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houbing金蟲 (初入文壇)
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[交流]
非線性方程組的迭代法(數(shù)值計算高手請進(jìn))
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| 我在用matlab求解一組非線性方程組的時候遇到了困難,因為初值選擇不合適,迭代幾乎都不收斂,由于數(shù)據(jù)量較大,沒有辦法對每個初值進(jìn)行調(diào)整,有沒有一種迭代算法可以對初值沒有要求,我目前使用的是幾個教科書上的算法,牛頓法,不動點迭代,弦割法。期待有高手可以指點迷津,先行謝過! |
版主 (知名作家)
木蟲 (正式寫手)
清靜的女孩

金蟲 (初入文壇)
金蟲 (初入文壇)
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為了方便向大家請教,我把我的程序貼了出來,第一次使用matlab,對著手冊編了一周,有不夠簡潔的地方還望見諒:) 基本問題就是求解kesai afa gama J0afa J1afa J0gama J1gama(分別為afa gama的零階和一階bessel函數(shù))七個變量的非線性方程組;共有5328個數(shù)據(jù)點,每個點都需要求解這樣一個方程組,初值只給了kesai的初值,其它變量有顯式的關(guān)系可以通過kesai求解,實際上是利用迭代法求fkesai=0; j=1,j=2都是收斂的,j=3就不收斂了 % 不動點迭代 %define constant clear; E=3000000000; rou=1200; K=2500000000; a=0.015; ita=1000000; sampling_rate=10000000; f=(1:5238)*sampling_rate/5238; im=i; %calculate parameters for j=1:5238 Estar(j)=-im*E*ita*f(j)/(E-im*ita*f(j)); end for j=1:5238 kesai0(j)=sqrt(rou*f(j)^2/Estar(j)); end for j=1:5238 miu(j)=3*K*f(j)*ita*im/(9*K*(1+im*f(j)*ita/E)-im*f(j)*ita); lamda(j)=K-2/3*miu(j); end %initial value of variables for j=1:5238 kesai(j)=kesai0(j); afa(j)=sqrt(rou*f(j)^2/(lamda(j)+2*miu(j))-kesai(j)^2); gama(j)=sqrt(rou*f(j)^2/miu(j)-kesai(j)^2); J0afa(j)=besselj(0,afa(j)*a); J1afa(j)=besselj(1,afa(j)*a); J0gama(j)=besselj(0,gama(j)*a); J1gama(j)=besselj(1,gama(j)*a); fkesai(j)=2*afa(j)/a*(gama(j)^2+kesai(j)^2)*J1afa(j)*J1gama(j)-(gama(j)^2-kesai(j)^2)*J0afa(j)*J1gama(j)-4*kesai(j)*afa(j)*gama(j)*J1afa(j)*J0gama(j); j %iterative n=1; while abs(fkesai(j))>0.0001&(n<=10000) %不動點迭代from fkesai=0 kesai(j)=(2*afa(j)/a*(gama(j)^2+kesai(j)^2)*J1afa(j)*J1gama(j)-(gama(j)^2-kesai(j)^2)*J0afa(j)*J1gama(j))/(4*afa(j)*gama(j)*J1afa(j)*J0gama(j)); afa(j)=sqrt(rou*f(j)^2/(lamda(j)+2*miu(j))-kesai(j)^2); gama(j)=sqrt(rou*f(j)^2/miu(j)-kesai(j)^2); J0afa(j)=besselj(0,afa(j)); J1afa(j)=besselj(1,afa(j)); J0gama(j)=besselj(0,gama(j)); J1gama(j)=besselj(1,gama(j)); fkesai(j)=2*afa(j)/a*(gama(j)^2+kesai(j)^2)*J1afa(j)*J1gama(j)-(gama(j)^2-kesai(j)^2)*J0afa(j)*J1gama(j)-4*kesai(j)*afa(j)*gama(j)*J1afa(j)*J0gama(j); n=n+1; abs(fkesai) end end |
金蟲 (小有名氣)
金蟲 (小有名氣)
鐵桿木蟲 (職業(yè)作家)
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