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 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.

Theorem

Let f and g be differentiable on [a,b].  If Wronskian W(f,g)(t0)  is nonzero for some t0 in [a,b] then f and g are linearly independent on [a,b]If f and g are linearly dependent then the Wronskian is zero for all t in [a,b].

 

 

Example 

Show that the functions f(t)  =  t   and g(t)  =  e2t are linearly independent.

Solution

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.

 

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.

Abel's Theorem

Let y1 and y2 be solutions on the differential equation 

          L(y)  =  y'' + p(t)y' + q(t)y  =  0

where p and q are continuous on [a,b].  Then the Wronskian is given by 

         

where c is a constant depending on only y1 and y2, but not on t.  The Wronskian is either zero for all t in [a,b] or no t in [a,b].

 

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)  =  cesin 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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