math111_logo Application

3. Linear differential system

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.

The general solution of a system x' = A(t)x of n first order linear differential equations in n dependent variables x = (x1(t), x2(t), ..., xn(t)) is

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 xx' - 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:

(eλtv)'
= (eλt)'v (v is a constant vector)
= eλtλv
= eλtAv (v is an eigenvector with eigenvalue λ)
= A(eλtv). (eλt is a scalar (number))

Thus we conclude the following.

Suppose v1, v2, ..., vn is a basis of eigenvectors of A with eigenvalues λ1, λ2, ..., λn. Then the general solution of the system x' = Ax of first order linear differential equations is

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

[ 13 -4 ],
-4 7

we conclude that the system

x' = 13x - 4y
y' = - 4x + 7y

has two (linearly independent) solutions

[ x1(t) ] = e5t[ 1 ], [ x2(t) ] = e15t[ 2 ],
y1(t) -2 y2(t) 1

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 + and λ = μ - , 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 + )t(u + iw) + c2e(μ - )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.


[exercise]
[previous topic] [part 1] [part 2] [part 3] [next topic]