The Wronskian

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We return to the homogeneous system of linear differential equations

$\begin{equation} \label{eq:linihsys2} \frac{dX}{dt} = A(t)X \end{equation}$ where $A(t)=(a_{ij}(t))$ is an $n\times n$ matrix. Let $X_1,\ldots ,X_n$ be a linearly independent set of solutions to (??) and let $\begin{equation} \label{E:Y(t)2} Y(t) = \left(X_1(t)|\cdots |X_n(t)\right) \end{equation}$ Using existence and uniqueness of solutions to the initial problem for (??), we showed in Lemma ?? that $Y(t)$ is an invertible matrix for every time $t$ . In this section we prove this same result by explicitly showing that the determinant of $Y(t)$ is always nonzero.

Define the Wronskian to be $W(t) = \det Y(t).$

Let $X_1,\ldots ,X_n$ be solutions to (??) such that the vectors $X_j(0)$ form a basis of $\R ^n$ . Let $Y(t)$ be the matrix defined in (??). Then $\begin{equation} \label{L:Wronskian} W(t) = e^{\int_0^t\trace(A(\tau))d\tau}W(0). \end{equation}$

It follows directly from (??) that the determinant of $Y(t)$ is nonzero when the determinant of $Y(0)$ is nonzero. But $\det Y(0)\neq 0$ since the vectors $X_j(0)$ form a basis of $\R ^n$ .

We prove Theorem ?? in two important special cases: linear constant coefficient systems and linear nonconstant $2\times 2$ systems. The proof for constant coefficient systems is based on Jordan normal forms, while the proof for $2\times 2$ systems is based on solving a separable differential equation for the Wronskian itself. It is this latter proof that generalizes to a proof of the theorem.

Wronskians for Constant Coefficient Systems

First, we interpret the Wronskian directly in terms of the constant coefficient matrix $A$ . Note that $X_j(t)=e^{tA}e_j$ is just the $j^{th}$ column of the matrix $e^{tA}$ . It follows that $W(t) = \det e^{tA}.$

Let $A$ be an $n\times n$ matrix. Then $\det e^A = e^{\trace (A)}.$

Proof

This result is proved using Jordan normal forms. To see why normal form theory is relevant, suppose that $A$ and $B$ are similar matrices. Then $e^A$ and $e^B$ are similar matrices, and $\trace (A)=\trace (B)$ and $\det e^A = \det e^B$ . So if we can show that the lemma is valid for matrices in Jordan normal form, then the lemma is valid for all matrices.

Suppose that the matrix $J$ is a $k\times k$ Jordan block matrix associated to the eigenvalue $\lambda$ . Then $J$ is upper triangular and the diagonal entries of $J$ all equal $\lambda$ . It follows that $\trace (J)=k\lambda$ . It also follows that $e^J$ is an upper triangular matrix whose diagonal entries all equal $e^\lambda$ . Hence $\det e^J = \left (e^\lambda \right )^k = e^{k\lambda } = e^{\trace (J)}.$ So the lemma is valid for Jordan block matrices.

Next suppose that $A$ is in block diagonal, that is $A=\mattwo {B}{0}{0}{C}.$ We claim that if the lemma is valid for matrices $B$ and $C$ , then it is valid for the matrix $A$ . To see this observe that $\trace (A) = \trace (B) + \trace (C),$ and that $e^A = \mattwo {e^B}{0}{0}{e^C}.$ Hence $\det e^A = \det e^B \det e^C = e^{\trace (B)}e^{\trace (C)},$ by assumption. It follows that $\det e^A = e^{\trace (B)+\trace (C)} = e^{\trace (A)},$ as desired. By induction, the lemma is valid for Jordan normal form matrices and hence for all matrices. $\blacksquare$

Wronskians for Planar Systems

In the time dependent case we verify Theorem ?? only for $2\times 2$ systems, as this substantially simplifies the discussion. Let $A(t) = \mattwo {a_{11}(t)}{a_{12}(t)}{a_{21}(t)}{a_{22}(t)},$ and let $X_1(t) = \vectwo {x_1(t)}{y_1(t)} \AND X_2(t) = \vectwo {x_2(t)}{y_2(t)}$ be solutions of (??). It follows that

$\begin{equation} \label{E:xyderiv} \begin{array}{rcl} \dot{x}_1 & = & a_{11}x_1 + a_{12}y_1 \\ \dot{y}_1 & = & a_{21}x_1 + a_{22}y_1 \\ \dot{x}_2 & = & a_{11}x_2 + a_{12}y_2 \\ \dot{y}_2 & = & a_{21}x_2 + a_{22}y_2. \end{array} \end{equation}$

In this notation $Y(t) = \mattwo {x_1(t)}{x_2(t)}{y_1(t)}{y_2(t)},$ and $W(t) = x_1(t)y_2(t) - x_2(t)y_1(t).$ We claim that $\dot {W} = \trace (A) W.$ If so, we can use separation of variables to solve this differential equation obtaining $\ln |W| = \int \trace (A(t))dt$ from which the proof of Theorem ?? follows.

Use the product rule to compute $\dot {W} = \dot {x}_1y_2 + x_1\dot {y}_2 - \dot {x}_2y_1 - x_2\dot {y}_1.$ Now substitute (??) to see that $\dot {W} = (a_{11}x_1 + a_{12}y_1)y_2 + x_1(a_{21}x_2 + a_{22}y_2) - (a_{11}x_2 + a_{12}y_2)y_1 - x_2(a_{21}x_1 + a_{22}y_1)$ from which it follows that $\dot {W} = (a_{11}+a_{22})W = \trace (A)W,$ as claimed.

Exercises

In Exercises ?? – ?? compute the Wronskian $W(t)$ for the specified matrices:

$Y(t) = \mattwo {t}{1}{-1}{t}$ .

$Y(t) = \mattwo {\cos t}{-\sin t}{\sin t}{\cos t}$ .

$Y(t) = \left (\begin {array}{ccc} 0 & t & e^t\\ 1 & t & te^t \\ t & t & t^2 e^t \end {array}\right )$ .

$Y(t) = \left (\begin {array}{ccc} 1 & 0 & 1\\ 1 + \sin t & 1 & e^t \\ -4 + \cos t & t & te^t \end {array}\right )$ .

Use (??) to verify that the Wronskian for solutions to the linear constant coefficient system of ODE $\dot {X}=AX$ is $\begin{equation} \label{E:ccw} W(t) = e^{\trace(A)t}W(0). \end{equation}$

In Exercises ?? – ?? verify (??) for the given matrix $A$ . Hint: First, you need to find linearly independent solutions to the system $\dot {X}=AX$ .

$A = \mattwo {0}{-1}{1}{0}$ .

$A = \mattwo {2}{3}{3}{2}$ .

$A = \mattwo {2}{-2}{4}{-3}$ .