Finding an unknown ordinary differential equation - matlab

Question

Given
d²x/dt² + a·dx/dt + 7.9·x³ = 3.2·sin(xt)
with initial conditions
x(0) = +1.2
dx/dt(0) = −3.3
x(2.3) = −0.6
Find numerically all the possible values of a, each accurate to at least 3 significant digits.
Is there any method other than brute force for solving this?

Answer 1

As far as I can see, it is not possible to solve this problem as stated.
Here is what I did. I implemented your problem in a reasonably general way:
%{
Find all 'a' for which
d²x/dt² + a·dx/dt + 7.9·x³ - 3.2·sin(xt) = 0
with initial conditions
x(0) = +1.2
dx/dt(0) = −3.3
x(2.3) = −0.6
%}
function odetest
% See how the function search_a(a) behaves around a = 0:
test_as = 0 : 0.1 : 10;
da = zeros(size(test_as));
for ii = 1:numel(test_as)
da(ii) = search_a(test_as(ii)); end
figure(100), clf, hold on
plot(test_as, da)
axis tight
xlabel('a')
ylabel('|x(2.3) - 0.6|')
% Roughly cherry-pick some positive values, improve the estimate, and
% plot the solutions
opt = optimset('tolfun',1e-14, 'tolx',1e-12);
plot_x(fminsearch(#search_a, 0.0, opt), 1)
plot_x(fminsearch(#search_a, 1.4, opt), 2)
plot_x(fminsearch(#search_a, 3.2, opt), 3)
% Plot single solution
function plot_x(a,N)
[xt, t] = solve_ode(a);
figure(N), clf, hold on
plot(t,xt)
plot(2.3, -0.6, 'rx', 'markersize', 20)
title (['x(t) for a = ' num2str(a)])
xlabel('t')
ylabel('x(t)')
end
end
% Solve the problem for a value a, and return the difference between the
% actual value and desired value (-0.6)
function da = search_a(a)
a_desired = -0.6;
xt = solve_ode(a);
da = abs(xt(end) - a_desired);
end
% Solve the problem for any given value of a
function [xt, t] = solve_ode(a)
y0 = [1.2 -3.3];
tfinal = 2.3;
opt = odeset('AbsTol',1e-12, 'RelTol',1e-6);
[t,yt] = ode45(#(y,t) odefun(y,t,a), [0 tfinal], y0, opt);
xt = yt(:,1); % transform back to x(t)
end
% Most ODE solvers solve first-order systems. This is not a problem for a
% second-order system, because if we make the transformation
%
% y(t) = [ x (t)
% x'(t) ]
%
% Then we can solve for
%
% y'(t) = [ x' (t)
% x''(t) ] <- the second-order homogeneous DE
%
function dydt = odefun(t,y,a)
dydt = [y(2)
-a*y(2) - 7.9*y(1)^3 + 3.2*sin(y(1)*t)];
end
The first part gave me this figure:
Some further investigation suggests that this only grows for larger a.
This figure gave rise to the initial estimates a = [0, 1.4, 3.2], which I improved via fminsearch() and plotted the solutions of:
So, that probably enables you to hand in your homework :)
However, why I say it's impossible to answer the question as stated, is because this is what the first plot looks like for negative a:
The oscillatory behavior seems to continue indefinitely, and the spacing in between the zeros seems to decrease in a non-predictable way.
Now, my university days are long behind me, and I'm not so well-versed in ODE theory anymore. Perhaps there is a pattern to it, that just doesn't show because of numerical problems. Or perhaps the oscillation stops after some value, never to return again. Or perhaps another zero turns up at a = +1053462664212.25.
I can't prove any of these things, I just know how to brute-force it; the rest is up to you.