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Suppose I have two linearly independent solution vectors \begin{bmatrix}x_1,_1(t)\\x_1,_2(t)\end{bmatrix} and \begin{bmatrix}x_2,_1(t)\\x_2,_2(t)\end{bmatrix}

If I take the Wronskian of these 2 solution vectors, it comes out to a nonzero number since they are stated to be Linearly Independent. My question is, if you take the Wronskian of the same solution vectors but their derivative:

\begin{bmatrix}x'_{1,1(t)}&x'_{2,1(t)}\\x''_{1,1(t)}&x''_{2,1(t)}\end{bmatrix}

Would it still be linearly independent? (Would the Wronskian still be a nonzero number?

nmasanta
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focoquachy
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2 Answers2

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You've got a couple things going on at once here. First, you are asking about the Wronskian of two vectors of functions: $$\mathbf{x}_1(t) = \left[\begin{array}{c} x_{1,1}(t) \\ x_{1,2}(t) \end{array}\right] \quad\text{and}\quad \mathbf{x}_2(t) = \left[\begin{array}{c} x_{2,1}(t) \\ x_{2,2}(t) \end{array}\right]$$
which is defined as: $$W(\mathbf{x}_1,\mathbf{x}_2)(t) = \left|\begin{array}{cc} x_{1,1}(t) & x_{1,2}(t) \\ x_{1,2}(t) & x_{2,2}(t) \end{array}\right|.$$ Next you are asking about the Wronskian of two differentiable functions $x_{1,1}(t)$ and $x_{2,1}(t)$, which is defined as: $$W(x_{1,1},x_{2,1})(t) = \left|\begin{array}{cc} x_{1,1}(t) & x_{2,1}(t) \\ x_{1,1}'(t) & x_{2,1}'(t) \end{array}\right|.$$ So be careful there. To answer your question (for functions at least), take: $$f(t) = t\quad\text{and}\quad g(t) = -1.$$ Then, $$W(f,g) = \left|\begin{array}{cc} t & -1 \\ 1 & 0 \end{array}\right| = 1$$ but $$W(f',g') = \left|\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right| =0.$$ Also, this link does a great job at explaining the Wronskian.

Community
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Any two linearly independent constant vectors have dependent (identically zero) derivatives. Thus, the answer to your question is "not necessarily".

nullUser
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