Math 115 ( ) Yum-Tong Siu 1. General Variation Formula and Weierstrass-Erdmann Corner Condition. J = x 1. F (x,y,y )dx.

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1 Math Yum-Tong Siu 1 General Variation Formula and Weierstrass-Erdmann Corner Condition General Variation Formula We take the variation of the functional F x,y,y dx with the two end-points,y 1, x 2,y 2 allowed freely to vary Suppose we have a family curves y = yx,t parametrized by t so that the extremal corresponds to t = 0 The new setting in our situation is that the end-points and x 2 are not independent of t In fact, = t and x 2 = x 2 t are both functions of t So the functional J as a function of t can be written as Jt = t t F x,yx,t, yx,t dx Before we differentiate the functional Jt with respect to t, we first discuss the two variations of y with respect to t, one with x fixed and the other with x also varying as a function of t By the chain rule applied to y = yxt,t, we get d dt yxt,t = t yxt,t + yxt,t d dt xt We denote the left-hand side by δy, which represents the actual variation of an end-point so that if we require the end-point to lie on a prescribed curve it is this variation δy which will be used The second factor d xt of the dt second term on the right-hand side is δx The first term on the right-hand we will denote by t y so that t y = δy y δx By applying the Fundamental Theorem of Calculus to the variation of the upper limit and the lower limit of the integral functional, we get δ F x,y,y δx t F x,y,y dx + The second term on the right-hand side can be dealt with in the same way as the simpler situation of the end-points being fixed and we obtain t F x,y,y dx = Fy t y + F y t y dx

2 Math Yum-Tong Siu 2 which by integration by parts gives Fy t y + F y t y dx = F y d dx F y t ydx+f y t y Now we combine together,, and to get δ F y d x=x dx F 2 y t y dx + F y δy + F y F y δx This is known as the general variation formula Condition for End-Point to be on Prescribed Curve Suppose the end-point x j,y j j = 1, 2 is constrained to be on the prescribed curve g j x,y First by considering the variation with the two end-points fixed we get the Euler- Lagrange equation to conclude from the general variation formula that δ F y δy + F y F y δx By differentiating the equation g j x j,y j = 0 with respect to the parameter t of the family of curves, we get δx x=xj + δy x=xj = 0 which can be rewritten as δy x=xj = δx x=xj When we put this into, from the vanishing of δj we get 0 = F y δx F y δx + F y F y δx x=x2 x=x1 g 2 g 2 g 1 g 1 Since δx is allowed to vary freely both at x = and x = x 2, it follows that F y + F y F y = 0 at x = x j for j = 1, 2, which can be written as at x = x j for j = 1, 2 F y = F y F y

3 Math Yum-Tong Siu 3 Weierstrass-Erdmann Corner Condition Suppose our functional J is given as the sum of two integrals F x,y,y dx + x=x2 F x,y,y dx so that the two end-points y 1 = y and y 2 = y x 2 but the ordinate η = y ξ of the middle point with abscissa x = ξ is allowed to vary freely Then we can first fix the middle point and vary the two integrals separately to get one Euler-Lagrange equation on each of the two intervals [,ξ] and [ξ,x 2 ] Then we consider the condition imposed by the free variation of the middle point ξ,η to get δ F y δy + F y F y δx + F y δy + F y F y δx which, on account of the fixing of the two end-points,y 1 and x 2,y 2, becomes +0 0 δ F y δy + F y F y δx 0 +0 If both δy and δx are allowed to vary freely, the vanishing of δj implies the following two statements which are known as the Weierstrass-Erdmann corner condition F y = F y, 0 +0 F y F y = F y F y 0 +0 In other words, the two canonical coordinates p = F y and H = y F y F are continuous at the corner if the middle point is allowed to vary freely If we have the free variation of only δy, then we have only the partial Weierstrass-Erdmann corner condition F y = F y 0 +0

4 Math Yum-Tong Siu 4 On the other hand, If we have the free variation of only δx, then we have only the partial Weierstrass-Erdmann corner condition F y F y = F y F y 0 +0 In general, if we have the free variation of only δx and δy on a prescribed curve gx,y = 0, then we have δx g x + δy g y = 0 and the partial Weierstrass-Erdmann corner condition g x F y + g y F y F y = g x F y + g y F y F y 0 +0 Snell s Law as Application of the Weierstrass-Erdmann Corner Condition Consider the problem of light traveling in two different media where the velocity of light is c 1 and c 2 respectively The light trajectory is the extremal of the functional where F 1 x,y,y dx + F j = x=x2 2 F 2 x,y,y dx, for j = 1, 2 First we consider the case of fixed the middle point where the corner occurs to get the two Euler-Lagrange equations F j y d dx F j y = 0 Since each F j is independent of x, we have the first integral which is the analog of the conservation of energy Explicit computation gives H j = y F j y F j = constant H j = y F j y F j = 1 y = 1 1 2

5 Math Yum-Tong Siu 5 This means that light travels in a straight line in each of the two media We now assume that the interface is given by x = ξ with ξ fixed Then only δy is allowed to vary freely at x = ξ At the interface, the canonical coordinate p is continuous That is, F 1 y = F 2 y, 0 +0 which means 1 y c 1 2 = 1 y 0 c Introduce the angle of incidence θ 1 and the angle of refraction θ 2 given by y 0 = tanθ 1 and y +0 = tanθ 2 Then we can write the above condition in the form of Snell s law for the refraction of light which states that sin θ 1 c 1 = sin θ 2 c 2