4 1-D Boundary Value Problems Heat Equation

Transcription

1 4 -D Boundary Vaue Probems Heat Equation The main purpose of this chapter is to study boundary vaue probems for the heat equation on a finite rod a x b. u t (x, t = ku xx (x, t, a < x < b, t > u(x, = ϕ(x The main new ingredient is that physica constraints caed boundary conditions must be imposed at the ends of the rod. The two main conditions are u(a, t =, u(b, t = Dirichet Conditions u x (a, t =, u x (b, t = Neumann Conditions We can aso have any combination of these conditions, i.e., we coud have a Dirichet condition at x = a and Neumann condition at x = b. The is one additiona important boundary condition u x (a, t k u(a, t =, u x (b, t + k u(b, t = Robin Conditions Finay we wi a more genera probem invoving two extra terms that correspond to heat conduction and convection. u t (x, t = k ( u xx (x, t 2au(x, t x + bu(x, t, < x <, t > u(x, = ϕ(x. The basic methodoogy presented here is the idea of eigenvaues and eigenvector expansions as presented in a inear agebra or differentia equations cass when studying inear systems of ODEs. The basic idea can be described by an exampe. Let X be an n-vector vaued function and A an n n matrix. Aso et X denote a constant initia condition n-vector. Then to sove the initia vaue probem d dt X = AX, X( = X we find the eigenvaues {λ j } n j= and eigenvectors {Φ j } n j= of A (i.e. AΦ j = λ j Φ j where we aso assume { Φ j, Φ k = Φ j = k Φ k = δ jk j k. Then, assuming there are n ineary independent eigenvectors, the soution can be written n X(t = e λjt X, Φ j Φ j. j= The generaization of this idea to the one dimensiona heat equation invoves the generaized theory of Fourier series. This method due to Fourier was deveop to sove the heat equation and it is one of the most successfu ideas in mathematics. We wi begin our study with cassica Fourier series and then turn to the heat equation and Fourier s idea of separation of variabes.

2 4. Fourier Series A series of the functions φ = 2, φ( n = cos(nx, φ (2 n = sin(nx, for n written in a series 2 a + ( an cos(nx + b n sin(nx is known as a Fourier series. (We choose φ = so a of the functions have the same norm. 2 A fairy genera cass of functions can be expanded in Fourier series. Let f(x be a function defined on < x < π. Assume that f(x can be expanded in a Fourier series f(x 2 a + ( an cos(nx + b n sin(nx. (4. Here the means has the Fourier series. We have not said if the series converges yet. For now et s assume that the series converges uniformy so we can repace the with an =. We integrate Equation 4. from to π to determine a. f(x dx = 2 a = πa + = πa dx + a n cos(nx + b n sin(nx dx (a n cos(nx dx + b n sin(nx dx So that a = π f(x dx. Mutipying by cos(mx and integrating wi enabe us to sove for a m. f(x cos(mx dx = 2 a + cos(mx dx (a n cos(nx cos(mx dx + b n sin(nx cos(mx dx A but one of the terms on the right side vanishes due to the orthogonaity of the functions. f(x cos(mx dx = a m cos(mx cos(mx dx = a m = πa m ( + cos(2mx dx 2 2

3 So that a m = π f(x cos(mx dx m =,, 2,... Simiary, we can mutipy by sin(mx and integrate to sove for b m. The resut is b m = π a n and b n are caed Fourier coefficients. f(x sin(mx dx m =, 2, 3,.... Athough we wi not show it, Fourier series converge for a fairy genera cass of functions. Let f(x denote the eft imit of f(x and f(x + denote the right imit. Exampe 4.. For the function defined { for x <, f(x = x + for x, the eft and right imits at x = are f( =, f( + =. Theorem 4.. Let f(x be a 2π-periodic function for which Fourier coefficients a n = π f(x cos(nx dx, b n = π f(x dx exists. Define the f(x sin(nx dx. If x is an interior point of an interva on which f(x has imited tota fuctuation, then the Fourier series of f(x a 2 + ( an cos(nx + b n sin(nx, converges to 2 (f(x + f(x +. If f is continuous at x, then the series converges to f(x. Exampe 4.2. Consider the function defined by { x for π x < f(x = π 2x for x < π. The Fourier series converges to the function defined by for x = x for π < x < ˆf(x = π/2 for x = π 2x for < x < π. The function ˆf(x is potted in the foowing figure. 3

4 Figure : Graph of ˆf(x. A ot can be earned about the Fourier coefficients from the geometry of the function. For exampe, if f(x is an even function, (f( x = f(x, then there wi not be any sine terms in the Fourier series for f(x. The Fourier sine coefficient is b n = π f(x sin(nx dx. Since f(x is an even function and sin(nx is odd, f(x sin(nx is odd. b n is the integra of an odd function from to π and is thus zero. Aso we can simpify the cosine coefficients, a n = π = 2 π f(x cos(nx dx f(x cos(nx dx. Exampe 4.3. Consider the function defined on [, π by { x for x < π/2 f(x = π x for π/2 x < π. The Fourier cosine coefficients for this function are /2 a n = 2 x cos(nx dx + 2 π π π 4 = 8 ( ( πn cos sin π/2 (π x cos(nx dx for n =, for n. In Figure 3 the even periodic extension of f(x is potted in a dashed ine and the sum of the first five nonzero terms in the Fourier cosine series are potted in a soid ine. 4

5 Figure 3: Fourier Cosine Series. If, on the other hand, f(x is an odd function, (f( x = f(x, then there wi not be any cosine terms in the Fourier series. Since f(x cos(nx is an odd function, the cosine coefficients wi be zero. Since f(x sin(nx is an even function,we can rewrite the sine coefficients b n = 2 π f(x sin(nx dx. Exampe 4.4. Consider the function defined on [, π by { x for x < π/2 f(x = π x for π/2 x < π. The Fourier sine coefficients for this function are /2 b n = 2 x sin(nx dx + 2 (π x sin(nx dx π π π/2 = 6 ( ( πn cos sin In Figure 4 the odd periodic extension of f(x is potted in a dashed ine and the sum of the first five nonzero terms in the Fourier sine series are potted in a soid ine Figure 4: Fourier Sine Series. 5

6 4.2 Fourier series for arbitrary interva The above formuas can easiy be extended to periodic functions with an arbitrary period 2 defined on [, ] and, using even and odd extensions, we can even write Fourier series expansions for functions defined on an interva [, ]. We present these formuas beow.. For f(x = f(x + 2 and f is piecewise smooth on x. Then for a x R where f(x+ + f(x 2 = a + { ( x ( x } a n cos + b n sin, a = 2 a n = b n = f(x dx, ( x f(x cos dx ( x f(x sin dx. If f is continuous (at x then f(x+ + f(x 2 = f(x. 2. Fourier Cosine: Given f on [, ] a = f(x dx, f(x+ + f(x 2 a n = 2 = a + ( x f(x cos dx ( x a n cos 3. Fourier Sine: Given f on [, ] b n = 2 ( x f(x sin dx f(x+ + f(x 2 = ( x b n sin 6

7 4.3 Heat Equation Dirichet Boundary Conditions u t (x, t = ku xx (x, t, < x <, t > (4.2 u(, t =, u(, t = u(x, = ϕ(x. Separate Variabes Look for simpe soutions in the form u(x, t = ϕ(xψ(t. Substituting into (4.4 and dividing both sides by ϕ(xψ(t gives ψ(t kψ(t = ϕ (x ϕ(x Since the eft side is independent of x and the right side is independent of t, it foows that the expression must be a constant: ψ(t kψ(t = ϕ (x ϕ(x = λ. (Here ψ means the derivative of ψ with respect to t and ϕ means means the derivative of ϕ with respect to x. We seek to find a possibe constants λ and the corresponding nonzero functions ϕ and ψ. We obtain ϕ λϕ =, ψ kλψ =. The soution of the second equation is ψ(t = Ce kλt (4.3 where C is an arbitrary constant. Furthermore, the boundary conditions give ϕ(ψ(t =, ϕ(ψ(t = for a t. Since ψ(t is not identicay zero we obtain the desired eigenvaue probem ϕ (x λϕ(x =, ϕ( =, ϕ( =. ( Find Eigenvaues and Eignevectors The next main step is to find the eigenvaues and eigenfunctions from (4.6. There are, in genera, three cases: (a If λ = then ϕ(x = ax + b so appying the boundary conditions we get Zero is not an eigenvaue. = ϕ( = b, = ϕ( = a a = b =. 7

8 (b If λ = µ 2 > then ϕ(x = a cosh(µx + b sinh(µx. Appying the boundary conditions we have = ϕ( = a a = = ϕ( = b sinh(µ b =. Therefore, there are no positive eigenvaues. Consider the foowing aternative argument: If ϕ (x = λϕ(x then mutipying by ϕ we have ϕ(xϕ (x = λϕ(x 2. Integrate this expression from x = to x =. We have λ ϕ(x 2 dx = ϕ(xϕ (x dx = ϕ (x 2 dx + ϕ(xϕ (x. Since ϕ( = ϕ( = we concude λ = ϕ (x 2 dx ϕ(x 2 dx and we see that λ must be ess than or equa to zero. (c So, finay, consider λ = µ 2 so that ϕ(x = a cos(µx + b sin(µx. Appying the boundary conditions we have = ϕ( = a a = = ϕ( = b sin(µ. From this we concude sin(µ = which impies µ = and therefore ( λ n = µ 2 n = 2, ϕn (x = b n sin(µ n x, n =, 2,.. (4.5 From (4.5 we aso have the associated functions ψ n (t = e kλnt. 3. Write Forma Sum From the above considerations we can concude that for any integer N and constants {b n } N n= u N (x, t = N b n ψ n (tϕ n (x = N ( x b n e kλnt sin. satisfies the differentia equation in (4.4 and the boundary conditions. 8

9 4. Use Fourier Series to Find Coefficients The ony probem remaining is to somehow pick the constants b n so that the initia condition u(x, = ϕ(x is satisfied. To do this we consider what we earned from Fourier series. In particuar we ook for u as an infinite sum ( x u(x, t = b n e kλnt sin and we try to find {b n } satisfying ϕ(x = u(x, = ( x b n sin. But this nothing more than a Sine expansion of the function ϕ on the interva (,. b n = 2 ( x ϕ(x sin dx. (4.6 Exampe 4.5. As an expicit exampe for( the initia condition consider =, k = / and ϕ(x = x( x. Let us reca that µ n = which in this case reduces to. b n = 2 = 2 = 2 [ x( x sin (x dx ( cos (x x( x dx x( x cos(x + ] ( 2x cos(x dx µ n = 2 = 2 [ = 4 ( 2 ( sin (x ( 2x dx ( 2x sin(x sin(x dx = 4 ( 2 ( 2 sin (x dx [ cos(x ] ] = 4 [ ( n ] ( 3 We arrive at the soution where u(x, t = 4 π 3 x( x = 4 π 3 [ ( n ] n 3 e n2 π2 t/ sin (x. (4.7 [ ( n ] n 3 sin (x. 9

10 As an exampe with N = 3 we have x( x 8 ( sin(πx + sin(3πx. π 3 27 In the foowing figure we pot the eft and right hand side of the above x Finay we pot the approximate soution at times t =, t =, t = 2, t = 3 u(x, t = 4 3 [ ( n ] e n2π2t/ sin (x. π 3 n x 4.4 Heat Equation Neumann Boundary Conditions u t (x, t = u xx (x, t, < x <, t > (4.8 u x (, t =, u x (, t = u(x, = ϕ(x

11 . Separate Variabes Look for simpe soutions in the form u(x, t = ϕ(xψ(t. Substituting into (4.8 and dividing both sides by ϕ(xψ(t gives ψ(t ψ(t = ϕ (x ϕ(x Since the eft side is independent of x and the right side is independent of t, it foows that the expression must be a constant: ψ(t ψ(t = ϕ (x ϕ(x = λ. (Here ψ means the derivative of ψ with respect to t and ϕ means means the derivative of ϕ with respect to x. We seek to find a possibe constants λ and the corresponding nonzero functions ϕ and ψ. We obtain The soution of the second equation is ϕ λϕ =, ψ λψ =. ψ(t = Ce λt (4.9 where C is an arbitrary constant. Furthermore, the boundary conditions give ϕ (ψ(t =, ϕ (ψ(t = for a t. Since ψ(t is not identicay zero we obtain the desired eigenvaue probem ϕ (x λϕ(x =, ϕ ( =, ϕ ( =. (4. 2. Find Eigenvaues and Eignevectors The next main step is to find the eigenvaues and eigenfunctions from (4.. There are, in genera, three cases: (a If λ = then ϕ(x = ax + b so appying the boundary conditions we get = ϕ ( = a, = ϕ ( = a a =. Notice that b is sti an arbitrary constant. We concude that λ = is an eigenvaue with eigenfunction ϕ (x =. (b If λ = µ 2 > then and ϕ(x = a cosh(µx + b sinh(µx ϕ (x = aµ sinh(µx + bµ cosh(µx. Appying the boundary conditions we have = ϕ ( = bµ b = = ϕ ( = aµ sinh(µ a =.

12 Therefore, there are no positive eigenvaues. Consider the foowing aternative argument: If ϕ (x = λϕ(x then mutipying by ϕ we have ϕ(xϕ (x = λϕ(x 2. Integrate this expression from x = to x =. We have λ ϕ(x 2 dx = ϕ(xϕ (x dx = ϕ (x 2 dx + ϕ(xϕ (x. Since ϕ ( = ϕ ( = we concude λ = ϕ (x 2 dx ϕ(x 2 dx and we see that λ must be ess than or equa to zero ( zero ony if ϕ =. (c So, finay, consider λ = µ 2 so that and ϕ(x = a cos(µx + b sin(µx ϕ (x = aµ sin(µx + bµ cos(µx. Appying the boundary conditions we have = ϕ ( = bµ b = = ϕ ( = aµ sin(µ. From this we concude sin(µ = which impies µ = and therefore ( λ n = µ 2 n = 2, ϕn (x = cos(µ n x, n =, 2,.. (4. From (4.9 we aso have the associated functions ψ n (t = e λnt. 3. Write Forma Infinite Sum From the above considerations we can concude that for any integer N and constants {a n } N n= u n (x, t = a N 2 + a n ψ n (tϕ n (x = a + N ( x a n e λnt cos. satisfies the differentia equation in (4.8 and the boundary conditions. 4. Use Fourier Series to Find Coefficients The ony probem remaining is to somehow pick the constants a n so that the initia condition u(x, = ϕ(x is satisfied. To do this we consider what we earned from Fourier series. In particuar we ook for u as an infinite sum ( x u(x, t = a + a n e λnt cos 2

13 and we try to find {a n } satisfying ϕ(x = u(x, = a + ( x a n cos. But this nothing more than a Cosine expansion of the function ϕ on the interva (,. Our work on Fourier series showed us that a = 2 ϕ(x dx, a n = 2 ( x ϕ(x cos dx. (4.2 As an expicit exampe for the initia condition consider = and ϕ(x = x( x. In this case (4.2 becomes We have and a n = 2 a = 2 a = 2 ϕ(x dx, a n = 2 ϕ(x dx = 2 [ ] x 2 = 2 2 x3 = 3 3. ϕ(x cos (x dx = 2 ϕ(x cos (x dx. x( x dx x( x cos (x dx = 2 = 2 [ = 2 = 2 ( sin (x x( x dx x( x sin (x ( cos (x ( 2x dx [ cos (x ( 2x ( 2 = 2 [ cos ( ] = 2 ( 2 (( n + = ] sin (x ( 2x dx cos (x 4 (, n even 2., n odd dx ] 3

14 In order to eiminate the odd terms in the expansion we introduce a new index, k by n = 2k where k =, 2,. So finay we arrive at the soution u(x, t = 6 π 2 k= k 2 e 4k2 π2t cos(2kπx. (4.3 As an exampe with N = 4 we have x( x 6 π 2 ( 4 cos(2kπx. k 2 Notice that as t the infinite sum converges to zero uniformy in x. Indeed, π2t k 2 e 4k2 cos(2kπx t e 4π2 k = π2 2 6 e 4π2t. k= So the soution converges to a nonzero steady state temperature which is exacty the average vaue of the initia temperature distribution. k= im u(x, t = t 6 = ϕ(x dx. In the foowing figure we pot the eft and right hand side of the above x Finay we pot the approximate soution at times t =, t = /, t = 2/, t = 3/. 4

15 x 4.5 Heat Equation Dirichet-Neumann Boundary Conditions u t (x, t = u xx (x, t, < x <, t > (4.4 u(, t =, u x (, t = u(x, = ϕ(x. Separate Variabes Look for simpe soutions in the form u(x, t = ϕ(xψ(t. Substituting into (4.4 and dividing both sides by ϕ(xψ(t gives ψ(t ψ(t = ϕ (x ϕ(x Since the eft side is independent of x and the right side is independent of t, it foows that the expression must be a constant: ψ(t ψ(t = ϕ (x ϕ(x = λ. (Here ψ means the derivative of ψ with respect to t and ϕ means means the derivative of ϕ with respect to x. We seek to find a possibe constants λ and the corresponding nonzero functions ϕ and ψ. We obtain ϕ λϕ =, ψ λψ =. The soution of the second equation is ψ(t = Ce λt (4.5 5

16 where C is an arbitrary constant. Furthermore, the boundary conditions give ϕ(ψ(t =, ϕ (ψ(t = for a t. Since ψ(t is not identicay zero we obtain the desired eigenvaue probem ϕ (x λϕ(x =, ϕ( =, ϕ ( =. ( Find Eigenvaues and Eignevectors The next main step is to find the eigenvaues and eigenfunctions from (4.6. There are, in genera, three cases: (a If λ = then ϕ(x = ax + b so appying the boundary conditions we get Zero is not an eigenvaue. (b If λ = µ 2 > then and = ϕ( = b, = ϕ ( = a a = b =. ϕ(x = a cosh(µx + b sinh(µx ϕ (x = aµ sinh(µx + bµ cosh(µx. Appying the boundary conditions we have = ϕ ( = aµ a = = ϕ ( = bµ cosh(µ b =. Therefore, there are no positive eigenvaues. Consider the foowing aternative argument: If ϕ (x = λϕ(x then mutipying by ϕ we have ϕ(xϕ (x = λϕ(x 2. Integrate this expression from x = to x =. We have λ ϕ(x 2 dx = ϕ(xϕ (x dx = ϕ (x 2 dx + ϕ(xϕ (x. Since ϕ( = ϕ ( = we concude λ = ϕ (x 2 dx ϕ(x2 dx and we see that λ must be ess than or equa to zero. (c So, finay, consider λ = µ 2 so that and ϕ(x = a cos(µx + b sin(µx ϕ (x = aµ sin(µx + bµ cos(µx. Appying the boundary conditions we have = ϕ( = aµ a = = ϕ ( = bµ cos(µ. 6

17 From this we concude cos(µ = which impies and therefore µ = (2n π 2 ( 2 (2n π λ n = µ 2 n =, ϕ n (x = sin(µ n x, n =, 2,.. (4.7 2 From (4.5 we aso have the associated functions ψ n (t = e λnt. 3. Write Forma Infinite Sum From the above considerations we can concude that for any integer N and constants {b n } N n= u n (x, t = N b n ψ n (tϕ n (x = N ( (2n πx b n e λnt sin. 2 satisfies the differentia equation in (4.4 and the boundary conditions. 4. Use Fourier Series to Find Coefficients The ony probem remaining is to somehow pick the constants b n so that the initia condition u(x, = ϕ(x is satisfied. To do this we consider what we earned from Fourier series. In particuar we ook for u as an infinite sum ( (2n πx u(x, t = b n e λnt sin 2 and we try to find {b n } satisfying ϕ(x = u(x, = ( (2n πx b n sin. 2 But this nothing more than a Sine type expansion of the function ϕ on the interva (,. Using ( (2n πx ϕ n (x = sin 2 we have ϕ(x = b k ϕ k (x. k= We proceed as usua by mutipying both sides by ϕ n (x and integrating from to and using the orthogonaity (described beow in (4.9, (4.2. ϕ n (xϕ(x dx = b k ϕ n (xϕ k (x dx k= which impies b n = 2 ϕ(xϕ n (x dx. (4.8 7

18 Orthogonaity: ϕ n (xϕ k (x dx = 2, n = k., n k To see this reca that ϕ j = λ j ϕ j and ϕ j ( =, ϕ j( =. First consider n k so λ n λ k and therefore λ n ϕ n (xϕ k (x dx = = = ϕ n(xϕ k (x dx (4.9 ϕ n(xϕ k(x dx + [ϕ n(xϕ k (x] ϕ n (xϕ k(x dx + [ϕ n(xϕ k (x ϕ n (xϕ k(x] Therefore = λ k ϕ n (xϕ k (x dx. (4.2 (λ n λ k For n = k we have ϕ 2 n(x dx = ϕ n (xϕ k (x dx = = 2 ϕ n (xϕ k (x dx =. ( (2n πx sin 2 dx (4.2 2 ( ( (2n πx cos dx = 2 ( ( (2n πx sin 2 (2n π = 2. (4.22 As an( expicit exampe for the initia condition consider ϕ(x = x. (2n π µ n = 2 Let us reca that 8

19 b n = 2 = 2 = 2 [ ϕ(xϕ n (x dx = 2 ( x cos (µ nx dx µ n x cos(µ nx µ n + [ = 2 cos(µ n + sin(µ ] nx µ n = 2 µ 2 n [ cos((2n π/2 + µ n = 8( n+ (2n 2 π 2 x sin (µ n x dx ] cos(µ n x dx µ n ] sin((2n π/2 µ 2 n We arrive at the soution u(x, t = 8 π 2 ( n+ (2n 2 eλnt sin (µ n x. ( Heat Equation Genera Boundary Conditions: Sturm Liouvie Theory We consider the heat equation on the interva (a, b with genera boundary conditions. These notes use the materia found in the book in Sections u t (x, t = u xx (x, t q(xu(x, t, a < x < b, t > (4.24 α u(a, t α 2 u x (a, t =, (4.25 β u(b, t + β 2 u x (b, t =, (4.26 u(x, = ϕ(x (4.27 where we assume that α 2 + α 2 2 and β 2 + β 2 2. If we carry out our standard procedure of separation of variabes we seek u(x, t = ϕ(xψ(t. Substituting into (4.24 and dividing both sides by ϕ(xψ(t gives ψ(t ψ(t = ϕ (x q(xϕ(x ϕ(x 9

20 Since the eft side is independent of x and the right side is independent of t, it foows that the expression must be a constant: ψ(t ψ(t = ϕ (x q(xϕ(x ϕ(x (Here ψ means the derivative of ψ with respect to t and ϕ means means the derivative of ϕ with respect to x. We seek to find a possibe constants λ and the corresponding nonzero functions ϕ and ψ. = λ. We obtain ϕ q(xϕ(x λϕ =, ψ λψ =. The soution of the second equation is ψ(t = Ce kλt (4.28 where C is an arbitrary constant. Furthermore, the boundary conditions give (α ϕ(a kϕ(aψ(t =, (β ϕ(b + β 2 ϕ(bψ(t = for a t. Since ψ(t is not identicay zero we obtain the desired eigenvaue probem ϕ (x q(xϕ(x λϕ(x =, α ϕ(a α 2 ϕ(a =, β ϕ(b + β 2 ϕ(b = (4.29 This is an eigenvaue probem which is referred to as a Reguar Sturm-Liouvie probem. Theorem 4.2. The probem (4.29 has infinitey many eigenpairs {λ n, ϕ n (x} which satisfy the foowing properties:. The eigenvaues are a simpe, i.e. they are eigenvaues of mutipicity one which means that λ j λ k for j k. 2. The eigenvaues are a rea and a but a finite number are negative. If α, α 2, β, β 2 then a the eigenvaues are ess than or equa to zero. 3. If we order the eigenvaues in decreasing order by λ n < λ n < < λ 2 < λ then λ n as n. 4. The nth eigenfunction ϕ n (x is rea and has exacty (n zeros in the interva (a, b 5. The eigenfunctions are orthogona in the foowing sense b a ϕ n (xϕ m (x dx = for n m. 2

21 6. If ϕ(x is piecewise smooth (P C ( (a, b in my notation in cass, then (ϕ(x+ + ϕ(x 2 = c n ϕ n (x, a < x < b, where c n = b f(xϕ a n(x dx b a ϕ2 n(x dx. At east for continuous ϕ, Theorem 4.2 aows us to concude that the soution to (4.24-(4.27 is given by u(x, t = c n e λnt ϕ n (x. 4.7 Heat Equation with Conduction and Convection We consider the heat equation on the interva (, with two extra terms that correspond to heat conduction and convection. u t (x, t = k ( u xx (x, t 2au(x, t x + bu(x, t, < x <, t > (4.3 u(, t =, u(, t =, (4.3 (4.32 u(x, = ϕ(x. (4.33 There are many different ways to approach this probem and one woud be to appy separation of variabe directy. The dissadvantange to this is that one gets a more compicated ode for ϕ(x and there is a more difficut anaysis of the eigenvaues and eigenvectors. We wi take a different approach which aows us to use our earier work after a change of dependent variabes. So to this end et us define v(x, t via u(x, t = e ax+βt v(x, t, β = k(b a 2. (4.34 Thus we have and we can compute v(x, t = e (ax+βt u(x, t v t kv xx = e (ax+βt ( βu + u t k [ e (ax+βt ( au + u x ] x = e (ax+βt {( βu + u t k [ a( au + u x + ( au x + u xx ]} = e (ax+βt [ u t k(u xx 2au x + a 2 u + βu ] = e (ax+βt [ u t k(u xx 2au x + a 2 u + (b a 2 u ] = e (ax+βt [u t k(u xx 2au x + bu] =. 2

22 Furthermore and Therefore, v(x, t is the soution of v(, t = e βt u(, t =, v(, t = e (a+βt u(, t = We have eigenvaues and eigenfunctions v(x, = e ax u(x, = e ax ϕ(x. v t = kv xx v(, t =, v(, t = v(x, = e ax ϕ(x. ( λ n = and we obtain the soution to this probem as v(x, t = ( b n e λnt sin x 2, sin ( x with b n = 2 ( e ax ϕ(x sin x dx. Finay our soution to (4.3-(4.33 can be written as u(x, t = e ax+βt ( b n e λnt sin x. 4.8 Assignment Eigenvaues and Heat Equation. Fourier Series Exampes: The Fourier series for f(x = x 2 on x π gives x 2 π ( n n 2 cos(nx, x π Find vaues of x and give a justification for using them aong with the above information to show that (a (b n = π2 2 6 ( (n+ n 2 = π Find the Fourier Cosine series expansion of f(x = sin(x on x π. 3. Find the Fourier series expansion of f(x = cos 2 (x on x π. (THINK 4. Determine a soutions 22

23 (a y y =, < x <, y( =, y( = 4. (b y + 4y =, < x < π, y( =, y (π =. (c y + y =, < x < 2π, y( =, y(2π =. (d y 2y + y =, < x <, y( =, y( = Find a eigenvaues λ and eigenvectors y (i.e., y nonzero (a y + λy =, < x < π, y( =, y(π =. (b y + λy =, < x < π, y ( =, y (π =. (c y + λy =, < x < π, y( =, y (π =. (d y + λy =, < x < 2π, y( = y(2π, y ( = y (2π. 6. Sove the heat probem u t = u xx with (a BC: u(, t =, u(π, t =, IC: u(x, = sin(x (b BC: u(, t =, u(π, t =, IC: u(x, = x(π x (c BC: u(, t =, u(π, t = IC: u(x, = sin(x 7 sin(3x (d BC: u x (, t =, u x (π, t =, IC: u(x, = cos(x (e BC: u(, t =, u(π, t =, IC: u(x, = sin(x cos(x (f BC: u x (, t =, u x (π, t =, IC: u(x, = cos 2 (x (Hint: haf-ange formua, x < π/2 (g BC: u(, t =, u(π, t =, IC: u(x, =, π/2 x < π (h BC: u x (, t =, u x (π, t =,, x < π/2 IC: u(x, =, π/2 x < π 7. Sove the initia boundary vaue probem u t (x, t = u xx (x, t, < x < π, t > u(, t =, u(π, t = u(x, = x 8. Sove the initia boundary vaue probem u t (x, t = u xx (x, t, < x < π, t > u x (, t =, u x (π, t = u(x, = x 9. Sove the initia boundary vaue probem u t (x, t = u xx (x, t, < x < π, t > u(, t =, u x (π, t = u(x, = x 23

24 . Sove the initia boundary vaue probem u t (x, t = u xx (x, t 2u x (x, t, < x <, t > u(, t =, u(, t = u(x, = e x sin(πx. Sove the initia boundary vaue probem u t (x, t = 2(u xx (x, t 3u(x, t, < x <, t > u(, t =, u(, t = u(x, = 2 sin(πx 3 sin(2πx 24