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.

Example

The functions f(t)  =  2sin² t  and  g(t)  =  1 - cos²(t) are linearly dependent since 

        (1)(2sin² t) + (-2)(1 - cos²(t))  =  0

Example

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.

Example 

Show that the functions f(t)  =  t   and g(t)  =  e^2t are linearly independent.

Solution

We compute the Wronskian. 

        f '(t)  =  1        g '(t)  =  2e^2t 

The Wronskian is 

        (t)(2e^2t) - (e^2t)(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.

Proof

If

        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.

Proof

First the Wronskian 

         W  =  y₁y₂' - y₁'y₂ 

has derivative 

        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.

Example

Find the Wronskian (up to a constant) of the differential equations

        y'' + cos(t) y'  =  0

Solution

We just use Abel's theorem, the integral of cos t is sin t hence the Wronskian is 

        W(t)  =  ce^{-sin t} 

A corollary of Abel's theorem is the following 

Corollary

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.

Example

Prove that 

        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].

Solution

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.

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