Mixed Boundary Value Problems Episode 4 doc

Historical Background 77 or x 3/2  ∞ 0 A(k) J 3 2 (kx) √ k dk = 2 √ 2 ωx 3 3 √ π (2.6.33) after using integral tables. Differentiating both sides of Equation 2.6.33 with respect to x,  ∞ 0 √ kA(k)J 1 2 (kx) dk = 2ω √ 2x √ π , 0 ≤ x<a. (2.6.34) We now turn to Equation 2.6.31. Multiplying this equation by (2x/π) 1/2 / √ r 2 − x 2 and integrating the resulting equation over r from x>ato ∞,we find that  ∞ 0 k(1 + aλ 0 k)A(k)   2 π  ∞ x √ xJ 1 (kr) √ r 2 − x 2 dr  dk =0; (2.6.35) or  ∞ 0 √ k(1 + aλ 0 k)A(k)J 1 2 (kx) dk =0,a<x<∞. (2.6.36) Let us now replace x with r in Equation 2.6.34 and Equation 2.6.36 and express J 1 2 (z)=  2/(πz)sin(z). This yields  ∞ 0 A(k)sin(kr) dk =2ωr, 0 ≤ r<a, (2.6.37) and  ∞ 0 (1 + aλ 0 k)A(k)sin(kr) dk =0,a<r<∞. (2.6.38) We can rewrite Equation 2.6.38 as  ∞ 0 A(k)sin(kr) dk = −λ 0  ∞ 0 akA(k)sin(kr) dk, a < r < ∞. (2.6.39) From the theory of Fourier integrals, A(k)= 4ω π  a 0 r sin(kr) dr − 2λ 0 π  ∞ a   ∞ 0 akA(k)sin(kr) dk  sin(kr) dr (2.6.40) =2ωa 2  2 πak J 3 2 (ak) − 2λ 0 π  ∞ 0   ∞ 0 aξA(ξ)sin(ξr) dξ  sin(kr) dr + 2λ 0 π  a 0   ∞ 0 aξA(ξ)sin(ξr) dξ  sin(kr) dr. (2.6.41) © 2008 by Taylor & Francis Group, LLC 78 Mixed Boundary Value Problems The second term in Equation 2.6.41 equals aλ 0 kA(k). Therefore, (1 + aλ 0 k)A(k)=2ωa 2  2 πak J 3 2 (ak) + 2λ 0 π  a 0   ∞ 0 aξA(ξ)sin(ξr) dξ  sin(kr) dr. (2.6.42) Interchanging the order of integration, we finally have that (1 + aλ 0 k)A(k)=2ωa 2  2 πak J 3 2 (ak)+λ 0  a 0 aξA(ξ)K(k, ξ) dξ, (2.6.43) where K(y, ξ)= 2 π  a 0 sin(ry)sin(rξ) dr = 1 π  sin[a(y − ξ)] y − ξ − sin[a(y + ξ)] y + ξ  . (2.6.44) One of the intriguing aspects of Equation 2.6.43 is that the unknown is the Hankel transform A(k). In general, the integral equations that we will see will involve an unknown which is related to A(k)viaanintegral definition. See Equation 2.5.61, Equation 2.5.62 and Equation 2.5.65. In Goodrich’s paper he solved the integral equation as a variational problem and found an approximate solution by the optimization of suitable solutions. However, we shall shortly outline an alternative method for any value of λ 0 . In 1978 Shail 27 reexamined Goodrich’s paperfortworeasons. First, the solution for the µ =0case,Equation2.6.16, creates a divergent integral when it is substituted back into dual equations, Equation 2.6.8 and Equation 2.6.9. Second, the governing Fredholm integral equation is over an infinite range and appears to be unsuitable for asymptotic solution as λ 0 → 0orλ 0 →∞. Shail’s analysis for the µ =0casebeginsby noting that ∂ 2 u(r, 0) ∂r 2 + 1 r ∂u(r, 0) ∂r − u(r, 0) r 2 =0,a≤ r<∞ (2.6.45) from Equation 2.6.1 and Equation 2.6.4. The general solution to Equation 2.6.45 is u(r, 0) = Cr + D/r, a < r < ∞. (2.6.46) The values of C and D follow from the limits that u(r, 0) → 0asr →∞and continuity at u(r, 0) at r = a.Thisyields u(r, 0) =  ωr, 0 ≤ r<a, ωa 2 /r, a < r < ∞. (2.6.47) 27 Takenfrom Shail, R., 1978: The torque on a rotating disk in the surface of a liquid with an adsorbed film. J. Engng. Math., 12, 59–76 with kind permission from Springer Science and Business Media. © 2008 by Taylor & Francis Group, LLC Historical Background 79 If we use Equation 2.6.47 in place of Equation 2.6.4, then the solution Equa- tion 2.6.17 follows directly. Shail also examined the general case and solved it using the method of complementary representations forgeneralized axially symmetric poten- tial functions. This method is very complicated and we introduce a greatly simplified version of an analysis first done by Chakrabarti. 28 We begin by introducing the function g(x)suchthat (1 + aλ 0 k)A(k)= 2 π  a 0 g(x)sin(kx) dx. (2.6.48) Turning to Equation 2.6.31 first, direct substitution yields  ∞ 0 k(1 + aλ 0 k)A(k)J 1 (kr) dk =  a 0 g(x)   ∞ 0 k sin(kx)J 1 (kr) dk  dx (2.6.49) = −  a 0 g(x) d dr   ∞ 0 sin(kx)J 0 (kr) dk  dx (2.6.50) =0, (2.6.51) because 0 ≤ x ≤ a<r<∞.Thus,ourchoiceofA(k)satisfiesEquation 2.6.31 identically. Turning to Equation 2.6.30 next, direct substitution gives 2 π  a 0 g(x)   ∞ 0 sin(kx)J 1 (kr) 1+aλ 0 k dk  dx = ωr, 0 ≤ r<a;(2.6.52) or  a 0 g(x)   ∞ 0 sin(kx)J 1 (kr) dk  dx (2.6.53) +  a 0 g(x)   ∞ 0 1 −aλ 0 k 1+aλ 0 k sin(kx)J 1 (kr) dk  dx = ωπr, 0 ≤ r<a. From tables, 29 the integral within the square brackets of the first integral in Equation 2.6.53 can be evaluated and Equation 2.6.53 simplifies to  r 0 xg(x) √ r 2 − x 2 dx +  a 0 g(τ)   ∞ 0 1 −aλ 0 k 1+aλ 0 k sin(kτ) rJ 1 (kr) dk  dτ = ωπr 2 (2.6.54) 28 Chakrabarti, A., 1989: On some dual integral equations involving Bessel functions of order one. Indian J. Pure Appl. Math., 20, 483–492. 29 Gradshteyn and Ryzhik, op. cit., Formula 6.671.1. © 2008 by Taylor & Francis Group, LLC 80 Mixed Boundary Value Problems for 0 ≤ r<a.UsingEquation 1.2.13 and Equation 1.2.14, we can solve for xg(x)andfind that xg(x)+ 2 π  a 0 g(τ)   ∞ 0 1 −aλ 0 k 1+aλ 0 k sin(kτ) d dx   x 0 ξ 2 J 1 (kξ)  x 2 − ξ 2 dξ  dk  dτ =2ω d dx   x 0 ξ 3  x 2 − ξ 2 dξ  , 0 ≤ x<a. (2.6.55) From integral tables, 30 d ds   s 0 ξ 2 J 1 (kξ)  s 2 − ξ 2 dξ  = s sin(ks), (2.6.56) and Equation 2.6.55 becomes g(x)+ 2 π  a 0 g(τ)   ∞ 0 1 −aλ 0 k 1+aλ 0 k sin(kτ)sin(kx) dk  dτ =4ωx, 0 ≤ x<a. (2.6.57) Equation 2.6.57 is identical to Chakrabarti’s equations (50) and (52). Once we solve Equation 2.6.57 numerically, its values of g(x)canbesubstituted into Equation 2.6.48. Finally the solution u(r, z)follows from Equation 2.6.5. The numerical solution of Equation 2.6.57 is nontrivial due to the nature of the integration over k.Tosolveit,weuse a spectral method. If we take g(τ)tobeanoddfunction over (−a, a), we have that g(τ)= ∞  n=1 A n sin  nπτ a  , (2.6.58) sin(kτ)=2 ∞  n=1 (−1) n nπ k 2 a 2 − n 2 π 2 sin(ka)sin  nπτ a  , (2.6.59) and x = −2a ∞  n=1 (−1) n nπ sin  nπx a  . (2.6.60) Substitution of Equation 2.6.58 through Equation 2.6.60 into Equation 2.6.57 givesthe infinite set of equations A m + ∞  n=1 H mn A n = C m ,m=1, 2, 3, , (2.6.61) 30 Ibid., Formula 6.567.1 with ν =1andµ = − 1 2 . © 2008 by Taylor & Francis Group, LLC Historical Background 81 0 0.5 1 1.5 2 0 0.5 1 1.5 2 0 0.2 0.4 0.6 0.8 1 r/a z/a u(r,z)/(aω) Figure 2.6.3:Thisissimilar to Figure 2.6.1 except that µ, η =0andλ 0 =5. where H mn =4a(−1) n+m nmπ  ∞ 0 1 −aλ 0 k 1+aλ 0 k sin 2 (ka) (k 2 a 2 − n 2 π 2 )(k 2 a 2 − m 2 π 2 ) dk, (2.6.62) and C m = 8ωa mπ (−1) m+1 . (2.6.63) The system of equations is then truncated to, say, N spectral components and the system is inverted to yield A m for m =1, 2, ,N.Next,A(k)can be found via (1 + aλ 0 k) A(ak) a =2 N  n=1 n(−1) n k 2 a 2 − n 2 π 2 sin(ka)A n . (2.6.64) The larger the value of N,thegreaterthe accuracy. Finally, u(r, z)=  ∞ 0 A(ξ) a e −ξz/a J 1 (ξr/a) dξ. (2.6.65) Figure 2.6.3 illustrates this solution when λ 0 =5. © 2008 by Taylor & Francis Group, LLC 0 0.5 1 1.5 2 −2 −1 0 1 2 3 4 −1 −0.5 0 0.5 1 r z u(r,z) Chapter 3 Separation of Variables Separation of variables is the most commonly used technique for solving boundary value problems. In the case of mixed boundary value problems they lead to dual or higher-numbered Fourier series which yield the Fourier coefficients. In this chapter we examine dual Fourier series in Section 3.1 and Section 3.2, while dual Fourier-Besselseriesaretreated in Section 3.3 and dual Fourier-Legendre series in Section 3.4. Finally Section 3.5 treats triple Fourier series. In Example 1.1.1 we showed that the method of separation of variables led to the dual cosine series: ∞  n=1 a n n − 1 2 cos  n − 1 2  x  =1, 0 ≤ x<c, (3.0.1) and ∞  n=1 a n cos  n − 1 2  x  =0,c<x≤ π. (3.0.2) Equations 3.0.1 and 3.0.2 are examples of a larger class of dual trigonometric 83 © 2008 by Taylor & Francis Group, LLC 84 Mixed Boundary Value Problems equations. The general form of these dual series can be written            ∞  n=1 n p a n sin(nx)=f(x), 0 ≤ x<c, ∞  n=1 a n sin(nx)=g(x),c<x≤ π, (3.0.3)            ∞  n=1  n − 1 2  p a n cos  n − 1 2  x  = f(x), 0 ≤ x<c, ∞  n=1 a n cos  n − 1 2  x  = g(x),c<x≤ π, (3.0.4)            ∞  n=1  n − 1 2  p a n sin  n − 1 2  x  = f (x), 0 ≤ x<c, ∞  n=1 a n sin  n − 1 2  x  = g(x),c≤ x ≤ π, (3.0.5) and            1 2 αa 0 + ∞  n=1 n p a n cos(nx)=f (x), 0 ≤ x<c, 1 2 a 0 + ∞  n=1 a n cos(nx)=g(x),c<x≤ π, (3.0.6) where −1 ≤ p ≤ 1. Comparing Equation 3.0.1 and Equation 3.0.2 with Equation 3.0.4, they are identical if we set p = −1. The purpose of this chapter is to focus on those mixed value problems that lead to these dual equations and solve them. 3.1 DUAL FOURIER COSINE SERIES Tranter 1 examined dual trigonometric series of the form            ∞  n=1  n − 1 2  p a n cos  n − 1 2  x  = F (x), 0 ≤ x<c, ∞  n=1 a n cos  n − 1 2  x  = G(x),c<x≤ π, (3.1.1) where p = ±1. The most commonly encountered case is when G(x)=0. When p =1,heshowed that a n = 2 π  c 0 h(x)cos  n − 1 2  x  dx, (3.1.2) 1 Tranter, C. J., 1960: Dual trigonometrical series. Proc. Glasgow Math. Assoc., 4, 49–57. © 2008 by Taylor & Francis Group, LLC Separation of Variables 85 where h(x)=  1 ξ χ(η)  η 2 − ξ 2 dη, (3.1.3) χ(η)= 4η π sin(c/2)  η 0 F {2arcsin[x sin(c/2)]}  (η 2 − x 2 )[1 −x 2 sin 2 (c/2)] dx, (3.1.4) and ξ =sin(x/2) csc(c/2). When p = − 1, a n =2χ(1) sin(c/2)P n−1 [cos(c)] − 2sin(c/2)  1 0 χ  (η)P n−1  1 −2η 2 sin 2 (c/2)  dη, (3.1.5) where χ(η)= 2 π  η 0 xF  {2arcsin[x sin(c/2)]}  η 2 − x 2 dx + C. (3.1.6) Here, C is a constant whose value is determined by substituting Equation 3.1.6 into Equation 3.1.5 and then choosing C so that ∞  n=1  n − 1 2  −1 a n = F(0). (3.1.7) • Example 3.1.1 To illustrate Tranter’s solution, let us assume that F(x)=1if0≤ x<c. From Equation 3.1.6, we have that χ(η)=C;fromEquation 3.1.5, a n = 2C sin(c/2)P n−1 [cos(c)]. To evaluate C,wesubstitute a n into Equation 3.1.7 and find that 2C sin(c/2) ∞  n=1  n − 1 2  −1 P n−1 [cos(c)] = 1. (3.1.8) From the generation formula for Legendre polynomials, ∞  n=1 x 2n−2 P n−1 [cos(c)] = [1 − 2x 2 cos(c)+x 4 ] −1/2 . (3.1.9) Integrating Equation 3.1.9 from 0 to 1, we have ∞  n=1  n − 1 2  P n−1 [cos(c)] = 2  1 0 dx  1 −2x 2 cos(c)+x 4 (3.1.10) =  π/2 0 dθ  1 −cos 2 (c/2) sin 2 (θ) (3.1.11) = K[cos 2 (c/2)], (3.1.12) © 2008 by Taylor & Francis Group, LLC 86 Mixed Boundary Value Problems 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x/π y u(x,y) Figure 3.1.1:Thesolutionu(x, y)tothe mixed boundary value problem posed in Example 1.1.1 when c = π/2. where K(·)denotes the complete elliptic integral 2 and sin(θ)=2x/(1 + x 2 ). Therefore, 2C sin(c/2)K[cos 2 (c/2)] = 1, and a n = P n−1 [cos(c)]/K[cos 2 (c/2)] is the solution to the dual Fourier cosine series Equation 3.0.1 and Equation 3.0.2. Recall that Equation 3.0.1 and Equation 3.0.2 arose from the separation of variables solution of Equation 1.1.1 through Equation 1.1.4. Therefore, the solution to this particular mixed boundary value problem is u(x, y)= ∞  n=1 P n−1 [cos(c)] K[cos 2 (c/2)] exp  −  n − 1 2  y  n − 1 2 cos  n − 1 2  x  . (3.1.13) This solution is illustrated in Fig. 3.1.1 when c = π/2. • Example 3.1.2 Let us solve 3 ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 =0, 0 <x<π, 0 <y<π/2, (3.1.14) 2 See Milne-Thomson, L. M., 1965: Elliptic integrals. Handbook of Mathematical Func- tions,M.Abromowitz and I. A. Stegun, Eds., Dover, 587–626. See Section 17.3. 3 TakenfromWhiteman, J. R., 1968: Treatment of singularities in a harmonic mixed boundary value problem by dual series methods. Quart. J. Mech.Appl.Math., 21, 41–50 with permission of Oxford University Press. © 2008 by Taylor & Francis Group, LLC Separation of Variables 87 subject to the boundary conditions u x (0,y)=0,u(π, y)=1, 0 <y<π/2, (3.1.15)  u(x, 0) = 1 2 , 0 ≤ x<π/2, u y (x, 0) = 0,π/2 <x≤ π, (3.1.16) and u y (x, π/2) = 0, 0 <x<π. (3.1.17) Using separation of variables, a solution to Equation 3.1.14 which also satisfies Equation 3.1.15 and Equation 3.1.17 is u(x, y)=1− ∞  n=0 B n n + 1 2 cosh  n + 1 2  y − π 2  sinh  n + 1 2  π/2  cos  n + 1 2  x  . (3.1.18) If we then substitute Equation 3.1.18 into the mixed boundary condition Equation 3.1.17, we obtain the dual series ∞  n=0 B n 2n +1 coth  n + 1 2  π/2  cos  n + 1 2  x  = 1 4 , 0 ≤ x<π/2, (3.1.19) and ∞  n=0 B n cos  n + 1 2  x  =0,π/2 <x≤ π. (3.1.20) The remaining challenge is to solve this dual series. Fortunately, in the 1960s Tranter 4 showed that the dual trigonometrical series ∞  n=0 A n 2n +1 cos  n + 1 2  x  = f(x), 0 <x<c, (3.1.21) and ∞  n=0 A n cos  n + 1 2  x  =0,c<x<π, (3.1.22) has the solution A n = A 0 P n [cos(c)] −  c 0 F (θ)P  n [cos(θ)] sin(θ) dθ, n =1, 2, 3, , (3.1.23) where F (θ)= 2 √ 2 π  θ 0 f  (x)sin(x)  cos(x) −cos(θ) dx, (3.1.24) 4 Tranter, C. J., 1964: An improved method for dual trigonometrical series. Proc. Glasgow Math. Assoc., 6, 136–140. © 2008 by Taylor & Francis Group, LLC [...]... (3.1 .40 ) and ˜ D(n) = −Pn (0) − C(n, 0) (3.1 .41 ) Having found these Bm ’s, we use Equation 3.1.19 with x = π /4 to compute B0 via B0 = 1 4 ∞ Bn /B0 cosh n + 2n + 1 n=0 © 2008 by Taylor & Francis Group, LLC 1 2 π/2 cos n + 1 2 x (3.1 .42 ) 90 Mixed Boundary Value Problems 1 u(x,y) 0.9 0.8 0.7 0.6 0.5 0 .4 0.5 0 .4 1 0.3 0.8 0.6 0.2 0 .4 0.1 y/π 0.2 0 x/π 0 Figure 3.1.2: The solution u(x, y) to the mixed boundary. .. Appl Math., 21, 245 –257 © 2008 by Taylor & Francis Group, LLC Separation of Variables 93 1 u(x,y) 0.8 0.6 0 .4 0.2 0 1 0.5 0 0.2 0 0 .4 x/b 0.6 −0.5 0.8 −1 1 y/b Figure 3.1 .4: The solution u(x, y) to the mixed boundary value problem given by Equation 3.1 .43 through Equation 3.1 .46 1 Figure 3.1 .4 illustrates the solution u(x, y) when b = h = 3, c = w = 1, = 1 and 2 = 2 • Example 3.1 .4 Let us solve6 ∂... 1 −b < x < b, |x| < w, w < |x| < b, (3.1 .44 ) (3.1 .45 ) and u(−b, y) = u(b, y) = 0, 0 < y < h (3.1 .46 ) Figure 3.1.3 illustrates the geometry for this problem If we use separation of variables, the solution to Equation 3.1 .43 is ∞ u(x, y) = An ekn (y−c) − e−kn (y+c) cos(kn x), n=1 © 2008 by Taylor & Francis Group, LLC 0 < y < c, (3.1 .47 ) 92 Mixed Boundary Value Problems Kiyono and Shimasaki5 developed... that the solution to Equation 3.0.6 is c π 2 π √ a0 = h(t) dt + g(t) dt , (3.1. 74) π 2 0 c 7 Sbragaglia, M., and A Prosperetti, 2007: A note on the effective slip properties for microchannel flows with ultrahydrophobic surfaces Phys Fluids, 19, Art No 043 603 8 See Section 5 .4. 3 in Sneddon, I N., 1966: Mixed Boundary Value Problems in Potential Theory North Holland, 283 pp © 2008 by Taylor & Francis Group,... these equations and find approximate values for the An ’s The potential then follows from Equation 3.1.110 Figure 3.1.7 illustrates this solution when ϕ = π/3 and N = 100 Problems 1 Solve Laplace’s equation 1 ∂ ∂u r r ∂r ∂r + © 2008 by Taylor & Francis Group, LLC 1 ∂2u = 0, r2 ∂θ2 0 ≤ r < ∞, 0 < θ < 2π, 102 Mixed Boundary Value Problems Problem 1 Step 2 : Using the final boundary condition and the results... substituting Equation 3.2.10 into Equation 3.2.6, 1 π ∞ nπx 1 sin n L n=1 p(t) sin − nπt L dt = U0 (x), |x| < (3.2.11) 14 Williams, W E., 19 64: The solution of dual series and dual integral equations Proc Glasgow Math Assoc., 6, 123–129 © 2008 by Taylor & Francis Group, LLC 1 04 Mixed Boundary Value Problems Let us assume that we can Chebyshev express p(t) by the expansion p(t) = √ 1 2 − t2 ∞ B2m+1 T2m+1 (t/... Press, Formula 7.355.1 and 7.355.2 © 2008 by Taylor & Francis Group, LLC Separation of Variables 105 1 u(x,y) 0.8 0.6 0 .4 0.2 0 0.2 0 1 0 .4 0.8 0.6 0.6 0 .4 0.8 0.2 x/L 0 1 y/L Figure 3.2.1: The solution u(x, y) to the mixed boundary value problem given by Equation 3.2.1 through Equation 3.2 .4 when U0 (x) = x/ and L/ = 3 For a given U0 (x), we compute b2n+1 Equation 3.2.17 then yields B2m+1 while Equation... © 2008 by Taylor & Francis Group, LLC 94 Mixed Boundary Value Problems If we use separation of variables, the solution to Equation 3.1.63 is ∞ An e−ny cos(nx), u(x, y) = y + A0 + 0 < y < ∞, (3.1.68) −∞ < y < 0 (3.1.69) n=1 and ∞ u(x, y) = y − A0 − An eny cos(nx), n=1 Equation 3.1.68 and Equation 3.1.69 satisfy not only Equation 3.1.63, but also Equation 3.1. 64 through Equation 3.1.66 Finally, we substitute... 2κ V, 2(κ + 1) D2 = − V , 2(κ + 1) and 1 · 3 · · · · (2n − 3)V , 2 · 4 · · · 2n where n = 2, 3, 4, The figure entitled Problem 1 illustrates the solution u(r, θ)/V when κ = 6 D2n−1 = −D2n = (−1)n+1 3.2 DUAL FOURIER SINE SERIES Dual Fourier sine series arise in the same manner as dual Fourier cosine series in mixed boundary value problems in rectangular domains However, for some reason, they do not...88 Mixed Boundary Value Problems and A0 is found by substituting Equation 3.1.23 into Equation 3.1.21 Can we apply Tranter’s results, Equation 3.1.21 through Equation 3.1. 24, to solve Equation 3.1.19 and Equation 3.1.20? We begin by rewriting these equations as follows: ∞ Bn cos n + 2n + 1 n=0 1 2 ∞ Bm 1 + 1 − coth m + 4 m=0 2m + 1 x = 1 2 × cos m + and ∞ Bn cos . Group, LLC 86 Mixed Boundary Value Problems 0 0.2 0 .4 0.6 0.8 1 0 0.5 1 1.5 2 0 0.1 0.2 0.3 0 .4 0.5 0.6 0.7 0.8 0.9 1 x/π y u(x,y) Figure 3.1.1:Thesolutionu(x, y)tothe mixed boundary value problem. (3.1. 24) 4 Tranter, C. J., 19 64: An improved method for dual trigonometrical series. Proc. Glasgow Math. Assoc., 6, 136– 140 . © 2008 by Taylor & Francis Group, LLC 88 Mixed Boundary Value Problems and. + 1 2  π/2  cos  n + 1 2  x   . (3.1 .42 ) © 2008 by Taylor & Francis Group, LLC 90 Mixed Boundary Value Problems 0 0.2 0 .4 0.6 0.8 1 0 0.1 0.2 0.3 0 .4 0.5 0 .4 0.5 0.6 0.7 0.8 0.9 1 x/π y/π u(x,y) Figure