i)*(x(i)+y(i))+g(i)*((u(i)-1)+expdata; f2=u(i)*y(i)+t(i)*y(i)+y(i)^(2*xexpdata); f=[f1,f2]';......分量方程組(每一分量均有此關(guān)系) Y=sum(x(i)-y(i))/xexpdata.....計(jì)算yexpdata Op=sum(Y-yexpdata)^2......優(yōu)化函數(shù) |
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