The 1-D Heat Equation (Solutions to Problems)

Solutions to Problems for The 1-D Heat Equation 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 2006 1. A bar with initial temperature profile f (x) > 0, with ends held at 0o C, will cool as t → ∞, and approach a steady-state temperature 0o C. However, whether or not all parts of the bar start cooling initially depends on the shape of the initial temperature profile. The following example may enable you to discover the relationship. (a) Find an initial temperature profile f (x), 0 ≤ x ≤ 1, which is a linear
df df (0) = 0 = dx (1), f 12 = combination of sin πx and sin 3πx, and satisfies dx 2. Solution: A linear combination of sin πx and sin 3πx is f (x) = a sin 3πx + b sin πx Imposing the conditions gives df (0) = π (3a + b) dx df 0 = (1) = −π (3a + b) dx  1 2 = f = −a + b 2 0 = The first two equations yield the same thing, 3a = −b. Substituting this into the last equation gives 1 a=− , 2 Thus 3 b= . 2 1 3 f (x) = − sin 3πx + sin πx 2 2 1 (1) (b) Solve the problem ut = uxx ; u (0, t) = 0 = u (1, t) ; u (x, 0) = f (x) . Note: you can just write down the solution we had in class, but make sure you know how to get it! Solution: This is the basic heat problem we considered in class, with solution ∞ X 2 2 u (x, t) = Bn sin (nπx) e−n π t (2) n=1 where Bn = 2 Z 1 f (x) sin (nπx) dx (3) 0 and f (x) is given in (1). The form of (1) is already a sine series, with B1 = 3/2, B3 = −1/2 and Bn = 0 for all other n. You can check this for yourself by computing integrals in (3) for f (x) given by (1), from the orthogonality of sin nπx. Therefore, u (x, t) = 1 3 2 2 sin (πx) e−π t − sin (3πx) e−9π t 2 2 (4) (c) Show that for some x, 0 ≤ x ≤ 1, ut (x, 0) is positive and for others it is negative. How is the sign of ut (x, 0) related to the shape of the initial temperature profile? How is the sign of ut (x, t), t > 0, related to subsequent temperature profiles? Graph the temperature profile for t = 0, 0.2, 0.5, 1 on the same axis (you may use Matlab). Solution: Differentiating u (x, t) in time gives   3 9 2 −π 2 t −9π 2 t ut (x, t) = −π − sin (3πx) e sin (πx) e 2 2 Setting t = 0 gives 3 ut (x, 0) = − π 2 (sin (πx) − 3 sin (3πx)) 2 Note that ut   15 1 , 0 = π 2 > 0, 6 4 ut   1 , 0 = −6π 2 < 0 2 Thus at x = 1/6, ut is positive and for x = 1/2, ut is negative. From the PDE, ut = uxx 2 2 1.8 t=0 1.6 1.4 0 u(x,t ) 1.2 1 0.8 0.6 t=0.5,1 0.4 t=0.2 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x Figure 1: Plots of u(x, t0 ) for t0 = 0, 0.2, 0.5, 1. and hence the sign of ut gives the concavity of the temperature profile u (x, t0 ), t0 constant. Note that for uxx (x, t0 ) > 0, the profile u (x, t0 ) is concave up, and for uxx (x, t0 ) < 0, the profile u (x, t0 ) is concave down. At t0 = 0, the sign of ut (x, 0) give the concavity of the initial temperature profile u (x, 0) = f (x). In Figure 1, u(x, t0 ) is plotted for t0 = 0, 0.2, 0.5, 1. 2. Initial temperature pulse. Solve the inhomogeneous heat problem with Type I boundary conditions: ∂u ∂ 2u = ; u (0, t) = 0 = u (1, t) ; u (x, 0) = Pw (x) ∂t ∂x2 where t > 0, 0 ≤ x ≤ 1, and   if 0 < x < 21 − w2  0 u0 Pw (x) = if 21 − w2 < x < 12 + w2 w   0 if 12 + w2 < x < 1 (5) Note: we derived the form of the solution in class. You may simply use this and replace Pw (x) with f (x). (a) Show that the temperature at the midpoint of the rod when t = 1/π 2 (dimensionless) is approximated by     1 1 2u0 sin (πw/2) ≈ , u 2 π2 e πw/2 3 Can you distinguish between a pulse with width w = 1/1000 from one with
w = 1/2000, say, by measuring u 21 , π12 ? (b) Illustrate the solution qualitatively by sketching (i) some typical temperature profiles in the u − t plane (i.e. x = constant) and in the u − x plane (i.e. t = constant), and (ii) some typical level curves u (x, t) = constant in the x − t plane. At what points of the set D = {(x, t) : 0 ≤ x ≤ 1, t ≥ 0} is u (x, t) discontinuous? Solution: This is the Heat Problem with Type I homogeneous BCs. The solution we derived in class is, with f (x) replaced by Pw (x), u (x, t) = ∞ X un (x, t) = n=1 ∞ X 2 π2 t Bn sin (nπx) e−n (6) n=1 where the Bn ’s are the Fourier coefficients of f (x) = Pw (x), given by Z 1 Bn = 2 Pw (x) sin (nπx) dx 0 Breaking the integral into three pieces and substituting for Pε (x) from (5) gives Z 1/2−w/2 Z 1/2+w/2 Bn = 2 Pw (x) sin (nπx) dx + 2 Pw (x) sin (nπx) dx 0 +2 1/2−w/2 Z 1 Pw (x) sin (nπx) dx 1/2+w/2 Z 1/2+w/2 u0 = 0+2 w 1/2−w/2 2u0 = w = u0  cos cos (nπx) − nπ nπ 2 sin (nπx) dx + 0 1/2+w/2 1/2−w/2
(1 − w) − cos wnπ/2 nπ 2
(1 + w) We apply the cosine rule cos (r − s) − cos (r + s) = 2 sin r sin s with r = nπ/2, s = nπw/2 to Eq. (7), Bn = 4u0 nπ nπw sin sin wnπ 2 2 When n is even (and nonzero), i.e. n = 2m for some integer m, B2m = 2u0 sin mπ sin mπw = 0 wmπ 4 (7) When n is odd, i.e. n = 2m − 1 for some integer m, B2m−1 = 2u0 (−1)m+1 sin ((2m − 1) πw/2) . (2m − 1) πw/2 (8) (a) The temperature at the midpoint of the rod, x = 1/2, at scaled time t = 1/π 2 is, from (6) and (8), u (x, t) = = ∞ X m=1 ∞ X m=1  π  −(2m−1)2 sin ((2m − 1) πw/2) sin (2m − 1) e (2m − 1) πw/2 2   sin ((2m − 1) πw/2) . (2m − 1) πw/2 2u0 (−1)m+1 2u0 e(2m−1) 2 For t ≥ 1/π 2 , the first term gives a good approximation to u (x, t),       1 1 1 1 2u0 sin (πw/2) u . ≈ u1 = , , 2 π2 2 π2 e πw/2 To distinguish between pulses with ε = 1/1000 and w = 1/2000, note that πw/2 limw→0 sinπw/2 = 1, and so for smaller and smaller w, the corresponding
temperature u 21 , π12 gets closer and closer to 2u0 /e,       π 2 ε2 1 1 1 1 2u0 1− ≈ u1 = , , + ··· , w ≪ 1. u 2 π2 2 π2 e 2 · 3! In particular,     1 1 1 1 1 1 u1 − u1 , ;w = , ;w = 2 π2 1000 2 π2 2000   2u0 sin (π/2000) sin (π/4000) − = e π/2000 π/4000 2u0 ≈ − × 3.1 × 10−7 e Thus it is hard to distinguish these two temperature distributions, at least by measuring the temperature at the center of the rod at time t = 1/π 2 . By this time, diffusion has smoothed out some of the details of the initial condition. (b) The solution u (x, t) is discontinuous at t = 0 at the points x = (1 ± w) /2. That said, u (x, t) is piecewise continuous on the entire interval [0, 1]. Thus, the Fourier series for u (x, 0) converges everywhere on the interval and equals u (x, 0) at all points except x = (1 ± w) /2. The temperature profiles (u−t plane, u−x plane), 3D solution and level curves are shown. 5 u−t plot 10 u(x0,t)/u0 8 6 4 2 0 0 0.02 0.04 0.06 0.08 0.1 t Figure 2: Time temperature profiles u (x0 , t) at x0 = 0.5, 0.4 and 0.1 (from top to bottom). The t-axis is the time profile corresponding to x0 = 0, 1. Recall that to draw the level curves, it is easiest to already have drawn the spatial temperature profiles. Draw a few horizontal broken lines across your u vs. x plot. Suppose you draw a horizontal line u = u1 . Suppose this line u = u1 crosses one of your profiles u (x, t0 ) at position x = x1 . Then (x1 , t0 ) is a point on the level curve u (x, t) = u1 . Now plot this point in your level curve plot. By observing where the line u = u1 crosses your various spatial profiles, you fill in the level curve u (x, t) = u1 . Repeat this process for a few values of u1 to obtain a few representative level curves. Plot also the special case level curves: u (x, t) = u0 /w, u (x, t) = 0, etc. Come and see me if you still have problems. 3. Consider the homogeneous heat problem with type II BCs: ∂u ∂ 2u = ; ∂t ∂x2 ∂u ∂u (0, t) = 0 = (1, t) ; ∂x ∂x u (x, 0) = f (x) (9) where t > 0, 0 ≤ x ≤ 1 and f is a piecewise smooth function on [0, 1]. (a) Find the eigenvalues λn and the eigenfunctions Xn (x) for this problem. Write the formal solution of the problem (a), and express the constant coefficients as integrals involving f (x). 6 u−x plot 12 10 0 u(x,t )/u 0 8 6 4 2 0 0 0.2 0.8 0.6 0.4 1 x Figure 3: Spatial temperature profiles u (x, t0 ) at t0 = 0 (dash), 0.001, 0.01, 0.1. The x-axis from 0 to 1 is the limiting temperature profile u (x, t0 ) as t0 → ∞. 3D plot of u(x,t) u(x,t)/u0 10 5 0 0 0.02 0.04 0.06 0.08 t 1 0.5 0.1 0 x Figure 4: 7 Level curves u(x,t)=const 0.1 0.08 t 0.06 1 0.04 0.25 0.5 0.25 0.02 0 0 0.5 2 4 0.2 0.6 0.4 0.8 1 x Figure 5: Level curves u (x, t) /u0 = C for various values of the constant C. Numbers adjacent to curves indicate the value of C. The line segment (1 − w) /2 ≤ x ≤ (1 + w) /2 at t = 0 is the level curve with C = 1/w = 10. The lines x = 0 and x = 1 are also level curves with C = 0. (b) Find a series solution in the case that f (x) = u0 , u0 a constant. Find an approximate solution good for large times. Sketch temperature profiles (u vs. x) for different times. (c) Evaluate limt→∞ u (x, t) for the solution (a) when f (x) = Pw (x) with Pw (x) defined in (5). Illustrate the solution qualitatively by sketching temperature profiles and level curves as in Problem 2(b). It is not necessary to find the complete formal solution. Solution: (a) To find a series solution, we use separation of variables, u (x, t) = X (x) T (t) (10) The PDE in (9) gives the usual T′ X ′′ = −λ = T X where λ is constant since the left hand side is a function of x only and the middle is a function of t only. Substituting (10) into the BCs in (9) gives X ′ (0) = X ′ (1) = 0 8 The Sturm-Liouville boundary value problem for X (x) is thus X ′′ + λX = 0; X ′ (0) = X ′ (1) = 0 (11) Let us try λ < 0. Then the solutions are √ √ X (x) = c1 e− |λ|x + c2 e |λ|x and imposing the BCs gives p p 0 = X ′ (0) = − |λ|c1 + |λ|c2 √ √ p p − |λ| ′ 0 = X (1) = − |λ|c1 e + |λ|c2 e |λ| The first equation gives c1 = −c2 and substituting this into the second, we have  √ √  p − |λ| + e |λ| 0 = |λ|c2 e Since λ < 0, the bracketed expression is positive. Hence c1 = c2 = 0, i.e. X (x) must be the trivial solution, and we discard the case λ < 0. For λ = 0, X (x) = c1 x + c2 and both BCs are satisfied by taking c1 = 0. Thus X (x) = c2 = A0 (we’ll use A0 by convention - it’s just another way to name the constant). Hence, the case λ = 0 is allowed and yields a non-trivial solution. For λ > 0, we have √ √ X = c1 sin λx + c2 cos λx The BC X ′ (0) = 0 implies c1 = 0. The other BC implies √ √ 0 = X ′ (1) = −c2 λ sin λ For a non-trivial solution, c2 must be nonzero. Since λ > 0 then we must √ have sin λ = 0, which implies the eigenvalues are λ n = n2 π 2 , n = 1, 2, 3, ... and the eigenfunctions are Xn (x) = cos (nπx) For each n, the solution for T (t) is Tn (t) = e−λn t . Hence the series solution for u (x, t) is u (x, t) = A0 + ∞ X n=1 9
An cos (nπx) exp −n2 π 2 t (12) At t = 0, the initial condition gives f (x) = u (x, 0) = A0 + ∞ X An cos (nπx) (13) n=1 The orthogonality conditions are found using the identity 2 cos (nπx) cos (mπx) = cos ((m − n) πx) + cos ((m + n) πx) Note also that for m, n = 1, 2, 3..., we have ( Z 1 0 cos ((m − n) πx) dx = Z 1 m=n= 6 0 0 m 6= n 1 cos ((m + n) πx) dx = 0 0 The last integral follows since m+n cannot be zero for any positive integers m, n. Combining the three previous equations gives the orthogonality conditions ( Z 1 1/2 m = n 6= 0 cos (nπx) cos (mπx) dx = (14) 0 m= 6 n 0 Multiplying each side of (13) by cos (mπx), integrating from x = 0 to 1, and applying the orthogonality condition (14) gives Z 1 A0 = f (x) dx (15) 0 Z 1 Am = 2 cos (mπx) f (x) dx (16) 0 (b) Substituting f (x) = u0 into (16) and (15) gives Z 1 An = 2u0 cos (nπx) dx = 0, n > 0 (17) 0 A0 = Z 1 u0 dx = u0 0 It is no surprise that An = 0 for n > 0 since the IC f (x) = u0 is one of the eigenfunctions, X0 (x) = A0 . The series solution is simply u (x, t) = u0 = f (x) . Temperature profiles are simply a plot of f (x) = u0 , i.e. the temperature along the rod does not change. This is reasonable, since the rod is initially 10 u−t plot 10 u(x0,t)/u0 8 6 4 2 0 0 0.02 0.04 0.06 0.08 0.1 t Figure 6: Time temperature profiles u (x0 , t) at x0 = 0.5, 0.4 and 0.1 (from top to bottom). at a constant temperature and is completely insulated - so nothing will happen. (c) Taking the limit t → ∞ of (12) and using (15) gives Z 1/2+w/2 Z 1 Z 1 u0 dx = u0 lim u (x, t) = A0 = f (x) dx = Pw (x) dx = t→∞ 0 0 1/2−w/2 w Thus the temperature along rod eventually becomes the constant u0 . Lastly, we illustrate the solution qualitatively by sketching temperature profiles and level curves in Figures 6 to 9. 4. Consider the homogeneous heat problem with type III (mixed) BCs: ∂u ∂u ∂ 2u = ; (0, t) = 0 = u (1, t) ; u (x, 0) = f (x) 2 ∂t ∂x ∂x where t > 0, 0 ≤ x ≤ 1 and f is a piecewise smooth function on [0, 1]. (a) Find the eigenvalues λn and the eigenfunctions Xn (x) for this problem. Write the formal solution of the problem (a), and express the constant coefficients as integrals involving f (x). (b) Find a series solution in the case that f (x) = u0 , u0 a constant. Find an approximate solution good for large times. Sketch temperature profiles (u vs. x) for different times. 11 u−x plot 12 10 0 u(x,t )/u 0 8 6 4 2 0 0 0.2 0.8 0.6 0.4 1 x Figure 7: Spatial temperature profiles u (x, t0 ) at t0 = 0 (dash), 0.001, 0.03, 0.1. The line u(x, t0 )/u0 = 1 is the limiting temperature profile as t0 → ∞. Note that the ends of the rod heat up! 3D plot of u(x,t) u(x,t)/u0 10 5 0 0 0.02 0.04 0.06 0.08 t 1 0.5 0.1 0 x Figure 8: 12 Level curves u(x,t)=const 0.1 0.08 1 t 0.06 0.04 1 0.02 0.5 0.5 2 0.25 0.25 4 0 0 0.2 0.6 0.4 0.8 1 x Figure 9: Level curves u (x, t) /u0 = C for various values of the constant C. Numbers adjacent to curves indicate the value of C. The line segment (1 − w) /2 ≤ x ≤ (1 + w) /2 at t = 0 is the level curve with C = 1/w = 10. (c) Evaluate limt→∞ u (x, t) for the solution (a) when f (x) = Pw (x) with Pw (x) defined in (5). Illustrate the solution qualitatively by sketching temperature profiles and level curves as in Problem 2(b). It is not necessary to find the complete formal solution. Solution: (a) To find a series solution, we use separation of variables as before (Eq. (10)), but now obtain the Sturm Liouville problem X ′′ + λX = 0; X ′ (0) = X (1) = 0 (18) Let us try λ < 0. Then the solutions are √ √ X (x) = c1 e− |λ|x + c2 e |λ|x and imposing the BCs gives p p 0 = X ′ (0) = − |λ|c1 + |λ|c2 √ √ 0 = X (1) = c1 e− |λ| + c2 e |λ| The first equation gives c1 = −c2 and substituting this into the second, we have √  √  0 = c2 e− |λ| 1 − e2 |λ| 13 √ Since λ < 0, then e2 |λ| > 1 and the bracketed expression is negative. Hence c2 = c1 = 0, i.e. X (x) must be the trivial solution, and we discard the case λ < 0. For λ = 0, X (x) = c1 x + c2 and imposing the BCs gives c1 = c2 = 0. Thus we discard the λ = 0 case. For λ > 0, we have X = c1 sin √ λx + c2 cos √ λx The BC X ′ (0) = 0 implies c1 = 0. The other BC implies √ 0 = X (1) = c2 cos λ For a non-trivial solution, c2 must be nonzero. Since λ > 0 then we must √ have cos λ = 0, which implies the eigenvalues are (2n − 1)2 π 2 λn = , 4 n = 1, 2, 3, ... and the eigenfunctions are Xn (x) = cos  2n − 1 πx 2  For each n, the solution for T (t) is Tn (t) = e−λn t . Hence the series solution for u (x, t) is !   ∞ X (2n − 1)2 π 2 2n − 1 πx exp − t (19) u (x, t) = An cos 2 4 n=1 At t = 0, f (x) = u (x, 0) = ∞ X An cos n=1  (2n − 1) π πx 2  (20) The orthogonality conditions are found using the identity     2m − 1 2n − 1 πx cos πx = cos ((m − n) πx)+cos ((m + n − 1) πx) 2 cos 2 2 Note also that for m, n = 1, 2, 3..., we have ( Z 1 0 cos ((m − n) πx) dx = Z 0 1 m=n= 6 0 0 m 6= n 1 cos ((m + n − 1) πx) dx = 0 14 The last integral follows since m+n−1 cannot be zero for any positive integers m, n. Combining the three previous equations gives the orthogonality conditions (     Z 1 1/2 m = n 6= 0 2n − 1 2m − 1 cos πx cos πx dx = (21) 2 2 0 m 6= n 0 Multiplying each side of (20) by cos ((2m − 1) πx/2), integrating from x = 0 to 1, and applying the orthogonality condition (21) gives   Z 1 2m − 1 πx f (x) dx Am = 2 cos 2 0 (b) Substituting f (x) = u0 into (16) and (15) gives An "
#1  πx sin 2n−1 2n − 1 2 = 2u0 cos πx dx = 2u0 2n−1 2 π 0 2 0   n+1 4u0 (−1) 4u0 2n − 1 = sin π = (2n − 1) π 2 (2n − 1) π Z 1  Thus the series solution is !   ∞ 2n − 1 (2n − 1)2 π 2 4u0 X (−1)n+1 cos πx exp − t u (x, t) = π n=1 2n − 1 2 4 After t ≥ 4/π 2 , we may approximate the series by the first term,  2   πx  4u0 π u (x, t) ≈ u1 (x, t) = cos exp − t π 2 4 The temperature profiles (u vs. x) for different times are given below in Figures 10 to 13. (c) Taking the limit t → ∞ of (19) gives lim u (x, t) = 0 t→∞ Thus the temperature along rod eventually goes to zero. Lastly, we illustrate the solution qualitatively by sketching temperature profiles and level curves (Figures 14 to 17). 15 u−t plot 1 u(x0,t)/u0 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 t Figure 10: Time temperature profiles u (x0 , t) at x0 = 0.01, 0.4 and 0.9 (from top to bottom). u−x plot 1 u(x,t0)/u0 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 x Figure 11: Spatial temperature profiles u (x, t0 ) at t = 0.001, 0.01, 0.1, 0.5 and 1 from top to bottom. The line u(x, t0 )/u0 = 1 is the initial temperature profile at t = 0. 16 3D plot of u(x,t) u(x,t)/u 0 1 0.5 0 0 0.5 t 1 0 0.4 x 0.2 0.8 0.6 1 Figure 12: Level curves u(x,t)=const 1 0.8 0.1 0.6 t 0.25 0.4 0.5 0.2 0 0 0.9 0.99 0.2 0.6 0.4 0.8 1 x Figure 13: Level curves u (x, t) /u0 = C for various values of the constant C. Numbers adjacent to curves indicate the value of C. 17 u−t plot 10 u(x0,t)/u0 8 6 4 x =0.5 0 2 0.4 0.1 0 0 0.1 0.2 0.3 0.4 0.5 t Figure 14: Time temperature profiles u (x0 , t) at x0 = 0.5, 0.4 and 0.1 (from top to bottom). u−x plot 12 10 t=0 0 u(x,t )/u 0 8 6 4 0.001 2 0 0 0.1 0.03 0.5 0.2 0.4 0.6 0.8 1 x Figure 15: Spatial temperature profiles u (x, t0 ) at t0 = 0 (dash), 0.001, 0.03, 0.1 and 0.5. 18 3D plot of u(x,t) u(x,t)/u0 10 5 0 0 0.1 0.2 1 0.3 0.5 0.4 t 0.5 0 x Figure 16: Level curves u(x,t)=const 0.5 0.4 0.3 0.5 t 0.25 0.2 0.1 1 0.5 0 0 0.25 4 0.2 2 0.6 0.4 0.8 1 x Figure 17: Level curves u (x, t) /u0 = C for various values of the constant C. Numbers adjacent to curves indicate the value of C. The line segment (1 − w) /2 ≤ x ≤ (1 + w) /2 at t = 0 is the level curve with C = 1/w = 10. 19