Continuous dynamical systems are described by systems of ordinary differential equations:
x' = f(t, x),
where t is the independent variable (usually considered as the time for evolution), and
x = x(t) = (x1(t), x2(t), ..., xn(t)),
f(t, x) = (f1(t, x1, x2, ..., xn), f2(t, x1, x2, ..., xn), ..., fn(t, x1, x2, ..., xn)).
The system is linear if f(t, x) is a linear transformation of x:
|x'1 =||a11(t)x1||+ a12(t)x2||+ ... + a1n(t)xn|
|x'2 =||a21(t)x1||+ a22(t)x2||+ ... + a2n(t)xn|
|x'n =||an1(t)x1||+ an2(t)x2||+ ... + ann(t)xn|
and may be abbreviated as x' = A(t)x. The following result tells us that the key to solving the system is to find n linearly independent solutions.
x(t) = c1v1(t) + c2v2(t) + ... + cnvn(t),
where v1(t), v2(t), ..., vn(t) are n special linearly independent solutions of the system and c1, c2, ..., cn are arbitrary constants.
Since x → x' - A(t)x is a linear transformation, the system x' = A(t)x may be considered as homogeneous. The result above can be compared with the structure of the solutions of Ax = 0.
In the special case the coefficient functions are constants: aij(t) = aij, we have a system x' = Ax of linear differential equations with constant coefficients. Suppose v is an eigenvector of A with eigenvalue λ, then x(t) = eλtv satisfies the system:
= (eλt)'v (v is a constant vector)
= eλtAv (v is an eigenvector with eigenvalue λ)
= A(eλtv). (eλt is a scalar (number))
Thus we conclude the following.
x(t) = c1eλ1tv1 + c2eλ2tv2 + ... + cneλntvn.
Example From the eigenvalues 5, 15 and the eigenvectors (1, -2), (2, 1) found in this example for the matrix
we conclude that the system
|x' =||13x||- 4y|
|y' =||- 4x||+ 7y|
has two (linearly independent) solutions
|[||x1(t)||] = e5t[||1||], [||x2(t)||] = e15t[||2||],|
and the general solution for the system is
|[||x(t)||] = c1e5t[||1||] + c2e15t[||2||] = [||c1e5t + 2c2e15t||].|
|y(t)||-2||1||- 2c1e5t + c2e15t|
Example We try to find the solution of the system
|x' =||x||+ 5y|
|y' =||- 2x||+ 3y|
satisfying the initial condition
x(0) = -1, y(0) = 14.
Based on this example, we find the general solution
x(t) = (1 - 3i)c1e(2 + 3i)t + (1 + 3i)c2e(2 - 3i)t,
y(t) = 2c1e(2 + 3i)t + 2c2e(2 - 3i)t
of the system. By (see this note)
e(2 + 3i)t = e2t(cos3t + isin3t)
and the similar formula for e(2 - 3i)t, we may rewrite the general solution as
x(t) = ae2t(cos3t + 3sin3t) + be2t(- 3cos3t + sin3t),
y(t) = ae2t(2cos3t) + be2t(2sin3t),
where a = c1 + c2, b = i(c1 -c2).
To find the solution satisfying the initial condition, we substitute the initial condition to the general solution and get
a - 3b = -1,
2a + 2b = 12.
Then we get a = 5 and b = 2 and the special solution
x(t) = e2t(- cos3t + 17sin3t),
y(t) = e2t(10cos3t + 4sin3t),
satisfying the initial condition.
In the example above, we saw that if we take a1 and a2 to be real numbers, then we get the real-valued general solution in case of complex eigenvalues. In general, suppose a real matrix A has a conjugate pair of truly complex eigenvalues λ = μ + iν and λ = μ - iν, where ν ≠ 0. Let v = u + iw and v = u - iw be the corresponding eigenvectors. Then we get a linear combination (a = c1 + c2, b = i(c1 -c2))
c1e(μ + iν)t(u + iw) + c2e(μ - iν)t(u - iw)
= aeμt((cosνt)u - (sinνt)w) + beμt((sinνt)u + (cosνt)w)
as part of the general solution of the system x' = Ax. If we take a and b to be real numbers for all the conjugate pairs of eigenvalues, then we get the general real-valued solutions.