The Heat and Wave Equations in 2D and 3D (Solutions to Problems)

Solutions to Problems for 2D & 3D Heat and Wave Equations 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 2006 1 Problem 1 A rectangular metal plate with sides of lengths L, H and insulated faces is heated to a uniform temperature of u0 degrees Celsius and allowed to cool with three of its edges maintained at 0o C and the other insulated. You may use dimensional coordinates, with PDE ut = κ∇2 u, 0 ≤ x ≤ L, 0 ≤ y ≤ H. The BCs are u (0, y, t) = 0 = u (L, y, t) , u (x, 0, t) = 0, ∂u (x, H, t) = 0. ∂y (1) (i) Solve for u (x, y, t) subject to an initial condition u (x, y, 0) = 100. (ii) Find the smallest eigenvalue λ and the first term approximation (i.e. the term with e−λκt ). (iii) For fixed t = t0 ≫ 0, sketch the level curves u = constant as solid lines and the heat flow lines as dotted lines, in the xy-plane. (iv) Of all rectangular plates of equal area, which will cool the slowest? Hint: for each type of plate, the smallest eigenvalue gives the rate of cooling. (v) Does a square plate, side length L, subject to the BCs (1) cool more or less rapidly than a rod of length L, with insulated sides, and with ends maintained at 0o C? You may use the results we derived in class for the rod, without derivation. Solution: (i) We separate variables as u (x, y, t) = v (x, y) T (t) 1 to obtain ∇2 v T′ = = −λ (2) κT v where λ is constant since the l.h.s. depends only on t and the middle depends only on (x, y). We so obtain the Sturm-Liouville problem ∇2 v + λv = 0 (3) The BCs are v (0, y) = 0 = v (L, y) , v (x, 0) = 0, ∂v (x, H) = 0. ∂y (4) Separating variables again for v (x, y) gives v (x, y) = X (x) Y (y) (5) and substituting into the PDE (3) gives X ′′ Y + XY ′′ + λ = 0 Rearranging gives X ′′ Y ′′ +λ=− =µ (6) Y X where µ is constant because the l.h.s. depends only on y and the middle on x. Introducing (5) into the BCs (4) gives 0 = v (0, y) = X (0) Y (y) 0 = v (L, y) = X (L) Y (y) ⇒ ⇒ X (0) = 0 X (L) = 0 0 = v (x, 0) = X (x) Y (0) ⇒ Y (0) = 0 ∂v 0 = (x, H) = X (x) Y ′ (H) Y ′ (H) = 0 ⇒ ∂y We begin with the problem for X (x), since we have a complete set (two) of homogeneous Type I BCs, the simplest to deal with. From (6), the ODE for X (x) is X ′′ + µX = 0; X (0) = 0 = X (L) We’ve seen before that the eigenfunctions and eigenvalues are Xn (x) = sin � nπx � L , n2 π 2 µn = 2 , L n = 1, 2, 3... From (6), with µ = µn , the ODE for Y (y) is Y ′′ + νY = 0; Y (0) = 0 = Y ′ (H) 2 where ν = λ − µn . Again, we’ve seen ODEs like this before, that ν > 0 and Y (y) = A cos Imposing the BCs gives �√ � �√ � vy + B sin vy 0 = Y (0) = A �√ � √ 0 = Y ′ (H) = B v cos vH √ Note that B cannot be zero since we desire a non-trivial solution. Hence cos ( vH) = 0 and � � √ 1 π, m = 1, 2, 3... vH = m + 2 Thus the eigenvalues and eigenfunctions are � �2 2 1 π , m = 1, 2, 3... vm = m− 2 H2 �� � � 1 y Ym (y) = sin m− π 2 H The general solution to the Sturm-Liouville problem for v (x, y) is � � �� � nπx � y 1 π , n, m = 1, 2, 3... vmn (x, y) = sin sin m− L 2 H From (2), the solution for T (t) is −κλt T =e −κ(νm +µn )t =e � �� � � �2 1 1 n2 2 m− = exp − + κπ t 2 H 2 L2 The general solution to the PDE and BCs for u (x, y, t) is � �� � � �2 �� � � � nπx � 1 1 y 1 n2 m− π exp − + 2 κπ 2 t , umn (x, y, t) = sin sin m− 2 L 2 H 2 H L for n, m = 1, 2, 3... To satisfy the initial condition, we use superposition and sum over all m, n to obtain u (x, y, t) = ∞ � ∞ � m=1 n=1 ∞ � ∞ � umn (x, y, t) � � �� y 1 m− π = Amn sin sin L 2 H m=1 n=1 � �� � � �2 1 1 n2 m− + 2 κπ 2 t · exp − 2 2 H L � nπx � 3 To find the constants Amn , we impose the IC: 100 = u (x, y, 0) = ∞ � ∞ � Amn sin m=1 n=1 � nπx � L sin �� 1 m− 2 � y π H � Using orthogonality relations for sin, we have �� � � � L� H � nπx � 1 y 4 100 sin sin m− π dydx Amn = LH 0 0 L 2 H �� � � � � nπx � � H 1 y 400 L sin dx sin m− π dy = LH 0 L 2 H 0 � �� � � � 400 L 2H 1 = (cos (nπ) − 1) cos m− π −1 LH nπ (2m − 1) π 2 800 = (1 − (−1)n ) 2 n (2m − 1) π Thus Am2n = 0 and Am2n−1 = and u (x, y, t) = ∞ � ∞ � m=1 n=1 1600 (2n − 1) (2m − 1) π 2 � � � � (2m − 1) πy (2n − 1) πx 1600 sin sin (2n − 1) (2m − 1) π 2 L 2H � � � � (2m − 1)2 (2n − 1)2 · exp − + κπ 2 t 4H 2 L2 (ii) The smallest eigenvalue λ is λ11 = � 1 1 + 2 2 4H L � π2 and the first term approximation is � � � � � πx � � πy � 1600 1 1 2 u (x, y, t) ≈ u11 (x, y, t) = 2 sin κπ t + sin exp − π L 2H 4H 2 L2 (iii) To draw the level curves, note that the isotherms (lines of constant temp, i.e. level curves) will intersect the insulated side (y = H) at right angles, and cannot intersect the other sides, which are also isotherms (u = 0). Using these constraints and continuity allow you to sketch the curves. The level curves and heat flow lines are plotted as solid and dotted lines, respectively, in Figure 1. [extra] The heat flow lines are found by finding the orthogonal trajectories to the level curves � πy � � πx � sin =c sin L 2H 4 1 0.95 0.8 0.6 0.4 y/H 0.8 0.6 0.4 0.2 0.2 0.05 0 0 0.2 0.4 0.6 x/L 0.8 1 1.2 Figure 1: Sketch of level curves (solid) and heat flow lines (broken) of u(x, y, t), scaled by its maximum. Differentiating implicitly with respect to x gives � πx � π � πy � � πx � � πy � π dy cos sin =0 + sin cos L L 2H L 2H 2H dx Solving for dy/dx yields � πx � � πy � 2H dy =− cot tan dx L L 2H The slope of the orthogonal trajectories are the negative reciprocal, � πx � � πy � dy L = tan cot dx 2H L 2H Thus tan Integrating gives − Thus � πy � 2H dy = � πx � L tan dx 2H L �� � πx ��� 2H ��� � πy ��� L L � � � ln �cos ln �cos � =− � +c 2H π 2H L π � � � �� 2 2 � πx �L /4H 2H � −1 A �cos cos y= � π L 5 In Figure 1, we chose H = L = 1 and various values of A to obtain the dashed lines. (iv) The rate of cooling is κλ11 , and hence given a material (constant κ) of a certain fixed area A = LW , we can write λ11 as � 2 � L 1 λ11 = + 2 π2 2 4A L The rectangle of area A with the smallest rate of cooling is found by minimizing λ11 with respect to L: � � 2L 2 dλ11 π2 = 0 = − dL 4A2 L3 so that 4 2 L = 4A ⇒ L= √ 2A ⇒ A H= = L Thus � A 2 � A/2 1 H = √ = 2 L 2A Thus the rectangle with these BCs (zero at all edges except the top) that cools the slowest is one with H = L/2, height half the width. [optional] This makes sense, since when the sides are all held at zero the rectangle that cools the slowest is the square, and this can be constructed by joining two of the rectangles in this problem along their insulated sides. (v) The smallest eigenvalue for the rod of length L with zero BCs at the ends is π2 λ1 = 2 L and for the square of side length L subject to BCs (1) is λ11 = 5 π2 > λ1 4 L2 Thus the rod cools more slowly, which makes sense since the square can cool from its lower horizontal side, unlike the rod which is insulated on its horizontal sides, and can only cool through its ends. 6 2 Problem 2 Haberman Problem 7.3.3, p. 287. Heat equation on a rectangle with different diffusivities in the x- and y-directions. Solution: We solve the heat equation where the diffusivity is different in the x and y directions: ∂ 2u ∂u ∂ 2u = k1 2 + k2 2 ∂t ∂x ∂y on a rectangle {0 < x < L, 0 < y < H} subject to the BCs ∂u (x, 0, t) = 0, ∂y ∂u (x, H, t) = 0, ∂y u (0, y, t) = 0, u (L, y, t) = 0, and the IC u (x, y, 0) = f (x, y) . We separate variables as before, u (x, y, t) = X (x) Y (y) T (t) so that the PDE becomes X (x) Y (y) T ′ (t) = k1 X ′′ (x) Y (y) T (t) + k2 X (x) Y ′′ (y) T (t) Dividing by X (x) Y (y) T (t) gives T ′ (t) Y ′′ (y) X ′′ (x) = k1 + k2 T (t) X (x) Y (y) Since the l.h.s. depends on t and the r.h.s. on (x, y), both sides must equal a constant −λ (we call this the separation constant): Y ′′ (y) X ′′ (x) T ′ (t) = k1 + k2 = −λ T (t) X (x) Y (y) Thus T (t) = ce−λt Separating the BCs gives u (0, y, t) = 0 u (L, y, t) = 0 ∂u (x, 0, t) = 0 ∂y ∂u (x, H, t) = 0 ∂y 7 ⇒ X (0) = 0 ⇒ X (L) = 0 ⇒ Y ′ (0) = 0 ⇒ Y ′ (L) = 0 Rearranging the other part of the equation gives X ′′ (x) Y ′′ (y) k2 + λ = −k1 Y (y) X (x) Since the l.h.s. depends on y and the r.h.s. on x, both sides must equal a constant, say µ, Y ′′ (y) X ′′ (x) k2 + λ = −k1 =µ Y (y) X (x) The problem for X (x) is now X ′′ (x) + µ X (x) = 0; k1 X (0) = 0 = X (L) The solution is, as we found in class, � nπ �2 � nπx � , µn = k1 , Xn (x) = sin L L n = 1, 2, 3... The problem for Y (y) is Y ′′ (y) + λ − µn Y (y) = 0; k2 Y ′ (0) = 0 = Y ′ (H) . We’ve found the solution to this problem before, � mπy � λmn − µn � mπ �2 Ym (y) = cos = , , H k2 H m = 1, 2, 3... Thus λmn = k2 � mπ �2 H + µn = π 2 � n2 m2 k2 2 + k1 2 H L � , m, n = 1, 2, 3... and the solutions to the PDE and BCs are � nπx � � mπy � umn (x, y, t) = sin cos e−λmn t , L H m, n = 1, 2, 3... Using superposition, we sum over all m, n to obtain the general solution, u (x, y, t) = = ∞ � ∞ � m=1 n=1 ∞ � ∞ � Amn umn (x, y, t) Amn sin m=1 n=1 � nπx � L cos � mπy � H e−λmn t where the constants Amn are found by imposing the IC: u (x, y, 0) = f (x, y) = ∞ � ∞ � m=1 n=1 8 Amn sin � nπx � L cos � mπy � H (7) Using the orthogonality of sines and cosines, we multiply by sin (kπx/L) cos (lπy/H) and integrate in x from 0 to L, and in y from 0 to H, to obtain � � � � � L � H kπx lπy dydx f (x, y) sin cos L H x=0 y=0 � � � � � L � H ∞ � ∞ � mπy � � nπx � � kπx lπy dydx = sin sin cos cos Amn L L H H x=0 y=0 m=1 n=1 � � H � � � � L ∞ � ∞ � mπy � � nπx � � lπy kπx dx cos cos dy = sin sin Amn L L H H y=0 x=0 m=1 n=1 = ∞ � ∞ � L H Amn δnk δml 2 2 m=1 n=1 = Alk LH 4 Hence Amn 3 4 = LH � L x=0 � H f (x, y) sin y=0 � nπx � L cos � mπy � H dydx Problem 3 Haberman Problem 7.7.4 (a), p. 316. The pie-shaped membrane problem. Solution: We consider the displacement on a pie shaped membrane of radius a and angle π/3 radians that satisfies ∂2u = c 2 ∇2 u ∂t2 Due to the circular geometry, we use polar coordinates and separate variables as u = R (r) H (θ) T (t) Recall that in polar coords the Laplacian is given by � � 1 ∂ 1 ∂2u ∂u 2 ∇ u= r + 2 2 r ∂r ∂r r ∂θ and hence the PDE becomes T ′′ (t) 1 d = 2 c T (t) rR (r) dr � dR (r) r dr � 1 d2 H (θ) + 2 r H (θ) dθ2 The l.h.s. depends on t and the r.h.s. on (r, θ), so that both sides equal a constant −λ (the separation constant) � � T ′′ (t) 1 d 1 dR (r) d2 H (θ) r + = = −λ c2 T (t) rR (r) dr dr r2 H (θ) dθ2 9 Recall that we showed in class that if the BCs are Type I or Type II then the eigenvalues are positive. In all the cases in the problem, the boundary either has Type I or Type II BCs, so that λ > 0. Thus � √ � � √ � T (t) = a cos c λt + b sin c λt The BCs are u (r, 0, t) = 0 ⇒ u (r, π/3, t) = 0 ∂u (a, θ, t) = 0 ∂r H (0) = 0, ⇒ H (π/3) = 0, ⇒ R′ (a) = 0. Since we have a complete set of homogeneous BCs for H (θ), we solve for H first. Notice that going from (x, y) to (r, θ), the sides of the membrane don’t touch, unlike the disc, and hence H (θ) does not need to be periodic. The equation for R (r) and H (θ) can be rearranged: � � r d dR (r) 1 d2 H (θ) r + λr2 = − R (r) dr dr H (θ) dθ2 Since the l.h.s. depends on r and the r.h.s. on θ, both sides must equal a constant µ: � � 1 d2 H (θ) r d dR (r) r + λr2 = − =µ H (θ) dθ2 R (r) dr dr Thus the problem for H (θ) is d2 H (θ) + µH (θ) = 0; dθ2 H (0) = 0 = H (π/3) . This is just the 1D Sturm Liouville problem with Type I homogeneous BCs. We have shown before that µ > 0 and the eigenfunctions and eigenvalues are � �2 � � nπ nπθ µn = = 9n2 , n = 1, 2, 3... Hn (θ) = sin = sin (3nθ) , π/3 π/3 The equation for R is then, on rearranging, � �� 2 √ �2 dR 2 2d R r + λr − (3n) R = 0 +r dr2 dr This is the Bessel Equation of order 3n. The general solution is �√ � �√ � Rn (r) = an J3n λr + bn Y3n λr , n = 1, 2, 3... 10 We must impose that the temperature at r = 0 (the origin) remains bounded, and hence must take bn = 0. Thus �√ � λr , n = 1, 2, 3... Rn (r) = an J3n Imposing the BC at r = a gives √ ′ �√ � λa 0 = Rn′ (a) = an λJ3n Thus the eigenvalues are given by solving �� � ′ J3n λnm a = 0, n, m = 1, 2, 3... Just like the zeros of the Bessel function, there are infinitely many local max and mins, and hence J3′ n has infinitely many zero. We enumerate the zeros with the index m, and the order of the Bessel function with the index n. The solution to the PDE and BCs is thus �� � � � � � � � �� unm = J3n λnm r sin (3nθ) anm cos c λnm t + bnm sin c λnm t , n, m = 1, 2, 3... Using superposition, we sum over all n, m to obtain the general solution u= ∞ � ∞ � n=1 m=1 J3n �� � � � � � � �� � λnm r sin (3nθ) anm cos c λnm t + bnm sin c λnm t where the anm and bnm are found from the ICs. The natural frequencies of oscillation are √ c λnm , n, m = 1, 2, 3... fnm = 2π 4 Problem 4 Find the eigenvalue λ and corresponding eigenfunction v for the 30o -60o -90o right triangle (i.e. a right triangle that has these angles); v and λ satisfy ∇2 v + λv = 0 in D, v = 0 on ∂D, √ � � where D = (x, y) : 0 < y < 3x, 0 < x < 1 . Hint: combine the eigenfunctions on the rectangle � √ � D = (x, y) : 0 < x < 1, 0 < y < 3 11 to obtain an eigenfunction on D that is positive on D. We know that the first eigenfunction can be characterized (up to a non-zero multiplicative constant) as the eigenfunction that is of one sign. You may use the eigenfunctions derived in-class for the rectangle, without derivation. Be sure to sketch the region correctly before solving the problem. Solution: Continuing where the notes left off, the trial function is vT = Av31 + Bv24 + Cv15 where vmn � y = sin (mπx) sin nπ √ 3 � and A, B, C are constants determined to make vT = 0 on y = √ Av31 + Bv24 + Cv15 = 0, y = 3x. √ 3x, Note that vT = 0 on x = 1 and y = 0 since v31 , v24 , v15 are eigenfunctons for the √ � � rectangle 0 < x < 1, 0 < y < 3 . We will use the fact that 1 (cos (a − b) − cos (a + b)) 2 cos (−a) = cos (a) sin a sin b = Thus, on y = √ 3x, 0 = Av31 + Bv24 + Cv15 = A sin (3πx) sin (πx) + B sin (2πx) sin (4πx) + C sin (πx) sin (5πx) A B = (cos (2πx) − cos (4πx)) + (cos (−2πx) − cos (6πx)) 2 2 C + (cos (−4πx) − cos (6πx)) 2 1 1 1 (A + B) cos (2πx) + (−A + C) cos (4πx) + (−B − C) = 2 2 2 Since cos (2πx), cos (4πx), cos (6πx) are linearly independent, each coefficient must be zero: A + B = 0, −A + C = 0, −B − C = 0 There are only two independent equations in these 3 (i.e you can get any one of the equations by combining the other two). Thus we are left with a multiplicative constant C, which is fine because eigen-functions are only defined up to a multiplicative constant (we often just take the constant to be 1, for simplicity). In summary, the function vT = Av31 + Bv24 + Cv15 = C (v31 − v24 + v15 ) 12 2 1 0 −1 −2 2 1.5 1 0.8 1 0.6 0.4 0.5 y 0 0.2 0 x Figure 2: Plot of vT (x, y). is zero on the boundary of DT . Also, since each of v31 , v24 , v15 has eigenvalue λ31 = λ24 = λ15 = 28π 2 /3, we have � � 28 28 ∇2 vT = C ∇2 v31 − ∇2 v24 + ∇2 v15 = C π 2 (v31 − v24 + v15 ) = π 2 vT 3 3 in the triangle. Thus vT is an eigen-function on the triangle with eigenvalue 28π 2 /3. [extra] Actually vT does equal zero on the interior (see Figure 2). So other methods would have to be used to find the smallest eigen-value. We can get an upper bound on v by using the Rayleigh quotient, and associated theorems. Consider the function � √ � v = y (x − 1) y − 3x √ Note that v = 0 on the boundary (y = 0, x = 1, y = 3x), v is nonzero on the interior, and � �2 � �2 ∂v ∂v 2 + |∇v| = ∂y ∂x � � √ �2 √ �2 √ 2 2 = y y − 2 3x + 3 + (x − 1) 2y − 3x 13 The Rayleigh quotient is √ �2 � � √ √ �2 � 1 �� √3x � 2 � 2� − 1) 2y − 3x dy dx − 2 3x + 3 + (x y y 0 0 � � √ R (v) = √ �2 � � 1 � 3x y 2 (x − 1)2 y − 3x dy dx 0 0 √ 1 3 112 √ = = 15 ≈ 37.3 1 3 3 560 Thus by our theorem, the smallest eigenvalue on the triangle is less than 37.3. In particular, the eigenvalue associated with vT was λT = 28 2 π ≈ 92.1 3 Clearly, this is NOT the smallest eigenvalue. 5 Problem 5 Consider the boundary value problem on the isosceles right angled triangle of side length 1, ∇2 v = 0, 0 < y < x, 0< x< 1 subject to the BCs ∂v (1, y) = 0, ∂x ∂v (x, 0) = 0, ∂y v (x, x) = 0, v (x, x) = 50, 0< y< 1 0< x< 1 0 < x < 1/2 1/2 < x < 1 Give a symmetry argument to show that v (x, 1 − x) = 25 for 0 < x < 1. Sketch the level curves and heat flow lines of v. Be sure to sketch the region correctly before solving the problem. Solution: Take any point P = (x0 , 1 − x0 ) on the line y = 1 − x inside the triangle. Let v1 = v (x0 , 1 − x0 ) be its steady-state temperature. Rotate the triangle about the line y = 1−x. When you do this, the position of the point P doesn’t change (see Figure 3(top) for a schematic of this argument). Add the solutions on the two plates to obtain a triangular plate insulated on the horizontal and vertical side, and kept at u = 50 on the hypotenuse. Thus the steady state temperature of this new plate is 50. Hence v1 + v1 = 50, or v1 = 25. Since the point we chose was arbitrary, then all points on the line y = 1 − x have steady state temp v (x, 1 − x) = 25 for 14 50 0 u1 0 + = u1 50 Problem 1 50 Problem 2 50 2u1 Problem 3 50 37.5 0 25 12.5 Figure 3: Top: schematic of symmetry argument. Bottom: sketch of level curves (solid) and heat flow lines (broken) of u(x, y). 0 < x < 1 (and inside the triangle, or 1/2 < x < 1). The level curves and heat flow lines are given as solid and dashed lines, respectively, in Figure 3(bottom). [extra] Note also that v (x, 0.5) = 75/2 for 1/2 < x < 1 and v (0.5, y) = 25/2 for 0 < y < 1/2. 15