THÔNG TIN TÀI LIỆU
288 Mixed Boundary Value Problems where a<1. Using transform methods or separation of variables, the general solution to Equation 4.4.27, Equation 4.4.28, and Equation 4.4.29 is u(r, z)= ∞ 0 A(k)J 0 (kr)e −kz dk. (4.4.31) Substituting Equation 4.4.31 into Equation 4.4.30, we find that ∞ 0 A(k)J 0 (kr) dk = V, 0 ≤ r<a, (4.4.32) ∞ 0 kA(k)J 0 (kr) dk =0,a<r<1, (4.4.33) and ∞ 0 A(k)J 0 (kr) dk =0, 1 <r<∞. (4.4.34) To solve this set of integral equations, we let A(k)=B(k)+D(k). Then, Equation 4.4.32 through Equation 4.4.34 can be rewritten ∞ 0 B(k)J 0 (kr) dk = f(r), 0 ≤ r<a, (4.4.35) ∞ 0 kB(k)J 0 (kr) dk =0,a<r<∞, (4.4.36) ∞ 0 kD(k)J 0 (kr) dk =0, 0 ≤ r<1, (4.4.37) and ∞ 0 D(k)J 0 (kr) dk = g(r), 1 <r<∞, (4.4.38) where f(r)=V − ∞ 0 D(k)J 0 (kr) dk, (4.4.39) and g(r)=− ∞ 0 B(k)J 0 (kr) dk. (4.4.40) Equation 4.4.36 and Equation 4.4.37 are automatically satisfied if we define B(k)andD(k)asfollows: B(k)= a 0 φ(t)cos(kt) dt, D(k)= ∞ 1 ψ(τ)sin(kτ) dτ. (4.4.41) © 2008 by Taylor & Francis Group, LLC Transform Methods 289 If we substitute Equation 4.4.41 into Equation 4.4.35, we have that a 0 φ(t) ∞ 0 cos(kt)J 0 (kr) dk dt = f(r)(4.4.42) after we interchange the order of integration. Using Equation 1.4.14, Equation 4.4.42 simplifies to r 0 φ(t) √ r 2 − t 2 dt = f(r). (4.4.43) From Equation 1.2.13 and Equation 1.2.14, we obtain φ(t)= 2 π d dt t 0 rf(r) √ t 2 − r 2 dr . (4.4.44) In a similar manner, we find that ψ(τ)=− 2 π d dτ ∞ τ rg(r) √ r 2 − τ 2 dr . (4.4.45) Next, we substitute for D(k)inEquation4.4.39 and find that f(r)=V − ∞ 0 ∞ 1 ψ(τ)sin(kτ) dτ J 0 (kr) dk (4.4.46) = V − ∞ 1 ψ(τ) ∞ 0 sin(kτ)J 0 (kr) dk dτ (4.4.47) = V − ∞ 1 ψ(τ) √ τ 2 − r 2 dτ (4.4.48) for 0 ≤ r<a.HerewehaveusedEquation1.4.13. In a similar manner, it is readily shown that g(r)=− a 0 φ(t) √ r 2 − t 2 dt, 1 <r<∞. (4.4.49) Finally, we substitute Equation 4.4.48 into Equation 4.4.44 and find that φ(t)= 2 π d dt t 0 rV √ t 2 − r 2 dr (4.4.50) − 2 π d dt t 0 ∞ 1 ψ(τ) √ τ 2 − r 2 dτ rdr √ t 2 − r 2 = 2V π − 2 π ∞ 1 ψ(τ) d dt t 0 rdr √ τ 2 − r 2 √ t 2 − r 2 dτ (4.4.51) = 2V π − 2 π ∞ 1 τψ(τ) t 2 − τ 2 dτ. (4.4.52) © 2008 by Taylor & Francis Group, LLC 290 Mixed Boundary Value Problems 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 z u(r,z)/V Figure 4.4.1:The solution to Equation 4.4.27 subject to the mixed boundary conditions given by Equation 4.4.28 through Equation 4.4.30 when a =0.5. In a similar manner, we also find that ψ(τ)=− 2 π a 0 τφ(t) τ 2 − t 2 dt. (4.4.53) If we introduce t = ax, τ = ay,Φ(x)=πφ(t)/(2V ), and Ψ(y)=πψ(t)/(2V ), Equation 4.4.52 and Equation 4.4.53 can be combined to yield Φ(x)=1+ 2 π 2 1 0 Φ(ξ)K(x, ξ) dξ, 0 <x<1, (4.4.54) where K(x, ξ)=2a ∞ 1 η 2 (η 2 − a 2 x 2 )(η 2 − a 2 ξ 2 ) dη (4.4.55) = 1 ξ 2 − x 2 ξ ln 1+aξ 1 − aξ − x ln 1+ax 1 − ax . (4.4.56) For the special case of ξ = x,weuseL’Hospital rule and find that K(x, x)= 1 2x ln 1+ax 1 − ax + a 1 − a 2 x 2 . (4.4.57) Ourcomputations begin by finding φ(t)fromEquation 4.4.54. The function ψ(τ)follows from Equation 4.4.53. With φ(t)andψ(τ), we compute B(k), D(k)andA(k). Finally, Equation 4.4.31 gives the potential u(r, z). Figure 4.4.1 illustrates this potential when a =0.5. © 2008 by Taylor & Francis Group, LLC Transform Methods 291 Finally Selvadurai 86 solved this problem as a system of integral equations. In the present case we have ∂ 2 u ∂r 2 + 1 r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<∞, 0 <z<∞, (4.4.58) subject to the boundary conditions lim r→0 |u(r, z)| < ∞, lim r→∞ u(r, z) → 0, 0 <z<∞, (4.4.59) lim z→∞ u(r, z) → 0, 0 ≤ r<∞, (4.4.60) and u(r, 0) = 1, 0 ≤ r<a, u z (r, 0) = 0,a<r<b, u(r, 0) = 0,b<r<∞, (4.4.61) where b>a. Using transform methods or separation of variables, the general solution to Equation 4.4.58, Equation 4.4.59, and Equation 4.4.60 is u(r, z)= ∞ 0 A(k)J 0 (kr)e −kz dk. (4.4.62) Substituting Equation 4.4.62 into Equation 4.4.61, we find that ∞ 0 A(k)J 0 (kr) dk =1, 0 ≤ r<a, (4.4.63) ∞ 0 kA(k)J 0 (kr) dk =0,a<r<b, (4.4.64) and ∞ 0 A(k)J 0 (kr) dk =0,b<r<∞. (4.4.65) To solve this set of integral equations, we introduce two new functions f(r)andg(r)suchthat ∞ 0 kA(k)J 0 (kr) dk = f(r), 0 ≤ r<a, (4.4.66) and ∞ 0 kA(k)J 0 (kr) dk = g(r),b<r<∞. (4.4.67) 86 Reprinted from Mech. Res. Commun., 23,A.P.S.Selvadurai,Ontheproblem of an electrified disc located at the central opening of a coplanar earthed sheet, 621–624, c 1996, with permission from Elsevier. © 2008 by Taylor & Francis Group, LLC 292 Mixed Boundary Value Problems Then, because A(k)= a 0 rf(r)J 0 (kr) dr + ∞ b rg(r)J 0 (kr) dr, (4.4.68) Equation 4.4.63 and Equation 4.4.65 become a 0 τf(τ)L(τ, r) dτ + ∞ b τg(τ)L(τ,r) dτ =1, 0 ≤ r<a, (4.4.69) and a 0 τf(τ)L(τ, r) dτ + ∞ b τg(τ)L(τ,r) dτ =0,b≤ r<∞ (4.4.70) after interchanging the order of integration, where L(τ,r)= ∞ 0 J 0 (kτ)J 0 (kr) dk. (4.4.71) Because 87 L(τ,r)= min(τ,r) 0 ds (τ 2 − s 2 )(r 2 − s 2 ) (4.4.72) = ∞ max(τ,r) ds (s 2 − τ 2 )(s 2 − r 2 ) , (4.4.73) Equation 4.4.69 can be rewritten 2 π r 0 τf(τ) τ 0 ds (τ 2 − s 2 )(r 2 − s 2 ) dτ + 2 π a r τf(τ) r 0 ds (τ 2 − s 2 )(r 2 − s 2 ) dτ (4.4.74) + 2 π ∞ b τg(τ) ∞ τ ds (s 2 − τ 2 )(s 2 − r 2 ) dτ =1, or, after interchanging the order of integration, 2 π r 0 a s τf(τ) √ τ 2 − s 2 dτ ds √ r 2 − s 2 + 2 π ∞ b s b τg(τ) √ s 2 − τ 2 dτ ds √ s 2 − r 2 =1. (4.4.75) 87 Cooke, op. cit. © 2008 by Taylor & Francis Group, LLC Transform Methods 293 Setting F (s)= a s τf(τ) √ τ 2 − s 2 dτ, 0 ≤ s ≤ a, (4.4.76) and G(s)= s b τg(τ) √ s 2 − τ 2 dτ, b ≤ s ≤∞, (4.4.77) Equation 4.4.75 simplifies to r 0 F (s) √ r 2 − s 2 ds = π 2 − ∞ b G(s) √ s 2 − r 2 ds. (4.4.78) Applying Equation 1.2.13 and Equation 1.2.14 to Equation 4.4.78, we have that F (s)=− d ds s 2 − r 2 s 0 − 2 π ∞ b G(t) d ds s 0 dy t 2 − s 2 + y 2 dt, (4.4.79) where weinterchanged the order of integration and set s 2 −r 2 = y 2 .Carrying out the integration in y and simplifying, we finally obtain F (r)+ 2 π ∞ b tG(t) t 2 − r 2 dt =1, 0 ≤ r<a. (4.4.80) In a similar manner, for Equation 4.4.70, we have that a 0 τf(τ) τ 0 ds (τ 2 − s 2 )(r 2 − s 2 ) dτ + r b τg(τ) ∞ r ds (s 2 − τ 2 )(s 2 − r 2 ) dτ (4.4.81) + ∞ r τg(τ) ∞ τ ds (s 2 − τ 2 )(s 2 − r 2 ) dτ =0, or a 0 a s τf(τ) √ τ 2 − s 2 dτ ds √ r 2 − s 2 + ∞ r s b τg(τ) √ s 2 − τ 2 dτ ds √ s 2 − r 2 =0. (4.4.82) Equation 4.4.82 simplifies to ∞ r G(s) √ s 2 − r 2 ds = − a 0 F (s) √ r 2 − s 2 ds. (4.4.83) Applying Equation 1.2.15 and Equation 1.2.16 to Equation 4.4.83, we have that G(s)= 2 π a 0 F (t) d ds ∞ s r (r 2 − t 2 )(r 2 − s 2 ) dr dt. (4.4 .84) © 2008 by Taylor & Francis Group, LLC 294 Mixed Boundary Value Problems Carrying out the integration and differentiation within the wavy brackets, we obtain G(r)+ 2s π a 0 F (t) s 2 − t 2 dt =0,b<r<∞. (4.4.85) Upon solving the dual Fredholm integral equations, Equation 4.4.80 and Equa- tion 4.4.85, we have F (r)andG(r). Next, we invert Equation 4.4.76 and Equation 4.4.77 to find f(r)andg(r). The Fourier coefficient A(k)follows from Equation 4.4.68 while Equation 4.4.62 yields u(r, z). • Example 4.4.2 Let us now solve Laplace’s equation 88 ∂ 2 u ∂r 2 + 1 r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<∞, 0 <z<∞, (4.4.86) when the boundary conditions are lim r→0 |u(r, z)| < ∞, lim r→∞ u(r, z) → 0, 0 <z<∞, (4.4.87) lim z→∞ u(r, z) → 0, 0 ≤ r<∞, (4.4.88) and u z (r, 0) = 0, 0 ≤ r<a, u(r, 0) = 1,a<r<b, u z (r, 0) = 0,b<r<∞, (4.4.89) where b>a>0. Using transform methods or separation of variables, the general solution to Equation 4.4.86, Equation 4.4.87, and Equation 4.4.88 is u(r, z)= ∞ 0 A(k)J 0 (kr)e −kz dk k . (4.4.90) Substituting Equation 4.4.90 into Equation 4.4.89, we find that ∞ 0 A(k)J 0 (kr) dk =0, 0 ≤ r<a, (4.4.91) ∞ 0 A(k)J 0 (kr) dk k =1,a<r<b, (4.4.92) and ∞ 0 A(k)J 0 (kr) dk =0,b<r<∞. (4.4.93) The mixed boundary condition, Equation 4.4.89, has led to three integral equations involving Fourier-Bessel integrals. Our remaining task is to find the Fourier-Bessel coefficient A(k). Can we find some general result that might assist us in solving these triple Fourier-Bessel equations? In 1963 Cooke 89 studied how to find A(k)governedbythe following 88 Similar to a problem by Borodachev, N. M., and F. N. Borodacheva, 1966: Pentration of an annular stamp into an elastic half-space. Mech. Solids, 1(4), 101–103. 89 Cooke, op. cit. © 2008 by Taylor & Francis Group, LLC Transform Methods 295 integral equations: ∞ 0 A(k)J ν (kr) dk =0, 0 ≤ r<a, (4.4.94) ∞ 0 k p A(k)J ν (kr) dk = f(r),a<r<b, (4.4.95) and ∞ 0 A(k)J ν (kr) dk =0,b<r<∞, (4.4.96) where p = ±1, ν>− 1 2 ,andb>a>0. For p = −1, he proved that A(k)=k b a rg(r)J ν (kr) dr, (4.4.97) where g(r)=− 2 π r ν−1 d dr b r ηh(η) η 2 − r 2 dη ,a<r<b, (4.4.98) η 2ν h(η)= d dη η a x ν+1 f(x) η 2 − x 2 dx − 4 π 2 η η 2 − a 2 b a th(t) √ t 2 − a 2 K(η, t) dt, (4.4.99) K(η, t)= a 0 y 2ν (a 2 − y 2 ) (η 2 − y 2 )(t 2 − y 2 ) dy, (4.4.100) and a<η<b. We can use Co oke’s results if we set ν =0. Then, from Equation 4.4.97 through Equation 4.4.100, we have that A(k)=k b a rg(r)J 0 (kr) dr, (4.4.101) where g(r)=− 2 πr d dr b r ηh(η) η 2 − r 2 dη , (4.4.102) h(η)= d dη η a x η 2 − x 2 dx − 4 π 2 η η 2 − a 2 b a th(t) √ t 2 − a 2 K(η, t) dt (4.4.103) = η η 2 − a 2 − 4 π 2 η η 2 − a 2 b a th(t) √ t 2 − a 2 K(η, t) dt, (4.4.104) © 2008 by Taylor & Francis Group, LLC 296 Mixed Boundary Value Problems and K(η, t)= a 0 a 2 − y 2 (η 2 − y 2 )(t 2 − y 2 ) dy (4.4.105) = 1 2(η 2 − t 2 ) η 2 − a 2 η ln η + a η − a − t 2 − a 2 t ln t + a t −a . (4.4.106) In the special case when t = η,weemployL’Hospitalrule and find that K(η, η)=− a 2η 2 + η 2 + a 2 4η 3 ln η + a η − a . (4.4.107) If we introduce ηh(η)= η 2 − a 2 χ(η), (4.4.108) then Equation 4.102 and Equation 4.4.104 become g(r)=− 2 πr d dr b r η 2 − a 2 η 2 − r 2 χ(η) dη , (4.4.109) and η 2 − a 2 η 2 χ(η)=1− 4 π 2 b a K(η, t)χ(t) dt. (4.4.110) To compute u(r, z), we solve Equation 4.4.110 by replacing the integral with its representation fromthemidpoint rule. Setting d eta = (b-a) / N, the M ATLAB code for computing χ(η)is: for j = 1:N xi(j) = (j-0.5)*d eta+a; eta(j) = (j-0.5)*d eta + a; end for m = 1:N for K = 0:K max k=K*dK; factor(K+1,m) = bessel(0,k*eta(m)); end; end for n = 1:N % rows loop (top to bottom in the matrix) x=xi(n); b(n) = 1; % right side of the integral equation for m = 1:N % columns loop (left to right in the matrix) t=eta(m); % start setting up Equation 4.4.110 if (n==m) AA(n,m) = (x-a)*(x+a)/(x*x); % first term on left side else AA(n,m) = 0; end © 2008 by Taylor & Francis Group, LLC Transform Methods 297 % introduce the integral in Equation 4.4.110 temp1 = (x-a)*(x+a)/x; temp2 = (t-a)*(t+a)/t; if (t == x) integrand = -a/(2*x*x)+(x*x+a*a)*log((x+a)/(x-a))/(4*x*x*x); else integrand = temp1*log((x+a)/(x-a))-temp2*log((t+a)/(t-a)); integrand = integrand/(2*(x*x-t*t)); end AA(n,m) = AA(n,m) + 4*integrand*d eta/(pi*pi); end end % compute χ(η) and call it f f=AA\b’ Equation 4.4.109 gives g(r). First use the midpoint rule to compute the integral and put it in F(n).Thencompute the derivative to find g(n).The M ATLAB code for this is: for n = 1:N r=a+(n-1)*d eta; F(n) = 0; for m = n:N sq = xi(m)*xi(m); temp1 = sqrt((sq-a*a)/(sq-r*r)); F(n) = F(n) + temp1*f(m)*d eta; end; end F(N+1) = 0; for n = 1:N g(n) = -2*(F(n+1)-F(n))/(pi*xi(n)*d eta); end Finally,combining Equation 4.4.90 and Equation 4.4.101, u(r, z)= b a ηg(η) ∞ 0 J 0 (kη)J 0 (kr)e −kz dk dη, (4.4.111) after the order of integration is interchanged. The integral within the square brackets is evaluated using Simpson’s rule. The M ATLAB code is: for j = 1:21 z=0.1*(j-1); for K = 0:K max k=K*dK; Z(K+1) = exp(-k*z); end © 2008 by Taylor & Francis Group, LLC [...]... by Taylor & Francis Group, LLC 300 Mixed Boundary Value Problems We begin our solution of Equation 4.4 .117 through Equation 4.4 .119 by introducing a F (k) = b µ1 (ξ) sin(kξ) dξ + 0 µ2 (ξ) sin(kξ) dξ (4.4.120) a Note that µ1 (ξ) is defined over the interval [0, a] while µ2 (ξ) is defined over the interval [a, b] Substituting Equation 4.4.120 into Equation 4.4 .119 and interchanging the order of integration,... 2(τ 2 τ +a 1 τ (τ 2 − a2 ) ln 2) −ξ τ −a Cooke, op cit © 2008 by Taylor & Francis Group, LLC − ξ(ξ 2 − a2 ) ln 308 Mixed Boundary Value Problems 4 3.5 u(r,z)/a2 3 2.5 2 1.5 1 0.5 0 0 0 0.5 0.5 1 1 z/a 1.5 2 1.5 2.5 2 r/a 3 Figure 4.4.5: The solution to Equation 4.4.161 subject to the mixed boundary conditions given by Equation 4.4.162 through Equation 4.4.164 when b/a = 2 F (ξ) and A(k) follow from Equation... when a = 1, b = 2, and h = 1 Problems 1 If 0 < a < 1, solve97 ∂ 2 u 1 ∂u ∂ 2 u + 2 = 0, + ∂r2 r ∂r ∂z 0 ≤ r < ∞, −∞ < z < ∞, 97 See Davis, A M J., 1991: Slow viscous flow due to motion of an annular disk; pressuredriven extrusion through an annular hole in a wall J Fluid Mech., 231, 51–71 © 2008 by Taylor & Francis Group, LLC 316 Mixed Boundary Value Problems subject to the boundary conditions lim |u(r,... Figure 4.4.4 illustrates this potential when a = 1 and b = 2 © 2008 by Taylor & Francis Group, LLC 306 Mixed Boundary Value Problems • Example 4.4.5 Kim and Kim solved Laplace’s equation93 ∂ 2 u 1 ∂u ∂ 2 u + 2 = 0, + ∂r2 r ∂r ∂z 0 ≤ r < ∞, 0 < z < ∞, (4.4.161) 0 < z < ∞, (4.4.162) subject to the boundary conditions lim |u(r, z)| < ∞, r→0 lim u(r, z) → 0, r→∞ lim u(r, z) → 0, 0 ≤ r < ∞, z→∞ urz... (4.4 .112 ) 0 < z < ∞, (4.4 .113 ) subject to the boundary conditions lim |u(r, z)| < ∞, r→0 lim u(r, z) → 0, r→∞ lim u(r, z) → 0, z→∞ 0 ≤ r < ∞, (4.4 .114 ) 90 Taken with permission from Sibgatullin, N R., I N Sibgatullin, A A Garcia, and V S Manko, 2004: Magnetic fields of pulsars surrounded by accretion disks of finite extension Astron Astrophys., 422, 587–590 © 2008 by Taylor & Francis Group, LLC 300 Mixed. .. (4.4.201) 96 See Kuz’min, Yu N., 1972: Electrostatic field of a circular disk near a plane containing an aperture Sov Tech Phys., 17, 473–476 © 2008 by Taylor & Francis Group, LLC 312 Mixed Boundary Value Problems subject to the boundary conditions lim |u(r, z)| < ∞, r→0 lim u(r, z) → 0, −∞ < z < ∞, r→∞ lim u(r, z) → 0, 0 ≤ r < ∞, |z|→∞ u(r, 0− ) = u(r, 0+ ), uz (r, 0− ) = uz (r, 0+ ), u(r, 0) = 0, 0 ≤ r... (4.4.141) Transform Methods 303 0.6 2 a u(r,z)/A 0.5 0.4 0.3 0.2 0.1 0 −0.1 −0.2 0 0 1 0.5 2 1 z/a r/a 3 1.5 2 4 Figure 4.4.3: The solution to Equation 4.4 .112 subject to the mixed boundary conditions given by Equation 4.4 .113 through Equation 4.4 .115 when b/a = 2 and a2 K/A = 0.5 We evaluate numerically the integrals inside of the square brackets (except for the case when z = 0 where there is an exact...298 Mixed Boundary Value Problems for i = 1:31 r = 0.1*(i-1); u(i,j) = 0; for K = 0:K max k = K*dK; R(K+1) = besselj(0,k*r); end for m = 1:N integral = 0; for K = 0:K max k = K*dK; integrand = factor(K+1,m)*R(K+1)*Z(K+1);... 2 u + 2 = 0, + ∂r2 r ∂r ∂z 0 ≤ r < ∞, 0 < z < h, (4.4.179) 0 < z < h, (4.4.180) subject to the boundary conditions lim |u(r, z)| < ∞, r→0 lim u(r, z) → 0, r→∞ and u(r, 0) = f (r), uz (r, h) = g(r), uz (r, 0) = u(r, h) = 0, 0 ≤ r < 1, 1 < r < ∞ (4.4.181) 95 Taken from Dhaliwal, R S., 1966: Mixed boundary value problem of heat conduction for infinite slab Appl Sci Res., 16, 228–240 with kind permission... ψ1 (t) cos(kt) dt, (4.4.189) and A(k) sinh(kh) + B(k) cosh(kh) = 1 cosh(kh) + 1− 0 −kh 1 ψ1 (t) cos(kt) dt e cosh(kh) 1 0 ψ2 (t) sin(kt) dt (4.4.190) © 2008 by Taylor & Francis Group, LLC 310 Mixed Boundary Value Problems Introducing Equation 4.4.187 and Equation 4.4.189 into Equation 4.4.183 and Equation 4.4.184, multiplying the resulting equations by dη/ r2 − η 2 and integrating from 0 to r, we obtain . Francis Group, LLC 290 Mixed Boundary Value Problems 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 z u(r,z)/V Figure 4.4.1:The solution to Equation 4.4.27 subject to the mixed boundary conditions given. 587–590. © 2008 by Taylor & Francis Group, LLC 300 Mixed Boundary Value Problems We begin our solution of Equation 4.4 .117 through Equation 4.4 .119 by introducing F (k)= a 0 µ 1 (ξ)sin(kξ) dξ. LLC 308 Mixed Boundary Value Problems 0 0.5 1 1.5 2 2.5 3 0 0.5 1 1.5 2 0 0.5 1 1.5 2 2.5 3 3.5 4 r/a z/a u(r,z)/a 2 Figure 4.4.5:The solution to Equation 4.4.161 subject to the mixed boundary
Ngày đăng: 13/08/2014, 05:22
Xem thêm: Mixed Boundary Value Problems Episode 11 docx