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huangmagudi鐵蟲 (小有名氣)
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[求助]
高手請進————matlab中的線性矩陣不等式問題
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利用matlab中Yalmip工具箱編的程序如下: %%%%%%%%%%%%%%%%%% d=57; A=[0.8 0;0 0.91]; Ad=[-0.1 0;-0.1 -0.1]; n=size(A,1); for i=1:3 e{i}=[]; for j=1:3 if j==i e{i}=[e{i} eye(n)]; else e{i}=[e{i} zeros(n)]; end end end f1=A*e{1}+Ad*e{2}; f2=f1-e{1}; f3=[e{1};(d+1)*e{3}-e{1}]; f4=[f1;(d+1)*e{3}-e{2}]; P1=sdpvar(2*n,2*n,'symmetric'); P2=sdpvar(n,n,'symmetric'); P3=sdpvar(n,n,'symmetric'); % N1=sdpvar(n,3*n); % N2=sdpvar(n,3*n); N1=1/d*[P2 -P2 zeros(n,n)]; N2=3/(d*(d+1))*[-P2 -P2 2*P2]; Delta=f4'*P1*f4-f3'*P1*f3+f2'*P2*f2-2/d*(e{1}'-e{2}')*N1+(2*d+2)/d*(e{1}'+e{2}'-2*e{3}')*N2+e{1}'*P3*e{1}-e{2}'*P3*e{2}; FAI=[Delta N1' (d+1)/sqrt(3)*N2';N1 -P2 zeros(n,n);(d+1)/sqrt(3)*N2 zeros(n,n) -P2]; con=[]; con1=[P1>=0]; con2=[P2>=0]; con3=[P3>=0]; con4=[FAI<=0]; con=[con,con1,con2, con3,con4]; sol=solvesdp(con,[],sdpsettings('solver','lmilab')) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 此時能找到可行解,如果我把 N1=1/d*[P2 -P2 zeros(n,n)]; N2=3/(d*(d+1))*[-P2 -P2 2*P2]; 這兩行程序換成如下程序 N1=sdpvar(n,3*n); N2=sdpvar(n,3*n); 則不能找到可行解了。按理說,第二種情況具有更少保守性,因為此時N1和N2都是自由變量。第一種情況是第二種的特殊情形,為啥換了兩行程序,就找不到可行解了?實在沒搞明白 |
鐵蟲 (小有名氣)
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