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Mathematics LibreTexts

3.6: Linear Independence and the Wronskian

  • Page ID
    401
  • [ "article:topic", "Wronskian", "Abel\'s theorem", "LINEAR DEPENDENCE", "Linear Independence", "authorname:green" ]

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    Recall from linear algebra that two vectors \(v\) and \(w\) are called linearly dependent if there are nonzero constants \(c_1\) and \(c_2\) with

    \[ c_1v + c_2w = 0. \]

    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 below.

    Definition: Linear Dependence and Independence

    Let \(f(t)\) and \(g(t)\) be differentiable functions. Then they are called linearly dependent if there are nonzero constants \(c_1\) and \(c_2\) with \( c_1f(t) + c_2g(t) = 0 \) for all \(t\). Otherwise they are called linearly independent.

    Example \(\PageIndex{1}\)

    The functions \(f(t) = 2\sin^2 t \) and \(g(t) = 1 - \cos^2(t)\) are linearly dependent since 

    \[ (1)(2\sin^2 t) + (-2)(1 - \cos^2(t)) = 0. \nonumber \]

    Example \(\PageIndex{1}\)

    The functions \(f(t) = t\) and \(g(t) = t^2\) are linearly independent since otherwise there would be nonzero constants \(c_1\) and \(c_2\) such that 

    \[ c_1t + c_2t^2 = 0 \nonumber\]

    for all values of \(t\). First let \(t = 1\). Then 

    \[ c_1 + c_2 = 0. \nonumber\]

    Now let \(t = 2\). Then 

    \[ 2c_1 + 4c_2 = 0\nonumber \]

    This is a system of 2 equations and two unknowns. The determinant of the corresponding matrix is 

    \[4 - 2 = 2.\nonumber\]

    Since the determinant is nonzero, the only solution is the trivial solution. That is 

    \[ c_1 = c_2 = 0 .\nonumber\]

    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 \(W(f,g)(t_0)\) is nonzero for some \(t_0\) 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 \(\PageIndex{3}\)

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

    Solution

    We compute the Wronskian. 

    \[f'(t) = 1 g'(t) = 2e^{2t}\nonumber\]

    The Wronskian is

    \[ (t)(2e^{2t}) - (e^{2t})(1)\nonumber\]

    Now plug in \(t=0\) to get

    \[ W(f, g )(0) = -1 \nonumber\]

    which is nonzero. We can conclude that \(f\) and \(g\) are linearly independent.

    Proof

    If

    \[ C_1 f(t) + C_2g(t) = 0 \nonumber\]

    Then we can take derivatives of both sides to get

    \[ C_1f"(t) + C_2g'(t) = 0 \nonumber\]

    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 \( t_0\), only the trivial solution exists. Hence they are linearly independent.

    \(\square\)

    There is a fascinating relationship between second order linear differential equations and the Wronskian. This relationship is stated below.

    Theorem: Abel's Theorem

    Let \(y_1\) and \(y_2\) be solutions on the differential equation 

    \[ L(y) = y'' + p(t)y' + q(t)y = 0 \nonumber\]

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

    \[ {W(y_1, y_2 )(t) = ce^{- \int p(t) dt}} \nonumber \]

    where \(c\) is a constant depending on only \(y_1\) and \(y_2\), but not on \(t\). The Wronskian is either zero for all \(t\) in \([a,b]\) or not in \([a,b]\).

    Proof

    First the Wronskian 

    \[ W = y_1y'_2 - y_1y_2 \nonumber\]

    has derivative

    \[W' = y_1y'_2 + y_1y''_2 - y''_1y_2 - y_1y'_2 = y_1y''_2 - y''_1y_2. \nonumber\]

    Since \(y_1\) and \(y_2\) are solutions to the differential equation, we have

    \[ y''_1 + p(t)y'_1 + q(t)y_1 = 0 \nonumber\]

    \[ y''_2 + p(t)y'_2 + q(t)y_2 = 0 . \nonumber\]

    Multiplying the first equation by \(-y_2\) and the second by \(y_1\) and adding gives

    \[ (y_1y''_2 - y''_1y_2) + p(t)(y_1y'_2 - y_1y_2) = 0. \nonumber\]

    This can be written as

    \[ W' + p(t)W = 0. \nonumber\] 

    This is a separable differential equation with 

    \[ \dfrac{dW}{W} = -p(t) dt.\nonumber \]

    Now integrate and Abel's theorem appears.

    \(\square\)

    Example \(\PageIndex{4}\)

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

    \[ y'' + cos(t) y = 0. \nonumber\]

    Solution

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

    \[ W(t) = ce^{ \sin t}.\nonumber \] 

    A corollary of Abel's theorem is the following 

    Corollary

    Let \( y_1\) and \( y_2\) be solutions to the differential equation

    \[ L(y) = y'' + p(t)y' + q(t)y = 0 \] 

    Then either \( W( y_1, y_2)\) is zero for all \(t\) or never zero.

    Example \(\PageIndex{5}\)

    Prove that 

    \[y_1(t) = 1 - t \;\;\; \text{and } y_2(t) = t^3\nonumber \]

    cannot both be solutions to a differential equation

    \[ y'' + p(t)y + q(t) = 0 \nonumber\]

    for \( p(t) \) and \(q(t)\) continuous on \(\left [ -1, 5 \right ] \).

    Solution

    We compute the Wronskian

    \[y'_1 = -1\;\;\; \text{and} \;\;\; y'_2 = 3t^2 \nonumber\]

    \[ W(y_1, y_2) = (1 - t)(3t^2) -(t^3)(-1) = 3t^2 - 2t^3. \nonumber \]

    Notice that the Wronskian is zero at \(t = 0\) but nonzero at \(t = 1\). By the above corollary, \(y_1\) and \(y_2\) cannot both be solutions.

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