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# Class Note for MATH 499 at UA

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Date Created: 02/06/15
MATH499H Summary Topic Calculus of Variations Professor Dr Bruce Bayly Student Jeffrey Truman A functional has a set of functions as its domain and a set of real numbers as its range For example CD f01yxdx is a functional Note that the functionals we will examine here will be of the form I I dx where I is the integrand Some of these have ranges that are bounded and the ones examined here will typically have a minimum or maximum that we are interested in finding such as F y 21 y2 2xyxdx and we will assume that they are continuous Next we will examine the difference quotient for a functional As one might expect the difference quotient for a functional is FLv Ay FIy Ay Keeping in mind that y is a function of x we evaluate FyAy 2 2 I y Ay2 2xy Aydx J y2 216 Ay2y 2x Ay2 dx 1 1 Now this will be at an extremum when the coef cient of the rst order of Ay is zero giving the function yxx Since the secondorder Ay term is always positive yxx gives a minimum We will call the rstorder coefficient the functional derivative and it can be derived that for integralform functionals whose integrand depends only on x yx and y39x the functional derivative where I is the integrand of the lnctional is 6F a1 d 01 E E 567 When this is set equal to zero we get the EulerLagrange equation BI d BI 0 By dx By However sometimes these problems are unsolvable such as 4 Kb39 f y2 2xyx5y39 y393 3 for which the EulerLagrange equation yields the nonlinear secondorder ODE 2y 2x 5x4 6y y 0 There are several applications of the EulerLagrange equation One simple example is the problem of finding the shortest path between two points This starts with the arc length integral Ly f 11 y392 dx Evaluating the functional derivative of this integral provides the result of 5L BI d 01 d a d 1 I2 I 2 6y 6y dxayl 0 dxay1y dxy1y2 d 1 1 d l 2 E 2 E I 3 y dx1y 1y dxm 3 1 yr 1 yll 2 Zylyll 1 ylz 2 3quot ylzyll1 ylz 2 yn1 ylz y 1 y 1 y z a 39 1 3 2 39 2 1 y 22 1 y 22 1 y 22 1 y 22 Now using the EulerLagrange equation the functional is at an eXtremum when this quantity is equal to zero for all X Note that this occurs when yquot 0 or when yX is linear Therefore we have shown that the shortest distance between two points is a straight line which does not come as a surprise but does give us some confidence in this method 12 3 y yu1 y127 However for more complex examples it is often useful to use the energy function In general for integralform functional dependent on y and 3139 but not X there is an energy for which the derivative with respect to X is zero at the eXtrema As it turns out this is the energy function 1500 y39a I 1 ayy We can use this to find the surface of revolution of smallest area between two rings also known as the soapbubble problem First of all the functional for this surface from Xr points l2 to 13 is the following 1 Sr 21rf r 1r 2 dx 1 Also the functional derivative for this is 65 HI 1 61 d Zm r 2 8y ay dxay39 2 1 r dx 1 r2 I Zn39rrquot 2nr 21 r 2 1 r zz rrquot r 21 r 2 ml 1 r 2 1 02132 2quot 1 02132 Applying the EulerLagrange equation to this yields 1 r zz rrquot r 21 r 2 0 and this reduces to the secondorder nonlinear ODE 1 r392 Trquot 0 On the other hand considering the energy function above gives rr392 rlr392 r V1 r392 W 39Y rltr 39Y r J l 4 1 We then evaluate the derivative of this with respect to X dE i 392 39 Iz Tll I2 III dx dxr1r r1r 2r1r er Ex r l r392 rl rrlrll quotIrquot rl1 112 7117quot 1 13 1 3 3 3 1 rrzE 1 rrzE 1 rd 1 rd When this is set equal to zero we get rr r r r 3 0 gt r rr 1 r z 0 Multiplying through by l and taking 1quot 0 this yields the same thing as the EulerLagrange equation Therefore the energy function must be a constant 2 2 r T 392 L 392 r L E W gtE 1r gtE2 1r gtr i E2 1 So this forms a firstorder separable ODE Considering only the positive 7quot and using the substitution cosh u we can determine that the solution to this problem is a hyperbolic cosine curve There are also more complex problems that we can examine such as geodesics Geodesics are 2D surfaces in 3D space such as the surface of a sphere Recall the equations for transforming Cartesian coordinates to spherical coordinates x RsimpcosB y Rsimpsina z Rcosq On a geodesic sphere R is fixed The change of the other two variables can be found as follows Ax Rsinp Ago cos9 A0 sintpcose Ay Rsin Acp sin9 A0 sincpsine A2 Rcosp Ap sincp Since distance d2 Ax2 A302 Az2 we can convert to p and 6 and ultimately show dz R2Ap2 Rerzsin2 This can be used to produce the functional for the distance between two points on a geodesic sphere using the arc length formula b b hp 2 f IR2dp2 stin2 d02 Rf sin2 d9 1 a dB This gives a functional dependent of p9 which can be examined through the methods developed earlier Additionally with the use of the functional version of Lagrange multipliers we can solve the isoperimetric problem ie the shape that encloses the most area for a given perimeter or has the least perimeter for a given area To start the functional for area and perimeter are Ayfbydx and Ly1fb1y392dx According to the functional version of Lagrange multipliers the way to accomplish our goal is to extremize ADI or Ly The first will have a maximum and the second will have a minimum at the same point We will maximize the first call it Fy by the method of the energy function EWFVEW 41y392y lly392 2 y392 y 1 y392 1 y392 39 1 39 y w P Finding the function yX for which this expression is equal to zero for all X should nd this shape Unfortunately however this marks the end of this independent study and we did not have the time to do this

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