The heat and wave equations in 2D and 3D

Transcription

1 The heat and wave equations in and Linear Partial ifferential Equations Matthew J. Hancock Fall and 3 Heat Equation Ref: Myint-U & ebnath.3.5 [Nov, 006] Consider an arbitrary 3 subregion V of R 3 (V R 3 ), with temperature u (x,t) defined at all points x = (x,y,z) V. We generalize the ideas of 1- heat flux to find an equation governing u. The heat energy in the subregion is defined as heat energy = cρudv Recall that conservation of energy implies rate of change heat energy into V from heat energy generated = + of heat energy boundaries per unit time in solid per unit time We desire the heat flux through the boundary S of the subregion V, which is the normal component of the heat flux vector φ, φ nˆ, where nˆ is the outward unit normal at the boundary S. Hats on vectors denote a unit vector, nˆ = 1 (length 1). If the heat flux vector φ is directed inward, then φ nˆ < 0 and the outward flow of heat is negative. To compute the total heat energy flowing across the boundaries, we sum φ nˆ over the entire closed surface S, denoted by a double integral ds. S Therefore, the conservation of energy principle becomes d cρudv = φ nˆ ds + dt V S V V QdV (1) 1

2 1.1 ivergence Theorem (a.k.a. Gauss s Theorem) For any volume V with closed smooth surface S, AdV = A nˆ ds V S where A is any function that is smooth (i.e. continuously differentiable) for x V. Note that =,, = ê x + ê y + ê z x y z x y z where ê x, ê y, ê z are the unit vectors in the x, y, z directions, respectively. The divergence of a vector valued function F = (F x,f y,f z ) is F = F x F y F z + +. x y z The Laplacian of a scalar function F is F = F F F x + y + z. Applying the ivergence theorem to (1) gives d cρudv = φ dv + dt V Since V is independent of time, the integrals can be combined as u cρ + φ Q dv = 0 t V V V QdV Since V is an arbitrary subregion of R 3 and the integrand is assumed continuous, the integrand must be everywhere zero, u cρ t + φ Q = 0 () 1. Fourier s Law of Heat Conduction The 3 generalization of Fourier s Law of Heat Conduction is φ = K 0 u (3) where K 0 is called the thermal diffusivity. Substituting (3) into () gives u t Q = κ u + cρ (4)

3 where κ = K 0 / (cρ). This is the 3 Heat Equation. Normalizing as for the 1 case, x κ x =, t = t, l l Eq. (4) becomes (dropping tildes) the non-dimensional Heat Equation, where q = l Q/ (κcρ) = l Q/K 0. u = u + q, (5) t and 3 Wave equation The 1 wave equation can be generalized to a or 3 wave equation, in scaled coordinates, u tt = u (6) This models vibrations on a membrane, reflection and refraction of electromagnetic (light) and acoustic (sound) waves in air, fluid, or other medium. 3 Separation of variables in and 3 Ref: Guenther & Lee 10., Myint-U & ebnath 4.10, 4.11 We consider simple subregions R 3. We assume the boundary conditions are zero, u = 0 on, where denotes the closed surface of (assumed smooth). The 3 Heat Problem is u t = u, x, t > 0, u (x,t) = 0, x, (7) u (x, 0) = f (x), x. The 3 wave problem is u tt = u, x, t > 0, u (x,t) = 0, x, (8) u (x, 0) = f (x), x, u t (x, 0) = g (x), x. We separate variables as u (x,t) = X (x)t (t) (9) 3

4 The 3 Heat Equation implies T T X = = λ = const (10) X where λ = const since the l.h.s. depends solely on t and the middle X /X depends solely on x. The 3 wave equation becomes On the boundaries, T T = X X X (x) = 0, = λ = const (11) x The Sturm-Liouville Problem for X (x) is X + λx = 0, x (1) X (x) = 0, x 4 Solution for T (t) Suppose that the Sturm-Liouville problem (1) has eigen-solution X n (x) and eigenvalue λ n, where X n (x) is non-trivial. Then for the 3 Heat Problem, the problem for T (t) is, from (10), with solution T = λ (13) T λ T n (t) = c n e nt and the corresponding solution to the PE and BCs is λ u n (x,t) = X n (x)t n (t) = X n (x)c n e nt. For the 3 Wave Problem, the problem for T (t) is, from (11), (14) T = λ (15) T with solution λn λn T n (t) = α n cos t + β n sin t (16) and the corresponding normal mode is u n (x,t) = X n (x)t n (t). 4

5 5 Uniqueness of the 3 Heat Problem Ref: Guenther & Lee 10.3 We now prove that the solution of the 3 Heat Problem u t = u, x u (x,t) = 0, u (x, 0) = f (x), x x is unique. Let u 1, u be two solutions. efine v = u 1 u. Then v satisfies v t = v, x v (x,t) = 0, v (x, 0) = 0, x x Let V (t) = v dv 0 V (t) 0 since the integrand v (x,t) 0 for all (x,t). ifferentiating in time gives dv (t) = vv t dv dt Substituting for v t from the PE yields dv (t) = v vdv dt By result (6) derived below, dv (t) = dt v v nˆds v dv (17) But on, v = 0, so that the first integral on the r.h.s. vanishes. Thus dv (t) = v dv 0 (18) dt Also, at t = 0, V (0) = v (x, 0) dv = 0 Thus V (0) = 0, V (t) 0 and dv/dt 0, i.e. V (t) is a non-negative, non-increasing function that starts at zero. Thus V (t) must be zero for all time t, so that v (x,t) must be identically zero throughout the volume for all time, implying the two solutions are the same, u 1 = u. Thus the solution to the 3 heat problem is unique. For insulated BCs, v = 0 on, and hence v v nˆ = 0 on. Thus we can still derive Eq. (18) from (17), and the uniqueness proof still holds. Thus the 3 Heat Problem with Type II homogeneous BCs also has a unique solution. 5

6 6 Sturm-Liouville problem Ref: Guenther & Lee 10., Myint-U & ebnath Both the 3 Heat Equation and the 3 Wave Equation lead to the Sturm-Liouville problem X + λx = 0, x, (19) X (x) = 0, x. 6.1 Green s Formula and the Solvability Condition Ref: Guenther & Lee 8.3, Myint-U & ebnath (Exercise 1) For the Type I BCs assumed here (u (x,t) = 0, for x ), we now show that all eigenvalues are positive. To do so, we need a result that combines some vector calculus with the ivergence Theorem. From vector calculus, for any scalar function G and vector valued function F, Using the divergence theorem, (GF) nˆds = S (GF) = G F + F G (0) V (GF)dV (1) Substituting (0) into (1) gives (GF) nˆds = G FdV + F GdV () S V V Result () applied to G = v 1 and F = v gives (v 1 v ) nˆds = v1 v + v v 1 dv (3) S Result () applied to G = v and F = v 1 gives (v v 1 ) nˆds = v v 1 + v 1 v dv (4) S Subtracting (3) and (4) gives Green s Formula (also known as Green s second identity): S (v 1 v v v 1 ) nˆds = V V V v1 v v v 1 dv (5) This is also known as a Solvability Condition, since the values of v 1 and v on the boundary of must be consistent with the values of v 1 and v on the interior of. Result (5) holds for any smooth function v defined on a volume V with closed smooth surface S. 6

7 6. Positive, real eigenvalues (for Type I BCs) Ref: Myint-U & ebnath 7. Choosing G = v and F = v, for some function v, in () gives v v nˆds = v v + v v dv S = V V v v + v dv. (6) Result (6) holds for any smooth function v defined on a volume V with closed smooth surface S. We now apply result (6) to a solution X (x) of the Sturm-Liouville problem (19). Letting v = X (x), S = and V =, Eq. (6) becomes X X nˆds = X XdV + X dv (7) Since X (x) = 0 for x, X X nˆds = 0 (8) Also, from the PE in (19), Substituting (8) and (9) into (7) gives 0 = λ X dv + X XdV = λ X dv (9) X dv (30) For non-trivial solutions, X = 0 at some points in and hence by continuity of X, X dv > 0. Thus (30) can be rearranged, X dv λ = 0 (31) X dv Since X is real, the the eigenvalue λ is also real. If X = 0 for all points in, then integrating and imposing the BC X (x) = 0 for x gives X = 0 for all x, i.e. the trivial solution. Thus X is nonzero at some points in, and hence by continuity of X, X dv > 0. Thus, from (31), λ > 0. 7

8 6.3 Orthogonality of eigen-solutions to Sturm-Liouville problem Ref: Myint-U & ebnath 7. Suppose v 1, v are two eigenfunctions with eigenvalues λ 1, λ of the 3 Sturm- Liouville problem v + λv = 0, v = 0, x x We make use of Green s Formula (5) with V =, S =, which holds for any functions v 1, v defined on a volume V with smooth closed connected surface S. Since v 1 = 0 = v on and v 1 = λ 1 v 1 and v = λ v, then Eq. (5) becomes 0 = (λ 1 λ ) v 1 v dv Thus if λ 1 = λ, v 1 v dv = 0, (3) and the eigenfunctions v 1, v are orthogonal. 7 Heat and Wave problems on a rectangle, homogeneous BCs Ref: Guenther & Lee 10., Myint-U & ebnath [Nov 7, 006] 7.1 Sturm-Liouville Problem on a rectangle We now consider the special case where the subregion is a rectangle = {(x,y) : 0 x x 0, 0 y y 0 } The Sturm-Liouville Problem (19) becomes v v x + λv = 0, y (x,y), (33) v (0,y) = v (x0,y) = 0, 0 y y0, (34) v (x, 0) = v (x,y 0 ) = 0, 0 x x 0. (35) 8

9 Note that in the PE (33), λ is positive a constant (we showed above that λ had to be both constant and positive). We employ separation of variables again, this time in x and y: substituting v (x,y) = X (x)y (y) into the PE (33) and dividing by X (x)y (y) gives Y X + λ = Y X Since the l.h.s. depends only on y and the r.h.s. only depends on x, both sides must equal a constant, say µ, Y X + λ = = µ (36) Y X The BCs (34) and (35) imply X (0)Y (y) = X (x 0 )Y (y) = 0, 0 y y 0, X (x)y (0) = X (x)y (y 0 ) = 0, 0 x x 0. To have a non-trivial solution, Y (y) must be nonzero for some y [0,y 0 ] and X (x) must be nonzero for some x [0,x 0 ], so that to satisfy the previous equations, we must have X (0) = X (x 0 ) = Y (0) = Y (y 0 ) = 0 (37) The problem for X (x) is the 1 Sturm-Liouville problem X + µx = 0, 0 x x 0 (38) X (0) = X (x 0 ) = 0 We solved this problem in the chapter on the 1 Heat Equation. We found that for non-trivial solutions, µ had to be positive and the solution is mπx mπ X m (x) = a m sin, µ m =, x 0 x 0 m = 1,, 3,... (39) The problem for Y (y) is Y + νy = 0, 0 y y 0, (40) Y (0) = Y (y 0 ) = 0 where ν = λ µ. The solutions are the same as those for (38), with ν replacing µ: nπy nπ Y n (y) = b n sin, ν n =, n = 1,, 3,... (41) y 0 y 0 The eigen-solution of the Sturm-Liouville problem (33) (35) is mπx nπy v mn (x,y) = X m (x) Y n (y) = c mn sin sin, m,n = 1,, 3,... (4) x 0 y 0 with eigenvalue m n λ mn = µ m + ν n = π +. x 0 y 0 9

10 7. Solution to heat equation on rectangle The heat problem on the rectangle is the special case of (7), u u u t = x +, (x,y), t > 0, y u (x,y,t) = 0, (x,y), u (x,y, 0) = f (x,y), (x,y), where is the rectangle = {(x,y) : 0 x x 0, 0 y y 0 }. We reverse the separation of variables (9) and substitute solutions (14) and (4) to the T (t) problem (13) and the Sturm-Liouville problem (33) (35), respectively, to obtain mπx nπy λ u mn (x,y,t) = A mn sin sin e mnt x 0 y 0 «m mπx nπy π + n t x y = A mn sin sin e 0 0 x 0 y 0 To satisfy the initial condition, we sum over all m, n to obtain the solution, in general form, mπx nπy u (x,y,t) = u mn (x,y,t) = A mn sin sin e x 0 y 0 m=1 n=1 m=1 n=1 Setting t = 0 and imposing the initial condition u (x,y, 0) = f (x,y) gives λ mnt mπx nπy f (x,y) = u (x,y, 0) = A mn v mn (x,y) = A mn sin sin x 0 y m=1 n=1 m=1 n=1 0 mπx nπy where v mn (x,y) = sin x 0 sin y 0 are the eigenfunctions of the Sturm Liouville problem on a rectangle, (33) (35). Multiplying both sides by vmnˆ ˆ (x,y) ( ˆm, ˆn = 1,, 3,...) and integrating over the rectangle gives (43) f (x,y) vmnˆ ˆ (x,y)da = A mn v mn (x,y) v mnˆ ˆ (x,y) da (44) m=1 n=1 where da = dxdy. Note that x 0 mπx mπx ˆ v mn (x,y) v mnˆ ˆ (x,y)da = sin sin dx 0 x 0 x 0 y0 nπy nπy ˆ sin sin dy 0 y 0 y 0 x 0 y 0 /4, m = ˆm and n = ˆn = 0, otherwise 10

11 Thus (44) becomes A ˆmnˆ f (x,y) v mˆ nˆ (x,y) da = x 0 y 0 4 Since ˆm, ˆn are dummy variables, we replace ˆm by m and ˆn by n, and rearrange to obtain A mn 4 = f (x,y) v mnˆ ˆ (x,y) da x 0 y 0 4 x0 y0 mπx nπy = f (x,y) sin sin dydx (45) x 0 y x 0 y Solution to wave equation on rectangle, homogeneous BCs The solution to the wave equation on the rectangle follows similarly. The general 3 wave problem (8) becomes u u u tt = x +, (x,y), t > 0, y u (x,y,t) = 0, (x,y), u (x,y, 0) = f (x,y), (x,y), u t (x,y, 0) = g (x,y), (x,y), where is the rectangle = {(x,y) : 0 x x 0, 0 y y 0 }. We reverse the separation of variables (9) and substitute solutions (16) and (4) to the T (t) problem (15) and the Sturm Liouville problem (33) (35), respectively, to obtain mπx nπy u mn (x,y,t) = sin sin α nm cos λnm t + β nm sin λnm t x 0 y 0 mπx nπy = sin sin x 0 y 0 m n m n α nm cos πt x 0 + y 0 + β nm sin πt x 0 + To satisfy the initial condition, we sum over all m, n to obtain the solution, in general form, m=1 n=1 Setting t = 0 and imposing the initial conditions u (x,y,t) = u mn (x,y,t) (46) y 0 u (x,y, 0) = f (x,y), u t (x,y, 0) = g (x,y) 11

12 gives f (x,y) = u (x,y, 0) = α mn v mn (x,y) m=1 n=1 mπx nπy = α mn sin sin m=1 n=1 x 0 g (x,y) = u t (x,y, 0) = m=1 n=1 λmn y0 β mn v mn (x,y) mπx nπy = λmn β mn sin sin x 0 y m=1 n=1 0 mπx nπy where v mn (x,y) = sin x 0 sin y 0 are the eigenfunctions of the Sturm Liouville problem on a rectangle, (33) (35). As above, multiplying both sides by vmnˆ ˆ (x,y) ( ˆm, ˆn = 1,, 3,...) and integrating over the rectangle gives 4 α mn = f (x,y) v mn (x,y) dxdy x 0 y 0 4 β mn = g (x,y) v mn (x,y) dxdy x 0 y 0 λmn 8 Heat and Wave equations on a circle, homogeneous BCs Ref: Guenther & Lee 10., Myint-U & ebnath 9.4, 9.13 (exercises) We now consider the special case where the subregion is the unit circle (we may assume the circle has radius 1 by choosing the length scale l for the spatial coordinates as the original radius): = (x,y) : x + y 1 The Sturm-Liouville Problem (19) becomes v v x + λv = 0, y (x,y), (47) v (x,y) = 0, x + y = 1, (48) where we already know λ is positive and real. It is natural to introduce polar coordinates via the transformation x = r cos θ, y = r sin θ, w (r,θ,t) = u (x,y,t) for 0 r 1, π θ < π. 1

13 You can verify that v = v v 1 w 1 w x + y = r + r r r r θ The PE becomes 1 w 1 w r r + + λw = 0, r r θ 0 r 1, π θ < π (49) The BC (48) requires w (1,θ) = 0, π θ < π. (50) We use separation of variables by substituting w (r,θ) = R (r) H (θ) (51) into the PE (49) and multiplying by r / (R (r) H (θ)) and then rearranging to obtain d dr 1 d H 1 r r + λr = dr dr R (r) dθ H (θ) Again, since the l.h.s. depends only on r and the r.h.s. on θ, both must be equal to a constant µ, d dr 1 d H 1 r r + λr = dr dr R (r) dθ H (θ) = µ (5) The BC (50) becomes w (1,θ) = R (1)H (θ) = 0 which, in order to obtain non-trivial solutions (H (θ) = 0 for some θ), implies R (1) = 0 (53) In the original (x,y) coordinates, it is assumed that v (x,y) is smooth (i.e. continuously differentiable) over the circle. When we change to polar coordinates, we need to introduce an extra condition to guarantee the smoothness of v (x,y), namely, that w (r, π) = w (r,π), w θ (r, π) = w θ (r,π). (54) Substituting (51) gives dh dh H ( π) = H (π), ( π) = (π). (55) dθ dθ The solution v (x,y) is also bounded on the circle, which implies R (r) must be bounded for 0 r 1. 13

14 The problem for H (θ) is d H dh dh + µh (θ) = 0; H ( π) = H (π), ( π) = (π). (56) dθ dθ dθ You can show that for µ < 0, we only get the trivial solution H (θ) = 0. For µ = 0, we have H (θ) = const, which works. For µ > 0, non-trivial solutions are found only when µ = m, H m (θ) = a m cos (mθ) + b m sin (mθ) Thus, in general, we may assume λ = m, for m = 0, 1,, 3,... The equation for R (r) in (5) becomes d dr 1 r r + λr = µ = m, m = 0, 1,, 3,... dr dr R (r) Rearranging gives d R m dr m r + r + λr m R m = 0; R dr m (1) = 0, R m (0) < (57) dr We know already that λ > 0, so we can let so that (57) becomes s = λr, Rm (s) = R m (r) d R m dr m s + s + s m ds R m = 0; R m λ = 0, R m (0) < (58) ds The OE is called Bessel s Equation which, for each m = 0, 1,,... has two linearly independent solutions, J m (s) and Y m (s), called the Bessel functions of the first and second kinds, respectively, of order m. The function J m (s) is bounded at s = 0; the function Y m (s) is unbounded at s = 0. The general solution to the OE is R m (s) = c m1 J m (s) + c m Y m (s) where c mn are constants of integration. Our boundedness criterion R m (0) < at s = 0 implies c m = 0. Thus [Nov 9, 006] R m (s) = c m J m (s), R m (r) = c m J m λr. The Bessel Function J m (s) of the first kind of order m has power series ( 1) k s k+m J m (s) =. (59) k! (k + m)! k+m k=0 J m (s) can be expressed in many ways, see Handbook of Mathematical Functions by Abramowitz and Stegun, for tables, plots, and equations. The fact that J m (s) is 14

15 expressed as a power series is not a drawback. It is like sin (x) and cos (x), which are also associated with power series. Note that the power series (59) converges absolutely for all s 0 and converges uniformly on any closed set s [0,L]. To see this, note that each term in the sum satisfies ( 1) k s k+m L k+m k! (k + m)! k+m k! (k + m)! k+m Note that the sum of numbers L k+m k! (k + m)! k+m k=0 converges by the Ratio Test, since the ratio of successive terms in the sum is L (k+1)+m (k+1)!(k+1+m)! (k+1)+m L L L = = L k+m (k + 1) (k m) 4 (k + 1) 4 (k + 1) k!(k+m)! k+m Thus for k > N = L/, L < < 1 L k+m (N + 1) L (k+1)+m (k+1)!(k+1+m)! (k+1)+m k!(k+m)! k+m Since the upper bound is less than one and is independent of the summation index k, then by the Ratio test, the sum converges absolutely. By the Weirstrass M-Test, the infinite sum in (59) converges uniformly on [0,L]. Since L is arbitrary, the infinite sum in (59) converges uniformly on any closed subinterval [0,L] of the real axis. Each Bessel function J m (s) has an infinite number of zeros (roots) for s > 0. Let j m,n be the n th zero of the function J m (s). Note that 1 3 J 0 (s) j 0,1 =.4048 j 0, = j 0,3 = J 1 (s) j 1,1 = 3.85 j 1, = j 1,3 = The second BC requires R m (1) = R m λ = J m λ = 0 This has an infinite number of solutions, namely λ = jm,n for n = 1,, 3,... Thus the eigenvalues are λ j mn = m,n, m,n = 1,, 3,... with corresponding eigenfunctions J m (rj m,n ). The separable solutions are thus J 0 (rj 0,n ) n = 1,, 3,... v mn (x,y) = w mn (r,θ) = J m (rj m,n ) (α mn cos (mθ) + +β mn sin (mθ)) m,n = 1,, 3,... (60) 15

16 8.1 Solution to heat equation on the circle, homogeneous BCs The heat problem on the circle is the special case of (7), u u u t = x +, (x,y), t > 0, y u (x,y,t) = 0, (x,y), u (x,y, 0) = f (x,y), (x,y), where is the circle = {(x,y) : x + y 1}. We reverse the separation of variables (9) and substitute solutions (14) and (4) to the T (t) problem (13) and the Sturm Liouville problem (47) (48), respectively, to obtain λ u mn (x,y,t) = v mn (x,y) e mnt = v mn (x,y) e Jm,n t where v mn is given in (60). To satisfy the initial condition, we sum over all m, n to obtain the solution, in general form, u (x,y,t) = u mn (x,y,t) = = m=1 n=1 m=1 n=1 m=1 n=1 λ vmn (x,y) e mnt λ Jm (rj m,n) (α mn cos (mθ) + β mn sin (mθ)) e mnt (61) where r = x + y and tanθ = y/x. Setting t = 0 and imposing the initial condition u (x,y, 0) = f (x,y) gives f (x,y) = u (x,y, 0) = v mn (x,y) m=1 n=1 We can use orthogonality relations to find α mn and β mn. 9 The Heat Problem on a square with inhomogeneous BC We now consider the case of the heat problem on the unit square = {(x,y) : 0 x,y 1}, (6) 16

17 where a hot spot exists on the left side, u t = u, (x,y) u 0 /ε, {x = 0, y y 0 < ε/} u (x,y,t) = 0, otherwise on u (x,y, 0) = f (x,y) (63) where the hot spot is confined to the left side: 0 y 0 ε/ y y 0 + ε/ Equilibrium solution and Laplace s Eq. on a rectangle Ref: Guenther & Lee 8.1, 8., Myint-U & ebnath 6.6, 8.7 As in the 1 case, we first find the equilibrium solution u E (x,y), which satisfies the PE and the BCs, u E = u E (x,y) = 0, (x,y) u 0 /ε {x = 0, y y 0 < ε/} 0 otherwise on The PE for u E is called Laplace s equation. Laplace s equation is an example of an elliptic PE. The wave equation is an example of a hyperbolic PE. The heat equation is a parabolic PE. These are the three types of second order (i.e. involving double derivatives) PEs: elliptic, hyperbolic and parabolic. We proceed via separation of variables: u E (x,y) = X (x)y (y), so that the PE becomes X Y = = λ X Y where λ is constant since the l.h.s. depends only on x and the middle only on y. The BCs are and Y (0) = Y (1) = 0, X (1) = 0 u 0 /ε y y 0 < ε/ X (0)Y (y) = 0 otherwise We first solve for Y (y), since we have easy BCs: Y + λy = 0; Y (0) = Y (1) = 0 The non-trivial solutions, as we have found before, are Y n = sin (nπy) with λ n = n π, for each n = 1,, 3,... Now we consider X (x): X n π X = 0 17

18 and hence nπx nπx X (x) = c 1 e + c e An equivalent and more convenient way to write this is X (x) = c 3 sinh nπ (1 x) + c 4 cosh nπ (1 x) Imposing the BC at x = 1 gives X (1) = c 4 = 0 and hence X (x) = c 3 sinh nπ (1 x) Thus the equilibrium solution to this point is u E (x,y) = An sinh (nπ (1 x)) sin (nπy) n=1 You can check that this satisfies the BCs on x = 1 and y = 0, 1. Also, from the BC on x = 0, we have u E (0,y) = An sinh (nπ) sin (nπy) = n=1 Multiplying both sides by sin (mπy) an integrating in y gives u 0 /ε y y 0 < ε/ 0 otherwise 1 y0 +ε/ u 0 An sinh (nπ) sin (nπy) sin (mπy)dy = sin (mπy)dy ε n=1 0 y 0 ε/ From the orthogonality of sin s, we have Thus, 1 y0 +ε/ u 0 A m sinh (mπ) = sin (mπy)dy y 0 ε/ ε y0 +ε/ u 0 A m = sin (mπy)dy ε sinh (mπ) y 0 ε/ u 0 cos (mπy) y 0 +ε/ = ε sinh (mπ) mπ = = y 0 ε/ u 0 εmπ sinh (mπ) (cos (mπ (y 0 ε/)) cos (mπ (y 0 + ε/))) 4u 0 sin (mπy 0 ) sin mπε εmπ sinh (mπ) 18

19 Thus 4u 0 u E (x,y) = επ n=1 sin (nπy 0 ) sin nπε sinh (nπ (1 x)) sin (nπy) n sinh (nπ) To solve the transient problem, we proceed as in 1- by defining the function so that v (x,y,t) satisfies v (x,y,t) = u (x,y,t) u E (x,y) v t = v v = 0 on v (x,y, 0) = f (x,y) u E (x,y) 9. First term approximation To approximate the equilibrium solution u E (x,y), note that For sufficiently large n, we have sinh nπ (1 x) e nπ(1 x) e nπ(1 x) = sinh nπ e nπ e nπ sinh nπ (1 x) e nπ(1 x) nπx = e sinh nπ e nπ Thus the terms decrease in magnitude (x > 0) and hence u E (x,y) can be approximated the first term in the series, 4u 0 sin (πy 0 ) sin πε u E (x,y) sinh (π (1 x)) sin (πy) επ sinh (π) A plot of sinh (π (1 x)) sin (πy) is given below. The temperature in the center of the square is approximately 1 1 4u0 sin (πy 0 ) sin πε π π u E, sinh sin επ sinh (π) 9.3 Easy way to find steady-state temperature at center For y 0 = 1/ and ε = 1, we have 1 1 4u 0 sinh π u 0 u E,. π sinh (π) 4 19

20 z y x Figure 1: Plot of sinh (π(1 x)) sin(πy). It turns out there is a much easier way to derive this last result. Consider a plate with BCs u = u 0 on one side, and u = 0 on the other 3 sides. Let α = u E 1, 1. Rotating the plate by 90 o will not alter u E 1, 1, since this is the center of the plate. Let u Esum be the sums of the solutions corresponding to the BC u = u 0 on each of the four different sides. Then by linearity, u Esum = u 0 on all sides and hence u Esum = u 0 across the plate. Thus 1 1, = 4α u 0 = u Esum Hence α = u 1 E, 1 = u 0 / Placement of hot spot for hottest steady-state center Note that 1 1 u E, = sin (πy 0 ) sin πε 4u 0 π sinh επ sinh (π) sinh π sin (πy0 ) sin πε 4u 0 π sinh (π) ε u 0 sin πε sin (πy 0 ) 4 π u 0 sin (πy 0 ) 4 π u 0 = sin (πy 0 ) π 0 πε

21 for small ε. Thus the steady-state center temperature is hottest when the hot spot is placed in the center of the side, i.e. y 0 = 1/. 9.5 Solution to inhomogeneous heat problem on square We now use the standard trick and solve the inhomogeneous Heat Problem (63), u t = u, (x,y) u 0 /ε, {x = 0, y y 0 < ε/} u (x,y,t) = 0, otherwise on u (x,y, 0) = f (x,y) (64) using the equilibrium solution u E. We define the transient part of the solution as v (x,y,t) = u (x,y,t) u E (x,y) where u (x,y,t) is the solution to (64). The problem for v (x,y,t) is therefore v t = v, (x,y) v (x,y,t) = 0, (x,y) v (x,y, 0) = f (x,y) u E (x,y) This is the Heat Problem with homogeneous PE and BCs. We found the solution to this problem above in (43), (provided we use x 0 = 1 = y 0 in (43); don t mix up the side length y 0 in (43) with the location y 0 of the hot spot in the homogeneous problem): v (x,y,t) = m=1 n=1 A mn sin (mπx) sin (nπy)e π (m +n )t where 1 1 A mn = 4 (f (x,y) u E (x,y)) sin (mπx) sin (nπy)dydx 0 0 Thus we have found the full solution u (x,y,t). 10 Heat problem on a circle with inhomogeneous BC Consider the heat problem u t = u, (x,y) u (x,y,t) = g (x,y), (x,y) (65) u (x,y, 0) = f (x,y) 1

22 where = {(x,y) : x + y 1} is a circle (disc) of radius Equilibrium solution and Laplace s Eq. on a circle Ref: Guenther & Lee 8.1, 8., Myint-U & ebnath 8.4, 8.5 To solve the problem, we must first introduce the steady-state u = u E (x,y) which satisfies the PE and BCs, u E = 0, (x,y), u E (x,y) = g (x,y), As before, switch to polar coordinates via (x,y). x = r cos θ, y = r sin θ, w E (r,θ) = u E (x,y) for 0 r 1, π θ < π. The problem for u E becomes 1 we 1 w E r + = 0, r r r r θ 0 r 1, π θ < π (66) w (r, π) = w (r,π), w θ (r, π) = w θ (r,π), (67) w (0,θ) < (68) w (1,θ) = ĝ (θ), π θ < π (69) where ĝ (θ) = g (x,y) for (x,y) and θ = arctan (y/x). We separate variables w (r,θ) = R (r) H (θ) and the PE becomes r d dr d H 1 r = R (r) dr dr dθ H (θ) Since the l.h.s. depends only on r and the r.h.s. on θ, both must be equal to a constant µ, r d dr d H 1 r = R (r) dr dr dθ H (θ) = µ The problem for H (θ) is, as before, d H dh dh + µh (θ) = 0; H ( π) = H (π), ( π) = dθ dθ dθ (π),

23 with eigen-solutions H m (θ) = a m cos (mθ) + b m sin (mθ), m = 0, 1,,... The problem for R (r) is d dr R d R dr 0 = r r m = r + r m dr dr dr dr Try R (r) = r α to obtain the auxiliary equation α (α 1) + α m = 0 whose solutions are α = ±m. Thus for each m, the solution is R m (r) = c 1 r m +c r m. For m > 0, r m blows up as r 0. Our boundedness criterion (68) implies c = 0. Hence m R m (r) = c m r where c m are constants to be found by imposing the BC (69). The separable solutions satisfying the PE (66) and conditions (67) to (68), are m w m (r,θ) = r (A m cos (mθ) + B m sin (mθ)), m = 0, 1,,... The full solution is the infinite sum of these over m, m u E (x,y) = w E (r,θ) = r (A m cos (mθ) + B m sin (mθ)) (70) We still need to find the A m, B m. m=0 Imposing the BC (69) gives (Am cos (mθ) + B m sin (mθ)) = ĝ (θ) (71) R m=0 The orthogonality relations are π cos (mθ) sin (nθ) dθ = 0 π π cos (mθ) cos (nθ) π, m = n dθ = π sin (mθ) sin (nθ) 0, m = n, (m > 0). Multiplying (71) by sin nθ or cosnθ and applying these orthogonality relations gives π 1 A 0 = ĝ (θ)dθ π π π 1 A m = ĝ (θ) cos (mθ) dθ (7) π π π 1 B m = ĝ (θ) sin (mθ)dθ π π 3

24 y u=u 0 /(ST 0 1 T 0 u=0 x Figure : Setup for hot spot problem on circle Hot spot on boundary Suppose u 0 θ ĝ (θ) = 0 π θ θ +π 0 0 otherwise which models a hot spot on the boundary. The Fourier coefficients in (7) are thus Thus the steady-state solution is u 0 u 0 A 0 =, A m = sin (mθ 0 ) π mπ (θ 0 + π) u0 B m = mπ (θ + π) (cos (mθ 0) ( 1) m ) 0 u 0 + r m u0 sin (mθ 0 ) u 0 (cos (mθ 0 ) ( 1) m ) u E = cos (mθ) sin (mθ) π mπ (θ 0 + π) mπ (θ 0 + π) m= Interpretation The convergence of the infinite series is rapid if r 1. If r 1, many terms are required for accuracy. The center temperature (r = 0) at equilibrium (steady-state) is π u 0 1 u E (0, 0) = w E (0,θ) = = ĝ (θ)dθ π π π 4

25 i.e., the mean temperature of the circumference. This is a special case of the Mean Value Property of solutions to Laplace s Equation u = 0. We now consider plots of u E (x,y) for some interesting cases. We draw the level curves (isotherms) u E = const as solid lines. Recall from vector calculus that the gradient of u E, denoted by u E, is perpendicular to the level curves. Recall also from the physics that the flux of heat is proportional to u E. Thus heat flows along the lines parallel to u E. Note that the heat flows even though the temperature is in steady-state. It is just that the temperature itself at any given point does not change. We call these lines the heat flow lines or the orthogonal trajectories, and draw these as dashed lines in the figure below. Note that lines of symmetry correspond to (heat) flow lines. To see this, let n l be the normal to a line of symmetry. Then the flux at a point on the line is u n l. Rotate the image about the line of symmetry. The arrow for the normal to the line of symmetry is now pointing in the opposite direction, i.e. n l, and the flux is u n l. But since the solution is the same, the flux across the line must still be u n l. Thus u n l = u n l which implies u n l = 0. Thus there is no flux across lines of symmetry. Equivalently, u is perpendicular to the normal to the lines of symmetry, and hence u is parallel to the lines of symmetry. Thus the lines of symmetry are flow lines. Identifying the lines of symmetry help draw the level curves, which are perpendicular to the flow lines. Also, lines of symmetry can be thought of as an insulating boundary, since u n l = 0. (i) θ 0 = 0. Then u E = u 0 u 0 n 1 sin ((n 1) θ) r π π n 1 m=1 Use the BCs for the boundary. Note that the solution is symmetric with respect to the y-axis (i.e. even in x). The solution is discontinuous at {y = 0,x = ±1}, or {r = 1,θ = 0,π}. See plot. (ii) π < θ 0 < π/. The sum for u E is messy, so we use intuition. We start with the boundary conditions and use continuity in the interior of the plate to obtain a qualitative idea of the level curves and heat flow lines. See plot. (iii) θ 0 π + (a heat spot). Again, use intuition to obtain a qualitative sketch of the level curves and heat flow lines. Note that the temperature at the hot point is infinite. See plot. The heat flux lines must go from a point of hot temperature to a point of low temperature. Since the temperature along boundary is zero except at 5

26 u=0 u=0 u=0 u 0 u=u 0 /S u S T 0 u of T 0 T 0 S T 0 o S Figure 3: Plots of steady-state temperature due to a hot segment on boundary. θ = π, then all the heat flux lines must leave the hot spot and extend to various points on the boundary. 10. Solution to inhomogeneous heat problem on circle We now use the standard trick and solve the inhomogeneous Heat Problem (65), u t = u, (x,y) u (x,y,t) = g (x,y), (x,y) (73) u (x,y, 0) = f (x,y) using the equilibrium solution u E. We define the transient part of the solution as v (x,y,t) = u (x,y,t) u E (x,y) where u (x,y,t) is the solution to (73). The problem for v (x,y,t) is therefore v t = v, (x,y) v (x,y,t) = 0, (x,y) v (x,y, 0) = f (x,y) u E (x,y) This is the Heat Problem with homogeneous PE and BCs. We found the solution to this problem above in (61), v (x,y,t) = Jm (rj m,n ) (α mn cos (mθ) + β mn sin (mθ))e m=1 n=1 λ mnt where α mn and β mn are found from orthogonality relations and f (x,y) u E (x,y). Thus we have found the full solution u (x,y,t). 6

27 11 Mean Value Property for Laplace s Eq. Ref: Guenther & Lee 8.4, Myint-U & ebnath 8.4 Theorem [Mean Value Property] Suppose v (x,y) satisfies Laplace s equation in a domain, v = 0, (x,y). (74) Then at any point (x 0,y 0 ) in, v equals the mean value of the temperature around any circle centered at (x 0,y 0 ) and contained in, π 1 v (x 0,y 0 ) = v (x 0 + R cos θ,y 0 + R sin θ) dθ. (75) π π Note that the curve {(x 0 + R cos θ,y 0 + R sin θ) : π θ < π} traces the circle of radius R centered at (x 0,y 0 ). Proof: To prove the Mean Value Property, we first consider Laplace s equation (74) on the unit circle centered at the origin (x,y) = (0, 0). We already solved this problem, above, when we solved for the steady-state temperature u E that took the value ĝ (θ) on the boundary. The solution is Eqs. (70) and (7). Setting r = 0 in (70) gives the center value π 1 u E (0, 0) = A 0 = ĝ (θ) dθ (76) π π On the boundary of the circle (of radius 1), (x,y) = (cosθ, sin θ) and the BC implies that u E = ĝ (θ) on that boundary. Thus, ĝ (θ) = u E (cosθ, sin θ) and (76) becomes We now consider the region π 1 u E (0, 0) = A 0 = u E (cos θ, sin θ)dθ. (77) π π B (x0,y 0 ) (R) = (x,y) : (x x0 ) + (y y 0 ) R, which is a disc of radius R centered at (x,y) = (x 0,y 0 ). Since B (x0,y 0 ) (R), Laplace s equation (74) holds on this disc. Thus We make the change of variable v = 0, (x,y) B (x0,y 0 ) (R). (78) x x 0 y y 0 xˆ =, ŷ =, u E (ˆx, ŷ) = v (x,y) (79) R R to map the circle B (x0,y 0 ) (R) into the unit disc {(ˆx, ŷ) : ˆx + ŷ 1}. Laplace s equation (78) becomes ˆ u E = 0, (ˆx, ŷ) (ˆx, ŷ) : ˆx + ŷ 1 7

28 where ˆ = ( / xˆ, / ŷ ). The solution is given by Eqs. (70) and (7), and we found the center value above in Eq. (77). Reversing the change of variable (79) in Eq. (77) gives v (x 0,y 0 ) = 1 π π π v (x 0 + R cosθ,y 0 + R sin θ)dθ as required. For the heat equation, the Mean Value Property implies the equilibrium temperature at any point (x 0,y 0 ) in equals the mean value of the temperature around any disc centered at (x 0,y 0 ) and contained in. 1 Maximum Principle for Laplace s Eq. Ref: Guenther & Lee 8.4, Myint-U & ebnath 8. Theorem [Maximum Principle] Suppose v (x,y) satisfies Laplace s equation in a domain, v = 0, (x,y). Then the function v takes its maximum and minimum on the boundary of,. Proof: Let (x 0,y 0 ) be an interior point in where v takes it s maximum. In particular, (x,y) is not on the boundary of. The Mean Value Property implies that for any disc B (x0,y 0 )(R) of radius R > 0 centered at (x 0,y 0 ), π 1 v (x 0,y 0 ) = v (x 0 + R cosθ,y 0 + R sin θ)dθ π π = Average of v on boundary of circle But the average is always between the minimum and maximum. Thus v (x 0,y 0 ) must be between the minimum and maximum value of v (x,y) on the boundary of the disc. But since v (x 0,y 0 ) is the maximum of v on, then the entire boundary of the circle must have value v (x 0,y 0 ). Since R > 0 is arbitrary, this holds for all values in the disc B (x0,y 0 )(R). Keep increasing R until the disc hits the boundary of. Then v (x 0,y 0 ) is a value of v along the boundary of. A similar argument holds for the minimum of v. Corollary If v = 0 everywhere on the boundary, then v must be zero at everywhere in. For the homogeneous heat equation (i.e. no sinks/sources), this implies the equilibrium temperature attains its maximum/minimum on the boundary. 8

29 If the boundary is completely insulated, v = 0 on, then the equilibrium temperature is constant. To see this, we apply result (6) to the case where v is the solution to Laplace s equation with v = 0 on, 0 = v v nˆds = v v + v dv = v dv Thus, by continuity, v = 0 and v = 0 everywhere in and hence v = const in. 13 Eigenvalues on different domains In this section we revisit the Sturm-Liouville problem v + λv = 0, v = 0, x x and consider the effect of the shape of the domain on the eigenvalues. efinition Rayleigh Quotient R (v) = v vdv v dv (80) Theorem Given a domain R 3 and any function v that is piecewise smooth on, non-zero at some points on the interior of, and zero on all of, then the smallest eigenvalue of the Laplacian on satisfies λ R (v) and R (h) = λ if and only if h (x) is an eigen-solution of the Sturm Liouville problem on. Sketch Proof: We use result (6) derived for any smooth function v defined on a volume V with closed smooth surface S. v v nˆds = v v + v v dv In the statement of the theorem, we assumed that v = 0 on, and hence v vdv = v vdv (81) Let {φ n } be an orthonormal basis of eigenfunctions on, i.e. all the functions φ n which satisfy φ n + λ n φ n = 0, x φ n = 0, x 9

30 and φ n φ m dv = We can expand v in the eigenfunctions, 1, m = n 0, m = n v (x) = A n φ n (x) where the A n are constants. Assuming we can differentiate the sum termwise, we have n=1 v = An φ n = λn A n φ n (8) n=1 n=1 The orthonormality property (i.e., the orthogonality property with φ n dv = 1) implies v dv = A A n A m φ n φ m dv = (83) and, from (8), n=1 m=1 n=1 v vdv = λ n A n A m φ n φ m dv = λ n A m=1 n=1 n=1 n n (84) Substituting (84) into (81) gives v vdv = v vdv = λn A n n=1 (85) Substituting (83) and (85) into (80) gives v vdv R (v) = v dv n=1 λ na n = A n=1 n We assume the eigenfunctions are arranged in increasing order. In particular, λ n λ 1. Thus λ 1 A n=1 n n=1 A n R (v) = λ 1 = λ 1. A n=1 n n=1 A n If v is an eigenfunction with eigenvalue λ 1, then λ 1 A 1 R (v) = = λ 1 A 1 If v is not an eigenfunction corresponding to λ 1, then there exists an n > 1 such that λ n > λ 1 and A n = 0, so that R (v) > λ 1. 30

31 Theorem If two domains and in R satisfy i.e., but =, then the smallest eigenvalues of the Sturm-Liouville problems on and, λ 1 and λˆ1, respectively, satisfy λ 1 < λ 1 In other words, the domain that contains the sub-domain is associated with a smaller eigenvalue. Proof: Note that the Sturm-Liouville problems are v + λv = 0, (x,y) v = 0, (x,y) v + λ v = 0, (x,y) v = 0, (x,y) Let v 1 be the eigenfunction corresponding to λ 1 on. Then, as we have proven before, λ 1 = R (v 1 ), (86) where R (v) is the Rayleigh Quotient. Extend the function v 1 continuously from to to obtain a function ˆv 1 on which satisfies v 1 (x,y), (x,y) vˆ1 = 0 (x,y), (x,y) / The extension is continuous, since v 1 is zero on the boundary of. Applying the previous theorem to the region and function ˆv 1 (which satisfies all the requirements of the theorem) gives λˆ1 R (ˆv 1 ). Equality happens only if ˆv 1 is the eigenfunction corresponding to λˆ1. Useful fact [stated without proof]: the eigenfunction(s) corresponding to the smallest eigenvalue λˆ1 on are nonzero in the interior of. From the useful fact, ˆv 1 cannot be an eigenfunction corresponding to λˆ1 on, since it is zero in the interior of (outside ). Thus, as the previous theorem states, λˆ1 < R (ˆv 1 ). (87) 31

32 Since ˆv 1 = 0 outside, the integrals over in the Rayleigh quotient reduce to integrals over, where ˆv 1 = v 1, and hence Combining (86), (87), and (88) gives the result, R (ˆv 1 ) = R (v 1 ). (88) λˆ1 < λ 1. Example: Consider two regions, 1 is a rectangle of length x 0, height y 0 and is a circle of radius R. Recall that the smallest eigenvalue on the rectangle 1 is 1 1 λ 11 = π + x 0 y 0 The smallest eigenvalue on the circle of radius 1 is λ 01 = J 0,1 where the first zero of the Bessel function J 0 (s) of the first kind is J 0,1 = Since λ multiplied r in the Bessel function, then for a circle of radius R, we d rescale by the change of λr λ variable ˆr = r/r, so that J m = J m R rˆ where ˆr goes from 0 to 1. Thus on the circle of radius R, the smallest eigenvalue is J0,1 λ 01 =, J 0,1 = R Suppose the rectangle is actually a square of side length R. Then π J0, λ 11 = =, λ 01 = = R R R R Thus, λ 11 < λ 01, which confirms the second theorem, since is contained inside the square. Now consider the function r v (r) = 1 R 1, i.e., the circle You can show that dv v v =, dr and R (v) = v vda = v da π R dv rdrdθ π 0 dr π R π 0 v rdrdθ 6 = R > λ 01 This confirms the first theorem, since v (r) is smooth on, v (R) = 0 (zero on the boundary of ), and v is nonzero in the interior. 3

33 13.1 Faber-Kahn inequality Thinking about the heat problem on a plate, what shape of plate will cool the slowest? It is a geometrical fact that of all shapes of equal area, the circle (disc) has the smallest circumference. Thus, on physical grounds, we expect the circle to cool the slowest. Faber and Kahn proved this in the 190s. Faber-Kahn inequality For all domains R of equal area, the disc has the smallest first eigenvalue λ 1. Example. Consider the circle of radius 1 and the square of side length π. Then both the square and circle have the same area. The first eigenvalues for the square and circle are, respectively, 1 1 λ 1SQ = π + = π = 6.8, λ 1CIRC = (J 0,1 ) = π π and hence λ 1SQ > λ 1CIRC, as the Faber-Kahn inequality states. 14 Nodal lines Consider the Sturm-Liouville problem v + λv = 0, x (89) v = 0, x Nodal lines are the curves where the eigenfunctions of the Sturm-Liouville problem are zero. For the solution to the vibrating membrane problem, the normal modes u nm (x,y,t) are zero on the nodal lines, for all time. These are like nodes on the 1 string. Here we consider the nodal lines for the square and the disc (circle) Nodal lines for the square For the square, the eigenfunctions and eigenvalues are a special case of those we found for the rectangle, with side length x 0 = y 0 = a, mπx mπy π v mn (x,y) = sin sin, λ mn = m + n, n,m = 1,, 3,... a a a The nodal lines are the lines on which v mn (x,y) = 0, and are ka la x =, y =, 1 k m 1, 1 l n 1 m n for m,n. Note that v 11 (x,y) has no nodal lines on the interior - it is only zero on boundary. Since λ mn = λ nm, then the function f nm = Av mn +Bv nm is also an 33

34 eigenfunction with eigenvalue λ mn, for any constants A, B. The nodal lines for f nm can be quite interesting. Examples: we draw the nodal lines on the interior and also the lines around the boundary, where v nm = 0. (i) m = 1, n = 1. πx πy v 11 = sin sin a a This is positive on the interior and zero on the boundary. Thus the nodal lines are simply the square boundary. (ii) m = 1, n =. πx πy v 1 = sin sin a a The nodal lines are the boundary and the horizontal line y = a/. (iii) m = 1, n = 3 πx 3πy v 13 = sin sin a a The nodal lines are the boundary and the horizontal lines y = a/3, a/3. (iv) m = 3, n = 1 3πx πy v 31 = sin sin a a The nodal lines are the boundary and the vertical lines x = a/3, a/3. (v) Consider v 13 v 31. Since λ 31 = λ 13 = 10π /a, this is a solution to SL problem (89) on with λ = λ 13. To find the nodal lines, we use the identity sin 3θ = (sin θ) 3 4 sin θ to write πx πy πy v sin 3 4 sin 13 = sin a a a πx πy πx v sin 3 4 sin 31 = sin a a a πx πy πx πy v sin sin sin 13 v 31 = 4 sin a a a a The nodal lines are the boundary of the square,, and lines such that πx πy πx πy πx πy 0 = sin sin = sin sin sin + sin a a a a a a i.e., Thus πx a πx πy sin = sin ± a a πy = ± + kπ, any integer k. a 34

35 Hence the nodal lines are y = ±x + ka for all integers k. We re only concerned with the nodal lines that intersect the interior of the square plate: y = x, y = x + a Thus the nodal lines of v 13 v 31 are the sides and diagonals of the square. Let T be the isosceles right triangle whose hypotenuse lies on the bottom horizontal side of the square. The function v 13 v 31 is zero on the boundary T, positive on the interior of T, and is thus the eigenfunction corresponding to the first (smallest) eigenvalue λ = λ 13 of the Sturm-Liouville problem (89) on the triangle T. In this case, we found the eigenfunction without using separation of variables, which would have been complicated on the triangle. (vi) With v 13, v 31 given above, adding gives πx πy 3 πx πy sin v + sin 13 + v 31 = 4 sin sin a a a a The nodal lines for v 13 + v 31 are thus the square boundary and the closed nodal line defined by πx πy 3 sin + sin =. a a Let c be the region contained within this closed nodal line. The function (v 13 + v 31 ) is zero on the boundary c, positive on the interior of c, and thus (v 13 + v 31 ) is the eigenfunction corresponding to the first (smallest) eigenvalue λ 13 of the Sturm- Liouville problem (89) on c. (vii) Find the first eigenvalue on the right triangle T = 0 y 3x, 0 x 1. Note that separation of variables is ugly, because you d have to impose the BC 3x X (x) Y = 0 We proceed by placing the triangle inside a rectangle of horizontal and vertical side lengths 1 and 3, respectively. The sides of the rectangle coincide with the perpendicular sides of the triangle. Thus, all eigenfunctions v mn for the rectangle are already zero on two sides of the triangle. However, any particular eigenfunction v mn will not be zero on the triangle s hypotenuse, since all the nodal lines of v mn are horizontal or vertical. Thus we need to add multiple eigenfunctions. However, to satisfy the Sturm-Liouville problem, all the eigenfunctions must be associated with the same 35

36 eigenvalue. For the rectangle with side lengths 3 and 1, the eigenvalues are given by m n π λ mn = π + = m + n We create a table of 3m + n and look for repeated values: n, m The smallest repeated value of 3m + n is 8: λ 31 = λ 4 = λ 15 = 8π /3 The linear combination of the corresponding eigenfunctions is itself an eigenfunction of the SL problem (89) on the rectangle, v = Av 31 + Bv 4 + Cv 15. We wish v to also be zero along the hypotenuse of T, and seek A, B, C which satisfy Av 31 + Bv 4 + Cv 15 = 0 on y = 3x. If we can find such constants A, B, C then we have found an eigenfunction of the SL problem (89) on the triangle T. If this eigenfunction is positive on the interior of T then we have also found the smallest eigenvalue (in this case λ 31 ). If we cannot find such A, B, C, then we seek the next largest repeated value of 3m + n and continue as above. If we can find (A, B, C) such that v = 0 on y = 3x, but v is zero on the interior of T, then λ 31 is not the smallest eigenvalue and we must pursue another method of finding the smallest eigenvalue. The Rayleigh quotient can help us identify an upper bound. 14. Nodal lines for the disc (circle) For the disc of radius 1, we found the eigenfunctions and eigenvalues to be v mns = J m (rj m,n ) sin mθ, v mnc = J m (rj m,n ) cos mθ 36

37 with λ mn = π jm,n, n,m = 1,, 3,... The nodal lines are the lines on which v mnc = 0 or v mns = 0. Examples. (i) m = 0, n = 1. v 01 = J 0 (rj 0,1 ) The nodal lines are the boundary of the disc (circle of radius 1). (ii) m = 0, n =. v 0 = J 0 (rj 0, ) The nodal lines are two concentric circles, one of radius r = 1, the other or radius r = J 0,1 /J 0, < 1. (iii) m = 1, n = 1 and sine. v 11S = J 0 (rj 1,1 ) sin θ The nodal lines are the boundary (circle of radius 1) and the line θ = π, 0, π (horizontal diameter). 15 Steady-state temperature in a 3 cylinder Suppose a 3 cylinder of radius a and height L has temperature u (r,θ,z,t). We assume the axis of the cylinder is on the z-axis and (r,θ,z) are cylindrical coordinates. Initially, the temperature is u (r,θ,z, 0). The ends are kept at a temperature of u = 0 and sides kept at u (a,θ,z,t) = g (θ,z). The steady-state temperature u E (r,θ,z) in the 3 cylinder is given by u E = 0, π θ < π, 0 r a, 0 z L, (90) u E (r,θ, 0) = u E (r,θ,l) = 0, π θ < π, 0 r a, (91) u E (a,θ,z) = g (θ,z), π θ < π, 0 z L. In cylindrical coordinates (r,θ,z), the Laplacian operator becomes 1 v 1 v v v = r + r r r r θ + z We separate variables as v (r,θ,z) = R (r)h (θ)z (z) 37

38 so that (90) becomes 1 d dr (r) 1 d H(θ) 1 d Z (z) r + + = 0. rr (r) dr dr r H(θ) dθ Z (z) dz Rearranging gives 1 d dr (r) 1 d H(θ) 1 d Z (z) r + = = λ (9) rr (r) dr dr r H(θ) dθ Z (z) dz where λ is constant since the l.h.s. depends only on r, θ while the middle depends only on z. The function Z (z) satisfies The BCs at z = 0, L imply d Z dz + λz = 0 0 = u (r,θ, 0) = R (r) H (θ) Z (0) 0 = u (r,θ,l) = R (r)h (θ)z (L) To obtain non-trivial solutions, we must have Z (0) = 0 = Z (L). As we ve shown many times before, the solution for Z (z) is, up to a multiplicative constant, nπz nπ Z n (z) = sin, λ n =, n = 1,, 3,... L L Multiplying Eq. (9) by r gives r d dr (r) 1 d H(θ) r λ n r = = µ (93) R (r) dr dr H(θ) dθ where µ is constant since the l.h.s. depends only on r and the middle only on θ. We assume g (θ,z) is smooth and π-periodic in θ. Hence H ( π) = H (π), H ( π) = H (π) We solved this problem above and found that µ = m for m = 0, 1,,..., H 0 (θ) = const and H m (θ) = A m cos mθ + B m sin mθ, m = 1,, 3,... Multiplying (93) by R (r) gives d r r dr (r) λ n r + m R (r) = 0 dr dr 38

39 This is the Modified Bessel Equation with linearly independent solutions I m r λn and K m r λn called the modified Bessel functions of order m of the first and second kinds, respectively. Thus R mn (r) = c 1m I m r λ n + c m K m r λ n Since the I m s are regular (bounded) at r = 0, while the K m s are singular (blow up), and since R mn (r) must be bounded, we must have c m = 0, or R mn (r) = c 1m I m r nπr λ n = c 1m I m Thus the general solution is, by combining constants, nπr nπz u E (r,θ,z) = I m sin [A mn cos (mθ) + B m sin (mθ)] L L m=0 n=1 In theory, we can now impose the condition u (a,θ,z) = g (θ,z) and find A mn, B mn using orthogonality of sin, cos. L 39