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I've been reading about the wronskian and I got stuck in the following:

Suppose we are given a multivariate function describing e.g. a plane: $z = m_1 x + m_2 y + b$. How is the wronskian computed? We have here to two variables ($x,y$). How is this case dealt with the wronskian?

Best regards

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For checking linear dependency, you must need at least two functions in your question.

Now for two functions of two variables we can follow the follwing theorem given by "Green, G. M., Trans. Amer. Math. Soc., New York, 17, 1916,(483-516)".

Theorem: Let $y_{1}$ and $y_{2}$ be functions of two independence variables $x_{1}$ and $x_{2}$ i.e., $y_{1} = y_{1}(x_{1} ,x_{2}) $ and $y_{2} = y_{1}(x_{1} ,x_{2}) $ for which all partial derivatives of $1^{st}$ order, $\frac{\partial y_{1}}{\partial x_{k}}$, $\frac{\partial y_{2}}{\partial x_{k}}$, $(k = 1,2)$ exists throughout the region $A$. Suppose, farther, that one of the functions, say $y_{1}$, vanishes at no point of $A$. Then if all the two rowed determinants in the matrix \begin{pmatrix} y_{1} & y_{2} \\ \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{2}}{\partial x_{1}} \\ \frac{\partial y_{1}}{\partial x_{2}} & \frac{\partial y_{2}}{\partial x_{2}} \end{pmatrix} vanish identically in $A$, $y_{1}$ and $y_{2}$ are linearly dependent in $A$, and in fact $y_{2}=c y_{1}$.

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  • $\begingroup$ Do you mean the following three determinants? $$\begin{vmatrix} y_{1} & y_{2} \\ \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{2}}{\partial x_{1}} \end{vmatrix} , \quad \begin{vmatrix} y_{1} & y_{2} \\ \frac{\partial y_{1}}{\partial x_{2}} & \frac{\partial y_{2}}{\partial x_{2}} \end{vmatrix} ,\quad \text{and}\quad \begin{vmatrix} \frac{\partial y_{1}}{\partial x_{1}} & \frac{\partial y_{2}}{\partial x_{1}} \\ \frac{\partial y_{1}}{\partial x_{2}} & \frac{\partial y_{2}}{\partial x_{2}} \end{vmatrix}. $$ $\endgroup$
    – Michael Levy
    Commented Mar 30, 2022 at 3:51
  • $\begingroup$ Yes, @MichaelLevy $\endgroup$
    – nmasanta
    Commented Mar 30, 2022 at 13:05
  • $\begingroup$ I get the theorem you put and and I can apply it. Yet, returning to initial question, can we, in fact, define a Wronskian in such case? $\endgroup$
    – Michael Levy
    Commented Mar 15, 2023 at 1:34
  • $\begingroup$ @MichaelLevy we can't define Wronskian for a function only. We need atleast two functions... $\endgroup$
    – nmasanta
    Commented Mar 16, 2023 at 5:13
  • $\begingroup$ Hi. Yes, I know this. Suppose that I have two functions, where each is function is a function of the same two variables. What is the Wronskian? $\endgroup$
    – Michael Levy
    Commented Mar 17, 2023 at 1:01

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