The multiplier problem of the calculus of variations for scalar ordinary differential equations

Abstract

In the long-standing inverse problem of the calculus of variations one is asked to find a Lagrangian and a multiplier so that a given differential equation, after being multiplied with the multiplier, becomes the Euler–Lagrange equation for the Lagrangian. An answer to this problem for the case of a scalar ordinary differential equation of order $2n, n\ge 2,$ is proposed.

Proofs of some elementary results

1.1 Proof of Lemma 2.3

Proof

Let $m, n, m\ge n,$ be fixed. Note that this lemma is trivial for $k=0$. If it holds for some $k<n$, then

$$\begin{aligned} \partial _n D_m^{k+1} =\,&\sum _{j=0}^k \left( {\begin{array}{c}k\\ j\end{array}}\right) D_m^{k-j} \partial _{n-j} D_m \\ =\,&\sum _{j=0}^k \left( {\begin{array}{c}k\\ j\end{array}}\right) D_m^{k-j} (D_m \partial _{n-j} + \partial _{n-j-1}) \\ =\,&\sum _{j=0}^k \left( {\begin{array}{c}k\\ j\end{array}}\right) D_m^{k+1-j} \partial _{n-j} + \sum _{j=1}^{k+1} \left( {\begin{array}{c}k\\ j-1\end{array}}\right) D_m^{k+1-j} \partial _{n-j} \\ =\,&D^{k+1}_m \partial _n + \sum ^k_{j=1} \left[ \left( {\begin{array}{c}k\\ j\end{array}}\right) + \left( {\begin{array}{c}k\\ j-1\end{array}}\right) \right] D_m^{k+1-j} \partial _{n-j} + \partial _{n-k-1} \\ =\,&\sum ^{k+1}_{j=0} \left( {\begin{array}{c}k+1\\ j\end{array}}\right) D_m^{k+1-j} \partial _{n-j}\ . \end{aligned}$$

Hence the formula holds for all $k=0,1,\ldots ,n$. On the other hand, if it is true for some $k\ge n$, we have

$$\begin{aligned} \partial _n D_m^{k+1} =\,&\sum _{j=0}^n \left( {\begin{array}{c}k\\ j\end{array}}\right) D_m^{k-j} \partial _{n-j} D_m \\ =\,&\sum _{j=0}^{n-1} \left( {\begin{array}{c}k\\ j\end{array}}\right) D_m^{k-j} (D_m \partial _{n-j} + \partial _{n-j-1}) + \left( {\begin{array}{c}k\\ n\end{array}}\right) D^{k-n}_m \partial _0 D_m \\ =\,&\sum _{j=0}^{n-1} \left( {\begin{array}{c}k\\ j\end{array}}\right) D_m^{k+1-j} \partial _{n-j} + \sum _{j=0}^{n-1} \left( {\begin{array}{c}k\\ j\end{array}}\right) D_m^{k-j} \partial _{n-j-1} + \left( {\begin{array}{c}k\\ n\end{array}}\right) D^{k+1-n}_m \partial _0 \\ =\,&D^{k+1}_m \partial _n + \sum _{j=1}^n \left( {\begin{array}{c}k\\ j\end{array}}\right) D_m^{k+1-j} \partial _{n-j} + \sum _{j=1}^n \left( {\begin{array}{c}k\\ j-1\end{array}}\right) D_m^{k+1-j} \partial _{n-j} \\ =\,&D^{k+1}_m \partial _n + \sum _{j=1}^n \left( {\begin{array}{c}k+1\\ j\end{array}}\right) D_m^{k+1-j} \partial _{n-j} \\ =\,&\sum _{j=0}^n \left( {\begin{array}{c}k+1\\ j\end{array}}\right) D_m^{k+1-j} \partial _{n-j}, \end{aligned}$$

and the result follows. $\square$

1.2 Analytic Proof of Lemma 2.5

Proof

Let m be fixed and we proceed by induction on k. For $k=1$, if

$$\begin{aligned} \left\| {I}\right\| _{} = i_{m-1} + 2i_{m-2} + \cdots + mi_0 \le 1, \end{aligned}$$

the only possible multi-indices are $I^m = (0,0,\ldots ,0)$ and $I^m = (1,0,\ldots ,0)$. Then the right hand side reduces to $D_m$, agreeing with the left hand side.

Assume that the result holds for some $1\le {k}\le {m-2}$. Then,

$$\begin{aligned} D_m^{k+1} =\,&(D_{m-1} + p_m \partial _{m-1} ) \sum _{\left\| {I}\right\| _{} \le k} a_{I}^{(k)} p^{\left| {I}\right| _{}}_m D^{k-\left\| {I}\right\| _{}}_{m-1} \partial ^{I} \\ =\,&\sum _{\left\| {I}\right\| _{} \le k} a_{I}^{(k)} p^{\left| {I}\right| _{}}_m D^{k+1-\left\| {I}\right\| _{}}_{m-1} \partial ^{I} + \sum _{\left\| {I}\right\| _{} \le k} a_{I}^{(k)} p^{\left| {I}\right| _{}+1}_m \partial _{m-1} D^{k-\left\| {I}\right\| _{}}_{m-1} \partial ^{I}. \end{aligned}$$

The idea is simple — use Lemma 2.3 to move the single derivative $\partial _{m-1}$ in the second term to the far right and perform a shift in the “dummy” multi-index I. This is possible since the coefficient $a_I^{(k)}\ne 0$ only when $I\ge 0$ and $\left( {\begin{array}{c}n\\ r\end{array}}\right) \ne 0$ only in the case $n\ge {r}$, for integers n and r.

To carry out this argument, we observe that by Lemma 2.3,

$$\begin{aligned} \partial _{m-1}D_{m-1}^{k-\left\| {I}\right\| _{}}=\sum _{j=0}^{k-\left\| {I}\right\| _{}} \left( {\begin{array}{c}k-\left\| {I}\right\| _{}\\ j\end{array}}\right) D_{m-1}^{k-\left\| {I}\right\| _{}-j}\partial _{m-1-j} \end{aligned}$$

since $k-\left\| {I}\right\| _{}\le k\le m-1$. However, since $\displaystyle \left( {\begin{array}{c}k-\left\| {I}\right\| _{}\\ j\end{array}}\right) =0$ for $j>k-\left\| {I}\right\| _{}$, we also have

$$\begin{aligned} \partial _{m-1}D_{m-1}^{k-\left\| {I}\right\| _{}} =\,&\sum _{j=0}^{m-1} \left( {\begin{array}{c}k-\left\| {I}\right\| _{}\\ j\end{array}}\right) D_{m-1}^{k-\left\| {I}\right\| _{}-j}\partial _{m-1-j} \\ =\,&\sum _{j=1}^{m} \left( {\begin{array}{c}k-\left\| {I}\right\| _{}\\ j-1\end{array}}\right) D_{m-1}^{k+1-\left\| {I}\right\| _{}-j}\partial _{m-j}. \end{aligned}$$

It follows that

$$\begin{aligned} D_m^{k+1} =\,&\sum _{\left\| {I}\right\| _{}\le k} a_{I}^{(k)}p^{\left| {I}\right| _{}}_m D^{k+1-\left\| {I}\right\| _{}}_{m-1} \partial ^{I} + \sum _{j=1}^m\sum _{\left\| {I}\right\| _{}\le k} \left( {\begin{array}{c}k-\left\| {I}\right\| _{}\\ j-1\end{array}}\right) a_{I}^{(k)} p_m^{\left| {I}\right| _{}+1}D_{m-1}^{k+1-\left\| {I}\right\| _{}-j}\partial _{m-j}\partial ^{I}. \\ \end{aligned}$$

By writing $\partial _{m-j}=\partial ^{e_j}$ where $e_j=(0,\ldots ,1,\ldots ,0)$, with its “1” appears in the j-th slot from the left, we have

$$\begin{aligned} D_{m}^{k+1}=\,&\sum _{\left\| {I}\right\| _{}\le k} a_{I}^{(k)}p^{\left| {I}\right| _{}}_m D^{k+1-\left\| {I}\right\| _{}}_{m-1} \partial ^{I} \\&+ \sum _{j=1}^m\sum _{\left\| {I+e_{j}}\right\| _{}\le k+j} \left( {\begin{array}{c}k+j-\left\| {I+e_{j}}\right\| _{}\\ j-1\end{array}}\right) a^{(k)}_{(I+e_{j})-e_{j}} \cdot p_m^{\left| {I+e_{j}}\right| _{}}D_{m-1}^{k+1-\left\| {I+e_{j}}\right\| _{}}\partial ^{I+e_{j}}. \\ \end{aligned}$$

Since we have set $a_{I}^{(k)}=0$ whenever $i_{m-j}<0$ for some j with $1\le {j}\le {m}$, we may implicitly assume that $I-e_j\ge 0$ in the summands. Hence,

$$\begin{aligned} D_m^{k+1} =\,&\sum _{\left\| {I}\right\| _{}\le k} a_{I}^{(k)} p_m^{\left| {I}\right| _{}}D_{m-1}^{k+1-\left\| {I}\right\| _{}}\partial ^{I} + \sum _{j=1}^{m} \sum _{\left\| {I}\right\| _{}\le k+j} \left( {\begin{array}{c}k+j-\left\| {I}\right\| _{}\\ j-1\end{array}}\right) a_{I-e_{j}}^{(k)}p_m^{\left| {I}\right| _{}}D_{m-1}^{k+1-\left\| {I}\right\| _{}}\partial ^{I}. \end{aligned}$$

Using the fact that $\displaystyle \left( {\begin{array}{c}k+j-\left\| {I}\right\| _{}\\ j-1\end{array}}\right) =0$ for $\left\| {I}\right\| _{}>k+1$, we have

$$\begin{aligned} D_m^{k+1} =\,&\sum _{\left\| {I}\right\| _{}\le k} a_{I}^{(k)} p_m^{\left| {I}\right| _{}}D_{m-1}^{k+1-\left\| {I}\right\| _{}}\partial ^{I} + \sum _{j=1}^{m} \sum _{\left\| {I}\right\| _{}\le k+1} \left( {\begin{array}{c}k+j-\left\| {I}\right\| _{}\\ j-1\end{array}}\right) a_{I-e_{j}}^{(k)}p_m^{\left| {I}\right| _{}}D_{m-1}^{k+1-\left\| {I}\right\| _{}}\partial ^{I} \\ =\,&\sum _{\left\| {I}\right\| _{}\le k} \left[ a_{I}^{(k)}+\sum _{j=1}^{m}\left( {\begin{array}{c}k+j-\left\| {I}\right\| _{}\\ j-1\end{array}}\right) a^{(k)}_{I-e_{j}}\right] p_m^{\left| {I}\right| _{}}D_{m-1}^{k+1-\left\| {I}\right\| _{}}\partial ^{I}. \end{aligned}$$

Finally, we compute

$$\begin{aligned}&a_{I}^{(k)}+\sum _{j=1}^{m} \left( {\begin{array}{c}k+j-\left\| {I}\right\| _{}\\ j-1\end{array}}\right) a_{I-e_{j}}^{(k)} \\&\quad = \dfrac{k!}{\left\| {I}\right\| _{}^*I!(k-\left\| {I}\right\| _{})!} +\sum _{j=1}^{m} \dfrac{(k+j-\left\| {I}\right\| _{})!}{(j-1)!(k+1-\left\| {I}\right\| _{})!}\cdot \dfrac{k!}{\frac{\left\| {I}\right\| _{}^*}{j!}\cdot \frac{I!}{i_{m-j}}\cdot (k+j-\left\| {I}\right\| _{})!} \\&\quad = \dfrac{k!}{\left\| {I}\right\| _{}^*I!(k-\left\| {I}\right\| _{})!} + \sum _{j=1}^{m} \dfrac{ji_{m-j}k!}{\left\| {I}\right\| _{}^*I^m!(k+1-\left\| {I}\right\| _{})!} \\&\quad = \dfrac{k!}{\left\| {I}\right\| _{}^*I!(k+1-\left\| {I}\right\| _{})!} \left( k+1-\left\| {I}\right\| _{}+\sum _{j=1}^{m} ji_{m-j} \right) \\&\quad = \dfrac{(k+1)!}{\left\| {I}\right\| _{}^*I!(k+1-\left\| {I}\right\| _{})!} \\&\quad = a^{(k+1)}_{I}. \end{aligned}$$

Since the result is also true for $k+1$, it holds for all $k=1,\ldots ,m-1$. $\square$

1.3 Combinatorial Proof of Lemma 2.5

Proof

By Lemma 2.2(a), $D_m^k$ can be expanded as a binomial

$$\begin{aligned}\begin{aligned} D_m^k&=(D_{m-1}+p_m\partial _{m-1})^k\\ \end{aligned}\end{aligned}$$

which is equal to the sum of all possible strings of length k formed by the alphabet $\{D_{m-1},p_m\partial _{m-1}\}$. Each $\partial _{m-1}$ followed by $D_{m-1}$ gives rise to more terms by Lemma 2.2(b). In what follows, in the commutation of $p_m\partial _{m-1}D_{m-1}$, we call $p_mD_{m-1}\partial _{m-1}$ the higher order term and $p_m\partial _{m-2}$ the lower order one. Hence,

$$\begin{aligned} p_m\partial _{m-1}D_{m-1}^{j-1}=p_m\partial _{m-j}+\text {higher order terms}. \end{aligned}$$

Suppose a term has partial derivatives $\partial ^I=\partial _{m-j}^{i_{m-j}}\cdots \partial _0^{i_0}$. Two observations are in order:

• Each $\partial _{m-j}$ is associated to one and only one $p_m$. Hence $\partial ^I$ is multiplied by $p_m^{\left| {I}\right| _{}}$.

• Each $\partial _{m-j}$ is obtained as the lowest order term of $\partial _{m-1}D_{m-1}^{j-1}$. The resulting number of $D_{m-1}$ plus $\left\| {I}\right\| _{}$ is then the constant k.

It suffices to count the number of times $p_m^{\left| {I}\right| _{}}D_{m-1}^{k-\left\| {I}\right\| _{}}\partial ^I$ appears.

We observe that the coefficient

$$\begin{aligned} \dfrac{k!}{\left\| {I}\right\| _{}^*I!}=a_I^{(k)}(k-\left\| {I}\right\| _{})! \end{aligned}$$

counts the number of ways partitioning a k-element set into sub-classes whose numbers of elements are given by I, see, for example, 12(a) in Chapter 1 of Lovász [10]. By dividing it by $(k-\left\| {I}\right\| _{})!$, the number that $D_{m-1}$ appears, we do not distinguish between the $D_{m-1}$’s. Therefore, $a_I^{(k)}$ is the number of ways forming permutations of $i_{m-1}$ many $(p_m\partial _{m-1})$’s, $i_{m-2}$ many $(p_m\partial _{m-1}D_{m-1})$’s, $i_{m-3}$ many $(p_m\partial _{m-1}D_{m-1}^2)$’s, etc., as well as $(k-i_{m-1}-2i_{m-2}-\cdots -mi_0)$ many $D_{m-1}$’s.

In this setting, $I=(i_{m-1},\ldots ,i_{0})$ where $i_{m-j}$ is the number of $p_{m}\partial _{m-1}$’s to be commuted with its (immediately) following $(j-1)$ $D_{m-1}$’s. To fix the idea, let us look at an example. Suppose $k=10$ and $I=(i_{m-1},i_{m-2},i_{m-3})=(3,1,1)$, i.e. consider

$$\begin{aligned} p_m^5D_{m-1}^2\partial _{m-1}^3\partial _{m-2}\partial _{m-3}. \end{aligned}$$

(29)

This could be a result of re-ordering (always taking the highest order terms), say,

$$\begin{aligned} D_{m-1}(p_m\partial _{m-2})D_{m-1}(p_m\partial _{m-1})^3(p_m\partial _{m-3}). \end{aligned}$$

(30)

In order that it comes into place, the lower order $\partial _{m-2}$ and $\partial _{m-3}$ must be originally $\partial _{m-1}D_{m-1}$ and $\partial _{m-1}D_{m-1}^2$. Thus the corresponding binomial expansion is

$$\begin{aligned} D_{m-1}(p_m\partial _{m-1}D_{m-1})D_{m-1}(p_m\partial _{m-1})^3(p_m\partial _{m-1}D_{m-1}^2), \end{aligned}$$

(31)

or

$$\begin{aligned} D_{m-1}(p_{m}\partial _{m-1})D_{m-1}^2(p_{m}\partial _{m-1})^4D_{m-1}^2, \end{aligned}$$

one of the many terms in $(D_{m-1}+p_m\partial _{m-1})^{10}$.

Since (30) has a one-to-one correspondence to (31) when I is fixed, the total number of terms of the form (29) is given by $a_I^{(k)}$, as desired. $\square$