Thus the general solution for u is u(r, θ) = a 0 2 + ∞  n=1 r n [a n cos(nθ) + b n sin(nθ)] . For the boundary condition u(1, θ) = f(θ) we have the equation f(θ) = a 0 2 + ∞  n=1 [a n cos(nθ) + b n sin(nθ)] . If f(θ) has a Fourier series then the coefficients are a 0 = 1 π  2π 0 f(θ) dθ a n = 1 π  2π 0 f(θ) cos(nθ) dθ b n = 1 π  2π 0 f(θ) sin(nθ) dθ. For the boundary condition u r (1, θ) = g(θ) we have the equation g(θ) = ∞  n=1 n [a n cos(nθ) + b n sin(nθ)] . g(θ) has a series of this form only if  2π 0 g(θ) dθ = 0. 2054 The coefficients are a n = 1 nπ  2π 0 g(θ) cos(nθ) dθ b n = 1 nπ  2π 0 g(θ) sin(nθ) dθ. 47.3 Laplace’s Equation in an Annulus Consider the problem ∇ 2 u = 1 r ∂ ∂r  r ∂u ∂r  + 1 r 2 ∂ 2 u ∂θ 2 = 0, 0 ≤ r < a, −π < θ ≤ π, with the boundary condition u(a, θ) = θ 2 . So far this problem only has one boundary condition. By requiring that the solution be finite, we get the boundary condition |u(0, θ)| < ∞. By specifying that the solution be C 1 , (continuous and continuous first derivative) we obtain u(r, −π) = u(r, π) and ∂u ∂θ (r, −π) = ∂u ∂θ (r, π). We will use the method of separation of variables. We seek solutions of the form u(r, θ) = R(r)Θ(θ). 2055 Substituting into the partial differential equation, ∂ 2 u ∂r 2 + 1 r ∂u ∂r + 1 r 2 ∂ 2 u ∂θ 2 = 0 R  Θ + 1 r R  Θ = − 1 r 2 RΘ  r 2 R  R + rR  R = − Θ  Θ = λ Now we have the boundary value problem for Θ, Θ  (θ) + λΘ(θ) = 0, −π < θ ≤ π, subject to Θ(−π) = Θ(π) and Θ  (−π) = Θ  (π) We consider the following three cases for the eigenvalue, λ, λ < 0. No linear combination of the solutions, Θ = exp( √ −λθ), exp(− √ −λθ), can satisfy the boundary conditions. Thus there are no negative eigenvalues. λ = 0. The general solution solution is Θ = a + bθ. By applying the boundary conditions, we get Θ = a. Thus we have the eigenvalue and eigenfunction, λ 0 = 0, A 0 = 1. λ > 0. The general solution is Θ = a cos( √ λθ)+b sin( √ λθ). Applying the boundary conditions yields the eigenvalues λ n = n 2 , n = 1, 2, 3, . . . with the associated eigenfunctions A n = cos(nθ) and B n = sin(nθ). 2056 The equation for R is r 2 R  + rR  − λ n R = 0. In the case λ 0 = 0, this becomes R  = − 1 r R  R  = a r R = a log r + b Requiring that the solution be bounded at r = 0 yields (to within a constant multiple) R 0 = 1. For λ n = n 2 , n ≥ 1, we have r 2 R  + rR  − n 2 R = 0 Recognizing that this is an Euler equation and making the substitution R = r α , α(α − 1) + α − n 2 = 0 α = ±n R = ar n + br −n . requiring that the solution be bounded at r = 0 we obtain (to within a constant multiple) R n = r n The general solution to the partial differential equation is a linear combination of the eigenfunctions u(r, θ) = c 0 + ∞  n=1 [c n r n cos nθ + d n r n sin nθ] . 2057 We determine the coefficients of the expansion with the boundary condition u(a, θ) = θ 2 = c 0 + ∞  n=1 [c n a n cos nθ + d n a n sin nθ] . We note that the eigenfunctions 1, cos nθ, and sin nθ are orthogonal on −π ≤ θ ≤ π. Integrating the boundary condition from −π to π yields  π −π θ 2 dθ =  π −π c 0 dθ c 0 = π 2 3 . Multiplying the boundary condition by cos mθ and integrating gives  π −π θ 2 cos mθ dθ = c m a m  π −π cos 2 mθ dθ c m = (−1) m 8π m 2 a m . We multiply by sin mθ and integrate to get  π −π θ 2 sin mθ dθ = d m a m  π −π sin 2 mθ dθ d m = 0 Thus the solution is u(r, θ) = π 2 3 + ∞  n=1 (−1) n 8π n 2 a n r n cos nθ. 2058 Part VI Calculus of Variations 2059 Chapter 48 Calculus of Variations 2060 48.1 Exercises Exercise 48.1 Discuss the problem of minimizing  α 0 ((y  ) 4 − 6(y  ) 2 ) dx, y(0) = 0, y(α) = β. Consider both C 1 [0, α] and C 1 p [0, α], and comment (with reasons) on whether your answers are weak or strong minima. Exercise 48.2 Consider 1.  x 1 x 0 (a(y  ) 2 + byy  + cy 2 ) dx, y(x 0 ) = y 0 , y(x 1 ) = y 1 , a = 0, 2.  x 1 x 0 (y  ) 3 dx, y(x 0 ) = y 0 , y(x 1 ) = y 1 . Can these functionals have broken extremals, and if so, find them. Exercise 48.3 Discuss finding a weak extremum for the following: 1.  1 0 ((y  ) 2 − 2xy) dx, y(0) = y  (0) = 0, y(1) = 1 120 2.  1 0  1 2 (y  ) 2 + yy  + y  + y  dx 3.  b a (y 2 + 2xyy  ) dx, y(a) = A, y(b) = B 4.  1 0 (xy + y 2 − 2y 2 y  ) dx, y(0) = 1, y(1) = 2 Exercise 48.4 Find the natural boundary conditions associated with the following functionals: 1.  D F (x, y, u, u x , u y ) dx dy 2.  D  p(x, y)(u 2 x + u 2 y ) − q(x, y)u 2  dx dy +  Γ σ(x, y)u 2 ds Here D represents a closed boundary domain with boundary Γ, and ds is the arc-length differential. p and q are known in D, and σ is known on Γ. 2061 Exercise 48.5 The equations for water waves with free surface y = h(x, t) and bottom y = 0 are φ xx + φ yy = 0 0 < y < h(x, t), φ t + 1 2 φ 2 x + 1 2 φ 2 y + gy = 0 on y = h(x, t), h t + φ x h x − φ y = 0, on y = h(x, t), φ y = 0 on y = 0, where the fluid motion is described by φ(x, y, t) and g is the acceleration of gravity. Show that all these equations may be obtained by varying the functions φ(x, y, t) and h(x, t) in the variational principle δ  R   h(x,t) 0  φ t + 1 2 φ 2 x + 1 2 φ 2 y + gy  dy  dx dt = 0, where R is an arbitrary region in the (x, t) plane. Exercise 48.6 Extremize the functional  b a F (x, y, y  ) dx, y(a) = A, y(b) = B given that the admissible curves can not penetrate the interior of a given region R in the (x, y) plane. Apply your results to find the curves which extremize  10 0 (y  ) 3 dx, y(0) = 0, y(10) = 0 given that the admissible curves can not penetrate the interior of the circle (x −5) 2 + y 2 = 9. Exercise 48.7 Consider the functional  √ y ds where ds is the arc-length differen tial (ds =  (dx) 2 + (dy) 2 ). Find the curve or curves from a given vertical line to a given fixe d point B = (x 1 , y 1 ) which minimize this functional. Consider b oth the classes C 1 and C 1 p . Exercise 48.8 A perfectly flexible uniform rope of length L hangs in equilibrium with one end fixed at (x 1 , y 1 ) so that it passes over a frictionless pin at (x 2 , y 2 ). What is the position of the free end of the rope? 2062 [...]... (w − )(4 α 2 3 +3 β α β α 4 6 3 β − 12 ) α β α 2 We can find the stationary points of the excess function by examining its derivative (Let λ = β/α.) E (w) = 4w3 − 12w + 4λ (λ)2 − 3 = 0 √ √ 1 1 −λ − 4 − λ2 w3 = −λ + 4 − λ2 2 2 The excess function evaluated at these points is w1 = λ, w2 = E(w1 ) = 0, √ 3 E(w2 ) = 3 4 − 6 2 − 6 − 3 (4 − λ2 )3/ 2 , 2 √ 3 E(w3 ) = 3 4 − 6 2 − 6 + 3 (4 − λ2 )3/ 2 2 √ √ E(w2... 48.5 Hint 48 .6 Hint 48.7 Hint 48.8 Hint 48.9 Hint 48.10 2075 Hint 48.11 Hint 48.12 Hint 48. 13 Hint 48.14 Hint 48.15 Hint 48. 16 Hint 48.17 Hint 48.18 Hint 48.19 Hint 48.20 Hint 48.21 20 76 Hint 48.22 Hint 48. 23 Hint 48.24 Hint 48.25 Hint 48. 26 Hint 48.27 Hint 48.28 Hint 48.29 Hint 48 .30 Hint 48 .31 Hint 48 .32 2077 Hint 48 .33 Hint 48 .34 Hint 48 .35 Hint 48. 36 Hint 48 .37 Hint 48 .38 Hint 48 .39 Hint 48.40... y− = ± 3 and y+ = ˆ ˆ 3 p = q, p= √ √ Case 1, β = ± 3 Notice the the Lagrangian is minimized point-wise if y = √ ± 3 For this case the unique, strong global minimum is √ y = 3 sign(β)x ˆ √ Case 2, |β| < 3| α| For this case there are an infinite number of strong √ √ minima Any piecewise linear curve satisfying y− (x) = ± 3 and y+ (x) = ± 3 and y(0) = 0, y(α) = β is a strong minima √ Case 3, |β| > 3| α|... negative for −1 < λ < 3 and E(w3 ) is negative for √ 3 < λ < 1 This implies that the weak − minimum y = βx/α is not a strong local minimum for |λ| < 3| Since E(w1 ) = 0,√ cannot use the ˆ we Weierstrass excess function to determine if y = βx/α is a strong local minima for |β/α| > 3 ˆ 2080 1 Cp [0, α] Extremals Erdmann’s Corner Conditions Erdmann’s corner conditions require that ˆ F,y = 4(ˆ )3 − 12ˆ y y and. .. ˆ F − y F,y = (ˆ )4 − 6( ˆ )2 − y (4(ˆ )3 − 12ˆ ) y y ˆ y y are continuous at corners Thus the quantities (ˆ )3 − 3 y y and (ˆ )4 − 2(ˆ )2 y y ˆ are continuous Denoting p = y− and q = y+ , the first condition has the solutions ˆ √ 1 p = q, p = −q ± 3 4 − q 2 2 The second condition has the solutions, p = ±q, Combining these, we have p = ± 2 − q2 √ √ √ 3, q = − 3, p = − 3, q = 3 √ √ Thus we see that... the Euler equation, and satisfies the strengthened Jacobi and Legendre conditions Thus it is a weak minima (For |β/α| < 1 it is a weak maxima for the same reasons.) Weierstrass Excess Function The Weierstrass excess function is E(x, y , y , w) = F (w) − F (ˆ ) − (w − y )F,y (ˆ ) ˆ ˆ y ˆ y = w4 − 6w2 − (ˆ )4 + 6( ˆ )2 − (w − y )(4(ˆ )3 − 12ˆ ) y y ˆ y y 4 2 = w − 6w − β α 4 +6 β = w − 6w − w 4 α 4 2 β α... on K it follows ˆ that K is continuous and approaches zero at ±∞.) ˆ 2 For λ in the spectrum of T , show that λ is an eigenvalue if and only if K takes on the value λ on at least some interval of positive length and that every other λ in the spectrum belongs to the continuous spectrum 3 Find an explicit representation for (T − λI)−1 f for λ not in the spectrum, and verify directly that this result agrees... Weierstrass E function E(x, y, q, y ) is non-negative for arbitrary finite p and y at any point of D What is the implication of this for Fermat’s Principle? Exercise 48.17 2 Consider the integral 1+y2 dx between fixed limits Find the extremals, (hyperbolic sines), and discuss the Jacobi, (y ) Legendre, and Weierstrass conditions and their implications regarding weak and strong extrema Also consider the value of... differentiable functions f on [−1, 1] satisfying f (−1) = f (1) = 0, and d2 W = C[−1, 1] Let L : U → W be the operator dx2 Call λ in the spectrum of L if the following does not occur: There is a bounded linear transformation T : W → U such that (L − λI)T f = f for all f ∈ W and T (L − λI)f = f for all f ∈ U Determine the spectrum of L 2 067 Exercise 48. 23 Solve the integral equations 1 x2 y − y 2 φ(y) dy 1 φ(x)... ) Exercise 48 .33 Show that ∞ − −∞ (1 − t2 )1/2 log(1 + t) π dt = π x log 2 − 1 + (1 − x2 )1/2 − arcsin(x) t−x 2 Exercise 48 .34 Let C be a simple closed contour Let g(t) be a given function and consider 1 f (t) dt − = g(t0 ) ıπ C t − t0 (48.5) Note that the left side can be written as F + (t0 ) + F − (t0 ) Define a function W (z) such that W (z) = F (z) for z inside C and W (z) = −F (z) for z outside . with boundary Γ, and ds is the arc-length differential. p and q are known in D, and σ is known on Γ. 2 061 Exercise 48.5 The equations for water waves with free surface y = h(x, t) and bottom y =. approximate solution of the form y = x(1 −x) (a 0 + a 1 x + ··· + a n x n ) , 2 066 and carry out the solutions for n = 0 and n = 1. •  2 0  (y  ) 2 + y 2 + 2xy  dx, y(0) = 0 = y(2). •  2 1  x(y  ) 2 − x 2 −. transformation T : W → U such that (L −λI)T f = f for all f ∈ W and T (L −λI)f = f for all f ∈ U. Determine the spectrum of L. 2 067 Exercise 48. 23 Solve the integral equations 1. φ(x) = x + λ  1 0  x 2 y