Solution 48.7 C1 Extremals Without loss of generality, we take the vertical line to be the y axis We will consider x1 , y1 > 1 With ds = 1 + (y )2 dx we extremize the integral, x1 √ y 1 + (y )2 dx 0 Since the Lagrangian is independent of x, we know that the Euler differential equation has a first integral d Fy − Fy = 0 dx y Fy y + y Fy y − Fy = 0 d (y Fy − F ) = 0 dx y Fy − F = const For the given Lagrangian, this is y √ y y )2 − √ 1 + (y )2 = const, y 1 + (y √ √ (y )2 y − y(1 + (y )2 ) = const 1 + (y )2 , √ y = const 1 + (y )2 y = const is one solution To find the others we solve for y and then solve the differential equation y = a(1 + (y )2 ) y =± dx = y−a a a dy y−a 2094 ±x + b = 2 a(y − a) y= bx b2 x2 ± + +a 4a 2a 4a The natural boundary condition is √ Fy x=0 = yy 1 + (y )2 = 0, x=0 y (0) = 0 The extremal that satisfies this boundary condition is y= x2 + a 4a Now we apply y(x1 ) = y1 to obtain a= 1 2 y1 ± 2 y1 − x2 1 for y1 ≥ x1 The value of the integral is x1 0 x2 +a 4a x 1+ 2a By denoting y1 = cx1 , c ≥ 1 we have 2 x1 (x2 + 12a2 ) 1 dx = 12a3/2 √ 1 cx1 ± x1 c2 − 1 2 The values of the integral for these two values of a are √ 2 2 √ 3/2 −1 + 3c ± 3c c − 1 √ 2(x1 ) 3(c ± c2 − 1)3/2 a= 2095 √ The values are equal only when c = 1 These values, (divided by x1 ), are plotted in Figure 48.1 as a function of c The former and latter are fine and coarse dashed lines, respectively The extremal with a= 1 2 2 y1 − x2 1 y1 + has the smaller performance index The value of the integral is 2 x1 (x2 + 3(y1 + y1 − x2 )2 1 1 √ 2 2 3 3 2(y1 + y1 − x1 ) √ The function y = y1 is an admissible extremal for all x1 The value of the integral for this extremal is x1 y1 which is larger than the integral of the quadratic we analyzed before for y1 > x1 4 3.5 3 2.5 1.2 1.4 1.6 1.8 Figure 48.1: Thus we see that y= ˆ x2 + a, 4a a= 1 2 2096 y1 + 2 y1 − x2 1 2 is the extremal with the smaller integral and is the minimizing curve in C 1 for y1 ≥ x1 For y1 < x1 the C 1 extremum is, y = y1 ˆ C1 Extremals Consider the parametric form of the Lagrangian p t1 y(t) (x (t))2 + (y (t))2 dt t0 The Euler differential equations are d fx − fx = 0 and dt d fy − fy = 0 dt If one of the equations is satisfied, then the other is automatically satisfied, (or the extremal is straight) With either of these equations we could derive the quadratic extremal and the y = const extremal that we found previously We will find one more extremal by considering the first parametric Euler differential equation d fx − fx = 0 dt d dt y(t)x (t) (x (t))2 + (y (t))2 y(t)x (t) (x (t))2 + (y (t))2 =0 = const Note that x(t) = const is a solution Thus the extremals are of the three forms, x = const, y = const, x2 bx b2 y= + + + a 4a 2a 4a 2097 The Erdmann corner conditions require that √ Fy = F − y Fy = √ y yy , 1 + (y )2 √ y(y )2 2− 1 + (y ) = 1 + (y )2 √ y 1 + (y )2 are continuous at corners There can be corners only if y = 0 1 Now we piece the three forms together to obtain Cp extremals that satisfy the Erdmann corner conditions The only possibility that is not C 1 is the extremal that is a horizontal line from (0, 0) to (x1 , 0) and then a vertical line from (x1 , y1 ) The value of the integral for this extremal is y1 0 √ 2 t dt = (y1 )3/2 3 Equating the performance indices of the quadratic extremum and the piecewise smooth extremum, 2 x1 (x2 + 3(y1 + y1 − x2 )2 2 1 1 = (y1 )3/2 , √ 2 2 3 3 3 2(y1 + y1 − x1 ) √ 3±2 3 √ y1 = ±x1 3 The only real positive solution is √ 3+2 3 √ y1 = x1 ≈ 1.46789 x1 3 The piecewise smooth extremal has the smaller performance index for y1 smaller than this value and the quadratic extremal has the smaller performance index for y1 greater than this value 1 The Cp extremum is the piecewise smooth extremal for y1 ≤ x1 √ √ and is the quadratic extremal for y1 ≥ x1 3 + 2 3/ 3 2098 √ √ 3 + 2 3/ 3 Solution 48.8 The shape of the rope will be a catenary between x1 and x2 and be a vertically hanging segment after that Let the length of the vertical segment be z Without loss of generality we take x1 = y2 = 0 The potential energy, (relative to y = 0), of a length of rope ds in 0 ≤ x ≤ x2 is mgy = ρgy ds The total potential energy of the vertically hanging rope is m(center of mass)g = ρz(−z/2)g Thus we seek to minimize, x2 ρg 0 1 y ds − ρgz 2 , 2 y(0) = y1 , y(x2 ) = 0, subject to the isoperimetric constraint, x2 ds − z = L 0 Writing the arc-length differential as ds = x2 ρg y 0 1 + (y )2 dx we minimize 1 1 + (y )2 ds − ρgz 2 , 2 y(0) = y1 , y(x2 ) = 0, subject to, x2 1 + (y )2 dx − z = L 0 b Consider the more general problem of finding functions y(x) and numbers z which extremize I ≡ a F (x, y, y ) dx+ b f (z) subject to J ≡ a G(x, y, y ) dx + g(z) = L Suppose y(x) and z are the desired solutions and form the comparison families, y(x) + 1 η1 (x) + 2 η2 (x), z + 1 ζ1 + 2 ζ2 Then, there exists a constant such that ∂ (I + λJ) ∂ 1 ∂ (I + λJ) ∂ 2 1 , 2 =0 1 , 2 =0 2099 =0 = 0 These equations are b a and d H,y − Hy η1 dx + h (z)ζ1 = 0, dx b d H,y − Hy η2 dx + h (z)ζ2 = 0, dx a where H = F + λG and h = f + λg From this we conclude that d H,y − Hy = 0, dx h (z) = 0 with λ determined by b G(x, y, y ) dx + g(z) = L J= a 1 Now we apply these results to our problem Since f (z) = − 2 ρgz 2 and g(z) = −z we have −ρgz − λ = 0, z=− λ ρg It was shown in class that the solution of the Euler differential equation is a family of catenaries, y=− λ + c1 cosh ρg x − c2 c1 One can find c1 and c2 in terms of λ by applying the end conditions y(0) = y1 and y(x2 ) = 0 Then the expression for y(x) and z = −λ/ρg are substituted into the isoperimetric constraint to determine λ Consider the special case that (x1 , y1 ) = (0, 0) and (x2 , y2 ) = (1, 0) In this case we can use the fact that y(0) = y(1) to solve for c2 and write y in the form y=− λ + c1 cosh ρg 2100 x − 1/2 c1 Applying the condition y(0) = 0 would give us the algebraic-transcendental equation, y(0) = − λ + c1 cosh ρg 1 2c1 = 0, which we can’t solve in closed form Since we ran into a dead end in applying the boundary condition, we turn to the isoperimetric constraint 1 1 + (y )2 dx − z = L 0 1 cosh 0 x − 1/2 c1 2c1 sinh dx − z = L 1 2c1 −z =L With the isoperimetric constraint, the algebraic-transcendental equation and z = −λ/ρg we now have z = −c1 cosh z = 2c1 sinh 1 , 2c1 1 − L 2c1 For any fixed L, we can numerically solve for c1 and thus obtain z You can derive that there are no solutions unless L is greater than about 1.9366 If L is smaller than this, the rope would slip off the pin For L = 2, c1 has the values 0.4265 and 0.7524 The larger value of c1 gives the smaller potential energy The position of the end of the rope is z = −0.9248 Solution 48.9 Using the method of Lagrange multipliers, we look for stationary values of c ((y )2 + λy 2 ) dx = 0 δ 0 2101 c ((y )2 0 + λy 2 ) dx, The Euler differential equation is d F( , y ) − F,y = 0, dx d (2y ) − 2λy = 0 dx Together with the homogeneous boundary conditions, we have the problem y − λy = 0, y(0) = y(c) = 0, which has the solutions, nπ 2 nπx , , yn = an sin c c Now we determine the constants an with the moment of inertia constraint λn = − c a2 sin2 n 0 nπx c dx = n ∈ Z+ ca2 n =A 2 Thus we have the extremals, 2A nπx sin , c c yn = n ∈ Z+ The drag for these extremals is 2A D= c c 0 nπ c 2 2 cos nπx c An2 π 2 dx = c2 We see that the drag is minimum for n = 1 The shape for minimum drag is y= ˆ 2A nπx sin c c 2102 4 (ii) The continuous extremal that satisfies the boundary conditions is y = 7 − x Since F,y y = 2x2 ≥ 0 has a Taylor series representation for all y , this extremal provides a strong minimum (iii) The continuous extremal that satisfies the boundary conditions is y = 1 This is a strong minimum Solution 48.19 For identity (a) we take P = 0 and Q = φψx − ψφx For identity (b) we take P = φψy − ψφy and Q = 0 For identity 1 (c) we take P = − 1 (φψx − ψφx ) and Q = 2 (φψy − ψφy ) 2 D 1 1 (φψy − ψφy )x − − 2 2 D (φψx − ψφx )y Γ 1 1 − (φψx − ψφx ) dx + (φψy − ψφy ) dy 2 2 1 1 (φx ψy + φψxy − ψx φy − ψφxy ) + (φy ψx φψxy − ψy φx − ψφxy ) 2 2 1 1 =− (φψx − ψφx ) dx + (φψy − ψφy ) dy 2 Γ 2 Γ ψφxy dx dy − φψxy dx dy = D dx dy = D 1 2 (φψx − ψφx ) dx + Γ 1 2 dx dy (φψy − ψφy ) dy Γ The variation of I is t1 (−2(uxx + uyy )(δuxx + δuyy ) + 2(1 − µ)(uxx δuyy + uyy δuxx − 2uxy δuxy )) dx dy dt δI = t0 D From (a) we have −2(uxx + uyy )δuxx dx dy = D −2(uxx + uyy )xx δu dx dy D −2((uxx + uyy )δux − (uxx + uyy )x δu) dy + Γ 2118 From (b) we have −2(uxx + uyy )δuyy dx dy = D −2(uxx + uyy )yy δu dx dy D −2((uxx + uyy )δuy − (uxx + uyy )y δu) dy − Γ From (a) and (b) we get 2(1 − µ)(uxx δuyy + uyy δuxx ) dx dy D 2(1 − µ)(uxxyy + uyyxx )δu dx dy = D 2(1 − µ)(−(uxx δuy − uxxy δu) dx + (uyy δux − uyyx δu) dy) + Γ Using c gives us 2(1 − µ)(−2uxy δuxy ) dx dy = 2(1 − µ)(−2uxyxy δu) dx dy D D 2(1 − µ)(uxy δux − uxyx δu) dx + Γ − 2(1 − µ)(uxy δuy − uxyy δu) dy Γ Note that ∂u ds = ux dy − uy dx ∂n Using the above results, we obtain t1 t1 δI = 2 (− t0 4 ∂( u)δu dx dy dt + 2 D t0 Γ 2 ∂n u) δu + ( 2 u) ∂(δu) ∂n ds dt t1 + 2(1 − µ) (uyy δux − uxy δuy ) dy + (uxy δux − uxx δuy ) dx t0 Γ 2119 dt Solution 48.20 1 Exact Solution The Euler differential equation is d F,y = F,y dx d [2y ] = −2y − 2x dx y + y = −x The general solution is y = c1 cos x + c2 sin x − x Applying the boundary conditions we obtain, y= sin x − x sin 1 The value of the integral for this extremal is J sin x 2 − x = cot(1) − ≈ −0.0245741 sin 1 3 n = 0 We consider an approximate solution of the form y(x) = ax(1−x) We substitute this into the functional 1 (y )2 − y 2 − 2xy dx = J(a) = 0 The only stationary point is 3 1 J (a) = a − = 0 5 6 5 a= 18 2120 3 2 1 a − a 10 6 Since 5 18 J = 3 > 0, 5 we see that this point is a minimum The approximate solution is y(x) = 5 x(1 − x) 18 This one term approximation and the exact solution are plotted in Figure 48.5 The value of the functional is J =− 5 ≈ −0.0231481 216 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.2 0.4 0.6 0.8 1 Figure 48.5: One Term Approximation and Exact Solution n = 1 We consider an approximate solution of the form y(x) = x(1 − x)(a + bx) We substitute this into the functional 1 1 J(a, b) = (y )2 − y 2 − 2xy dx = 63a2 + 63ab + 26b2 − 35a − 21b 210 0 2121 We find the stationary points 1 (18a + 9b − 5) = 0 30 1 Jb = (63a + 52b − 21) = 0 210 7 71 , b= a= 369 41 Ja = Since the Hessian matrix H= Jaa Jab Jba Jbb = 3 5 3 10 3 10 26 105 , is positive definite, 41 3 > 0, det(H) = , 5 700 we see that this point is a minimum The approximate solution is y(x) = x(1 − x) 71 7 + x 369 41 This two term approximation and the exact solution are plotted in Figure 48.6 The value of the functional is J =− 136 ≈ −0.0245709 5535 2 Exact Solution The Euler differential equation is d F,y = F,y dx d [2y ] = 2y + 2x dx y − y = x 2122 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.2 0.4 0.6 0.8 1 Figure 48.6: Two Term Approximation and Exact Solution The general solution is y = c1 cosh x + c2 sinh x − x Applying the boundary conditions, we obtain, y= 2 sinh x − x sinh 2 The value of the integral for this extremal is J =− 2(e4 −13) ≈ −0.517408 3(e4 −1) Polynomial Approximation Consider an approximate solution of the form y(x) = x(2 − x)(a0 + a1 x + · · · an xn ) 2123 The one term approximate solution is y(x) = − 5 x(2 − x) 14 This one term approximation and the exact solution are plotted in Figure 48.7 The value of the functional is J =− 0.5 10 ≈ −0.47619 21 1 1.5 2 -0.05 -0.1 -0.15 -0.2 -0.25 -0.3 -0.35 Figure 48.7: One Term Approximation and Exact Solution The two term approximate solution is y(x) = x(2 − x) − 33 7 − x 161 46 This two term approximation and the exact solution are plotted in Figure 48.8 The value of the functional is J =− 416 ≈ −0.51677 805 2124 1 0.5 1.5 2 -0.05 -0.1 -0.15 -0.2 -0.25 -0.3 -0.35 Figure 48.8: Two Term Approximation and Exact Solution Sine Series Approximation Consider an approximate solution of the form y(x) = a1 sin πx πx + a2 sin (πx) + · · · + an sin n 2 2 The one term approximate solution is y(x) = − 16 πx sin 2 + 4) π(π 2 This one term approximation and the exact solution are plotted in Figure 48.9 The value of the functional is J =− 64 ≈ −0.467537 + 4) π 2 (π 2 The two term approximate solution is y(x) = − πx 2 16 sin + sin(πx) 2 + 4) 2 + 1) π(π 2 π(π 2125 1 0.5 1.5 2 -0.05 -0.1 -0.15 -0.2 -0.25 -0.3 -0.35 Figure 48.9: One Term Sine Series Approximation and Exact Solution This two term approximation and the exact solution are plotted in Figure 48.10 The value of the functional is J =− 4(17π 2 + 20) ≈ −0.504823 π 2 (π 4 + 5π 2 + 4) 3 Exact Solution The Euler differential equation is d F,y = F,y dx d x2 − 1 [2xy ] = −2 y − 2x2 dx x 1 1 y + y + 1 − 2 y = −x x x The general solution is y = c1 J1 (x) + c2 Y1 (x) − x 2126 0.5 1 1.5 2 -0.1 -0.2 -0.3 Figure 48.10: Two Term Sine Series Approximation and Exact Solution Applying the boundary conditions we obtain, y= (Y1 (2) − 2Y1 (1))J1 (x) + (2J1 (1) − J1 (2))Y1 (x) −x J1 (1)Y1 (2) − Y1 (1)J1 (2) The value of the integral for this extremal is J ≈ −0.310947 Polynomial Approximation Consider an approximate solution of the form y(x) = (x − 1)(2 − x)(a0 + a1 x + · · · an xn ) The one term approximate solution is y(x) = (x − 1)(2 − x) 2127 23 6(40 log 2 − 23) This one term approximation and the exact solution are plotted in Figure 48.11 The one term approximation is a surprisingly close to the exact solution The value of the functional is J =− 529 ≈ −0.310935 360(40 log 2 − 23) 0.2 0.15 0.1 0.05 1.2 1.4 1.6 1.8 2 Figure 48.11: One Term Polynomial Approximation and Exact Solution Solution 48.21 1 The spectrum of T is the set, {λ : (T − λI) is not invertible.} 2128 (T − λI)f = g ∞ K(x − y)f (y) dy − λf (x) = g −∞ ˆ ˆ ˆ K(ω)f (ω) − λf (ω) = g (ω) ˆ ˆ ˆ ˆ K(ω) − λ f (ω) = g (ω) ˆ ˆ ˆ We may not be able to solve for f (ω), (and hence invert T − λI), if λ = K(ω) Thus all values of K(ω) are in ˆ the spectrum If K(ω) is everywhere nonzero we consider the case λ = 0 We have the equation, ∞ K(x − y)f (y) dy = 0 −∞ Since there are an infinite number of L2 (−∞, ∞) functions which satisfy this, (those which are nonzero on a set of measure zero), we cannot invert the equation Thus λ = 0 is in the spectrum The spectrum of T is the range ˆ of K(ω) plus zero 2 Let λ be a nonzero eigenvalue with eigenfunction φ (T − λI)φ = 0, ∀x ∞ K(x − y)φ(y) dy − λφ(x) = 0, ∀x −∞ Since K is continuous, T φ is continuous This implies that the eigenfunction φ is continuous We take the Fourier transform of the above equation ˆ ˆ ˆ K(ω)φ(ω) − λφ(ω) = 0, ˆ ˆ K(ω) − λ φ(ω) = 0, 2129 ∀ω ∀ω ˆ ˆ If φ(x) is absolutely integrable, then φ(ω) is continous Since φ(x) is not identically zero, φ(ω) is not identically ˆ zero Continuity implies that φ(ω) is nonzero on some interval of positive length, (a, b) From the above equation ˆ we see that K(ω) = λ for ω ∈ (a, b) ˆ ˆ Now assume that K(ω) = λ in some interval (a, b) Any function φ(ω) that is nonzero only for ω ∈ (a, b) satisfies ˆ ˆ K(ω) − λ φ(ω) = 0, ∀ω By taking the inverse Fourier transform we obtain an eigenfunction φ(x) of the eigenvalue λ 3 First we use the Fourier transform to find an explicit representation of u = (T − λI)−1 f u = (T − λI)−1 f (T − λI)u = f ∞ K(x − y)u(y) dy − λu = f −∞ ˆ ˆˆ 2π K u − λˆ = f u ˆ f u= ˆ ˆ 2π K − λ ˆ 1 f u=− ˆ ˆ λ 1 − 2π K/λ ˆ For |λ| > |2π K| we can expand the denominator in a geometric series ∞ 1ˆ u=− f ˆ λ n=0 u=− 1 λ ∞ n=0 1 λn ˆ 2π K λ n ∞ Kn (x − y)f (y) dy −∞ 2130 Here Kn is the nth iterated kernel Now we form the Neumann series expansion u = (T − λI)−1 f 1 =− λ =− =− 1 λ 1 λ 1 =− λ −1 1 I− T λ ∞ n=0 ∞ n=0 ∞ n=0 f 1 n T f λn 1 n T f λn 1 λn ∞ Kn (x − y)f (y) dy −∞ The Neumann series is the same as the series we derived with the Fourier transform Solution 48.22 We seek a transformation T such that (L − λI)T f = f We denote u = T f to obtain a boundary value problem, u − λu = f, u(−1) = u(1) = 0 This problem has a unique solution if and only if the homogeneous adjoint problem has only the trivial solution u − λu = 0, u(−1) = u(1) = 0 This homogeneous problem has the eigenvalues and eigenfunctions, λn = − nπ 2 2 , un = sin 2131 nπ (x + 1) , 2 n ∈ N The inhomogeneous problem has the unique solution 1 u(x) = G(x, ξ; λ)f (ξ) dξ −1 where √ √  − sin( −λ(x< +1)) sin( −λ(1−x> )) , λ < 0, √ √   −λ sin(2 −λ)  1 G(x, ξ; λ) = − 2 (x< + 1)(1 − x> ), λ = 0,  sinh √λ(x +1) sinh √λ(1−x )  < ) √( > )  − ( √ , λ > 0, λ sinh(2 λ) for λ = −(nπ/2)2 , n ∈ N We set 1 Tf = G(x, ξ; λ)f (ξ) dξ −1 and note that since the kernel is continuous this is a bounded linear transformation If f ∈ W , then 1 (L − λI)T f = (L − λI) G(x, ξ; λ)f (ξ) dξ −1 1 (L − λI)[G(x, ξ; λ)]f (ξ) dξ = −1 1 δ(x − ξ)f (ξ) dξ = −1 = f (x) 2132 ... and the exact solution are plotted in Figure 48 .5 The value of the functional is J =− ≈ −0.023 148 1 2 16 0.07 0. 06 0.05 0. 04 0.03 0.02 0.01 0.2 0 .4 0 .6 0.8 Figure 48 .5: One Term Approximation and. .. differentiating (48 .6) at 2αh h x = yields 4c2 α = −2c2 Together with c2 = 4c2 (h − c2 ) we obtain c2 = 1+α2 and c2 = − 1+α2 Thus the equation 1 of the pencil (48 .6) will have the form + α2 y = αx + x (48 .7)... (48 .7) 4h α To find the envelope of this family we differentiate ( 48 .7) with respect to α to obtain = x + 2h x2 and eliminate α between this and ( 48 .7) to obtain x2 y = −h + 4h See Figure 48 .3 for