First Order Differential Equation

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I need to solve the below first ODE with a set time step say (Delt = 0.01s) and from 0 - 20s. Initial condiitons Y0 = [ 0 ; 0 ; 0 ]
dof = 3; %Three Story
I = 1.5796*10^(-2); %m^4
E = 0.209*10^9; %kN/m^2
m = 500000; %kg
k = 849116255; %N/m
m1 = 2*m;
m2 = 2*m;
m3 = m;
k1 = 3*k;
k2 = 3*k;
k3 = 2*k;
M = [ m1 0 0 ; 0 m2 0 ; 0 0 m3 ]; %Mass Matrix
K = [(k1+k2) -k 0 ; -k2 (k2+k3) -k3 ; 0 -k3 k3 ]; %Stiffness Matrix
C = [ 7 -3 0 ; -3 3.2 -1.4 ; 0 -1.4 1.4 ]*10^5; %Damping Matrix
o = [ 0 0 0 ; 0 0 0 ; 0 0 0 ]; %Zero Matrix
A = [ M o ; o eye(dof) ]
B = [ o K ; -eye(dof) o ]
P = 2*sin(4*(4*pi*t))
ft = [P ; 0 ; 0]
Yt+1 = Yt+ delt(inv(A)*(ft-B*Yt))
Once this has been computed, how does one store the results for each itteration of time and subsequently plot it?
  2 个评论
Alan Stevens
Alan Stevens 2020-8-4
Your matrices don't seem to be consistent. e.g. B is 6x6, so it has 6 columns, but you multiply it by Yt, which, if the initial value is to be believed, has just 3 rows.
What is/are the actual ODE's you are trying to solve?
Christopher Wijnberg
Hi Alan
My apologies -
ft = [ P ; 0 ; 0 ; 0 ; 0 ; 0]
I no longer have the original ODE's as they have been fransferred from second order set of n differential equations to a 2n system of first order differential equations by making use of the State Space Form.

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采纳的回答

Alan Stevens
Alan Stevens 2020-8-5
Well, the following shows how to progress through time. However, the explicit form you've adopted is unstable. I've included an implicit form as an alternative. Also, I notice that you haven't included the damping in the equations yet.
dof = 3; %Three Story
I = 1.5796*10^(-2); %m^4
E = 0.209*10^9; %kN/m^2
m = 500000; %kg
k = 849116255; %N/m
m1 = 2*m;
m2 = 2*m;
m3 = m;
k1 = 3*k;
k2 = 3*k;
k3 = 2*k;
M = [ m1 0 0 ; 0 m2 0 ; 0 0 m3 ]; %Mass Matrix
K = [(k1+k2) -k 0 ; -k2 (k2+k3) -k3 ; 0 -k3 k3 ]; %Stiffness Matrix
C = [ 7 -3 0 ; -3 3.2 -1.4 ; 0 -1.4 1.4 ]*10^5; %Damping Matrix
o = [ 0 0 0 ; 0 0 0 ; 0 0 0 ]; %Zero Matrix
A = [ M o ; o eye(dof) ];
B = [ o K ; -eye(dof) o ];
u = ones(6,1);
delt = 0.01;
Tend = 2;
t = 0:delt:Tend;
Y = zeros(6,length(t));
% Explicit - unstable
% for i = 1:length(t)-1
%
% P = 2*sin(4*(4*pi*t(i)));
% ft = [P ; 0 ; 0; 0; 0; 0];
%
% Y(:,i+1) = Y(:,i)+ delt*A\(ft-B*Y(:,i));
%
% end
% Implicit - stable
for i = 1:length(t)-1
P = 2*sin(4*(4*pi*t(i)));
ft = [P ; 0 ; 0; 0; 0; 0];
Y(:,i+1) = (Y(i) + delt*A\ft)./(u + delt*A\(B*u));
end
plot(t,Y(1,:))
  9 个评论
Alan Stevens
Alan Stevens 2020-8-5
There is no Y0. Try replacing
Y0(:,1) = [0 ; 0 ; 0 ; 0 ; 0 ; 0.005];
by either
Y(:,1) = [0 ; 0 ; 0 ; 0 ; 0 ; 0.005];
or
Y(6,1) = 0.005;
Christopher Wijnberg
Sorry Alan
I am just not getting the right outputs, while I should be getting an output for Y(t) at each floor/ dof I am getting a single plot... I am trying not be be useless here or waste your time but please forgive me this is my first time coding in any language let alone in Matlab..
This was a graph taken from a Youtube video for a similar example coded in Python.

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更多回答(1 个)

Alan Stevens
Alan Stevens 2020-8-5
Try plot(t, Y(3:6,:))
  1 个评论
Christopher Wijnberg
Thank you Alan, that works a treat!
And thanks for ALL your help prior!
Regards

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