Mathematical Problems : $b Lecture delivered before the International Congress of Mathematicians at Paris in 1900 — John Shaqi
Mathematical Problems : $b Lecture delivered before the International Congress of Mathematicians at Paris in 1900Hilbert, David
Science
Mathematical Problems : $b Lecture delivered before the International Congress of Mathematicians at Paris in 1900
Hilbert, David
Mathematics
which also admits these integral curves as solutions, then the function
is always an integral of the partial differential equation
(1*) of the first order; and conversely, if denotes any
solution of the partial differential equation (1*) of the first
order, all the non-singular integrals of the ordinary differential
equation (2) of the first order are at the same time integrals of
the differential equation (1) of the second order, or in short if
is an integral equation of the first order of the
differential equation (1) of the second order, represents
an integral of the partial differential equation (1*) and conversely;
[Pg 39]
the integral carves of the ordinary differential equation of the second
order are therefore, at the same time, the characteristics of the
partial differential equation (1*) of the first order.
In the present case we may find the same result by means of a simple
calculation; for this gives us the differential equations (1) and (1*)
in question in the form
where the lower indices indicate the partial derivatives with respect
to . The correctness of the affirmed relation
is clear from this.
The close relation derived before and just proved between the ordinary
differential equation (1) of the second order and the partial
differential equation (1*) of the first order, is, as it seems to me,
of fundamental significance for the calculus of variations. For, from
the fact that the integral is independent of the path of
integration it follows that
if we think of the left hand integral as taken along any path and
the right hand integral along an integral curve of
the differential equation
With the help of equation (3) we arrive at Weierstrass's formula
where designates Weierstrass's expression, depending upon
,
Since, therefore, the solution depends only on finding an integral
which is single valued and continuous in a certain
neighborhood of the integral curve , which we are
considering, the developments just indicated lead immediately—without
the introduction of the second variation, but only[Pg 40] by the application
of the polar process to the differential equation (1)—to the
expression of Jacobi's condition and to the answer to the question:
How far this condition of Jacobi's in conjunction with Weierstrass's
condition 0"> is necessary and sufficient for the occurrence of a
minimum.
The developments indicated may be transferred without necessitating
further calculation to the case of two or more required functions, and
also to the case of a double or a multiple integral. So, for example,
in the case of a double integral
to be extended over a given region , the vanishing of the
first variation (to be understood in the usual sense)
gives the well-known differential equation of the second order
for the required function of and .
On the other hand we consider the integral
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