Peano on Wronskians: A Translation - Appendix 2: Translations

Sur le déterminant Wronskien, by Giuseppe Peano

1. On the Wronskian determinant (by G. Peano, professor at the University of Turin). In nearly all papers, one finds the proposition: If the determinant formed with n functions of the same variable and their derivatives of orders $1,\dots,(n-1)$ is identically zero, there is between these functions a homogeneous linear relationship with constant coefficients.    

This wording is too general. We offer, in fact, $$x = t^2 [1+\phi(t)],\quad\quad y = t^2 [1-\phi(t)],$$ where $\phi(t)$ designates a function equal to zero for $t = 0$, equal to $1$ for positive $t$, to $-1$ for negative $t$ (*). One has for $$t\le 0,\quad \phi(t)=-1,\quad x=0,\quad {y=2t^2};$$ for $$t= 0,\quad \phi(t)=0,\quad x=0,\quad {y=0};$$ for $$t\ge 0,\quad \phi(t)=1,\quad {x=2t^2},\quad {y=0}.$$

(*) For example:  $$\phi (t) = \frac{2}{\pi} {\int_0^\infty} \frac{\sin tx}{x}\,dx$$ says Mr. Peano. If one wants a more elementary example, we offer $$\psi (t) = E\left({\frac{1}{1+t^2}}\right),\quad\quad \phi (t) = [1 - \psi (t)]\,{\frac{t+\psi(t)}{{\rm mod.}t+\psi(t)}},$$ where $E(u)$ designates the largest integer less than $u,\,{\rm mod.}\, t,$ the absolute value of $t$. (P.M.).

The functions $x$ and $y,$ and their derivatives $${x^\prime} = 2t [1+\phi(t)],\quad\quad {y^\prime} = 2t [1-\phi(t)],$$ are continuous, and one has, for all values of $t ,$ $$\left| \begin{array}{cc} x&y\\x^\prime&y^\prime\\ \end{array} \right| =0$$ But between $x$ and $y,$ there is no relation of the expressed form, because the coordinate point $(x, y)$  does not describe a straight line passing through the origin, but the two half-axes $ox$ and $oy.$

The proposal is true if one supposes that there exists no value of $t$ which cancels all the minors of the last line of the proposed determinant, and perhaps in other cases which would be interesting to investigate.

(*) One can correct the wording of the expressed theorem at the beginning of this note, as we have indicated in our Résume d’analyse, by adding the words: or one of the functions is identically zero. (P.M.).

Sur les Wronskiens, by Giuseppe Peano

6.  On the Wronskians (Excerpt from a letter by M.G. Peano). The note added to the end of my brief article on the Wronskians (Mathesis, p. 75-76), if it contributes to the clarity, it adds nothing to the rigor of the contested proposition; and I must return to the subject.

I. The proposition I contest is the following: If the determinant formed with $n$ functions $x_1,\dots, x_n$ of the same variable $t,$ and their derivatives of orders $1,\dots, (n-1),$ is zero for all values of $t,$  there is between these functions a homogeneous linear relationship with constant coefficients; in other words, one can determine $n$ constants, $C_1,\dots, C_n ,$ all of which cannot equal zero at the same time, such that $C_1 x_1+\cdots + C_n x_n =0$ for all values of $t.$

If one of the functions, for example $x_1,$ is identically zero, these functions are connected by the linear relation $x_1+0x_2 +\cdots +0x_n =0.$ Thus the conclusion of the proposition in question is not at all modified if one adds the words: or one of the functions is identically zero. 

This proposition may again be expressed in this way: If the Wronskian of the functions $x_1,\dots, x_n$ is identically zero, the determinant formed with the values of the functions, when one gives the variable $n$ ordinary values, is also zero.

A particular case of the general theorem is as follows: If the determinant formed with the derivatives of orders $1, 2, 3$ of the functions $x, y, z$ of the same variable $t$ is identically zero, the curve described by the coordinate point $x, y, z$ is planar.

II. To demonstrate that these propositions are not always true, I have indicated in my previous note, an example in which I consider only two functions $x$ and $y$ of the variable $t;$ the determinant $x{y^\prime} – {x^\prime} y$ is identically zero; but between them there exists no linear relation. Neither one nor the other of the two functions is identically zero, because $x>0$ if $t>0,$ and $y>0$ if $t<0.$

But these two functions present this unusual characteristic that, for all values of $t,$ one or the other is zero. To make this anomaly disappear, we propose $$X={t^2},\quad Y=t\,{{\rm mod}\,t}.$$ These two functions of $t$ satisfy the condition; they only cancel for $t=0,$ and between them there is no homogeneous linear relation. The sum and the difference of these functions are the functions of my first example; the coordinate point $(X,Y)$ now describes the two half-bisectors of the axes.

III. The given demonstrations of the proposition, including yours, only allow us to prove that it is true if one of the minor determinants of the last line is always equal to zero, or if it is never equal to zero.

For determinants of the second order, one can demonstrate the proposition, except where, for any value of the independent variable, the two given functions and their derivatives each cancel at the same time. 

It would be interesting to identify all the cases in which the proposition holds true.

I believed it necessary to publish this brief note on the Wronskians, because I have never seen the proposition clearly expressed. (See Hermite, Cours d’Analyse, p. 133; Jordan, Cours d’Analyse, III, p. 150; Laurent, Traité d’Analyse, I, p. 183.)

From Résumé du Cours d’Analyse Infinitésimale de l’Université de Gand (Survey of Infinitesimal Analysis, Ghent University), by Paul Mansion

III. A Wronskian $W(r,s,t,u)$ is identically zero if one of the functions $r,s,t,u$  is identically zero, or there exists between them a homogeneous linear relationship; and VICE VERSA.

If $r,$ for example, is identically zero, the same is true of the derivatives ${r^{\prime}}, {r^{\prime\prime}}, {r^{\prime\prime\prime}},$ and $W,$ having a column of zeros, is also identically zero. If one has $u=ar+bs,$ and as a result, ${{u^{\prime}}=a{r^{\prime}}+b{s^{\prime}}}, {{u^{\prime\prime}}=a{r^{\prime\prime}}+b{s^{\prime\prime}}}, {{u^{\prime\prime\prime}}=a{r^{\prime\prime\prime}}+b{s^{\prime\prime\prime}}},$ $W$ will have a column of zeros, when one takes out the fourth column, the first multiplied by  $a$ and the second multiplied by $b$.  

SIMILARLY, if $W$ is identically zero, one of the functions $r,s,t,u$ is equal to zero, or there exists between $r,s,t,u$ a linear relationship. We suppose first this reciprocal is established for Wronskians of three lines and we prove that it is true for Wronskians of four lines. Thus, define $k,m,n,p,$ the minors of $W$ with regard to ${r^{\prime\prime\prime}},{s^{\prime\prime\prime}},{t^{\prime\prime\prime}},{u^{\prime\prime\prime}}.$ These minors are themselves Wronskians of three lines. If one of them is identically zero, there exists a linear relationship between the functions involved (according to the hypothesis made for Wronskians of three lines), and the theorem is demonstrated.

If none of the minors $k,m,n,p$ is equal to zero, we consider the identical relations  $$kr+ms+nt+pu=0,\quad \left({{\mathcal{Z}}_1}\right)$$ $$k{r^{\prime}}+m{s^{\prime}}+n{t^{\prime}}+p{u^{\prime}}=0, \quad \left({{\mathcal{Z}}_2}\right)$$ $$k{r^{\prime\prime}}+m{s^{\prime\prime}}+n{t^{\prime\prime}}+p{u^{\prime\prime}}=0, \quad\left({{\mathcal{Z}}_3}\right)$$ $$k{r^{\prime\prime\prime}}+m{s^{\prime\prime\prime}}+n{t^{\prime\prime\prime}}+p{u^{\prime\prime\prime}}=0, \quad\left({{\mathcal{Z}}_4}\right)$$ obtained in expressing the properties of the minors of the zero determinant $W$. Because ${D_z}W=0,$ one still has, according to formula (2), $$k{r^{iv}}+m{s^{iv}}+n{t^{iv}}+p{u^{iv}}=0, \quad \left({{\mathcal{Z}}_5}\right)$$

Differentiating successively ${{ \mathcal{Z} }_1}, {{\mathcal{Z}}_2}, {{\mathcal{Z}}_3}, {{\mathcal{Z}}_4},$ we simplify the derivative of each of these equations, by means of the following, $${k^{\prime}}r+ {m^{\prime}}s+ {n^{\prime}}t+ {p^{\prime}}u=0,$$ $${k^{\prime}}{r^{\prime}}+{m^{\prime}}{s^{\prime}}+{n^{\prime}}{t^{\prime}}+{p^{\prime}}{u^{\prime}}=0,$$ $${k^{\prime}}{r^{\prime\prime}}+{m^{\prime}}{s^{\prime\prime}}+{n^{\prime}}{t^{\prime\prime}}+{p^{\prime}}{u^{\prime\prime}}=0,$$ $${k^{\prime}}{r^{\prime\prime\prime}}+{m^{\prime}}{s^{\prime\prime\prime}}+{n^{\prime}}{t^{\prime\prime\prime}}+{p^{\prime}}{u^{\prime\prime\prime}}=0.$$

These four relationships give immediately, according to the properties of homogeneous linear equations and the definition of $k,m,n,p ,$  $${\frac{k^\prime}{k} = \frac{m^\prime}{m} =\frac{n^\prime}{n} =\frac{p^\prime}{p} },\quad {\rm{or}}\quad {Dlk=Dlm=Dln=Dlp}.$$

As a result, according to numbers $221$ or $103$, $\alpha, \beta, \gamma$ being the constants, $$lm=lk+l\alpha,\quad ln=lk+l\beta,\quad lp=lk+l\gamma,$$

$$m={\alpha}k,\quad n={\beta}k,\quad p={\gamma}k.$$

Substituting these values of $m, n, p$ into the identity $kr+ms+nt+pu=0,$ it becomes, after division by $k,$ $r + {\alpha}s + {\beta}t + {\gamma}u = 0,$ a relationship which had to be demonstrated.

IV. REMARK. One can evidently establish analogous theorems to the previous ones on the Wronskians where the derivatives are replaced by partial derivatives or total differentials of the functions in question.

From Cours d’analyse de l’école polytechnique, tome troisieme, by Camille Jordan

122. One can note that the condition $$\left| {\begin{array}{*{20}c} {x_1 } & {} &  \cdots  & {} & {x_n }  \\ \vdots  & {} &  \ddots  & {} &  \vdots   \\ {x_1^{n - 1} } & {} &  \cdots  & {} & {x_n^{n - 1} }  \\ \end{array}} \right| \ne 0$$ expresses the necessary and sufficient condition for there to exist between the functions $x_1,\dots, x_n$ no linear relationship with constant coefficients, such that $${\alpha}_1 x_1+\cdots + {\alpha}_n x_n =0.$$

In fact, if there existed a relationship of this type, one would obtain, in differentiating it, $${\alpha}_1 {x_1^\prime} +\cdots + {\alpha}_n {x_n^\prime} =0,$$ $$\ \vdots$$ $${\alpha}_1 {x_1^{n-1}} +\cdots + {\alpha}_n {x_n^{n-1}} =0,$$ and, in eliminating the parameters ${\alpha}_1 ,\dots , {\alpha}_n ,$ it would become $$\left| {\begin{array}{*{20}c} {x_1 } & {} &  \cdots  & {} & {x_n }  \\ {x_1^\prime} & {} &  \cdots  & {} & {x_n^\prime}  \\ \vdots  & {} &  \cdots  & {} &  \vdots   \\ {x_1^{n - 1} } & {} &  \cdots  & {} & {x_n^{n - 1} }  \\ \end{array}} \right| = 0.$$

Similarly, if this determinant is zero, $x_1,\dots, x_n$ will be $n$ particular solutions of the linear equation of order $n-1$ $$\left| {\begin{array}{*{20}c} {X} & {x_2} &  \cdots  & {x_n }  \\ {X^\prime} & {x_2^{\prime}} &  \cdots  & {x_n^\prime}  \\ \vdots  & {\vdots} &  \cdots  &  \vdots   \\ {X^{n - 1} } & {x_2^{n-1}} &  \cdots  & {x_n^{n - 1} }  \\ \end{array}} \right| = 0.$$ One will have therefore, in designating by $X_1,\dots, X_n$ the independent solutions of this equation, and by $C_1^{\prime}, \dots , C_{n-1}^n$ the constants, $$x_1 ={C_1^\prime}X_1 +\cdots + {C_{n-1}^{\prime}}X_{n-1} ,$$ $$\ \vdots$$ $$x_n ={C_1^{n}}X_1 +\cdots + {C_{n-1}^{n}}X_{n-1}.$$ Eliminating $X_1,\dots, X_n$ between these equations, a linear relationship between $x_1,\dots, x_n$ will be deduced.