Solving two second order ODEs

Question

Jake Barlow 2021-12-25

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评论： Star Strider 2021-12-25

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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

.

Then I want to plot the solutions

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.

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Answer 1

Star Strider 2021-12-25

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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)]

vars =

[OTV,Subs] = odeToVectorField([eq1,eq2])

OTV =

Subs =

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')

.

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Star Strider 2021-12-25

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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.

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])

OTV =

Subs =

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

Hi Star Strider, thank you very much for your comment. It is clearer now!

Star Strider 2021-12-25

My pleasure!

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Answer 2

Paul 2021-12-25

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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)]

vars =

[OTV,S] = odeToVectorField([eq1,eq2]);

S

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])

sol = struct with fields:

V: sin(100*t)/100 - cos(100*t)/100 + 1/100 U: cos(100*t)/100 + sin(100*t)/100 - 1/100

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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Jake Barlow 2021-12-25

Hi Paul, thank you very much for your answer.

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Solving two second order ODEs

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