Solutions VI. MAE294A/SIO203A: Methods in Applied Mechanics Fall Quarter

Transcription

1 MAE94A/SIO3A: Methods in Applied Mechanics Fall Quarter 17 Solutions VI 1 We let Tx, t) = Xx)τt). We have two homogeneous conditions in x, so let X = λ X to obtain sinusoids, so we are left with τ = λ κτ. Our general solution is thus Tx, t) = A cos λx + B sin λx)e λ κt. The insulated endwalls provide the boundary condition T x x = and L) =. We apply this to our general solution. T x, ) = Bλ =, T x L, ) = Aλ sin λx =, where we took B = before applying the second condition. We now have the condition λ = nπ L. Next we apply the boundary condition so we solve and A n = L Tx, ) = n L A = 1 L A n cos nπx L = x L T, L x L T dx = T x L T cos nπx L dx = T n π [ 1)n 1], for n >. This will be zero for all even n, so our final solution is Tx, t) = T 4T cos π nπx L odd n n e κtn π /L. 1

2 We let ux, t) = Xx)Tt). We have two homogeneous conditions in x, so let X = λ X to obtain sinusoids, so we are left with T = c λ T. Our general solution is then ux, t) = A cos λx + B sin λx) C cos λct + D sin λct). Clamped ends mean that ux = ±a) =, which tells us λ = nπ/a. Next consider the boundary conditions ux, ) = C n A n cos nπx a + B n sin nπx ) = sin πx/a), a ) u t x, ) = nπ a D n A n cos nπx a + B n sin nπx =, a so D n = and we can absorb C n into our constants A n, B n. We then have ux, ) = A n cos nπx a + B n sin nπx a = sin πx/a) so by matching we see we need n = 4 and A 4 = while B 4 = 1. Our final solution is then ux, t) = sin πx/a) cos πct/a). 3 The general solution in polar coordinates is uρ, φ) = A log ρ + B + An ρ n + B n ρ n) C n cos nφ + D n sin nφ). n> Consider u a as satisfying the condition at ρ = a but zero at ρ = b and u b as satisfying the condition at ρ = b but zero at ρ = a. Applying the conditions to u a we find u a a, φ) = A log a + B + An a n + B n a n) C n cos nφ + D n sin nφ) n> = 1 = cosφ). By matching we see that we need n =, so the terms in the sum all vanish. We then have u a a, φ) = A log a + B = 1 u a b, φ) = A log b + B = So we find A = 1/log b log a) and B = log b/log b log a) and u a ρ, φ) = Next we apply the conditions to u b and find log ρ log b log b log a. u b b, φ) = A log b + B + An b n + B n b n) C n cos nφ + D n sin nφ) = sin φ. n>

3 Now by matching we see that the terms outside of the sum vanish and we pick out n = 1. We also have C 1 = to leave only the sin φ. We then have u b a, φ) = A 1 a + B 1 /a) sin φ = u b b, φ) = A 1 b + B 1 /b) sin φ = sin φ, so we find A 1 = 1/b a /b) and B 1 = a /b a /b). Thus we have solved for ρ a u b ρ, φ) = ) /ρ b a sin φ /b and so uρ, φ) = u a ρ, φ) + u b ρ, φ) = log ρ log b log b log a + ρ a ) /ρ b a sin φ. /b 4 In order to be single-valued, the dependence on φ must be periodic. Therefore choose a separable solution form of ψρ, θ) = Rρ)A cos nθ + B sin nθ). The biharmonic equation in polar coordinates is ψ ρρρρ + ρ ψ ρρρ 1 ρ ψ ρρ + 1 ρ 3 ψ ρ + ρ ψ ρρφφ + 1 ρ 4 ψ φφφφ ρ 3 ψ ρφφ + 4 ρ 4 ψ φφ =. We plug in our separable form of ψ, and for convenience use only one sinusoid to solve for Rρ). R ρρρρ + ρ R ρρρ 1 ρ R ρρ + 1 ρ 3 R ρ n ρ R ρρ + n4 ρ 4 R + n ρ 3 R ρ 4n ρ 4 R =. We can factorize this back to ψ ) and see d dρ + 1 ρ ) [ d dρ n d ρ dρ + 1 ) ] d ρ dρ n ρ R =. This is an equidimensional equation, so we look for the form R = ρ α. Within the square brackets, we will end up with powers ρ α, so the outer Laplacian will see this form instead of the ρ α that the inner Laplacian sees. We now have α ) n ) α n ) = with roots α = ±n, ± n. We will have two double valued roots when n = and one when n = 1, so our general solution is ψρ, φ) = A + B log ρ + C ρ + D ρ log ρ ) + A 1 ρ + B 1 ρ 1 + C 1 ρ 3 + D 1 ρ log ρ E 1 cos φ + F 1 sin φ) + A n ρ n + B n ρ n + C n ρ +n + D n ρ n) E n cos nφ + F n sin nφ). n>1 3

4 If we consider ψ Ux + C = Uρ cos φ + C, we know that the highest power of ρ must be 1 so immediately spot that we should only take n 1 and C n = for all n. We also need B = D = D 1 =. Furthermore, since our limit has only cos φ, we must have F 1 = and we can let E 1 be absorbed into the other constants. Now we have ψρ, φ) = A + A 1 ρ + B 1 ρ 1) cos φ. Applying the other boundaries ψa, φ) = ψ ρ a, φ) = we find A + A 1 a + B 1 /a) cos φ = A 1 B 1 /a = A + A 1 ρ cos φ = Uρ cos φ + C. We can find A 1 = U and A = C and B 1 = a U, but the first condition will not hold for all φ! Thus there is no solution and we have the paradox. 5 We have the general solution ur, θ) = A l r l + B l r l+1)) P l cos θ). l We want our solution to be finite at r =, so we must have B l =. Then we apply our other boundary condition ur, θ) = l A l R l P l cos θ) = cos θ. We can either multiply both sides by P m cos θ) and integrate in θ, or simply rewrite the boundary condition in terms of these Legendre polynomials. We find cos θ = [ 1 3 cos θ 1) ] = 3 P cos θ) P cos θ), so by matching with our solution we pick out l =, and find A = 1 3, A R = 3, so our final solution is ur, θ) = r R P cos θ). 4

5 6 First consider the PDE for n = 1. Choose our independent variable to be x. d dx Gx) k Gx) = δx). This is an ODE over the domain x, ), so we can solve it as we did before. Our homogeneous solutions are e ± k x and we split our domain in two. { Ae k x + Be k x for x < Gx) = Ce k x + De k x for x >. In order to satisfy the boundary conditions G x ), we take only e kx in both cases. Now { Ae k x for x < Gx) = Ce k x for x >. Now apply continuity and a jump condition at x = to solve for the constants. so A = C = 1/ k and our solution is A = C, C k A k = 1, Gx) = e kx k. For n =, 3 we can t do it as before because we only have r >. Instead, integrate the equation over a small circle. In D our equation d dr Gr) + 1 d r dr Gr) k Gr) = δr) has homogeneous solutions K k r), I k r). In order to satisfy the condition at r, we know that I k r) cannot contribute. If we integrate our equation over a circle of radius ɛ we find G k G ds = δr)ds π r<ɛ dgɛ) ɛdθ k π dr π dgɛ) ɛ k π dr ɛ ɛ r<ɛ Gr)rdr = 1, Gr)rdr = 1, where we have used the divergence theorem fin the second line. Now take ɛ. From the PDE we know that G δr) and therefore Gr) is smoother because derivatives are un-smoothing. Therefore the integral of G will vanish as ɛ. We then have lim ɛ dgɛ) π ɛ = 1. dr 5

6 In order for the LHS to be a constant, G r r) r 1, and therefore Gr) log r. Looking back at our homogeneous solutions, we have G = AK k r), and some research into this Bessel function tells us lim K k r) log r, r and therefore G r ɛ) Aɛ 1. Plugging this into our condition above gives πa = 1, and therefore Gr) = K k r) π. For n = 3 we have d dr Gr) + d r dr Gr) k Gr) = δr). This has homogeneous solutions e ± k r /r, and we again neglect the positive exponential that blows up at r. When we integrate the PDE on a sphere of radius ɛ we will find, as above, that the integral of Gr) will disappear because only G is not well behaved. We thus have dgɛ) lim 4π ɛ = 1, ɛ dr so we must have G r ɛ) r and Gɛ) r 1. By a Taylor Expansion we have ) k e k r lim G rr) = lim A e k r r ɛ r ɛ r r A 1 ɛ. Plugging this into the condition above gives 4πA = 1, so we have Gr) = e k r 4πr. 6