Physics 430: Lecture 14 Calculus of Variations. Dale E. Gary NJIT Physics Department. y. y 2 y 1. y = y ( x ). 2. 1. x. x 1 x 2. 6.1 Two Examples. The calculus of variations involves finding an extremum (maximum or minimum) of a quantity that is expressible as an integral.
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Dale E. Gary
NJIT Physics Department
y = y(x)
x1x26.1 Two Examples
y = y(x), and an element of length along the path is
this a lot in the next few chapters) as
which is valid because Thus, the length is
y = y(x) (right)
x1x26.2 Euler-Lagrange Equation-1
1 and 2, the “right” curve y(x), and a “wrong” curve
Y(x) that is a small displacement from the “right”
curve, as shown in the figure.
so dS/da = 0 gives
but the first term of this relation (the end-point term) is zero because h(x) is zero at the endpoints.
where is the function appropriate to your problem. Write down the Euler-Lagrange equation, and solve for the function y(x) that defines the required stationary path.
y(x) = mx + b. In other words, a straight line is the shortest path.
Give ds in the following situations:
yExample 6.2: The Brachistochrone
Notice that, since we are
using y as the independent
variable, we swap x and y
where the constant is renamed 1/2a for future convenience.
3Example 6.2: cont’d
for a cart to travel this path from 23 is the same,
no matter where 2 is placed, from 1 to 3! Thus,
oscillations of the cart along that path are exactly
isochronous (period perfectly independent of amplitude).