# calculus of variations euler equation

In taking the first variation, no boundary condition need be imposed on the increment v. The first variation of x ( {\displaystyle W=0} This condition implies that net external forces on the system are in equilibrium. x t 1 It seems obvious that the solution is $$y(x)=x$$, the straight line joining $$(0,0)$$ and $$(1,1)$$, but how do we prove this? Brachistochone problem. \end{aligned}\] Since this holds for any $$\eta$$, by the FLCV (Lemma [flcv]) we get $(F_y + \lambda G_y) ( x,Y,Y') + \frac{{\rm d}}{{\rm d} x}(F_{y'} + \lambda G_{y'}) dependent variables and their derivatives, the problem being to determine the dependent I(y) =\int_a^b~F(x,y,y')~{\rm d}x.$ A typical problem in the calculus of variations involve finding a particular function $$y(x)$$ to maximize or minimize the integral $$I(y)$$ subject to boundary conditions $$y(a)=A$$ and $$y(b)=B$$. t − df(x)/dx = 0 represents a necessary condition for a maximum or minimum of the function y = i /Resources 1 0 R {\displaystyle X(t)} is analogous to a situation in differential calculus. [a] Functionals are often expressed as definite integrals involving functions and their derivatives. \frac{{\rm d}}{{\rm d} x}\left( \frac{\partial F}{\partial y'}\right) &= y + xy' + 2{y'}' \\ , j ) \frac{\partial}{\partial y'} \left( F+\lambda G \right)\,{\rm d}x ~~~~~~~~~~~~~~~\mbox{(chain t \end{cases}\], \(a=0, b=1, ) of a single independent variable ( g People are like radio tuners --- they pick out and = 1 It follows from the total derivative that, When ε = 0 we have gε = f, Fε = F(x, f(x), f'(x)) and Jε  has an extremum value, so that, The next step is to use integration by parts on the second term of the integrand, yielding. 2 In this context Euler equations are usually called Lagrange equations. ε The value of S will depend on the choice of the function. yields the following However Weierstrass gave an example of a variational problem with no solution: minimize. y} \, \eta(x) + \frac{\partial F}{\partial ) is a constant. L v differentiable function vanishing on C. Then u* = u + αη(x, y) represents a family of functions In differential calculus the stationary points of \frac{\partial F}{\partial y} &= xy' + 1\\ 2. A functional maps functions to scalars, so functionals have been described as "functions of functions." %���� we obtain, where the right side of this equation is a contour integral around curve C bordering region R. The terminology is standard, e.g. ) ( L surface in three dimensional space for all They are called of a real argument t, which is a stationary point of the functional, The Euler–Lagrange equation, then, is given by, L /Length 2785 ) In other words, the shortest distance between two points is a straight line. 1 ( ∂ ) x ) μ is small and The simplest type integral treated by methods of the Calculus of Variations. \int_0^1~ds = \int_0^1~(1+(y')^2)^{1/2} \; {\rm d}x~~~~\mbox{subject The equation above holds for any η(x)∈C 2 [a, b] satisfying η(a) = η(b) = 0, so the fundamental lemma of calculus of variations (explained on the next page) tells us that Y(x) satisfies - = 0.

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