Linear Independence and the Wronskian
Recall from linear algebra that two vectors v and w are called linearly dependent if there are nonzero constants c1 and c2 with.
c1v + c2w = 0
We can think of differentiable functions f(t) and g(t) as being vectors in the vector space of differentiable functions. The analogous definition is
Let f(t) and g(t) be differentiable functions. Then they are called linearly dependent if there are nonzero constants c1 and c2 with
c1f(t) + c2g(t) = 0
for all t. Otherwise they are called linearly independent.
The functions f(t) = 2sin2 t and g(t) = 1 - cos2(t) are linearly dependent since
(1)(2sin2 t) + (-2)(1 - cos2(t)) = 0
The functions f(t) = t and g(t) = t2 are linearly independent since otherwise there would be nonzero constants c1 and c2 such that
c1t + c2t2 = 0
for all t. First let t = 1. Then
c1 + c2 = 0
Now let t = 2. Then
2c1 + 4c2 = 0
This is a system of 2 equations and two unknowns. The determinant of the corresponding matrix is
4 - 2 = 2
Since the determinant is nonzero, the only solution is the trivial solution. That is
c1 = c2 = 0
The two functions are linearly independent.
In the above example, we arbitrarily selected two values for t. It turns out that there is a systematic way to check for linear dependence. The following theorem states this way.
Show that the functions f(t) = t and g(t) = e2t are linearly independent.
We compute the Wronskian.
f '(t) = 1 g '(t) = 2e2t
The Wronskian is
(t)(2e2t) - (e2t)(1)
Now plug in 0 to get
W(f,g)(0) = -1
which is nonzero. We can conclude that f and g are linearly independent.
c1f(t) + c2g(t) = 0
Then we can take derivatives of both sides to get
c1f '(t) + c2g '(t) = 0
This is a system of two equations with two unknowns. The determinant of the corresponding matrix is the Wronskian. Hence, if the Wronskian is nonzero at some t0, only the trivial solution exists. Hence they are linearly independent.
There is a fascinating relationship between second order linear differential equations and the Wronskian. This relationship is stated below.
First the Wronskian
W = y1y2' - y1'y2
W' = y1'y2' + y1y2'' - y1''y2 - y1'y2' = y1y2'' - y1''y2
Since y1 and y2 are solutions to the differential equation, we have
y1'' + p(t)y1' + q(t)y1 = 0
y2'' + p(t)y2' + q(t)y2 = 0
Multiplying the first equation by -y2 and the second by y1 and adding gives
(y1y2'' - y1''y2) + p(t)(y1y2' - y1'y2) = 0
This can be written as
W' + p(t)W = 0
This is a separable differential equation with
dW/W = -p(t)dt
Now integrate and Abel's theorem appears.
Find the Wronskian (up to a constant) of the differential equations
y'' + cos(t) y = 0
We just use Abel's theorem, the integral of cos t is sin t hence the Wronskian is
W(t) = cesin t
A corollary of Abel's theorem is the following
Let y1 and y2 be solutions to the differential equation
L(y) = y'' + p(t)y' + q(t)y = 0
Then either W(y1,y2) is zero for all t or never zero.
y1(t) = 1 - t and y2(t) = t3
cannot both be solutions to a differential equation
y'' + p(t)y + q(t) = 0
for p(t) and q(t) continuous on [-1,5].
We compute the Wronskian
y1' = -1 y2' = 3t2
W(y1,y2) = (1 - t)(3t2) - (t3)(-1) = 3t2 - 2t3
Notice that the Wronskian is zero at t = 0 but nonzero at t = 1. By the above corollary, y1 and y2 cannot both be solutions.