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 [Nov, 005] Ref: Haberman 1.5, 7.1 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 = cρu dv 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 n, ˆ where n ˆ is the outward unit normal at the boundary S. Hats on vectors denote a unit vector, ˆ n = 1 (length 1. If the heat flux vector φ is directed inward, then φ n ˆ < 0 and the outward flow of heat is negative. To compute the total heat energy flowing across the boundaries, we sum φ n ˆ over the entire closed surface S, denoted by a double integral ds. S Therefore, the conservation of energy principle becomes d cρu dv = φ n ˆ ds + Q dv (1 dt V S V V 1

2 1.1 ivergence Theorem (a.k.a. Gauss s Theorem Ref: Haberman 1.5 For any volume V with closed smooth surface S, AdV = A n ˆ ds V where A is any function that is smooth (i.e. continuously differentiable for x V. Note that ( =,, = ˆ e x + ˆ e y + ˆe z x y z x y z where ˆe x, ˆy, e ˆ e 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 The Laplacian of a scalar function F is F x F y F z F = + +. x y z S F F F F = + +. x y z Applying the ivergence theorem to (1 gives d cρu dv = 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 Q dv Since V is an arbitrary subregion of R 3 and the integrand is assumed continuous, the integrand must be everywhere zero, u cρ + φ Q = 0 ( t 1. Fourier s Law of Heat Conduction The 3 generalization of Fourier s Law of Heat Conduction is (see Appendix to 1.5, pp 3, Haberman φ = K 0 u (3 where K 0 is called the thermal diffusivity. Substituting (3 into ( gives u t Q = κ u + (4 cρ

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 the dimensional Heat Equation (4 becomes (dropping tildes where q = l Q/ (κcρ = l Q/K 0. u = u + q, (5 t and 3 Wave equation Ref: Haberman 4.5, 4.6, 7.1 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 [Nov 4, 005] Ref: Haberman, 7., 7.3 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 The 3 wave problem is u t = u, x, t > 0, u (x, t = 0, x, (7 u (x, 0 = f (x, x. 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. 3

4 We separate variables as u (x, t = X (xt (t (9 The 3 Heat Equation implies T X = = λ = const (10 T 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 X = = λ = const (11 T X X (x = 0, 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 and the corresponding solution to the PE and BCs is T = λ (13 T T n (t = c n e λnt (14 u n (x, t = X n (xt n (t = X n (xc n e λnt. For the 3 Wave Problem, the problem for T (t is, from (11, with solution T = λ (15 T ( ( T n (t = α n cos λn t + β n sin λn t (16 and the corresponding normal mode is u n (x, t = X n (xt n (t. 4

5 5 Uniqueness of the 3 Heat Problem 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 = v v nds ˆ v dv (17 dt 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, 0dV = 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 n ˆ = 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: 7.4 Haberman 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: Haberman 7.4, parts of 7.5 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, (GF = G F + F G (0 Using the divergence theorem, (GF nds ˆ = (GF dv (1 S Substituting (0 into (1 gives (GF nds ˆ = G FdV + F GdV ( S V V Result ( applied to G = v 1 and F = v gives ( (v 1 v nds ˆ = v1 v + v v 1 dv (3 S Result ( applied to G = v and F = v 1 gives ( (v v 1 nds ˆ = v v 1 + v 1 v dv (4 S Subtracting (3 and (4 gives Green s Formula (also known as Green s second identity: ( (v 1 v v v 1 nds ˆ = v1 v v v 1 dv (5 S 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 V V V V

7 6. Positive, real eigenvalues (for Type I BCs Choosing G = v and F = v, for some function v, in ( gives ( v v nds ˆ = v v + v v dv S = V ( v v + v dv. (6 V 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 nds ˆ = X XdV + X dv (7 Since X (x = 0 for x, Also, from the PE in (19, X X nds ˆ = 0 (8 X XdV = λ X dv (9 Substituting (8 and (9 into (7 gives 0 = λ X dv + 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, λ = X dv X dv Since X is real, the the eigenvalue λ is also real. 0 (31 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: 7.4 Haberman Suppose v 1, v are two eigen-functions with eigenvalues λ 1, λ of the 3 Sturm- Liouville problem v + λv = 0, x v = 0, 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 eigen-functions v 1, v are orthogonal. 7 Heat and Wave problems on a rectangle, homogeneous BCs Ref: Haberman 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 + + λv = 0, x y (x, y, (33 v (0, y = v (x 0, y = 0, 0 y y 0, (34 v (x, 0 = v (x, y 0 = 0, 0 x x 0. (35 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 8

9 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 (0Y (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 y 0 0 See plots in Figure 7.3. (p 84 Haberman of v mn (x, y for various m, n. 9

10 7. Solution to heat equation on rectangle [Nov 7, 005] The heat problem on the rectangle is the special case of (7, u u u t = + (x, y, t > 0, x 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 ( ( «mπx π m nπy + 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, u (x, y, t = umn (x, y, t = m=1 n=1 m=1 n=1 ( ( mπx nπy A mn sin sin e λmnt (43 x 0 y 0 Setting t = 0 and imposing the initial condition u (x, y, 0 = f (x, y gives ( ( mπx nπy f (x, y = u (x, y, 0 = Amnv 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 sin are the eigen-functions of the Sturm Lix 0 y 0 ouville problem on a rectangle, (33 (35. Multiplying both sides by v mˆ ˆ n (x, y ( m, ˆ ˆn = 1,, 3,... and integrating over the rectangle gives f (x, y v ˆ n (x, yda = Amn v mn (x, y v ˆ n (x, y da (44 mˆ m=1 n=1 where da = dxdy. Note that x 0 ( ( mπx ˆmπx v mn (x, y v mˆ ˆ n (x, yda = sin sin dx 0 x ( 0 x ( 0 y0 nπy ˆnπy sin sin dy 0 y { 0 y 0 x 0 y 0 /4, m = m ˆ and n = n ˆ = 0, otherwise 10 mˆ

11 Thus (44 becomes A mˆ ˆ n mˆ 4 f (x, y v ˆ n (x, y da = x 0 y 0 Since m, ˆ n ˆ are dummy variables, we replace m ˆ by m and n ˆ by n, and rearrange to obtain 4 A mn = f (x, y v mˆ ˆ n (x, y da x 0 y 0 x 0 4 y0 ( ( mπx nπy = f (x, y sin sin dydx (45 x 0 y x 0 y Solution to wave equation on rectangle, homogeneous BCs Application: waves on a rectangular membrane ( 7.3 Haberman The solution to the wave equation on the rectangle follows similarly. The general 3 wave problem (8 becomes u u u tt = + (x, y, t > 0, x 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 + + β nm sin πt + x 0 y 0 x 0 y 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 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 x 0 y m=1 n=1 0 g (x, y = u t (x, y, 0 = λ mn β mn v mn (x, y = m=1 n=1 m=1 n=1 ( ( mπx nπy λ mn β mn sin sin x 0 y 0 ( ( mπx nπy where v mn (x, y = sin sin are the eigen-functions of the Sturm Lix 0 y 0 ouville problem on a rectangle, (33 (35. As above, multiplying both sides by v mˆ ˆ n (x, y ( m, ˆ n ˆ = 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 [Nov 9, 005] Ref: Haberman 7.7, 7.8 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 1

13 for You can verify that 0 r 1, π θ < π. v v 1 ( w 1 w v = + = r + x y r r r r θ The PE becomes 1 ( w 1 w r + + λw = 0, r r 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, The BC (50 becomes d ( dr 1 d H 1 r r (5 dr dr R (r + λr = dθ H (θ = µ w (1, θ = R (1H (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 [RAW (x, y circle, (r, θ domain] Substituting (51 gives w (r, π = w (r, π, w θ (r, π = w θ (r, π. (54 dh dh H ( π = H (π, ( π = (π. (55 dθ dθ 13

14 The solution v (x, y is also bounded on the circle, which implies R (r must be bounded for 0 r 1. The problem for H (θ is d H dh dh dθ + µh (θ = 0; H ( π = H (π, ( π = (π. (56 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 Rearranging gives d ( dr 1 r r + λr = µ = m, m = 0, 1,, 3,... dr dr R (r dr m r d R m ( + r + λr m R m = 0; R m (1 = 0, R m (0 < (57 dr 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 s + s dr m ( + s m Rm = 0; R m λ = 0, ds ds ( R m (s = c m J m (s, R m (r = c m J m λr. Rm (0 < (58 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 bound edness criterion R m (0 < at s = 0 implies c m = 0. Thus The Bessel Function J m (s of the first kind of order m has power series ( 1 k s k+m J m (s = k! (k + m! k+m k=0 (59 14

15 [RAW J m (s]. 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 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 k+m s 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 (k + 1 k!(k+m! k+m Thus for k > N = L/, L (k+1+m (k+1!(k+1+m! ( (k+1+m L < L k+m (N + 1 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 such zero for the function J m (s. Note that The second BC requires < 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 = ( λ ( λ R m (1 = R m = J m = 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,... 15

16 with corresponding eigen-functions 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,... ( 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, y, t > 0, x 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 where v mn is given in (60. u m,n t mn (x, y, t = v mn (x, y e λmnt = v mn (x, y e J 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 = v mn (x, y e = λ mnt m=1 n=1 m=1 n=1 Jm (rj m,n (α mn cos (mθ + β mn sin (mθ e λmnt (61 m=1 n=1 where r = x + y and tan θ = 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. 16

17 9 The Heat Problem on a square with inhomogeneous BC [Nov 14, 005] We now consider the case of the heat problem on the unit square ( = {(x, y : 0 x, y 1}, 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 (6 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: Haberman.5.1 As in the 1 case, we first find the equilibrium solution u E (x, y, which satisfies the PE and the BCs, u E = 0, (x, y { u 0 /ε {x = 0, y y 0 < ε/} u E (x, y = 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 Y (0 = Y (1 = 0, X (1 = 0 and X (0Y (y = { u 0 /ε y y 0 < ε/ 0 otherwise 17

18 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 and hence nπx X (x = c 1 e nπx + 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 = A n 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 0 /ε y y 0 < ε/ u E (0, y = A n sinh (nπ sin (nπy = 0 otherwise n=1 Multiplying both sides by sin (mπy an integrating in y gives 1 y0 +ε/ u 0 An sinh (nπ sin (nπy sin (mπydy = sin (mπydy ε n=1 0 y 0 ε/ From the orthogonality of sin s, we have y0 +ε/ 1 u 0 A m sinh (mπ = sin (mπydy y 0 ε/ ε 18

19 Thus, Thus y0 +ε/ u 0 A m = sin (mπydy 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 ( 4u 0 sin (nπy 0 sin nπε u E (x, y = sinh (nπ (1 x sin (nπy n sinh (nπ επ n=1 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 sinh nπ (1 x sinh nπ For sufficiently large n, we have e nπ(1 x e nπ(1 x = enπ 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 4u 0 sin (πy 0 sin πε u E, sinh ( π ( π sin επ sinh (π 19

20 z y x Figure 1: Plot of sinh (π(1 x sin(πy. 9.3 Easy way to find steady-state temperature at center For y 0 = 1/ and ε = 1, we have u E ( 1, 1 4u 0 π sinh ( π sinh (π u 0 4. 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 1 E, 1. ( Rotating the plate by 90 o will not alter u 1 E, 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 u 0 = u Esum, 1 = 4α Hence α = u E ( 1, 1 = u0 /4. 0

21 9.4 Placement of hot spot for hottest steady-state center Note that ( ( 1 1 4u 0 sin (πy 0 sin πε ( π u E, sinh επ sinh (π 4u 0 sinh ( ( π sin (πy 0 sin πε = π sinh (π ε [ ( u 0 sin πε ] sin (πy 0 π πε 4 u 0 sin (πy 0 4 π u 0 = sin (πy 0 π 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 Ref: Haberman We now use the standard trick and solve the inhomogeneous Heat Problem (6, 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 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 (63. 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: π (m +n t v (x, y, t = A mn sin (mπx sin (nπye m=1 n=1 1

22 where 1 A mn = Thus we have found the full solution u (x, y, t. 1 (f (x, y u E (x, y sin (mπx sin (nπydydx 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 (64 u (x, y, 0 = f (x, y where = {(x, y : x + y 1} is a circle (disc of radius Equilbrium solution and Laplace s Eq. on a circle Ref: Haberman.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 ( w E 1 w E r + = 0, r r r r θ 0 r 1, π θ < π (65 w (r, π = w (r, π, w θ (r, π = w θ (r, π, (66 w (0, θ < (67 w (1, θ = ĝ (θ, π θ < π (68

23 where ĝ (θ = g (x, y for (x, y and θ = arctan (y/x. We separate variables and the PE becomes w (r, θ = R (r H (θ 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, The problem for H (θ is, as before, r d ( dr d H 1 r = = µ R (r dr dr dθ H (θ d H dh dh + µh (θ = 0; H ( π = H (π, ( π = (π, dθ dθ dθ 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 d R dr 0 = r r m R = r + r m R dr dr dr dr Try R (r = r α to obtain the auxiliary equation α (α 1 + α m = 0 m whose solutions are α = ±m. Thus for each m, the solution is R m (r = c 1 r m +c r. For m > 0, r m blows up as r 0. Our boundedness criterion (67 implies c = 0. Hence R m (r = c m r m where c m are constants to be found by imposing the BC (68. The separable solutions satisfying the PE (65 and conditions (66 to (67, are w m (r, θ = r m (A m cos (mθ + B m sin (mθ, m = 0, 1,,... The full solution is the infinite sum of these over m, u E (x, y = w E (r, θ = r m m=0 (A m cos (mθ + B m sin (mθ (69 3

24 y u=u0/( ST0 1 T 0 u=0 x We still need to find the A m, B m. Figure : Setup for hot spot problem on circle. Imposing the BC (68 gives (Am cos (mθ + B m sin (mθ = ĝ (θ (70 m=0 The orthogonality relations are π cos (mθ sin (nθ dθ = 0 π { } { π cos (mθ cos (nθ π, m = n dθ =, (m > 0. π sin (mθ sin (nθ 0, m = n Multiplying (70 by sin nθ or cos nθ and applying these orthogonality relations gives A 0 = Hot spot on boundary 1 π π π π ĝ (θdθ 1 A m = ĝ (θ cos (mθ dθ (71 π π π 1 B m = ĝ (θ sin (mθdθ π π Suppose [RAW] { u 0 θ ĝ (θ = 0 π θ θ +π 0 0 otherwise 4

25 which models a hot spot on the boundary. The Fourier coefficients in (71 are thus u 0 u 0 A 0 =, A m = sin (mθ 0 π mπ (θ 0 + π u 0 B m = (cos (mθ 0 ( 1 m mπ (θ0 + π Thus the steady-state solution is u 0 ( u0 sin (mθ 0 + r mπ (θ 0 + π cos (mθ u 0 (cos (mθ 0 ( 1 m m u E = sin (mθ 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 (θdθ π π π 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 5

26 u=0 u=0 u=0 u=u u 0 u 0 /S u of S T 0 T T 0 S T o S 0 0 Figure 3: Plots of steady-state temperature due to a hot segment on boundary. lines. Also, lines of symmetry can be thought of as an insulating boundary, since u n l = 0. See Problem of Assnt 5. (i θ 0 = 0. Then ue = u 0 u 0 r n 1 sin ((n 1θ 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 π + 0 (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 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 Ref: Haberman We now use the standard trick and solve the inhomogeneous Heat Problem (64, u t = u, (x,y u (x,y,t = g (x,y, (x,y (7 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 6

27 where u (x, y, t is the solution to (7. 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 = J m (rj m,n (α mn cos (mθ + β mn sin (mθe 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. 7

28 11 Mean Value Property for Laplace s Eq. Ref: Haberman.5.4 Theorem [Mean Value Property] Suppose v (x, y satisfies Laplace s equation in a domain, v = 0, (x, y. (73 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. (74 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 (73 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. (69 and (71. Setting r = 0 in (69 gives the center value π 1 u E (0, 0 = A 0 = ĝ (θ dθ (75 π π On the boundary of the circle (of radius 1, (x, y = (cos θ, sin θ and the BC implies that u E = ĝ (θ on that boundary. Thus, ĝ (θ = u E (cos θ, sin θ and (75 becomes π 1 u E (0, 0 = A 0 = u E (cos θ, sin θdθ. (76 π π To prove the Mean Value Property, we consider the region B (x0,y 0 (R = { (x, y : (x x0 + (y y 0 R }, which is a circle of radius R centered at (x, y = (x 0, y 0. Since B (x0,y 0 (R, Laplace s equation (73 holds on this circle. Thus We make the change of variable x ˆ = v = 0, (x, y B (x0,y 0 (R. (77 x x 0 y y 0, ŷ =, ue (ˆ x, y ˆ = v (x, y (78 R R to map the circle B x, ˆ x (x0,y 0 (R into the unit circle {(ˆ y : ˆ + yˆ 1}. Laplace s equation (77 becomes ˆ ue = 0, (ˆ y { (ˆ y : ˆ + ˆ 1 } x, ˆ x, ˆ x y 8

29 where ˆ = ( / ˆ x, / ŷ. The solution is given by Eqs. (69 and (71, and we found the center value above in Eq. (76. Reversing the change of variable (78 in Eq. (76 gives v (x 0, y 0 = as required. 1 π π π v (x 0 + R cos θ, y 0 + R sin θdθ 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 circle centered at (x 0, y 0 and contained in. 1 Maximum Principle for Laplace s Eq. Ref: Haberman.5.4 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 any interior point of, i.e. (x, y is not on the boundary of. The Mean Value Property implies that for any circle of radius R centered at (x 0, y 0, 1 π v (x 0, y 0 = v (x 0 + R cos θ, y 0 + R sin θdθ π π = Average of v on boundary of circle But the the average of a set of numbers is always between the minimum and maximum of those numbers. Thus the average value of v (x, y on the boundary must be between the minimum and maximum value of v (x, y on the boundary, and hence v (x 0, y 0 is between the minimum and maximum values of v (x, y on the boundary. For the heat equation, this implies the equilibrium temperature cannot attain its maximum in the interior, unless temperature is constant everywhere. Also, if v = 0 everywhere on the boundary, then v must be zero at everywhere in. Now, 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 nds = v ˆ v + v dv = v dv Thus, by continuity, v = 0 everywhere in, hence v = 0 everywhere in. Thus v = const in. 9

30 13 Eigenvalues on different domains Ref: Haberman 7.6 In this section we revisit the Sturm-Liouville problem v + λv = 0, x v = 0, x and consider the effect of the shape of the domain on the eigenvalues. efinition Rayleigh Quotient v vdv R (v = (79 v dv 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 nds ˆ = v vdv + v vdv In the statement of the theorem, we assumed that v = 0 on, and hence v vdv = v vdv (80 Let {φ n } be an orthonormal basis of eigen-functions on, i.e. all the functions φ n which satisfy and φ n + λ n φ n = 0, x φ n = 0, x φ n φ m dv = We can expand v in the eigen-functions, { v (x = An φ n (x n=1 30 1, m = n 0, m n

31 where the A n are constants. Assuming we can differentiate the sum termwise, we have A n = λ n A n φ n (81 φ n v = n=1 n=1 The orthonormality property (i.e., the orthogonality property with φ dv = 1 n implies and, from (81, v dv = = A A n A m φ n φ m dv (8 v vdv = λ A n A n A m φ n φ m dv = λ n (83 Substituting (83 into (80 gives Substituting (8 and (84 into (79 gives v vdv λ n=1 n R (v = = v dv A n=1 m=1 n=1 m=1 n=1 n=1 v vdv = v vdv = λ n (84 n=1 We assume the eigen-functions 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. n=1 A n n=1 A n Also, equality holds (i.e., R (v = λ 1 if and only if v is an eigen-function for the eigenvalue λ 1. Otherwise there will be a λ n > λ 1 such that A n = 0 and R (v > λ 1. The above is not a complete proof, because we have not shown that the sums converge or that they can be differentiated termwise. Theorem If two domains and in R satisfy i.e., but =, then the smallest eigenvalues of the Sturm-Liouville problems on and, λ1 and ˆλ 1, respectively, satisfy λ 1 < λ 1 In other words, the domain that contains the sub-domain is associated with a smaller eigenvalue. 31 n=1 A n n A n n n

32 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 eigen-function corresponding to λ 1 on. Then, as we have proven before, λ 1 = R (v 1, (85 where R (v is the Rayleigh Quotient. Extend the function v 1 continuously from to to obtain a function vˆ 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 vˆ 1 (which satisfies all the requirements of the theorem gives λˆ 1 R (ˆ v 1. Equality happens only if vˆ 1 is the eigen-function corresponding to λ ˆ 1. Useful fact [stated without proof]: the eigen-function(s corresponding to the smallest eigenvalue λˆ 1 on are nonzero in the interior of. This is the crux of the proof. Haberman does not prove this in general, but states it on p From the useful fact, vˆ 1 cannot be an eigen-function corresponding to λˆ 1 on, since it is zero in the interior of (outside. Thus, as the previous theorem states, λˆ 1 < R (ˆ v 1. (86 Since vˆ 1 = 0 outside, the integrals over in the Rayleigh quotient reduce to integrals over, where ˆv 1 = v 1, and hence Combining (85, (86, and (87 gives the result, R (ˆv 1 = R (v 1. (87 ˆλ 1 < λ 1. 3

33 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 where the first zero 0,1 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 variable rˆ = r/r, so that J m rˆ where ˆr goes from 0 to 1. Thus λ = J m ( R 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 = = R, λ 01 = = R R R Thus, λ 11 < λ 01, which confirms the second theorem, since 1, i.e., the circle is contained inside the square. Now consider the function You can show that and R (v = ( r v (r = 1 R ( dv v v =, dr v vda v da π R ( dv π 0 dr rdrdθ 6 = π R = > λ 01 π 0 v rdrdθ R 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 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 Krahn proved this in the 190s. Faber-Kahn inequality For all domains R of equal area, the disc has the smallest first eigenvalue λ 1. 33

34 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 + = π = 6.8, λ 1CIRC = (J 0,1 1SQ = π = 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 v = 0, x Nodal lines are the curves where the eigen-functions 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 [Nov 17, 004] Ref: Haberman, p. 9. For the square, the eigen-functions 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 eigen-function 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. 34

35 (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 [RAW]. (ii m = 1, n =. ( ( πx πy v 1 = sin sin a a The nodal lines are the boundary and the horizontal line y = a/. [RAW] (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. [RAW] (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. [RAW] (v Consider v 13 v 31. Since λ 31 = λ 13 = 10π /a, this is a solution to 10π v + v = 0, a x v = 0, x To find the nodal lines, we use the identity sin 3θ = (sin θ ( 3 4 sin θ to write ( πx ( πy ( ( πy v 13 = sin sin 3 4 sin a a a ( πx ( πy ( ( πx v 31 = sin sin 3 4 sin a a a ( πx ( πy ( ( πx ( πy v 13 v 31 = 4 sin sin sin sin a a a a The nodal lines are the boundary of the square,, and lines such that ( πx ( 0 = sin πy ( ( πx ( πy ( ( πx ( πy sin = sin sin sin + sin a a a a a a i.e., ( πx ( πy sin = sin ± a a Note that sin ψ = sin ϕ if ψ ϕ = kπ or ψ + ϕ = (k 1π for any integer k. Thus the nodal lines are given by ( πy ( πx πy πx πy πx sin = sin = = kπ, + = (k 1π a a a a a a 35

36 ( πx ( πy sin = sin a a Hence the nodal lines are = πx ( a πy = kπ, a y = ±x + la πx ( a + πy = (k 1π a for all integers l. 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 [RAW]. Let T be the isosceles right triangle whose hypotenuse lies on the bottom horizontal side of the square [RAW]. The function v 13 v 31 is zero on the boundary T, positive on the interior of T, and thus satisfies the Sturm-Liouville problem 10π v + v = 0, a x T v = 0, x T Hence v 13 v 31 is the eigen-function corresponding to the first eigenvalue λ 13 of the Sturm-Liouville problem on the triangle T. In this case, we found the eigenfunction without using separation of variables, which would have been complicated on the triangle. NOTE: this is half the triangle considered in problem 4 of Assnt 5, but you use a similar solution method. (vi With v 13, v 31 given above, adding gives ( } πx ( πy { 3 ( ( πx ( πy v 13 + v 31 = 4 sin sin 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 area 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 satisfies the Sturm-Liouville problem 10π v + v = 0, a x c v = 0, x c Hence (v 13 + v 31 is the eigen-function corresponding to the first eigenvalue λ 13 of the Sturm-Liouville problem on c. (vii Find the first eigenvalue on the right triangle { } = 0 y 3x, 0 x 1. 36

37 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 eigen-functions v mn for the rectangle are already zero on two sides of the triangle. However, any particular eigen-function 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 eigen-functions. However, to satisfy the Sturm-Liouville problem, all the eigen-functions must be associated with the same eigenvalue. For the rectangle with side lengths 3 and 1, the eigenvalues are given by ( m n λ mn = π π ( + = 3m + n We create a table of 3m + n and look for eigenvalues that have multiple eigenfunctions. n, m The smallest value of 3m +n that is the same for multiple sets of (m, n is 8. Thus λ 31 = λ 4 = λ 15 = 8π /3 has multiple eigen-functions. We add the corresponding eigen-functions and set them to zero along y = 3x, Av 31 + Bv 4 + Cv 15 = 0. If we can find the constants A, B, C then we re done - we ve found the first (smallest eigenvalue 8π /3 and eigen-function Av 31 +Bv 4 +Cv 15 on the triangle. Otherwise, if we can t solve for A, B, C, we look for the next largest value of 3m + n that repeats, and try again. 14. Nodal lines for the disc (circle Ref: Haberman,

38 For the disc of radius 1, we found the eigen-functions and eigenvalues to be v mns = J m (rj m,n sin mθ, v mnc = J m (rj m,n cos mθ 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. [RAW] (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. [RAW] (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 [RAW]. 15 Steady-state temperature in a 3 cylinder Ref: Haberman 7.9 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, (88 u E (r, θ, 0 = u E (r, θ, L = 0, π θ < π, 0 r a, (89 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 38

39 We separate variables as v (r, θ, z = R (rh (θz (z so that (88 becomes Rearranging gives 1 d ( dr (r 1 d H 1 d Z (z r + + = 0. rr (r dr dr r dθ Z (z dz 1 d ( dr (r 1 d H 1 d Z (z r + = (90 rr (r dr dr r dθ Z (z dz = λ 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 + λz = 0 dz 0 = u (r, θ, 0 = R (r H (θ Z (0 0 = u (r, θ, L = R (rh (θz (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 (z = sin, λ n =, n = 1,, 3,... L L And this also shows that the constant λ = λ n. Multiplying Eq. (90 by r gives r d ( dr n (r λ n r d H n r = = µ (91 R n (r dr dr dθ where µ is constant since the l.h.s. depends only on r and the middle only on θ. Note: solving for H n (θ looks easier, until you realize that we don t have nice BCs on H n (θ. Multiplying (91 by R n (r gives r d ( r dr (r ( λr + µ R (r = 0 dr dr 39

40 Change variables to s = λn r, R n (s = R n (r, so that d ( dr n (s ( s + µ s s R n (r = 0 ds ds If µ = m for some m = 1,, 3,..., we have d ( dr nm (s s s ( s + m R mn (r = 0 ds ds which is the Modified Bessel Equation with tabulated solutions I m (s and K m (s called the modified Bessel functions of order m of the first and second kinds, respectively [RAW]. We assume µ = m, so that R mn (r = c 1m I m (s + c m K m (s Transforming back to r = s/ λ n gives ( nπr ( nπr R mn (r = c 1m I m + c m K m L L 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 ( nπr R mn (r = c 1m I m L The corresponding solutions for H (θ satisfy d H m + m H m = 0, m = 0, 1,, 3,... dθ and hence H m (θ = A m cos mθ + B m sin mθ. 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. 40