MA 201: Partial Differential Equations Lecture - 11

Transcription

1 MA 201: Partia Differentia Equations Lecture - 11 Heat Equation

2 Heat conduction in a thin rod The IBVP under consideration consists of: The governing equation: u t = αu xx, (1) where α is the therma diffusivity. The boundary conditions: for a t > 0 u(0, t) = 0, (2a) u(, t) = 0. (2b) The initia condition: for 0 x u(x, 0) = f (x). (3)

3 Heat conduction in a thin rod (Contd.) We use separation of variabe technique: assume a soution u(x, t) = X (x)t (t). The governing equation is converted to the foowing ODEs: X kx = 0, T kαt = 0. where k is a separation constant. The boundary conditions for this IBVP are the same as those in the one-dimensiona wave equation. We assume that a feasibe nontrivia soution exists for the negative vaues of k. The above equations give X + λ 2 X = 0, (4) T + λ 2 αt = 0. (5)

4 Heat conduction in a thin rod (Contd.) The soutions: X (x) = A cos(λx) + B sin(λx), (6) T (t) = Ce αλ2t. (7) Soution for u(x, t): u(x, t) = [A cos(λx) + B sin(λx)]ce αλ2t. (8) Using the boundary conditions (2a) and (2b) we get A = 0 and λ n = nπ, n = 1, 2, 3,.... Soution corresponding to each n: u n (x, t) = B n sin( nπx ) e α n2 π 2 2 t.

5 Heat conduction in a thin rod (Contd.) The genera soution: u(x, t) = u n (x, t) = n=1 n=1 B n sin( nπx ) e α n2 π 2 2 t where B n is obtained (refer to the soution of the one-dimensiona wave equation) by using the initia condition (3): B n = 2 0 f (x) sin( nπx ) dx, n = 1, 2, 3,... (9) Observe that what we have soved is a homogenous equation subject to homogenous (zero) conditions. Ideay the boundary conditions are very ikey to be non-homogenous for diffusion probems. (It wi be discussed ater.)

6 Heat conduction probem with fux and radiation conditions An insuation-type fux condition at, say, x = 0 is a condition on the derivative of the form u x (0, t) = 0, t > 0; because the fux is proportiona to the temperature gradient (Fourier s heat fow aw states fux = αu x, where α is the conductivity). Heat fux is defined as the rate of heat transfer per unit cross-sectiona area. The insuation condition requires that no heat fow across the boundary x = 0.

7 Heat conduction probem with fux and radiation conditions A radiation condition, on the other hand, is a specification of how heat radiates from the end of the rod, say at x = 0, into the environment, or how the end absorbs heat from its environment. Linear, homogeneous radiation condition take the form αu x (0, t) + bu(0, t) = 0, t > 0, where b is a constant. A typica probem in heat conduction may have a combination of Dirichet, insuation and radiation boundary conditions.

8 Heat conduction probem with fux conditions Consider the IBVP consisting of the foowing: u t = αu xx, (10) The boundary conditions for a t > 0 u x (0, t) = 0, (11a) u(, t) = 0. (11b) The initia condition for 0 x : u(x, 0) = f (x). (12) The boundary conditions (11) te us that the eft end of the rod is insuated and the right end is kept at zero degrees.

9 Heat conduction probem with fux conditions Assuming a soution of the form u(x, t) = X (x)t (t) the PDE is converted to the foowing ODEs: X kx = 0, T kαt = 0. Taking k = λ 2, the above equations become Giving us the soutions: X + λ 2 X = 0, (13) T + λ 2 αt = 0. (14) X (x) = A cos(λx) + B sin(λx), (15) T (t) = Ce αλ2t. (16) The soution: u(x, t) = [A cos(λx) + B sin(λx)]ce αλ2t. (17)

10 Heat conduction probem with fux conditions To use the boundary condition (11a) we have to differentiate (17) w.r.t. x to get u x = λ[ A sin(λx) + B cos(λx)]ce αλ2t. (18) Using boundary condition (11a) B = 0. The boundary condition (11b) wi give the eigenvaues as ( ) 2n + 1 π λ n =, n = 0, 1, 2, 3,.... 2

11 Heat conduction probem with fux conditions The soution to the IBVP wi be given by [ ( ) ] 2n π 2 u(x, t) = C n exp α 2 2 t (2n + 1)πx cos( ), 2 n=0 where C n can be obtained from IC (12): (2n + 1)πx f (x) = C n cos( ) 2 as C n = 2 0 f (x) cos( n=0 (19) (2n + 1)πx ) dx, n = 0, 1, 2, 3,... (20) 2 Note that here n = 0 aso contributes to the soution In other words, we can say that the eigenvaue λ 0, corresponding to n = 0 aso contributes.

12 Heat conduction probem with different fux conditions at ends Note the difference in soution when Dirichet condition at both ends of the rod are changed to one Neumann (at x = 0) and one Dirichet (at x = ) conditions We may have two other combinations of pairs of conditions, viz., 3. Dirichet condition at x = 0 and Neumann condition at x = 4. Neumann condition at both ends x = 0 and x =.

13 Heat conduction probems with different conditions at ends Probem I. Governing equation: u t = αu xx Boundary conditions: u(0, t) = 0 = u(, t), Initia Condition: u(x, 0) = f (x) Soution is with u(x, t) = A n = 2 n=1 0 A n sin( nπx ) e α n2 π 2 2 t. f (x) sin( nπx )dx.

14 Heat conduction probems with different conditions at ends Probem II. Governing equation u t = αu xx Boundary conditions: u x (0, t) = 0 = u(, t), Initia Condition: u(x, 0) = f (x) The soution is [ ( 2n + 1 u(x, t) = A n exp α 2 n=0 ) ] 2 π 2 2 t cos( (2n + 1)πx ). 2 with A n = 2 0 f (x) cos( (2n + 1)πx )dx. 2

15 Heat conduction probems with different conditions at ends Probem III. Governing equation u t = αu xx Boundary conditions: u(0, t) = 0 = u x (, t), Initia Condition: u(x, 0) = f (x) Soution is u(x, t) = [ ( 2n + 1 A n exp α 2 n=0 ) ] 2 π 2 2 t sin( (2n + 1)πx ) 2 with A n = 2 0 f (x) sin( (2n + 1)πx )dx. 2

16 Heat conduction probems with different conditions at ends Probem IV. Governing equation u t = αu xx Boundary conditions: u x (0, t) = 0 = u x (, t), Initia Condition: u(x, 0) = f (x) with u(x, t) = A n = 2 n=1 1 0 The soution is A n cos( nπx ) e α n2 π 2 2 t f (x) cos( nπx )dx. In each of the four soutions OBSERVE the function for x and aso the eigenvaues. TRY soving Probems III and IV yoursef.

17 A Non-homogeneous Heat Conduction Probem Consider the non-homogeneous PDE u t = αu xx + ax, (21) where a is a nonzero constant. Boundary conditions: u(0, t) = 0 = u(, t), Initia Condition: u(x, 0) = f (x) Seek a soution in the form: u(x, t) = v(x, t) + h(x), (22) where h(x) is an unknown function of x aone.

18 A Non-homogeneous Heat Conduction Probem (Contd.) Now substituting (22) into (21) v t = α[v xx + h (x)] + ax, (23) subject to v(0, t) + h(0) = 0, (24a) v(, t) + h() = 0, (24b) and v(x, 0) + h(x) = f (x), (25)

19 A Non-homogeneous Heat Conduction Probem (Contd.) Equations (23)-(25) can be convenienty spit into two probems: Probem I: v t = αv xx, v(0, t) = 0 = v(, t), v(x, 0) = f (x) h(x). Probem II: αh (x) = ax, h(0) = 0 = h().

20 A Non-homogeneous Heat Conduction Probem (Contd.) The soution of Probem I is known to us from the previous probem, which is: v(x, t) = Soution is n=1 A n sin( nπx ) e α n2 π 2 2 t. (26) with A n = 2 0 [f (x) h(x)] sin( nπx )dx.

21 A Non-homogeneous Heat Conduction Probem (Contd.) The soution for Probem II can be easiy found by integrating h (x) twice and using the boundary conditions: h(x) = ax 6α (x 2 2 ). (27) The coefficients A n in (26) can now be obtained by putting h(x) as in and (27). by the sum of (26) and (27). Hence the soution u(x, t) for our IBVP is given

22 Time-Independent Non-homogeneous BC We now turn to the situation where the BC are not both homogeneous, but are independent of time variabe t. The method of soution consists of the foowing steps: Step 1: Find a particuar soution of the PDE and BC. Step 2: Find the soution of a reated probem with homogeneous BC. Then, add this soution to that particuar soution obtained in Step 1. The procedure is iustrated in the foowing exampe: PDE: u t = αu xx, 0 x, t > 0, (28) BC: u(0, t) = a, u(, t) = b, t > 0, (29) IC: u(x, 0) = f (x), 0 x, (30) where a and b are arbitrary constants and f (x) is a given function.

23 Time-Independent Non-homogeneous BC (Contd.) Soution. Seek a particuar soution u p (x) of the form u p (x) = cx + d, where c and d are to be chosen so that the BC are satisfied: a = u p (0) = c 0 + d = d, b = u p () = c + d = c + a. = d = a and c = (b a)/. Thus u p (x) = (b a)x/ + a soves the PDE with the BC satisfied. Consider the reated homogeneous probem (i.e., with homogeneous PDE and BC) PDE: v t = αv xx 0 x, t > 0, BC: v(0, t) = 0, v(, t) = 0, t > 0, (31) IC: v(x, 0) = f (x) u p (x), 0 x.

24 Time-Independent Non-homogeneous BC (Contd.) If f (x) u p (x) = c n sin(nπx/), then its soution is given by n=1 v(x, t) = c n e (nπ/)2αt sin(nπx/). n=1 Now, set u(x, t) = v(x, t) + u p (x). Then it is easy to verify that u(x, t) soves (28). Indeed, u(x, t) soves (28) by the superposition principe. Further, we have BC: IC: u(0, t) = v(0, t) + u p (0) = 0 + a = a u(, t) = v(, t) + u p () = 0 + b = b u(x, 0) = v(x, 0) + u p (x) = f (x) u p (x) + u p (x) = f (x).

25 Time-Independent Non-homogeneous BC (Contd.) Exampe PDE: ut = 2u xx 0 x 1, t > 0, (32) BC: u x (0, t) = 1 u(1, t) = 1, t > 0, (33) IC: u(x, 0) = x + cos 2 (3πx/4) 5/2. (34) Soution. Take u p (x) = cx + d. The first BC u x (0, t) = 1 yieds c = 1, whie u p (1) = 1 + d yieds d = 2 by the second BC. Thus, u p (x) = x 2. The reated homogeneous probem is v t = 2v xx 0 x 1, t > 0, v x (0, t) = 0 v(1, t) = 0, t > 0 v(x, 0) = [x + cos 2 (3πx/4) 5/2] (x 2) = cos(3πx/2) 5/2 + 2 = 1 2 cos(3πx/2).

26 Time-Independent Non-homogeneous BC (Contd.) It is easy to obtain the soution of the reated homogeneous probem as v(x, t) = e 9π2 t/2 [ 1 2 cos(3πx/2)]. From the above exampes, we notice that the particuar soution is time independent, or in steady-state. Note: Any steady-state soution of the heat equation u t = αu xx is of the form cx + d. The soutions u(x, t) are sums of a steady-state particuar soution of the PDE and BC and the soution v(x, t) of the reated homogeneous probem which is transient in the sense that v(x, t) 0 as t. Thus u(x, t) = u p (x) + v(x, t) u p (x), as t. (steady-state soution) (transient soution) That is, the soution u approaches the steady-state soution as t.

27 However, for some types of BC there are no steady-state particuar soutions. Exampe PDE: u t = αu xx, 0 x, t > 0, (35) BC: u x (0, t) = a, u x (, t) = b, (36) IC: u(x, 0) = f (x), (37) where a and b are constants, and f (x) is a given function. Soution. Let u p (x) = cx + d. Then, using BC, we obtain c = a and c = b, which is impossibe uness a = b. NOTE: The boundary conditions state that heat is being drained out of the end x = 0 at a rate u x (0, t) = a and heat is fowing into the end x = at a rate u x (, t) = b. If b > a, then the heat energy is being added to the rod at a constant rate. If b < a, the rod oses heat at a constant rate. Thus, we cannot expect a steady-state soution of the PDE and BC, uness a = b.