Solving two second order ODEs
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Hi there!
I am trying to solve for U(t) and V(t) for the two second order ODEs
and 
, where a and w are constants and with the initial conditions 
and 
.
and . Then I want to plot the solutions 
against 
for a given time interval.
against for a given time interval. syms U(t) V(t)
%Constants definition
a = 1;
w = 100;
dU=diff(U,t);
dV=diff(V,t);
%Initial Conditions
y0 = [0 0 1 1];
eq1 = diff(U, t, 2) == -a*w*dV;
eq2 = diff(V,t, 2) == a*w*dU;
vars = [U(t); V(t)]
OTV = odeToVectorField([eq1,eq2])
M = matlabFunction(OTV,'vars', {'t','Y'});
interval = [0 5]; %time interval
ySol = ode45(M,interval,y0);
tValues = linspace(interval(1),interval(2),1000);
yValues1 = deval(ySol,tValues,1); %U(t) solution
yValues2 = deval(ySol,tValues,2); %V(t) solution
plot(yValues1,yValues2)
Does the above code correctly solve the system of differential equations and initial conditions and plot V(t) against U(t)?
If not, then please let me know what is incorrect. Also plese let me know if there is another way to solve the system of ODEs.
Thank you for your help. Much appreciated.
回答（2 个）
Essentially, yes.
However it could be made a bit more efficient —
syms U(t) V(t)
%Constants definition
a = 1;
w = 100;
dU=diff(U,t);
dV=diff(V,t);
%Initial Conditions
y0 = [0 0 1 1];
eq1 = diff(U, t, 2) == -a*w*dV;
eq2 = diff(V,t, 2) == a*w*dU;
vars = [U(t); V(t)]
[OTV,Subs] = odeToVectorField([eq1,eq2])
M = matlabFunction(OTV,'vars', {'t','Y'});
interval = [0 5]; %time interval
ySol = ode45(M,interval,y0);
tValues = linspace(interval(1),interval(2),1000);
yValues1 = deval(ySol,tValues,1); %U(t) solution
yValues2 = deval(ySol,tValues,2); %V(t) solution
plot(yValues1,yValues2)
% ---------- Slightly More Efficient: Solves Directly & Avoids The 'deval' Calls ----------
interval = [0 5]; %time interval
tValues = linspace(interval(1),interval(2),1000);
[t,y] = ode45(M,tValues,y0);
yValues1 = y(:,1); %U(t) solution
yValues2 = y(:,2); %V(t) solution
figure
plot(yValues1,yValues2)
xlabel('$V(t)$', 'Interpreter','latex')
ylabel('$\frac{dV(t)}{dt}$', 'Interpreter','latex')
.
4 个评论
Jake Barlow
2021-12-25
My pleasure!
It was not clear what the first plot represented.
That is where the ‘Subs’ output of odeToVectorField helps, and the reason I always include it in the requested outputs from odeToVEctorVield. According to it, the first two columns (in the second integration that I added) are V and 
. The U and 
results would be ‘y(:3)’ and ‘y(:,4)’ in order.
. The U and results would be ‘y(:3)’ and ‘y(:,4)’ in order. To plot V as a function of U would be —
syms U(t) V(t)
%Constants definition
a = 1;
w = 100;
dU=diff(U,t);
dV=diff(V,t);
%Initial Conditions
y0 = [0 0 1 1];
eq1 = diff(U, t, 2) == -a*w*dV;
eq2 = diff(V,t, 2) == a*w*dU;
[OTV,Subs] = odeToVectorField([eq1,eq2])
M = matlabFunction(OTV,'vars', {'t','Y'});
% ---------- Slightly More Efficient: Solves Directly & Avoids The 'deval' Calls ----------
interval = [0 5]; %time interval
tValues = linspace(interval(1),interval(2),1000);
[t,y] = ode45(M,tValues,y0);
yValues1 = y(:,3); %U(t) solution
yValues2 = y(:,1); %V(t) solution
figure
plot(yValues1,yValues2)
xlabel('$U(t)$', 'Interpreter','latex')
ylabel('$V(t)$', 'Interpreter','latex')
Check that to be certain, however it appears to be the desired result to me.
.
Jake Barlow
2021-12-25
Star Strider
2021-12-25
My pleasure!
I think there are a few mistakes in the code
syms U(t) V(t)
%Constants definition
a = 1;
w = 100;
dU=diff(U,t);
dV=diff(V,t);
%Initial Conditions
y0 = [0 0 1 1];
eq1 = diff(U, t, 2) == -a*w*dV;
eq2 = diff(V,t, 2) == a*w*dU;
vars = [U(t); V(t)]
[OTV,S] = odeToVectorField([eq1,eq2]);
S
Note that S is ordered [V dV U dU], so that should be the ordering of the solution of ode45
M = matlabFunction(OTV,'vars', {'t','Y'});
interval = [0 5]; %time interval
% note the IC's are in the same order as S
ySol = ode45(M,interval,[0 1 0 1],odeset('MaxStep',0.0001,'InitialStep',0.0001));
tValues = linspace(interval(1),interval(2),10000);
yValues1 = deval(ySol,tValues,1); % V(t) solution
yValues2 = deval(ySol,tValues,2); % Vdot(t) solution
yValues3 = deval(ySol,tValues,3); % U(t) solution
yValues4 = deval(ySol,tValues,4); % Udot(t) solution
figure
plot(yValues3,yValues1) % plot U vs V
xlabel('U');ylabel('V')
An exact solution can be computed using dsolve()
sol = dsolve([eq1; eq2],[U(0)==0; V(0)==0; dU(0)==1; dV(0)==1])
Ufunc = matlabFunction(sol.U);
Vfunc = matlabFunction(sol.V);
plot(Ufunc(tValues),Vfunc(tValues))
xlabel('U');ylabel('V')
% compare numerical and exact solutions
figure
plot(tValues,yValues3-Ufunc(tValues),tValues,yValues1-Vfunc(tValues))
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