1 Heat Equation Dirichlet Boundary Conditions

Transcription

1 Chapter 3 Heat Exampes in Rectanges Heat Equation Dirichet Boundary Conditions u t (x, t) = ku xx (x, t), < x <, t > (.) u(, t) =, u(, t) = u(x, ) = f(x). Separate Variabes Look for simpe soutions in the form u(x, t) = ϕ(x)ψ(t). Substituting into (.) and dividing both sides by ϕ(x)ψ(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. 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 kλψ =. ψ(t) = Ce kλt (.) 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() =. (.3). Find Eigenvaues and Eignevectors The next main step is to find the eigenvaues and eigenfunctions from (.3). 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 =.

2 (b) If λ = µ > then Appying the boundary conditions we have ϕ(x) = a cosh(µx) + b sinh(µx). = 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(x)ϕ (x) = λϕ(x). Integrate this expression from x = to x =. We have λ ϕ(x) = Since ϕ() = ϕ() = we concude ϕ(x)ϕ (x) = λ = ϕ (x) ϕ(x) and we see that λ must be ess than or equa to zero. (c) So, finay, consider λ = µ so that Appying the boundary conditions we have ϕ(x) = a cos(µx) + b sin(µx). ϕ (x) + ϕ(x)ϕ (x). = ϕ() = a a = = ϕ() = b sin(µ). From this we concude sin(µ) = which impies µ =. So we have eigenfunctions b n sin( x) and we choose the constant b n so that and therefore ϕ n (x) = b n = ( λ n = µ n = ), ϕn (x) = From (.) we aso have the associated functions ψ n (t) = e kλnt. sin(x), n =,,. (.4) 3. Write Forma Sum From the above considerations we can concude that for any integer N and constants {c n } N n= N u N (x, t) = c n ψ n (t)ϕ n (x). satisfies the differentia equation in (.) and the boundary conditions.

3 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, ) = f(x) is satisfied. To do this we consider what we earned from Fourier series. In particuar we ook for u as an infinite sum and we try to find {c n } satisfying u(x, t) = c n e kλnt ϕ n (x) f(x) = u(x, ) = c n ϕ n (x) But this nothing more than a Sine expansion of the function ϕ on the interva (, ). c n = f(x)ϕ n (x). (.5) Exampe.. As an expicit ( exampe for the initia condition consider =, k = / and f(x) = x( x). ) Let us reca that = which in this case reduces to. c n = = = [ x( x) sin (x) ( ) cos (x) x( x) x( x) cos(x) + ( x) cos(x) = = [ = () ( ) sin (x) ( x) ( x) sin(x) sin(x) = () ( ) sin (x) [ cos(x) = [ ( ) n () 3 We arrive at the soution u(x, t) = 4 π 3 [ ( ) n n 3 e n π t/ sin (x). (.6) where x( x) = 4 π 3 [ ( ) n n 3 sin (x). 3

4 As an exampe with N = 3 we have x( x) 8 ( sin(πx) + sin(3πx) ). π 3 7 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 =, t = 3 u(x, t) = 4 3 [ ( ) n e nπt/ sin (x). π 3 n x Heat Equation Neumann Boundary Conditions u t (x, t) = u xx (x, t), < x <, t > (.7) u x (, t) =, u x (, t) = u(x, ) = f(x) 4

5 . Separate Variabes Look for simpe soutions in the form u(x, t) = ϕ(x)ψ(t). Substituting into (.7) and dividing both sides by ϕ(x)ψ(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. 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 (.8) 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 () =. (.9). Find Eigenvaues and Eignevectors The next main step is to find the eigenvaues and eigenfunctions from (.9). 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 normaized eigenfunction ϕ (x) = /. (b) If λ = µ > then and Appying the boundary conditions we have ϕ(x) = a cosh(µx) + b sinh(µx) ϕ (x) = aµ sinh(µx) + bµ cosh(µx). = ϕ () = bµ b = = ϕ () = aµ sinh(µ) a =. Therefore, there are no positive eigenvaues. Consider the foowing aternative argument: If ϕ (x) = λϕ(x) then mutipying by ϕ we have ϕ(x)ϕ (x) = λϕ(x). Integrate this expression from x = to x =. We have λ ϕ(x) = ϕ(x)ϕ (x) = ϕ (x) + ϕ(x)ϕ (x). 5

6 Since ϕ () = ϕ () = we concude λ = ϕ (x) ϕ(x) and we see that λ must be ess than or equa to zero ( zero ony if ϕ = ). (c) So, finay, consider λ = µ so that and Appying the boundary conditions we have ϕ(x) = a cos(µx) + b sin(µx) ϕ (x) = aµ sin(µx) + bµ cos(µx). = ϕ () = bµ b = = ϕ () = aµ sin(µ). From this we concude sin(µ) = which impies µ =. So we have eigenfunctions a n cos( x) and we choose the constant a n so that and therefore ϕ n (x) = a n = ( ) λ n = µ n =, ϕn (x) = cos(x), n =,,.. (.) From (.8) 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= N u n (x, t) = c ϕ + c n ψ n (t)ϕ n (x) satisfies the differentia equation in (.7) 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, ) = f(x) is satisfied. To do this we consider what we earned from Fourier series. In particuar we ook for u as an infinite sum u(x, t) = c ϕ + c n e λnt ϕ n (x) and we try to find {a n } satisfying ϕ(x) = u(x, ) = c ϕ + c n ϕ n (x). But this nothing more than a Cosine expansion of the function ϕ on the interva (, ). Our work on Fourier series showed us that c n = 6 f(x)ϕ n (x). (.)

7 As an expicit exampe for the initia condition consider = (so ϕ (x) = ) and f(x) = x( x). In this case (.) becomes We have c n = f(x)ϕ n (x). and for n > c = f(x)ϕ (x) = [ x = x3 = 3 6. x( x) c n = = = [ ϕ(x) cos (x) = ( ) sin (x) x( x) x( x) sin (x) x( x) cos (x) ( x) sin (x) ( ) cos (x) = ( x) [ cos (x) = ( x) cos (x) ( ) = [ cos () = () (( )n + ) = (), n even, n odd. In order to eiminate the odd terms in the expansion we introduce a new index, k by n = k where k =,,. So finay we arrive at the soution u(x, t) = 6 π k= k e 4k πt cos(kπx). (.) As an exampe with N = 4 we have x( x) 6 π ( 4 7 ) cos(kπx). k

8 Notice that as t the infinite sum converges to zero uniformy in x. Indeed, πt k e 4k cos(kπx) t e 4π k = π 6 e 4πt. k= So the soution converges to a nonzero steady state temperature which is exacty the average vaue of the initia temperature distribution. im u(x, t) = t 6 = f(x). In the foowing figure we pot the eft and right hand side of the above. k= x Finay we pot the approximate soution at times t =, t = /, t = /, t = 3/ x 8

9 . Heat Equation Dirichet-Neumann Boundary Conditions u t (x, t) = u xx (x, t), < x <, t > u(, t) =, u x (, t) = u(x, ) = f(x) Appy Separation of Variabes to obtain the Sturm-Liouvie probem for {λ n, ϕ n (x)}. Find Eigenvaues and Eignevectors The next main step is to find the eigenvaues and eigenfunctions. There are, in genera, three cases:. If λ = then ϕ(x) = ax + b so appying the boundary conditions we get Zero is not an eigenvaue.. If λ = µ > then and Appying the boundary conditions we have = ϕ() = b, = ϕ () = a a = b =. ϕ(x) = a cosh(µx) + b sinh(µx) ϕ (x) = aµ sinh(µx) + bµ cosh(µx). = 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(x)ϕ (x) = λϕ(x). Integrate this expression from x = to x =. We have λ ϕ(x) = ϕ(x)ϕ (x) = Since ϕ() = ϕ () = we concude λ = ϕ (x) ϕ(x) and we see that λ must be ess than or equa to zero. 3. So, finay, consider λ = µ so that and Appying the boundary conditions we have ϕ(x) = a cos(µx) + b sin(µx) ϕ (x) = aµ sin(µx) + bµ cos(µx). ϕ (x) + ϕ(x)ϕ (x). = ϕ() = aµ a = = ϕ () = bµ cos(µ). 9

10 From this we concude cos(µ) = which impies µ = (n )π So we have eigenfunctions b n sin( x) and we choose the constant b n so that and therefore ϕ n (x) = b n = ( ) (n )π λ n = µ n =, ϕ n (x) = From (.) we aso have the associated functions ψ n (t) = e λnt. sin(x), n =,,. (.3) Write Forma Infinite Sum From the above considerations we can concude that for any integer N and constants {b n } N n= N u n (x, t) = c n ψ n (t)ϕ n (x). satisfies the heat equation and the boundary conditions. Use Fourier Series to Find Coefficients The ony probem remaining is to somehow pick the constants c n so that the initia condition u(x, ) = f(x) is satisfied. To do this we consider what we earned from Fourier series. In particuar we ook for u as an infinite sum and we try to find {b n } satisfying u(x, t) = c n e λnt ϕ n (x) f(x) = u(x, ) = c n ϕ n (x). But this nothing more than a Sine type expansion of the function ϕ on the interva (, ) and we have f(x) = c n = c n ϕ n (x). f(x)ϕ n (x). (.4) As an expicit exampe et = so that ϕ n (x) = ( ) sin( x) and for the initia condition consider (n )π f(x) = x. Let us reca that =

11 c n = = = [ ϕ(x)ϕ n (x) = ( x cos (µ ) nx) x cos(x) + = [ cos() + sin(µ nx) µ n = [ cos((n )π/) + = 4 ( ) n+ (n ) π x sin ( x) cos( x) sin((n )π/) µ n We arrive at the soution u(x, t) = 8 π ( ) n+ (n ) eλnt sin ( x). (.5)