Question

我正在尝试使用interp1将变量p作为时间的函数引入，但是它不起作用并且它给了我一个向量而不是为我的集成获得一个值，

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

Answer 1

您使用interp1时遇到问题。

p = interp1(t,p,T) - 这将返回一个向量'p = interp1(t,p,method,'pp') - 这将返回一个具有分段函数的结构。

Docs在这里。

此外，您可以使用ppval从分段函数中获取值。最后一点 - 在Matlab r2014a中有一个警告：

Warning: INTERP1(...,'PP') will be removed in a future release. Use GRIDDEDINTERPOLANT instead.

所以请准备好GRIDDEDINTERPOLANT。

有人可以帮我解决这个问题吗？

1 个答案: