- Please will you help me introduce the dependance on t of k(t), thanks!
MATLAB: solving a 2nd order ODE which has a parameter which evolves in time
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Mark Anslow
on 23 Feb 2020
Commented: Star Strider
on 24 Feb 2020
Hi, I'm trying to solve the equation: y'' + k(t)y = 0 where k(t) is an array of values and y(0) = 1/2 and y'(0) = 1 in the range t = 0 to 20.
I understand it is possible to solve this as 2 coupled first order differential equations using ode45 as follows when k is constant:
k = 2
syms y(t)
[System] = odeToVectorField(diff(y, 2) == -(k).*y);
M = matlabFunction(System,'vars', {'t','Y'});
sol = ode45(M,[0 20],[1/2 1]);
fplot(@(x)deval(sol,x,1), [0, 20])
Please will you help me introduce the dependance on t of k(t), thanks!
1 Comment
darova
on 24 Feb 2020
Can you be more specific? What is it? Where is dependance?
Accepted Answer
Star Strider
on 24 Feb 2020
Try this:
syms k(t) y(t) % Declare ‘k(t)’ Here
[System,Subs] = odeToVectorField(diff(y, 2) == -(k).*y); % Note ‘k(t)’ Included In Code
M = matlabFunction(System,'vars', {'t','Y'})
kv = rand(1,20); % ‘k’ Vector
tv = linspace(0, 20, numel(kv)); % Time Vector Matching ‘k’
k = @(t) interp1(tv, kv, t, 'linear', 'extrap'); % Function Interpolating ‘k’ To Specific Time
M = @(t,Y)[Y(2);-k(t).*Y(1)]; % Output Of 'matlabFunction’ (Needs To Be Copied, And Pasted In The Code After The 'k' Function Is Defined)
[T,Y] = ode45(M,[0 20],[1/2 1]);
figure
plot(T, Y)
grid
xlabel('T')
ylabel('Amplitude')
legend(string(Subs))
Use your own ‘kv’ vector. The rest of the code should adapt to it, whatever it is.
6 Comments
J. Alex Lee
on 24 Feb 2020
oops guess i need help with reading skills. i saw a similar question before, where "k(t)" in the function sense was being confused with "k(t)" in the index sense. Hope this isn't one of those cases and one really needs the discretized time-dependence.
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