0
我试图用interp1引入变量p作为时间的功能，但它不工作，它给了我一个载体，而不是得到一个值作为我的整合结果，任何人都可以帮我解决这个代码吗？
function RunLogisticOscilFisher
omega = 1;
k = 10;
N0 = 1;
A = 1;
p0 = .1;
tspan=(0:0.1:10);
% Finding the numerical solution for the function using ode45 solver
[t,p]=ode45(@logisticOscilfisher,tspan,p0,[],N0,k,omega);
% [t,p]=ode23s(@(t,p) N0*sin(omega*t)*p*(1-p./k),tspan,p0,odeset('AbsTol',1e-8,'RelTol',1e-10'));
% Plotting the function with time
figure(1)
plot(t,p)
% Finding the integral to get the fisher information with respect to p
f = @(p) ((A.*(((N0*sin(omega*t).^2.*(1-2*p./k))+(omega.*cos(omega*t))
).^2)./(N0.^2*sin(omega*t).^4.*(p-p.^2./k).^2)))
I1=integral(f, 11,20,'ArrayValued',true)
I2=integral(f,11,40,'ArrayValued',true)
I3=integral(f,11,60,'ArrayValued',true)
I4=integral(f,11,80,'ArrayValued',true)
I5=integral(f,11,101,'ArrayValued',true)
I=[I1,I2,I3,I4,I5]
II=[I1./20 I2./40 I3./60 I4./80 I5./101]
T=[20 40 60 80 101]';
%Plotting the Fisher Information
figure(2)
plot(T,I,'b-'), hold on
plot(T,II,'r-')
hold off
% Finding the integral to get the fisher information with respect to t
P = @(T) interp1(t,p,T);
ff = @(t) (A.*(((N0*sin(omega*t).^2.*(1-2*p./k))+(omega.*cos(omega*t))
).^2)./(N0.^2*sin(omega*t).^4.*(p-p.^2./k).^2))
J1=integral(ff, 0.1,2,'ArrayValued',true)
J2=integral(ff, 0.1,4,'ArrayValued',true)
J3=integral(ff, 0.1,6,'ArrayValued',true)
J4=integral(ff,0.1,8,'ArrayValued',true)
J5=integral(ff,0.1,10,'ArrayValued',true)
J=[J1,J2,J3,J4,J5]
JJ=[J1./2 J2./4 J3./6 J4./8 J5./10]
R=[2,4,6,8,10]';
%Plotting the Fisher Information
figure(3)
plot(R,J,'b-'), hold on
plot(R,JJ,'r-')
hold off
figure(4)
plot(t,f(t))
figure(5)
plot(log(t),log(f(t)))
P = @(T) interp1(t,p,T);
a=exp(1-cos(t));
fff = @(T) ( (A.* ((sin(t)).^2 .* (1-2.*(a./ (9.9 + 0.1 .* a))./10) + cos(t)).^2
) ./ ((sin(t)).^4 .* ((a ./ (9.9+(0.1.*a))) - ((a ./ (9.9 + (0.1 .* a))
).^2 ./10)).^2 ) )
K1=integral(fff, 0,2,'ArrayValued',true)
K2=integral(fff, 0,4,'ArrayValued',true)
K3=integral(fff, 0,6,'ArrayValued',true)
K4=integral(fff,0,8,'ArrayValued',true)
K5=integral(fff,0,10,'ArrayValued',true)
K=[K1,K2,K3,K4,K5]
KK=[K1./2 K2./4 K3./6 K4./8 K5./10]
%Plotting the Fisher Information
figure(6)
plot(R,K,'b-'), hold on
plot(R,KK,'r-')
hold off
1;
% function dpdt = logisticOscilfisher(t,p,N0,k,omega)
% dpdt = N0*sin(omega*t)*p*(1-p/k);
% end
尝试减少代码以实际显示您的问题。我们没有经历所有那些无法解释的代码。 –