The Heat Equation. Lectures INF2320 p. 1/88

Save this PDF as:
Size: px
Start display at page:

Download "The Heat Equation. Lectures INF2320 p. 1/88"

Transcription

1 The Heat Equation Lectures INF232 p. 1/88

2 Lectures INF232 p. 2/88 The Heat Equation We study the heat equation: u t = u xx for x (,1), t >, (1) u(,t) = u(1,t) = for t >, (2) u(x,) = f(x) for x (,1), (3) where f is a given initial condition defined on the unit interval (,1). We shall in the following study physical properties of heat conduction versus the mathematical model (1)-(3) separation of variables - a technique, for computing the analytical solution of the heat equation analyze the stability properties of the explicit numerical method

3 Lectures INF232 p. 3/88 Energy arguments We define the energy of the solution u at a time t by E 1 (t) = u 2 (x,t)dx for t. (4) Note that this is not the physical energy This energy is a mathematical tool, used to study the behavior of the solution We shall see that E 1 (t) is a non-increasing function of time

4 Lectures INF232 p. 4/88 Energy arguments If we multiply the left and right hand sides of the heat equation (1) by u it follows that u t u = u xx u for x (,1), t > By the chain rule for differentiation we observe that t u2 = 2uu t Hence 1 2 t u2 = u xx u for x (,1), t >

5 Lectures INF232 p. 5/88 Energy arguments By integrating both sides with respect to x, and applying the rule of integration by parts, we get 1 2 t u2 (x,t)dx = u xx (x,t)u(x,t)dx (5) = u x (1,t)u(1,t) u x (,t)u(,t) = u x (x,t)u x (x,t)dx u 2 x(x,t)dx for t >, where the last equality is a consequence of the boundary condition (2)

6 Lectures INF232 p. 6/88 Energy arguments We assume that u is a smooth solution of the heat equation, which implies that we can interchange the order of integration and derivation in (5), that is t u 2 (x,t)dx = 2 u 2 x(x,t)dx for t > (6) Therefore E 1(t) = 2 u 2 x(x,t)dx for t > This implies that E 1(t)

7 Lectures INF232 p. 7/88 Energy arguments Thus E 1 is a non-increasing function of time t, i.e., In particular E 1 (t 2 ) E 1 (t 1 ) for all t 2 t 1 u 2 (x,t)dx u 2 (x,)dx = f 2 (x)dx for t > (7)

8 Lectures INF232 p. 8/88 Energy arguments This means that the energy, in the sense of E 1 (t), is a non-increasing function of time The integral of u 2 x with respect to x, tells us how fast the energy decreases From a physical point of view it seems reasonable that a the energy will decrease in a system without any heat source In the following we study another energy function which can be analyzed in a similar manner

9 Lectures INF232 p. 9/88 Energy arguments Define the new energy function E 2 (t) = u 2 x(x,t)dx Similar to above, we can multiply by u t and integrate with respect to x and get u 2 t (x,t)dx = u xx (x,t)u t (x,t)dx

10 Lectures INF232 p. 1/88 Energy arguments Integration by parts leads to u 2 t (x,t)dx = [u x (x,t)u t (x,t)] 1 = u x (,t)u t (,t) u x (1,t)u t (1,t) u x (x,t)u xt (x,t)dx u x (x,t)u tx (x,t)dx By the chain rule we get t u2 x = 2u x u xt

11 Lectures INF232 p. 11/88 Energy arguments Hence u 2 t (x,t)dx = u x (,t)u t (,t) u x (1,t)u t (1,t) u x (x,t)u xt (x,t)dx = u x (,t)u t (,t) u x (1,t)u t (1,t) 1 2 t u2 x(x,t)dx We can interchange the order of integration and differentiation and thereby conclude that 1 2 t u 2 x(x,t)dx = u 2 t (x,t)dx u x (,t)u t (,t) + u x (1,t)u t (1,t)

12 Lectures INF232 p. 12/88 Energy arguments According to the boundary condition (2), u(,t) = u(1,t) = for all t > Since u(,t) and u(1,t) are constant with respect to time, we can conclude that u t (,t) = u t (1,t) = for t > (8)

13 Lectures INF232 p. 13/88 Energy arguments Thus, we get that E 2(t) = t u 2 x(x,t)dx = 2 u 2 t (x,t)dx, This means that E 2 is a non-increasing function of time, i.e. E 2 (t 2 ) E 2 (t 1 ) for all t 2 t 1 (9) u 2 x(x,t)dx u 2 x(x,)dx = f 2 x (x)dx, t >

14 Lectures INF232 p. 14/88 Stability We will now study how modifications of the initial condition (3) influence on the solution Consider the problem with a modified initial condition v t = v xx for x (,1), t > (1) v(,t) = v(1,t) = for t > (11) v(x,) = g(x) for x (,1) (12) If g is close to f, will v be approximately equal to u?

15 Lectures INF232 p. 15/88 Stability Let e denote the difference between u and v, i.e. e(x,t) = u(x,t) v(x,t) for x [,1], t From equations (1) and (1) we find that e t = (u v) t = u t v t = u xx v xx = (u v) xx = e xx Furthermore from (2)-(3) and (11)-(12) we get e(,t) = u(,t) v(,t) = = for t > e(1,t) = u(1,t) v(1,t) = = for t > e(x,) = u(x,) v(x,) = f(x) g(x) for x (,1)

16 Lectures INF232 p. 16/88 Stability Thus, e solves the heat equation with homogeneous Dirichlet boundary conditions at x =,1 and with initial condition h = f g From the above discussion, we therefore get that e 2 (x,t)dx h 2 (x)dx, t >, (u(x,t) v(x,t)) 2 dx ( f(x) g(x)) 2 dx, t > (13)

17 Lectures INF232 p. 17/88 Stability Thus, if g is close to f then the integral of (u v) 2, at any time t >, must be small Hence, we conclude that minor changes in the initial condition of (1)-(3) will not alter its solution significantly The problem is stable with respect to changes in the initial condition

18 Lectures INF232 p. 18/88 Uniqueness We will now prove that (1)-(3) can have at most one smooth solution Assume that both u and v are smooth solutions of this problem Using the above notation, this means that g(x) = f(x), x (,1) (13) implies that (u(x,t) v(x,t)) 2 dx =, t >

19 Lectures INF232 p. 19/88 Uniqueness Note that the function (u v) 2 is continuous, and furthermore (u(x,t) v(x,t)) 2 for all x [,1] and t We can therefore conclude that u(x,t) = v(x,t) for all x [,1] and t

20 Lectures INF232 p. 2/88 Maximum principles We know that the solution of (1)-(3), will be more and more smooth, and that it will approach zero, as time increases Physically it is reasonable that the maximum temperature must appear either initially or at the boundary We now study the initial-boundary value problem u t = u xx for x (,1), t > (14) u(,t) = g 1 (t) and u(1,t) = g 2 (t) for t > (15) u(x,) = f(x) for x (,1) (16) We shall see that the maximum value of the solution u(x,t) will occur initially or on the boundary

21 Lectures INF232 p. 21/88 Calculus First, we recall an important theorem from calculus Let q(x), x [,1] be a smooth function of one variable, which attains its maximum value at an interior point x, i.e. x (,1) and q(x) q(x ) for all x [,1] Then q must satisfy q (x ) = q (x )

22 Lectures INF232 p. 22/88 Calculus Consequently, if a smooth function h(x), defined on [,1], is such that or h (x) for all x (,1) h (x) > for all x (,1), then h must attain its maximum value at one of the endpoints

23 Lectures INF232 p. 23/88 Calculus Hence, we conclude that h() h(x) for all x [,1] (17) or h(1) h(x) for all x [,1] (18) Furthermore, if (17) hold, then h must satisfy h (), (19) and if (18) is the case, then it follows that h (1) (2)

24 Lectures INF232 p. 24/88 Calculus Let v = v(x,t) be a smooth function of space x [,1] and time t [,T] That is, we assume that the partial derivatives of all orders of v with respect x and t are continuous, and that where v : Ω T IR, Ω T = {(x,t) < x < 1 and < t < T }, (21) Ω T = {(x,) x 1} {(1,t) t T } {(x,t) x 1} {(,t) t T } (22) and Ω T = Ω T Ω T (23)

25 Lectures INF232 p. 25/88 Calculus Assume that (x,t ) Ω T, an interior point, is a maximum point for v in Ω T, i.e., v(x,t) v(x,t ) for all (x,t) Ω T Then, as in the single variable case, v must satisfy v x (x,t ) =, v t (x,t ) =, (24) v xx (x,t ) and v tt (x,t ) (25)

26 Lectures INF232 p. 26/88 Calculus Thus, a smooth function w = w(x,t), (x,t) Ω T such that either w x (x,t), w t (x,t), w xx (x,t) > or w tt (x,t) >, for all (x,t) Ω T, (26) must attain its maximum value at the boundary Ω T of Ω T We also have that if the maximum is achieved for t = T, say at (x,t), then w must satisfy the inequality w t (x,t) (27)

27 Lectures INF232 p. 27/88 Maximum principles Assume that u is a smooth solution of (14)-(16), and that u achieves its maximum value at an interior point (x,t ) Ω T From (14) it follows that By (24) we conclude that u t (x,t ) = u xx (x,t ) u xx (x,t ) = If there were strict inequalities in (25), we would have had a contradiction, and we could have concluded that u must achieve its maximum at Ω T

28 Lectures INF232 p. 28/88 Maximum principles Define a family of auxiliary functions {v ε } ε> by v ε (x,t) = u(x,t)+εx 2 for ε >, (28) where u solves (14)-(16) Note that v ε t (x,t) = u t (x,t), (29) v ε xx(x,t) = u xx (x,t)+2ε > u xx (x,t) Now, if v ε achieves its maximum at an interior point, say (x,t ) Ω T, then the property (24) implies that v ε t (x,t ) = u t (x,t ) =

29 Lectures INF232 p. 29/88 Maximum principles We can apply the heat equation (14) to conclude that Therefore u xx (x,t ) = u t (x,t ) = v ε xx(x,t ) = u xx (x,t )+2ε = 2ε > This violates the property (25), which must hold at a maximum point of v ε Hence, we conclude that v ε must attain its maximum value at the boundary Ω T of Ω T, i.e., v ε (x,t) max (y,s) Ω T v ε (y,s) for all (x,t) Ω T

30 Lectures INF232 p. 3/88 Maximum principles Can v ε reach its maximum value at time t = T and for x (,1)? Assume that (x,t), with < x < 1, is such that v ε (x,t) v ε (x,t) for all (x,t) Ω T Then, according to property (27), v ε t (x,t), which implies that u t (x,t) see (29) Consequently, since u satisfies the heat equation u xx (x,t) = u t (x,t)

31 Lectures INF232 p. 31/88 Maximum principles It follows that v ε xx(x,t) = u xx (x,t)+2ε +2ε > This contradicts the property (25) that must be fulfilled at such a maximum point This means that where v ε (x,t) max (y,s) Γ T v ε (y,s) for all (x,t) Ω T, (3) Γ T = {(,t) t T } {(x,) x 1} {(1,t) t T }

32 Maximum principles Let M = max ( max t g 1(t), max t g 2(t), max x (,1) ) f(x) And let Γ = {(,t) t } {(x,) x 1} {(1,t) t } Since Γ T Γ, (3) and the definition (28) of v ε, we have that v ε (x,t) max v ε (y,s) M + ε (y,s) Γ for all (x,t) Ω T By (28), it follows that u(x,t) v ε for all (x,t) Ω T Lectures INF232 p. 32/88

33 Lectures INF232 p. 33/88 Maximum principles Therefore u(x,t) M + ε for all (x,t) Ω T and all ε > This inequality is valid for all ε >, hence it must follow that u(x,t) M for all (x,t) Ω T (31) Since T > was arbitrary chosen, this inequality must hold for any t >, i.e., u(x,t) M for all x [,1], t > (32)

34 Lectures INF232 p. 34/88 Minimum principle The maximum principle above can be used to prove a similar minimum principle Let u(x,t) be a solution of the (14)-(16), and define w(x,t) = u(x,t) for all x [,1], t > Note that w satisfies the heat equation, since w t = ( u) t = u t = u xx = ( u) xx = w xx Moreover w(,t) = g 1 (t) and w(1,t) = g 2 (t) for t >, w(x,) = f(x) for x (,1)

35 Lectures INF232 p. 35/88 Minimum principle Thus, from the analysis presented above we find that ( w(x,t) max max ( g 1(t)), max ( g 2(t)), max t t ( = min min g 1(t), min g 2(t), min f(x) t t x (,1) We can write this u(x,t) for all x [,1], t > min x (,1) ), ( ) min g 1(t), min g 2(t), min f(x), t t x (,1) for all x [,1], t > ) ( f(x))

36 Lectures INF232 p. 36/88 Maximum principles A smooth solution of the problem (14)-(16) must satisfy the bound where m u(x,t) M for all x [,1], t >, (33) m = min M = max ( min ) f(x) x (,1) g 1(t), min g 2(t), min t t ( max g 1(t), max g 2(t), max t t x (,1) ) f(x)., (34) (35)

37 Lectures INF232 p. 37/88 Separation of variables Separation of variables is a technique for computing the analytical solution of the heat equation First, we study functions satisfying u t = u xx for x (,1), t >, (36) u(,t) = u(1,t) = for t > (37) We guess that we have solutions on the form u(x,t) = X(x)T(t), (38) where X(x) and T(t) only depend on x and t The boundary conditions imply that u(,t) = = X()T(t) and u(1,t) = = X(1)T(t)

38 Lectures INF232 p. 38/88 Separation of variables Consequently, X(x) must satisfy X() = and X(1) =, (39) provided that T(t) for t > By using (38) for u(x,t) in the heat equation (36) we find that X and T must satisfy the relation X(x)T (t) = X (x)t(t) for all x (,1),t > Thus, if X(x) for x (,1) and T(t) for t >, we get T (t) T(t) = X (x) for all x (,1), t > (4) X(x)

39 Lectures INF232 p. 39/88 Separation of variables Since the left hand side of (4) only depends on t and the right hand side only depends on x, we conclude that there must exist a constant λ such that T (t) T(t) = λ (41) X (x) X(x) = λ (42) We have studied problems on the form (41) earlier and have seen that a solution is given by where c is an arbitrary constant T(t) = ce λt, (43)

40 Lectures INF232 p. 4/88 Separation of variables We write (42) on the form X (x) = λx(x), (44) with the boundary conditions X() = and X(1) = (45) We have that: Therefore sin (kπx) = k 2 π 2 sin(kπx) X(x) = X k (x) = sin(kπx), (46) λ = λ k = k 2 π 2 (47) satisfy (44) and (45), for k =..., 2, 1,,1,2,...

41 Lectures INF232 p. 41/88 Separation of variables We can now summarize that u k (x,t) = c k e k2 π 2t sin(kπx) for k =..., 2, 1,,1,2,... (48) satisfy both the heat equation (36) and the boundary condition (37) Note that, {c k } are arbitrary constants

42 Lectures INF232 p. 42/88 Example 22 Consider the problem u t = u xx for x (,1), t >, (49) u(,t) = u(1,t) = for t >, (5) u(x,) = sin(πx) for x (,1). (51) We have that u(x,t) = e π2t sin(πx) satisfies (49) and (5). Furthermore, u(x,) = e π2 sin(πx) = sin(πx) and thus this is the unique smooth solution of this problem.

43 Lectures INF232 p. 43/88 Example 23 Our second example is u t = u xx for x (,1), t >, u(,t) = u(1,t) = for t >, u(x,) = 7sin(5πx) for x (,1). Putting k = 5 and c 5 = 7 in equation (48) we find that u(x,t) = 7e 25π2t sin(5πx) satisfies the heat equation and the boundary condition of this problem. Furthermore, u(x,) = 7e 25π2 sin(5πx) = 7sin(5πx) and hence the initial condition is also satisfied.

44 Lectures INF232 p. 44/88 Super-positioning Assume that both v 1 and v 2 satisfy equations (36) and (37), and let w = v 1 + v 2 Observe that w t = (v 1 +v 2 ) t = (v 1 ) t +(v 2 ) t =(v 1 ) xx +(v 2 ) xx = (v 1 +v 2 ) xx = w xx Furthermore and w(,t) = v 1 (,t)+v 2 (,t) = + = for t >, w(1,t) = v 1 (1,t)+v 2 (1,t) = + = for t >, hence w also satisfies (36) and (37)

45 Lectures INF232 p. 45/88 Super-positioning For an arbitrary constant a 1 consider the function p(x,t) = a 1 v 1 (x,t) Since a 1 is a constant and v 1 satisfy the heat equation it follows that p t = (a 1 v 1 ) t = a 1 (v 1 ) t = a 1 (v 1 ) xx = (a 1 v 1 ) xx = p xx Furthermore p(,t) = a 1 v 1 (,t) = for t >, p(1,t) = a 1 v 1 (1,t) = for t >. Thus we conclude that p satisfy both (36) and (37)

46 Lectures INF232 p. 46/88 Super-positioning We can now conclude that, if v 1 and v 2 satisfy (36)-(37) then any function on the form a 1 v 1 + a 2 v 2, also solves this problem for all constants a 1 and a 2 More generally, for any sequence of numbers c,c 1,c 2,... such that the series of functions c k e k2 π 2t sin(kπx) k=1 converges, this sum forms a solution of (36)-(37) This is called super-positioning

47 Lectures INF232 p. 47/88 Fourier series and the initial condition We now have that, for any sequence of numbers the function u(x,t) = c 1,c 2,... c k e k2 π 2t sin(kπx) (52) k=1 defines a formal solution of (36)-(37), provided that the series in (52) converges Therefore, if we drop the initial condition, we can find infinitely many functions satisfying the heat equation We shall now see how our model problem (1)-(3) (including the initial condition) can be solved

48 Lectures INF232 p. 48/88 Example 24 Consider the problem u t = u xx for x (,1), t >, (53) u(,t) = u(1,t) = for t >, (54) u(x,) = 2.3sin(3πx)+1sin(6πx) for x (,1). (55) This fit into (52) with c k = for k 3 and k 6, c 3 = 2.3 and c 6 = 1. Thus, the unique smooth solution is given by u(x,t) = 2.3e 9π2t sin(3πx)+1e 36π2t sin(6πx).

49 Lectures INF232 p. 49/88 Example 25 Let us determine a formula for the solution of the problem u t = u xx for x (,1), t >, u(,t) = u(1,t) = for t >, u(x,) = 2sin(πx)+8sin(3πx)+sin(67πx)+12sin(1 4 πx). By putting c k = for k 1, 3, 67, 1 4, c 1 = 2, c 3 = 8, c 67 = 1, c 1 4 = 12, in formula (52), we get the solution u(x,t) = 2e π2t sin(πx)+8e 9π2t sin(3πx) +e (67π)2t sin(67πx)+12e (14 π) 2t sin(1 4 πx).

50 Lectures INF232 p. 5/88 Sums of sine functions We can generalize as follows: Let S be any finite set of positive integers and let {c k } k S be arbitrary given constants Consider an initial condition on the form f(x) = c k sin(kπx) k S Then the solution of (1)-(3) is given by u(x,t) = c k e k2 π 2t sin(kπx) k S

51 Lectures INF232 p. 51/88 Sums of sine functions This is valid because u t (x,t) = k 2 π 2 c k e k2 π 2t sin(kπx) k S u xx (x,t) = k 2 π 2 c k e k2 π 2t sin(kπx) k S u(,t) = c k e k2 π 2t sin() = k S u(x,) = k S u(1,t) = k S c k e k2 π 2t sin(kπ) = c k e sin(kπx) = k S c k sin(kπx) = f(x)

52 Lectures INF232 p. 52/88 Sums of sine functions What about infinite series? For any initial condition on the form f(x) = c k sin(kπx) for x (,1), k=1 where the sum on the right hand side is an convergent infinite series. The solution of (1)-(1) is given by u(x,t) = c k e k2 π 2t sin(kπx). (56) k=1

53 Lectures INF232 p. 53/88 Computing Fourier sine series For a given function f(x), does there exist constants c 1, c 2, c 3,... such that f(x) = c k sin(kπx) for x (,1)? (57) k=1 If so, how can the constants c 1, c 2, c 3,... be determined? Let us first assume that the answer to the first question is yes

54 Lectures INF232 p. 54/88 Computing Fourier sine series Recall the trigonometric identity that, for any real numbers a and b sin(a)sin(b) = 1 2 (cos(a b) cos(a+b)) (58) Therefore it follows that sin(kπx) sin(lπx) dx = { k l, 1/2 k = l (59)

55 Lectures INF232 p. 55/88 Computing Fourier sine series Now f(x)sin(lπx)dx = = ( k=1 c k sin(kπx) ) sin(lπx) dx c k sin(kπx) sin(lπx) dx k=1 = 1 2 c l Therefore c k = 2 f(x)sin(kπx)dx for k = 1,2,... (6) We conclude that, if the function f can be written on the form (57), then the coefficients must satisfy (6)

56 Lectures INF232 p. 56/88 Computing Fourier sine series Usually, a function can be written on the form (57), and we refer to such functions as well-behaved. We can now summarize The Fourier coefficients for a "well-behaved" function f(x), x (,1), is defined by c k = 2 f(x)sin(kπx)dx for k = 1,2,..., (61) and the associated Fourier sine series by f(x) = c k sin(kπx) for x (,1). (62) k=1

57 Lectures INF232 p. 57/88 Computing Fourier sine series Furthermore, if f satisfies (62) then u(x,t) = c k e k2 π 2t sin(kπx) (63) k=1 defines a formal solution of the problem u t = u xx for x (,1), t >, (64) u(,t) = u(1,t) = for t >, (65) u(x,) = f(x) for x (,1). (66)

58 Lectures INF232 p. 58/88 Computing Fourier sine series For a given integer N, we can approximate u by the Nth partial sum u N of the Fourier series, i.e., u(x,t) u N (x,t) = N c k e k2 π 2t sin(kπx) (67) k=1

59 Lectures INF232 p. 59/88 Example 26 Determine the Fourier sine series of the constant function f(x) = 1 for x (,1). From formula (61) we find that c k = 2 = 2 f(x)sin(kπx)dx = 2 [ 1 kπ cos(kπx) ] 1 1 sin(kπx) dx = 2 (cos(kπ) 1) = kπ { if k is eve if k is odd 4 kπ Thus we find that the Fourier sine series of this function is f(x) = 1 = k=1 4 (2k 1)π sin((2k 1)πx) for x (,1) (68)

60 Lectures INF232 p. 6/ Figure 1: The first 2 (dashed line), 7 (dashed-dotted line) and 1 (solid line) terms of the Fourier sine series of the function f(x) = 1.

61 Lectures INF232 p. 61/88 Example 26 Solve u t = u xx for x (,1), t >, u(,t) = u(1,t) = for t >, u(x,) = 1 for x (,1). The formal solution is given by u(x,t) = k=1 4 (2k 1)π e (2k 1)2 π 2t sin((2k 1)πx) (69) In figures 2 and 3 we have graphed the function given by the 1th partial sum of the series (69) at time t =.5 and t = 1, respectively.

62 Figure 2: A plot of the function given by the sum of the 1 first terms of the series defining the formal solution of the problem studied in Example 26. The figure shows a snapshot of this function at time t =.5. Lectures INF232 p. 62/88

63 8 x Figure 3: A plot of the function given by the sum of the 1 first terms of the series defining the formal solution of the problem studied in Example 26. The figure shows a snapshot of this function at time t = 1. Lectures INF232 p. 63/88

64 Lectures INF232 p. 64/88 Note that A Fourier sine series, finite or infinite, g(x) = will have the property that c k sin(kπx), k=1 g() = g(1) = So, if a function f(x) is not zero at the endpoints, it can not be written as a sum of sine series on the closed interval [, 1] But the Fourier sine series of f, will in most cases still converge to f(x) in the open interval (,1)

65 Example 27 Compute the Fourier coefficients {c k } of the function According to formula (61) c k = 2 (x x 2 )sin(kπx)dx=2 We have x sin(kπx) dx = f(x) = x(1 x) = x x 2. [ x 1 ] 1 kπ cos(kπx) = 1 kπ cos(kπ)+ 1 kπ = 1 kπ x sin(kπx) dx 2 cos(kπ) = ( 1)k+1 kπ x 2 sin(kπx)dx. 1 kπ cos(kπx)dx [ 1 kπ sin(kπx). ] 1 Lectures INF232 p. 65/88

66 Example 27 x 2 sin(kπx)dx = [ x 2 1 ] 1 kπ cos(kπx) = 1 kπ cos(kπ)+ 2 kπ 2 kπ 1 kπ sin(kπx)dx = 1 kπ cos(kπ)+ 2 (kπ) 2 2x 1 kπ cos(kπx)dx [ x 1 kπ sin(kπx) ] 1 [ 1 kπ cos(kπx) ] 1 = 1 kπ cos(kπ)+ 2 (kπ) 3 cos(kπ) 2 (kπ) 3 = ( 1)k+1 kπ + 2( 1)k (kπ) 3 2 (kπ) 3. Lectures INF232 p. 66/88

67 Lectures INF232 p. 67/88 Example 27 Thus c k = 2 ( ) 2 (kπ) 3 2( 1)k (kπ) 3 = { if k is even, 8 (kπ) 3 if k is odd, and we find that f(x) = x x 2 ( ) 8 = k=1 ((2k 1)π) 3 sin((2k 1)πx).

68 Figure 4: The first (dashed line), 7 first (dashed-dotted line) and 1 first (solid line) terms of the Fourier sine series of the function f(x) = x x 2. It is impossible to distinguish between the figures representing the 7 first and 1 first terms. They are both accurate approximations of x x 2. Lectures INF232 p. 68/88

69 Lectures INF232 p. 69/88 Example 28 Solve the problem k=1 u t = u xx for x (,1), t >, u(,t) = u(1,t) = for t >, u(x,) = x(1 x) = x x 2 for x (,1). According to (63) the function ( ) 8 u(x,t) = ((2k 1)π) 3 e (2k 1)2 π 2t sin((2k 1)πx) (7) defines a formal solution of this problem.

70 Lectures INF232 p. 7/88 2 x Figure 5: A snapshot, at time t =.5, of the function given by the sum of the 1 first terms of the series defining the formal solution of the problem studied in Example 28.

71 Lectures INF232 p. 71/ x Figure 6: A snapshot, at time t = 1, of the function given by the sum of the 1 first terms of the series defining the formal solution of the problem studied in Example 28.

72 Lectures INF232 p. 72/88 Stability analysis of the num. sol. We shall now study the stability properties of the explicit finite difference scheme for heat equation presented earlier As above, the discretization parameters are defined by t = T m and x = 1 n 1, and functions are only defined in the gridpoints u l i = u(x i,t l ) = u((i 1) x,l t) for i = 1,...,n and l =,...,m

73 Lectures INF232 p. 73/88 Stability analysis of the num. sol. The numerical scheme is written u l+1 i = u l i + t x 2(ul i 1 2u l i + u l i+1) = αu l i 1 +(1 2α)u l i + αu l i+1 (71) for i = 2,...,n 1 and l =,...,m 1, where Boundary conditions are u l = ul 1 α = t x 2 (72) = for l = 1,...,m We shall see that this numerical scheme is only conditionable stable, and the stability depends on the parameter α

74 Lectures INF232 p. 74/88 Example 29 Consider the following problem with the analytical solution u t = u xx for x (,1), t >, u(,t) = u(1,t) = for t >, u(x,) = sin(3πx) for x (,1), u(x,t) = e π2t sin(3πx). In Figures 7-9 we have graphed this function and the numerical results generated by the scheme (71) for various values of the discretization parameters in space and time. Notice how the solution depends on α.

75 Lectures INF232 p. 75/ x Figure 7: The solid line represents the solution of the problem studied in Example 29. The dotted, dash-dotted and dashed lines are the numerical results generated in the cases of n = 1 and m = 17 (α =.4765), n = 2 and m = 82 (α =.442), n = 6 and m = 76 (α =.4931), respectively.

76 1.5 x Figure 8: The dashed line represents the results generated by the explicit scheme (71) in the case of n = 6 and m = 681, corresponding to α =.5112, in Example 29. The solid line is the graph of the exact solution of the problem studied in this example. Lectures INF232 p. 76/88

77 Lectures INF232 p. 77/ Figure 9: A plot of the numbers generated by the explicit scheme (71), with n = 6 and m = 675, in Example 29. Observe that α =.5157 >.5 and that, for these discretization parameters, the method fails to solve the problem under consideration!

78 Lectures INF232 p. 78/88 Example 3 We reconsider the problem analyzed Example 28, and study the performance of the explicit scheme (71) applied to u t = u xx for x (,1), t >, u(,t) = u(1,t) = for t >, u(x,) = x(1 x) = x x 2 for x (,1). We compare the numerical approximations generated by (71) with the formal solution (7) for various discretization parameters t and x

79 Figure 1: Numerical results from Example 3. The solid line represents the sum of the 1 first terms of the sine series of the formal solution of the problem studied in this example. The dotted, dash-dotted and dashed curves are the numerical results generated in the cases of n = 4 and m = 3 (α =.3), n = 7 and m = 8 (α =.45), n = 14 and m = 34 (α =.4971), respectively. Lectures INF232 p. 79/88

80 Lectures INF232 p. 8/ Figure 11: The dashed line represents the results generated by the explicit scheme (71) in the case of n = 14 and m = 29, corresponding to α =.5828, in Example 3. The solid line is the graph of the Fourier based approximation of the solution.

81 Lectures INF232 p. 81/ Figure 12: A plot of the numerical results produced by the explicit scheme (71), using n = 25 grid points in the spatial dimension and m = 25 time steps, in Example 3. In this case α =.676 >.5 and the method fails to solve the problem.

82 Lectures INF232 p. 82/88 Analysis We have observed that the explicit scheme (71) works fine, provided that α 1/2 For small discretization parameters t and x, it seems to produce accurate approximations of the solution of the heat equation However, for α > 1/2 the scheme tends to break down, i.e., the numbers produced are not useful. Our goal now is to investigate this property from a theoretical point of view We will derive, provided that α 1/2, a discrete analogue to the maximum principle Note that, for (1)-(3), the maximums principle implies u(x,t) max f(x) for all x (,1) and t x

83 Lectures INF232 p. 83/88 Analysis Assume that t and x satisfy α = t x Then We introduce 1 2α (73) ū l = max i u l i for l =,...,m Note that ū = max i f(x i )

84 Lectures INF232 p. 84/88 Analysis Recall that u l+1 i = αu l i 1 +(1 2α)ul i + αul i+1 It now follows from the triangle inequality that u l+1 i = αu l i 1 +(1 2α)u l i + αu l i+1 αu l i 1 + (1 2α)u l i + αu l i+1 = α u l i 1 +(1 2α) u l i +α u l i+1 αū l +(1 2α)ū l + αū l for i = 2,...,n 1 = ū l (74) Note that u l+1 1 = u l+1 n =

85 Lectures INF232 p. 85/88 Analysis Since (74) is valid for i = 2,...,n 1, we get or max u l+1 i i ū l+1 ū l ū l Finally, by a straightforward induction argument we conclude that ū l+1 ū = max i f(x i )

86 Lectures INF232 p. 86/88 Analysis Assume that the discretization parameters t and x satisfy α = t x (75) Then the approximations generated by the explicit scheme (71) satisfy the bound max i u l i max i f(x i ) for l =,...,m, (76) where f is the initial condition in the model problem (1)-(3).

87 Lectures INF232 p. 87/88 Consequences For a given n, m must satisfy m 2T(n 1) 2 Hence, the number of time steps, m, needed increases rapidly with the number of grid points, n, used in the space dimension If T = 1 and n = 11, then m must satisfy m 2, and in the case of n = 11 at least time steps must be taken! This is no big problem in 1D, but in 2D and 3D this problem may become dramatic

88 Lectures INF232 p. 88/88

MATH 425, PRACTICE FINAL EXAM SOLUTIONS.

MATH 425, PRACTICE FINAL EXAM SOLUTIONS. MATH 45, PRACTICE FINAL EXAM SOLUTIONS. Exercise. a Is the operator L defined on smooth functions of x, y by L u := u xx + cosu linear? b Does the answer change if we replace the operator L by the operator

More information

Second Order Linear Partial Differential Equations. Part I

Second Order Linear Partial Differential Equations. Part I Second Order Linear Partial Differential Equations Part I Second linear partial differential equations; Separation of Variables; - point boundary value problems; Eigenvalues and Eigenfunctions Introduction

More information

The one dimensional heat equation: Neumann and Robin boundary conditions

The one dimensional heat equation: Neumann and Robin boundary conditions The one dimensional heat equation: Neumann and Robin boundary conditions Ryan C. Trinity University Partial Differential Equations February 28, 2012 with Neumann boundary conditions Our goal is to solve:

More information

An Introduction to Partial Differential Equations

An Introduction to Partial Differential Equations An Introduction to Partial Differential Equations Andrew J. Bernoff LECTURE 2 Cooling of a Hot Bar: The Diffusion Equation 2.1. Outline of Lecture An Introduction to Heat Flow Derivation of the Diffusion

More information

College of the Holy Cross, Spring 2009 Math 373, Partial Differential Equations Midterm 1 Practice Questions

College of the Holy Cross, Spring 2009 Math 373, Partial Differential Equations Midterm 1 Practice Questions College of the Holy Cross, Spring 29 Math 373, Partial Differential Equations Midterm 1 Practice Questions 1. (a) Find a solution of u x + u y + u = xy. Hint: Try a polynomial of degree 2. Solution. Use

More information

Rolle s Theorem. q( x) = 1

Rolle s Theorem. q( x) = 1 Lecture 1 :The Mean Value Theorem We know that constant functions have derivative zero. Is it possible for a more complicated function to have derivative zero? In this section we will answer this question

More information

Høgskolen i Narvik Sivilingeniørutdanningen STE6237 ELEMENTMETODER. Oppgaver

Høgskolen i Narvik Sivilingeniørutdanningen STE6237 ELEMENTMETODER. Oppgaver Høgskolen i Narvik Sivilingeniørutdanningen STE637 ELEMENTMETODER Oppgaver Klasse: 4.ID, 4.IT Ekstern Professor: Gregory A. Chechkin e-mail: chechkin@mech.math.msu.su Narvik 6 PART I Task. Consider two-point

More information

EXISTENCE AND NON-EXISTENCE RESULTS FOR A NONLINEAR HEAT EQUATION

EXISTENCE AND NON-EXISTENCE RESULTS FOR A NONLINEAR HEAT EQUATION Sixth Mississippi State Conference on Differential Equations and Computational Simulations, Electronic Journal of Differential Equations, Conference 5 (7), pp. 5 65. ISSN: 7-669. UL: http://ejde.math.txstate.edu

More information

Inner Product Spaces

Inner Product Spaces Math 571 Inner Product Spaces 1. Preliminaries An inner product space is a vector space V along with a function, called an inner product which associates each pair of vectors u, v with a scalar u, v, and

More information

Nonlinear Algebraic Equations. Lectures INF2320 p. 1/88

Nonlinear Algebraic Equations. Lectures INF2320 p. 1/88 Nonlinear Algebraic Equations Lectures INF2320 p. 1/88 Lectures INF2320 p. 2/88 Nonlinear algebraic equations When solving the system u (t) = g(u), u(0) = u 0, (1) with an implicit Euler scheme we have

More information

BANACH AND HILBERT SPACE REVIEW

BANACH AND HILBERT SPACE REVIEW BANACH AND HILBET SPACE EVIEW CHISTOPHE HEIL These notes will briefly review some basic concepts related to the theory of Banach and Hilbert spaces. We are not trying to give a complete development, but

More information

No: 10 04. Bilkent University. Monotonic Extension. Farhad Husseinov. Discussion Papers. Department of Economics

No: 10 04. Bilkent University. Monotonic Extension. Farhad Husseinov. Discussion Papers. Department of Economics No: 10 04 Bilkent University Monotonic Extension Farhad Husseinov Discussion Papers Department of Economics The Discussion Papers of the Department of Economics are intended to make the initial results

More information

General Theory of Differential Equations Sections 2.8, 3.1-3.2, 4.1

General Theory of Differential Equations Sections 2.8, 3.1-3.2, 4.1 A B I L E N E C H R I S T I A N U N I V E R S I T Y Department of Mathematics General Theory of Differential Equations Sections 2.8, 3.1-3.2, 4.1 Dr. John Ehrke Department of Mathematics Fall 2012 Questions

More information

Application of Fourier Transform to PDE (I) Fourier Sine Transform (application to PDEs defined on a semi-infinite domain)

Application of Fourier Transform to PDE (I) Fourier Sine Transform (application to PDEs defined on a semi-infinite domain) Application of Fourier Transform to PDE (I) Fourier Sine Transform (application to PDEs defined on a semi-infinite domain) The Fourier Sine Transform pair are F. T. : U = 2/ u x sin x dx, denoted as U

More information

PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5. PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

More information

3. INNER PRODUCT SPACES

3. INNER PRODUCT SPACES . INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space.

More information

Class Meeting # 1: Introduction to PDEs

Class Meeting # 1: Introduction to PDEs MATH 18.152 COURSE NOTES - CLASS MEETING # 1 18.152 Introduction to PDEs, Fall 2011 Professor: Jared Speck Class Meeting # 1: Introduction to PDEs 1. What is a PDE? We will be studying functions u = u(x

More information

Systems with Persistent Memory: the Observation Inequality Problems and Solutions

Systems with Persistent Memory: the Observation Inequality Problems and Solutions Chapter 6 Systems with Persistent Memory: the Observation Inequality Problems and Solutions Facts that are recalled in the problems wt) = ut) + 1 c A 1 s ] R c t s)) hws) + Ks r)wr)dr ds. 6.1) w = w +

More information

Numerical methods for American options

Numerical methods for American options Lecture 9 Numerical methods for American options Lecture Notes by Andrzej Palczewski Computational Finance p. 1 American options The holder of an American option has the right to exercise it at any moment

More information

Pacific Journal of Mathematics

Pacific Journal of Mathematics Pacific Journal of Mathematics GLOBAL EXISTENCE AND DECREASING PROPERTY OF BOUNDARY VALUES OF SOLUTIONS TO PARABOLIC EQUATIONS WITH NONLOCAL BOUNDARY CONDITIONS Sangwon Seo Volume 193 No. 1 March 2000

More information

2.3 Convex Constrained Optimization Problems

2.3 Convex Constrained Optimization Problems 42 CHAPTER 2. FUNDAMENTAL CONCEPTS IN CONVEX OPTIMIZATION Theorem 15 Let f : R n R and h : R R. Consider g(x) = h(f(x)) for all x R n. The function g is convex if either of the following two conditions

More information

TOPIC 4: DERIVATIVES

TOPIC 4: DERIVATIVES TOPIC 4: DERIVATIVES 1. The derivative of a function. Differentiation rules 1.1. The slope of a curve. The slope of a curve at a point P is a measure of the steepness of the curve. If Q is a point on the

More information

Metric Spaces. Chapter 7. 7.1. Metrics

Metric Spaces. Chapter 7. 7.1. Metrics Chapter 7 Metric Spaces A metric space is a set X that has a notion of the distance d(x, y) between every pair of points x, y X. The purpose of this chapter is to introduce metric spaces and give some

More information

Taylor and Maclaurin Series

Taylor and Maclaurin Series Taylor and Maclaurin Series In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions

More information

4. Complex integration: Cauchy integral theorem and Cauchy integral formulas. Definite integral of a complex-valued function of a real variable

4. Complex integration: Cauchy integral theorem and Cauchy integral formulas. Definite integral of a complex-valued function of a real variable 4. Complex integration: Cauchy integral theorem and Cauchy integral formulas Definite integral of a complex-valued function of a real variable Consider a complex valued function f(t) of a real variable

More information

Sequences and Series

Sequences and Series Sequences and Series Consider the following sum: 2 + 4 + 8 + 6 + + 2 i + The dots at the end indicate that the sum goes on forever. Does this make sense? Can we assign a numerical value to an infinite

More information

The Method of Least Squares. Lectures INF2320 p. 1/80

The Method of Least Squares. Lectures INF2320 p. 1/80 The Method of Least Squares Lectures INF2320 p. 1/80 Lectures INF2320 p. 2/80 The method of least squares We study the following problem: Given n points (t i,y i ) for i = 1,...,n in the (t,y)-plane. How

More information

1 if 1 x 0 1 if 0 x 1

1 if 1 x 0 1 if 0 x 1 Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or

More information

OPTIMAL CONTROL OF A COMMERCIAL LOAN REPAYMENT PLAN. E.V. Grigorieva. E.N. Khailov

OPTIMAL CONTROL OF A COMMERCIAL LOAN REPAYMENT PLAN. E.V. Grigorieva. E.N. Khailov DISCRETE AND CONTINUOUS Website: http://aimsciences.org DYNAMICAL SYSTEMS Supplement Volume 2005 pp. 345 354 OPTIMAL CONTROL OF A COMMERCIAL LOAN REPAYMENT PLAN E.V. Grigorieva Department of Mathematics

More information

Fourth-Order Compact Schemes of a Heat Conduction Problem with Neumann Boundary Conditions

Fourth-Order Compact Schemes of a Heat Conduction Problem with Neumann Boundary Conditions Fourth-Order Compact Schemes of a Heat Conduction Problem with Neumann Boundary Conditions Jennifer Zhao, 1 Weizhong Dai, Tianchan Niu 1 Department of Mathematics and Statistics, University of Michigan-Dearborn,

More information

The two dimensional heat equation

The two dimensional heat equation The two dimensional heat equation Ryan C. Trinity University Partial Differential Equations March 6, 2012 Physical motivation Consider a thin rectangular plate made of some thermally conductive material.

More information

Differentiation of vectors

Differentiation of vectors Chapter 4 Differentiation of vectors 4.1 Vector-valued functions In the previous chapters we have considered real functions of several (usually two) variables f : D R, where D is a subset of R n, where

More information

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics Undergraduate Notes in Mathematics Arkansas Tech University Department of Mathematics An Introductory Single Variable Real Analysis: A Learning Approach through Problem Solving Marcel B. Finan c All Rights

More information

Numerical Analysis Lecture Notes

Numerical Analysis Lecture Notes Numerical Analysis Lecture Notes Peter J. Olver. Finite Difference Methods for Partial Differential Equations As you are well aware, most differential equations are much too complicated to be solved by

More information

15 Kuhn -Tucker conditions

15 Kuhn -Tucker conditions 5 Kuhn -Tucker conditions Consider a version of the consumer problem in which quasilinear utility x 2 + 4 x 2 is maximised subject to x +x 2 =. Mechanically applying the Lagrange multiplier/common slopes

More information

Parabolic Equations. Chapter 5. Contents. 5.1.2 Well-Posed Initial-Boundary Value Problem. 5.1.3 Time Irreversibility of the Heat Equation

Parabolic Equations. Chapter 5. Contents. 5.1.2 Well-Posed Initial-Boundary Value Problem. 5.1.3 Time Irreversibility of the Heat Equation 7 5.1 Definitions Properties Chapter 5 Parabolic Equations Note that we require the solution u(, t bounded in R n for all t. In particular we assume that the boundedness of the smooth function u at infinity

More information

4.5 Chebyshev Polynomials

4.5 Chebyshev Polynomials 230 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION 4.5 Chebyshev Polynomials We now turn our attention to polynomial interpolation for f (x) over [ 1, 1] based on the nodes 1 x 0 < x 1 < < x N 1. Both

More information

CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

More information

1 Completeness of a Set of Eigenfunctions. Lecturer: Naoki Saito Scribe: Alexander Sheynis/Allen Xue. May 3, 2007. 1.1 The Neumann Boundary Condition

1 Completeness of a Set of Eigenfunctions. Lecturer: Naoki Saito Scribe: Alexander Sheynis/Allen Xue. May 3, 2007. 1.1 The Neumann Boundary Condition MAT 280: Laplacian Eigenfunctions: Theory, Applications, and Computations Lecture 11: Laplacian Eigenvalue Problems for General Domains III. Completeness of a Set of Eigenfunctions and the Justification

More information

Lecture 13 Linear quadratic Lyapunov theory

Lecture 13 Linear quadratic Lyapunov theory EE363 Winter 28-9 Lecture 13 Linear quadratic Lyapunov theory the Lyapunov equation Lyapunov stability conditions the Lyapunov operator and integral evaluating quadratic integrals analysis of ARE discrete-time

More information

FIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.

FIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper. FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is

More information

TMA4213/4215 Matematikk 4M/N Vår 2013

TMA4213/4215 Matematikk 4M/N Vår 2013 Norges teknisk naturvitenskapelige universitet Institutt for matematiske fag TMA43/45 Matematikk 4M/N Vår 3 Løsningsforslag Øving a) The Fourier series of the signal is f(x) =.4 cos ( 4 L x) +cos ( 5 L

More information

Duality in General Programs. Ryan Tibshirani Convex Optimization 10-725/36-725

Duality in General Programs. Ryan Tibshirani Convex Optimization 10-725/36-725 Duality in General Programs Ryan Tibshirani Convex Optimization 10-725/36-725 1 Last time: duality in linear programs Given c R n, A R m n, b R m, G R r n, h R r : min x R n c T x max u R m, v R r b T

More information

CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e.

CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. This chapter contains the beginnings of the most important, and probably the most subtle, notion in mathematical analysis, i.e.,

More information

1 The 1-D Heat Equation

1 The 1-D Heat Equation The 1-D Heat Equation 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 006 1 The 1-D Heat Equation 1.1 Physical derivation Reference: Guenther & Lee 1.3-1.4, Myint-U & Debnath.1 and.5

More information

LECTURE 15: AMERICAN OPTIONS

LECTURE 15: AMERICAN OPTIONS LECTURE 15: AMERICAN OPTIONS 1. Introduction All of the options that we have considered thus far have been of the European variety: exercise is permitted only at the termination of the contract. These

More information

Numerical PDE methods for exotic options

Numerical PDE methods for exotic options Lecture 8 Numerical PDE methods for exotic options Lecture Notes by Andrzej Palczewski Computational Finance p. 1 Barrier options For barrier option part of the option contract is triggered if the asset

More information

The degree of a polynomial function is equal to the highest exponent found on the independent variables.

The degree of a polynomial function is equal to the highest exponent found on the independent variables. DETAILED SOLUTIONS AND CONCEPTS - POLYNOMIAL FUNCTIONS Prepared by Ingrid Stewart, Ph.D., College of Southern Nevada Please Send Questions and Comments to ingrid.stewart@csn.edu. Thank you! PLEASE NOTE

More information

Economics 2020a / HBS 4010 / HKS API-111 FALL 2010 Solutions to Practice Problems for Lectures 1 to 4

Economics 2020a / HBS 4010 / HKS API-111 FALL 2010 Solutions to Practice Problems for Lectures 1 to 4 Economics 00a / HBS 4010 / HKS API-111 FALL 010 Solutions to Practice Problems for Lectures 1 to 4 1.1. Quantity Discounts and the Budget Constraint (a) The only distinction between the budget line with

More information

n k=1 k=0 1/k! = e. Example 6.4. The series 1/k 2 converges in R. Indeed, if s n = n then k=1 1/k, then s 2n s n = 1 n + 1 +...

n k=1 k=0 1/k! = e. Example 6.4. The series 1/k 2 converges in R. Indeed, if s n = n then k=1 1/k, then s 2n s n = 1 n + 1 +... 6 Series We call a normed space (X, ) a Banach space provided that every Cauchy sequence (x n ) in X converges. For example, R with the norm = is an example of Banach space. Now let (x n ) be a sequence

More information

Section 3.7. Rolle s Theorem and the Mean Value Theorem. Difference Equations to Differential Equations

Section 3.7. Rolle s Theorem and the Mean Value Theorem. Difference Equations to Differential Equations Difference Equations to Differential Equations Section.7 Rolle s Theorem and the Mean Value Theorem The two theorems which are at the heart of this section draw connections between the instantaneous rate

More information

5.4 The Heat Equation and Convection-Diffusion

5.4 The Heat Equation and Convection-Diffusion 5.4. THE HEAT EQUATION AND CONVECTION-DIFFUSION c 6 Gilbert Strang 5.4 The Heat Equation and Convection-Diffusion The wave equation conserves energy. The heat equation u t = u xx dissipates energy. The

More information

Lectures 5-6: Taylor Series

Lectures 5-6: Taylor Series Math 1d Instructor: Padraic Bartlett Lectures 5-: Taylor Series Weeks 5- Caltech 213 1 Taylor Polynomials and Series As we saw in week 4, power series are remarkably nice objects to work with. In particular,

More information

Level Set Framework, Signed Distance Function, and Various Tools

Level Set Framework, Signed Distance Function, and Various Tools Level Set Framework Geometry and Calculus Tools Level Set Framework,, and Various Tools Spencer Department of Mathematics Brigham Young University Image Processing Seminar (Week 3), 2010 Level Set Framework

More information

In this section, we will consider techniques for solving problems of this type.

In this section, we will consider techniques for solving problems of this type. Constrained optimisation roblems in economics typically involve maximising some quantity, such as utility or profit, subject to a constraint for example income. We shall therefore need techniques for solving

More information

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization

More information

THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS

THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear

More information

Scientific Computing: An Introductory Survey

Scientific Computing: An Introductory Survey Scientific Computing: An Introductory Survey Chapter 10 Boundary Value Problems for Ordinary Differential Equations Prof. Michael T. Heath Department of Computer Science University of Illinois at Urbana-Champaign

More information

Microeconomic Theory: Basic Math Concepts

Microeconomic Theory: Basic Math Concepts Microeconomic Theory: Basic Math Concepts Matt Van Essen University of Alabama Van Essen (U of A) Basic Math Concepts 1 / 66 Basic Math Concepts In this lecture we will review some basic mathematical concepts

More information

9 More on differentiation

9 More on differentiation Tel Aviv University, 2013 Measure and category 75 9 More on differentiation 9a Finite Taylor expansion............... 75 9b Continuous and nowhere differentiable..... 78 9c Differentiable and nowhere monotone......

More information

Homework # 3 Solutions

Homework # 3 Solutions Homework # 3 Solutions February, 200 Solution (2.3.5). Noting that and ( + 3 x) x 8 = + 3 x) by Equation (2.3.) x 8 x 8 = + 3 8 by Equations (2.3.7) and (2.3.0) =3 x 8 6x2 + x 3 ) = 2 + 6x 2 + x 3 x 8

More information

x a x 2 (1 + x 2 ) n.

x a x 2 (1 + x 2 ) n. Limits and continuity Suppose that we have a function f : R R. Let a R. We say that f(x) tends to the limit l as x tends to a; lim f(x) = l ; x a if, given any real number ɛ > 0, there exists a real number

More information

3 Contour integrals and Cauchy s Theorem

3 Contour integrals and Cauchy s Theorem 3 ontour integrals and auchy s Theorem 3. Line integrals of complex functions Our goal here will be to discuss integration of complex functions = u + iv, with particular regard to analytic functions. Of

More information

Calculus 1: Sample Questions, Final Exam, Solutions

Calculus 1: Sample Questions, Final Exam, Solutions Calculus : Sample Questions, Final Exam, Solutions. Short answer. Put your answer in the blank. NO PARTIAL CREDIT! (a) (b) (c) (d) (e) e 3 e Evaluate dx. Your answer should be in the x form of an integer.

More information

MATH 381 HOMEWORK 2 SOLUTIONS

MATH 381 HOMEWORK 2 SOLUTIONS MATH 38 HOMEWORK SOLUTIONS Question (p.86 #8). If g(x)[e y e y ] is harmonic, g() =,g () =, find g(x). Let f(x, y) = g(x)[e y e y ].Then Since f(x, y) is harmonic, f + f = and we require x y f x = g (x)[e

More information

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS

SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS A second-order linear differential equation has the form 1 Px d y dx dy Qx dx Rxy Gx where P, Q, R, and G are continuous functions. Equations of this type arise

More information

On computer algebra-aided stability analysis of dierence schemes generated by means of Gr obner bases

On computer algebra-aided stability analysis of dierence schemes generated by means of Gr obner bases On computer algebra-aided stability analysis of dierence schemes generated by means of Gr obner bases Vladimir Gerdt 1 Yuri Blinkov 2 1 Laboratory of Information Technologies Joint Institute for Nuclear

More information

CITY UNIVERSITY LONDON. BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION

CITY UNIVERSITY LONDON. BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION No: CITY UNIVERSITY LONDON BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION ENGINEERING MATHEMATICS 2 (resit) EX2005 Date: August

More information

Notes on metric spaces

Notes on metric spaces Notes on metric spaces 1 Introduction The purpose of these notes is to quickly review some of the basic concepts from Real Analysis, Metric Spaces and some related results that will be used in this course.

More information

y cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx

y cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx Trigonometric Integrals In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE Evaluate cos 3 x dx.

More information

The continuous and discrete Fourier transforms

The continuous and discrete Fourier transforms FYSA21 Mathematical Tools in Science The continuous and discrete Fourier transforms Lennart Lindegren Lund Observatory (Department of Astronomy, Lund University) 1 The continuous Fourier transform 1.1

More information

Lecture 4: BK inequality 27th August and 6th September, 2007

Lecture 4: BK inequality 27th August and 6th September, 2007 CSL866: Percolation and Random Graphs IIT Delhi Amitabha Bagchi Scribe: Arindam Pal Lecture 4: BK inequality 27th August and 6th September, 2007 4. Preliminaries The FKG inequality allows us to lower bound

More information

A QUICK GUIDE TO THE FORMULAS OF MULTIVARIABLE CALCULUS

A QUICK GUIDE TO THE FORMULAS OF MULTIVARIABLE CALCULUS A QUIK GUIDE TO THE FOMULAS OF MULTIVAIABLE ALULUS ontents 1. Analytic Geometry 2 1.1. Definition of a Vector 2 1.2. Scalar Product 2 1.3. Properties of the Scalar Product 2 1.4. Length and Unit Vectors

More information

MATH 4330/5330, Fourier Analysis Section 11, The Discrete Fourier Transform

MATH 4330/5330, Fourier Analysis Section 11, The Discrete Fourier Transform MATH 433/533, Fourier Analysis Section 11, The Discrete Fourier Transform Now, instead of considering functions defined on a continuous domain, like the interval [, 1) or the whole real line R, we wish

More information

e.g. arrival of a customer to a service station or breakdown of a component in some system.

e.g. arrival of a customer to a service station or breakdown of a component in some system. Poisson process Events occur at random instants of time at an average rate of λ events per second. e.g. arrival of a customer to a service station or breakdown of a component in some system. Let N(t) be

More information

24. The Branch and Bound Method

24. The Branch and Bound Method 24. The Branch and Bound Method It has serious practical consequences if it is known that a combinatorial problem is NP-complete. Then one can conclude according to the present state of science that no

More information

RANDOM INTERVAL HOMEOMORPHISMS. MICHA L MISIUREWICZ Indiana University Purdue University Indianapolis

RANDOM INTERVAL HOMEOMORPHISMS. MICHA L MISIUREWICZ Indiana University Purdue University Indianapolis RANDOM INTERVAL HOMEOMORPHISMS MICHA L MISIUREWICZ Indiana University Purdue University Indianapolis This is a joint work with Lluís Alsedà Motivation: A talk by Yulij Ilyashenko. Two interval maps, applied

More information

1. Let X and Y be normed spaces and let T B(X, Y ).

1. Let X and Y be normed spaces and let T B(X, Y ). Uppsala Universitet Matematiska Institutionen Andreas Strömbergsson Prov i matematik Funktionalanalys Kurs: NVP, Frist. 2005-03-14 Skrivtid: 9 11.30 Tillåtna hjälpmedel: Manuella skrivdon, Kreyszigs bok

More information

y cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx y 1 u 2 du u 1 3u 3 C

y cos 3 x dx y cos 2 x cos x dx y 1 sin 2 x cos x dx y 1 u 2 du u 1 3u 3 C Trigonometric Integrals In this section we use trigonometric identities to integrate certain combinations of trigonometric functions. We start with powers of sine and cosine. EXAMPLE Evaluate cos 3 x dx.

More information

Algebra Unpacked Content For the new Common Core standards that will be effective in all North Carolina schools in the 2012-13 school year.

Algebra Unpacked Content For the new Common Core standards that will be effective in all North Carolina schools in the 2012-13 school year. This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these tools to better serve teachers. Algebra

More information

Metric Spaces. Chapter 1

Metric Spaces. Chapter 1 Chapter 1 Metric Spaces Many of the arguments you have seen in several variable calculus are almost identical to the corresponding arguments in one variable calculus, especially arguments concerning convergence

More information

Factoring Polynomials

Factoring Polynomials Factoring Polynomials Hoste, Miller, Murieka September 12, 2011 1 Factoring In the previous section, we discussed how to determine the product of two or more terms. Consider, for instance, the equations

More information

The Exponential Distribution

The Exponential Distribution 21 The Exponential Distribution From Discrete-Time to Continuous-Time: In Chapter 6 of the text we will be considering Markov processes in continuous time. In a sense, we already have a very good understanding

More information

Date: April 12, 2001. Contents

Date: April 12, 2001. Contents 2 Lagrange Multipliers Date: April 12, 2001 Contents 2.1. Introduction to Lagrange Multipliers......... p. 2 2.2. Enhanced Fritz John Optimality Conditions...... p. 12 2.3. Informative Lagrange Multipliers...........

More information

Numerical Methods for Differential Equations

Numerical Methods for Differential Equations Numerical Methods for Differential Equations Chapter 1: Initial value problems in ODEs Gustaf Söderlind and Carmen Arévalo Numerical Analysis, Lund University Textbooks: A First Course in the Numerical

More information

Math 55: Discrete Mathematics

Math 55: Discrete Mathematics Math 55: Discrete Mathematics UC Berkeley, Spring 2012 Homework # 9, due Wednesday, April 11 8.1.5 How many ways are there to pay a bill of 17 pesos using a currency with coins of values of 1 peso, 2 pesos,

More information

A REMARK ON ALMOST MOORE DIGRAPHS OF DEGREE THREE. 1. Introduction and Preliminaries

A REMARK ON ALMOST MOORE DIGRAPHS OF DEGREE THREE. 1. Introduction and Preliminaries Acta Math. Univ. Comenianae Vol. LXVI, 2(1997), pp. 285 291 285 A REMARK ON ALMOST MOORE DIGRAPHS OF DEGREE THREE E. T. BASKORO, M. MILLER and J. ŠIRÁŇ Abstract. It is well known that Moore digraphs do

More information

Introduction to Partial Differential Equations By Gilberto E. Urroz, September 2004

Introduction to Partial Differential Equations By Gilberto E. Urroz, September 2004 Introduction to Partial Differential Equations By Gilberto E. Urroz, September 2004 This chapter introduces basic concepts and definitions for partial differential equations (PDEs) and solutions to a variety

More information

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

More information

INTERPOLATION. Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y).

INTERPOLATION. Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y). INTERPOLATION Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y). As an example, consider defining and x 0 =0, x 1 = π 4, x

More information

N 1. (q k+1 q k ) 2 + α 3. k=0

N 1. (q k+1 q k ) 2 + α 3. k=0 Teoretisk Fysik Hand-in problem B, SI1142, Spring 2010 In 1955 Fermi, Pasta and Ulam 1 numerically studied a simple model for a one dimensional chain of non-linear oscillators to see how the energy distribution

More information

An important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate.

An important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate. Chapter 10 Series and Approximations An important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate 1 0 e x2 dx, you could set

More information

Rotation Rate of a Trajectory of an Algebraic Vector Field Around an Algebraic Curve

Rotation Rate of a Trajectory of an Algebraic Vector Field Around an Algebraic Curve QUALITATIVE THEORY OF DYAMICAL SYSTEMS 2, 61 66 (2001) ARTICLE O. 11 Rotation Rate of a Trajectory of an Algebraic Vector Field Around an Algebraic Curve Alexei Grigoriev Department of Mathematics, The

More information

tegrals as General & Particular Solutions

tegrals as General & Particular Solutions tegrals as General & Particular Solutions dy dx = f(x) General Solution: y(x) = f(x) dx + C Particular Solution: dy dx = f(x), y(x 0) = y 0 Examples: 1) dy dx = (x 2)2 ;y(2) = 1; 2) dy ;y(0) = 0; 3) dx

More information

Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

More information

Estimating the Average Value of a Function

Estimating the Average Value of a Function Estimating the Average Value of a Function Problem: Determine the average value of the function f(x) over the interval [a, b]. Strategy: Choose sample points a = x 0 < x 1 < x 2 < < x n 1 < x n = b and

More information

INDISTINGUISHABILITY OF ABSOLUTELY CONTINUOUS AND SINGULAR DISTRIBUTIONS

INDISTINGUISHABILITY OF ABSOLUTELY CONTINUOUS AND SINGULAR DISTRIBUTIONS INDISTINGUISHABILITY OF ABSOLUTELY CONTINUOUS AND SINGULAR DISTRIBUTIONS STEVEN P. LALLEY AND ANDREW NOBEL Abstract. It is shown that there are no consistent decision rules for the hypothesis testing problem

More information

1. Prove that the empty set is a subset of every set.

1. Prove that the empty set is a subset of every set. 1. Prove that the empty set is a subset of every set. Basic Topology Written by Men-Gen Tsai email: b89902089@ntu.edu.tw Proof: For any element x of the empty set, x is also an element of every set since

More information

A characterization of trace zero symmetric nonnegative 5x5 matrices

A characterization of trace zero symmetric nonnegative 5x5 matrices A characterization of trace zero symmetric nonnegative 5x5 matrices Oren Spector June 1, 009 Abstract The problem of determining necessary and sufficient conditions for a set of real numbers to be the

More information

MATH PROBLEMS, WITH SOLUTIONS

MATH PROBLEMS, WITH SOLUTIONS MATH PROBLEMS, WITH SOLUTIONS OVIDIU MUNTEANU These are free online notes that I wrote to assist students that wish to test their math skills with some problems that go beyond the usual curriculum. These

More information