Question about ODE45

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Zhukun Wang
Zhukun Wang 2021-2-24
回答： Walter Roberson 2021-2-24
function Math462hw2Q5partcmodel
kG=0.002;
kAV=0.004;
k0=0.3;
k12=1;
k21=0.7;
T=1500;
tspan=0:1:168; %We have to do this for the rest intervals of same length from 0 to 168.
x1=0;
x2=0;
T=0;
miut=15;
y=[x1,x2,T];
[t,soln]=ode45(@(t,y)dotfunctions(t,y,k0,k12,k21,kG,kAV),tspan,y);
%Plot solution
plot(t,soln);
xlabel('time');
ylabel('value');
title('Progress of treatment');
function ddt=dotfunctions(~,xt,k0,k12,k21,kG,kAV)
x1=xt(1);
x2=xt(2);
T=xt(3);
x1dot=miut-(k0+k12)*x1+k21*x2;
x2dot=k12*x1-k21*x2;
Tdot=kG*T-kAV*T*x1;
ddt=[x1dot;x2dot;Tdot];
end
end
For this code, how am I supposed to change the code if I only want miut to be 15 at t=0,24,48,72,96,120,144? Thanks!
  2 个评论
Walter Roberson
Walter Roberson 2021-2-24
What do you want it to be at other times?
Are you asking for an impulse equal to 15 at those particular times ?
Zhukun Wang
Zhukun Wang 2021-2-24
I want it to be zero at all other times Yeah, I want an impulse function.

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回答（1 个）

Walter Roberson
Walter Roberson 2021-2-24
The ode*() functions cannot handle impulses. More completely, they are only designed to handle objectives in which the second derivative of the function you supply is continuous. That rules out impulses, and also rules out most cases of abs() or min() or max() or if statements.
What you need to do is stop integrating and re-start from where you left off. That would look like
breakpoints = 0:24:24:144;
yinit = y0;
for K = 1 : length(breakpoints)
[t{K}, y{K}] = ode45(fun, breakpoints(K):breakpoints(K+1), yinit);
yinit = y{K}(end,:);
yinit(1) = yinit(1) + miut; %apply the impulse
end
Then afterwards, stitch together all of the t{:} and y{:}

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