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< >已知微分方程组<FONT face="Times New Roman"> </FONT></P>
< ><FONT face="Times New Roman">dS/dt=Y*y1-0.338*M*S</FONT></P>
< ><FONT face="Times New Roman">dI/dt=e*M-2.0265*I+0.01*Y</FONT></P>
<P ><FONT face="Times New Roman">dR/dt=2.0265*I</FONT></P>
<P ><FONT face="Times New Roman">dY/dt=-y1*Y-0.01*Y+0.338*M*S*a</FONT></P>
<P ><FONT face="Times New Roman">dM/dt=0.338*M*S*(1-a)-e*M</FONT></P>
<P ><FONT face="Times New Roman">y1</FONT>、<FONT face="Times New Roman">e</FONT>、<FONT face="Times New Roman">a</FONT>分别在[<FONT face="Times New Roman">0,0.5]</FONT>、[0<FONT face="Times New Roman">,1]</FONT>、<FONT face="Times New Roman">[0,1]</FONT>之间，还知道一组<FONT face="Times New Roman">S</FONT>和<FONT face="Times New Roman">I</FONT>的数值。现欲通过上述条件拟合出<FONT face="Times New Roman">y1</FONT>、<FONT face="Times New Roman">n</FONT>、<FONT face="Times New Roman">e</FONT>、<FONT face="Times New Roman">a</FONT>的最优值。我的思路如下：</P>
<P >先建立一个表示方程组的函数，其中代定系数用字母表示。之后建立一个<FONT face="Times New Roman">M</FONT>文件，利用多重循环给系数付值，每次根据所付值解出解析解，与所给<FONT face="Times New Roman">S</FONT>和<FONT face="Times New Roman">I</FONT>的数值相比较，选出最优的一组参数值。下面是我的程序，总调试不出来，系统提示是函数中代定系数不能用字母表示，请哪位大侠指点一下如何求出参数最优值，这关系到俺们建模队能否通过校内选拔，感激涕零！！！</P>
<P > </P><wrapblock><v:shapetype><v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path connecttype="rect" gradientshapeok="t" extrusionok="f"></v:path><lock aspectratio="t" v:ext="edit"></lock></v:shapetype>
<P ><v:shape><v:imagedata></v:imagedata><v:textbox style="mso-next-textbox: #_x0000_s1026"></v:textbox><w:wrap type="topAndBottom">ts=11:63;<BR>x0=[0.9995,2.3892e-004, 2.8000e-005,2.1769e-004, 1.1415e-004];<BR>sure=[1553,1636,1741,1803,1897,1960,2049,2136,2177,2227,2265,2304,2347,2370,2388,2405,2420,2434,2437,2444,2444,2456,2465,2490,2499,2504,2512,2514,2517,2520,2521,2521,2521,2521,2521,2521,2521,2521,2521,2522,2522,2522,2522,2522,2522,2522,2522,2522,2522,2522,2522,2523,2523];<BR>y1=0;e=0;a=0;<BR>[t,x]=ode45('after_control',ts,x0);<BR>sum=0;<BR>for i=1:53<BR> sum=sum+x(i,2)-sure(1,i);<BR>end<BR>min=sum;<BR>for y1=0:0.1:0.5<BR> for e=0:0.1:1<BR> for a=0:0.1:1<BR> [t,x]=ode45('after_control',ts,x0);<BR> sum=0;<BR> for i=1:53<BR> sum=sum+x(i,2)-sure(1,i)<BR> end<BR> if sum<min<BR> min=sum;<BR> yy1=y1;yy2=y2;ee=e;aa=a;<BR> end<BR> end<BR> end<BR>end<BR>y1=yy1;e=ee;a=aa;<BR>[t,x]=ode45('after_control',ts,x0);<BR>plot(t,x(:,2)*6500000,'r')<BR>hold on,plot(t,sure,'g')<BR></w:wrap></v:shape>function xdot=after_control(t,x)<BR>y1;y2=0.01;q=2.0265;n=0.338;a;<BR>xdot=[x(4)*y1-n*x(5)*e,e*x(5)-q*x(2)+y2*x(4),q*x(2),-y1*x(4)-y2*x(4)+n*x(5)*x(1)*a,n*x(5)*x(1)*(1-a)-e*x(5)]'</P>
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<P ></wrapblock><BR clear=all><FONT face="Times New Roman"> <p></p></FONT></P>
<P ><FONT face="Times New Roman"> <p></p></FONT></P> |
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发表于 2005-8-12 08:46:25