MA 201, Mathematics III, July-November 2016, Partial Differential Equations: 1D wave equation (contd.) and 1D heat conduction equation

Transcription

1 MA 201, Mathematics III, July-November 2016, Partial Differential Equations: 1D wave equation (contd.) and 1D heat conduction equation Lecture 12 Lecture 12 MA 201, PDE (2016) 1 / 24

2 Formal Solution of the Finite Vibrating String Problem Let us revisit the problem of vibrating string of length L. The solution is given by u(x,t) = n=1 sin nπx L [ A n cos nπct L +B nsin nπct ] L (1) with A n = 2 L B n = L 2 nπc 0 L φ(x)sin nπx dx, n = 1,2,3,... L 0 ψ(x)sin nπx L dx, n = 1,2,3,... Lecture 12 MA 201, PDE (2016) 2 / 24

3 Solution of the Finite Vibrating String Problem: Example Example: For a string of length L stretched between the points x = 0 and x = L, find the vibration in the string subject to the following initial conditions: u(x,0) = sin(πx/l)+1/2sin(3πx/l), u t (x,0) = 0. Solution: Here B n = 0, and A n = 2 L L 0 (sin πx L Note that in the solution only A 1 and A 3 will exist. A 1 = 1,A 3 = 1/2. 3πx nπx sin )sin L L dx. The solution is u(x,t) = sin πx πct cos L L πx 3πct sin cos L L. Lecture 12 MA 201, PDE (2016) 3 / 24

4 IBVP for vibrating string with gravity We consider the external force is due to the gravitational acceleration g only (consider a string oriented horizontally), the equation becomes u tt = c 2 u xx g, 0 < x < L, t > 0. (2) We seek to find the displacement of the string at any position and at any time subject to the following boundary condition (for t > 0) and initial conditions (0 x L): and u(0,t) = 0, u(l,t) = 0, (3a) (3b) u(x,0) = φ(x), u t (x,0) = ψ(x). (4a) (4b) Lecture 12 MA 201, PDE (2016) 4 / 24

5 IBVP for vibrating string with gravity (Contd.) Due to the presence of the term g in equation (2), the direct application of the method of separation of variables will not work. We intend to convert the given IBVP into two problems: one IBVP and one BVP, former resembling the problem we solved previously and the other taking care of the nonhomogeneous term in the governing equation. Seek a solution in the form: u(x,t) = v(x,t)+h(x), (5) where h(x) is an unknown function of x alone. Lecture 12 MA 201, PDE (2016) 5 / 24

6 IBVP for vibrating string with gravity (Contd.) Now substituting (5) into (2) v tt = c 2 [v xx +h (x)] g, (6) subject to v(0,t)+h(0) = 0, v(l,t)+h(l) = 0, (7a) (7b) and v(x,0)+h(x) = φ(x), v t (x,0) = ψ(x). (8a) (8b) Lecture 12 MA 201, PDE (2016) 6 / 24

7 IBVP for vibrating string with gravity (Contd.) Equations (6)-(8) can be conveniently split into two problems: Problem I: v tt = c 2 v xx, v(0,t) = 0 = v(l,t), v(x,0) = φ(x) h(x), v t (x,0) = ψ(x). Problem II: c 2 h (x) = g, h(0) = 0 = h(l). Lecture 12 MA 201, PDE (2016) 7 / 24

8 IBVP for vibrating string with gravity (Contd.) The solution of Problem I is known to us from the previous problem, which is: v(x,t) = n=1 sin nπx L where A n and B n are given, respectively, by A n = 2 L B n = L 2 nπc 0 L [ A n cos( nπct L )+B nsin( nπct ] L ), (9) [φ(x) h(x)]sin( nπx ) dx, n = 1,2,3,... (10) L 0 ψ(x)sin( nπx ) dx, n = 1,2,3,... (11) L Lecture 12 MA 201, PDE (2016) 8 / 24

9 IBVP for vibrating string with gravity (Contd.) The solution for Problem II can be easily found by integrating h (x) twice Upon using the boundary conditions h(x) = gx2 2c 2 +Ax+B. we get B = 0 and A = gl/(2c 2 ) and hence Hence the solution u(x,t) for our IBVP is given by the sum of (9) and (12). h(x) = g (L x)x 2c 2. (12) Lecture 12 MA 201, PDE (2016) 9 / 24

10 Separation of Variables for Heat conduction Problem Lecture 12 MA 201, PDE (2016) 10 / 24

11 Heat conduction in a thin rod One important problem is the heat conduction in a thin metallic rod of finite length. This transient problem represents a class of problem known as diffusion problems. Just like one-dimensional wave equation, separation of variables can be appropriately applied. The IBVP under consideration for heat conduction in a thin metallic rod of length L consists of: The governing equation where α is the thermal diffusivity. u t = αu xx, 0 < x < L, t > 0, (13) Lecture 12 MA 201, PDE (2016) 11 / 24

12 Heat conduction in a thin rod (Contd.) The boundary conditions for all t > 0 due to maintaining the temperature at both ends at zero degrees u(0,t) = 0, u(l,t) = 0. (14a) (14b) The initial condition for 0 x L, i.e., initial temperature distribution: u(x,0) = f(x). (15) The assumptions are: The rod is very thin so that it can be considered as one-dimensional. The curved surface is insulated. There is no external source of heat. Using the separation of variable technique by assuming a solution of the form u(x,t) = X(x)T(t) Lecture 12 MA 201, PDE (2016) 12 / 24

13 Heat conduction in a thin rod (Contd.) The governing equation is converted to the following ODEs: X kx = 0, where k is a separation constant. T kαt = 0. Observe that the boundary conditions for this IBVP are the same as those in the one-dimensional wave equation. We can immediately assume that a feasible nontrivial solution exists only for the negative values of k. Above equations X +λ 2 X = 0, (16) T +λ 2 αt = 0. (17) Lecture 12 MA 201, PDE (2016) 13 / 24

14 Heat conduction in a thin rod (Contd.) The solutions: X(x) = Acos(λx)+Bsin(λx), (18) T(t) = Ce αλ2t. (19) Solution for u(x, t): u(x,t) = [Acos(λx)+Bsin(λx)]Ce αλ2t. (20) Using the boundary conditions (14a) and (14b) A = 0 and λ n = nπ L, n = 1,2,3,... Solution corresponding to each n: u n (x,t) = B n sin( nπx L ) n 2 π 2 e α L 2 t. Lecture 12 MA 201, PDE (2016) 14 / 24

15 Heat conduction in a thin rod The general solution: u(x,t) = = u n (x,t) n=1 n=1 B n sin( nπx L ) n 2 π 2 e α L 2 t (21) where B n is obtained (refer to the solution of the one-dimensional wave equation) by using the initial condition (15): B n = 2 L L 0 f(x)sin( nπx ) dx, n = 1,2,3,... (22) L Observe that what we have solved here is a homogenous equation subject to homogenous (zero) conditions. Ideally the boundary conditions are very likely to be non-homogenous for diffusion problems. Lecture 12 MA 201, PDE (2016) 15 / 24

16 Heat conduction in a thin rod Case II. With nonhomogeneous boundary conditions Consider the heat conduction in a thin metal rod of length L with insulated sides with the ends x = 0 and x = L kept at temperatures u = u 0 0 C and u = u L 0 C, respectively, for all time t > 0. Suppose that the temperature distribution at t = 0 is u(x,0) = φ(x), 0 < x < L. To determine the temperature distribution in the rod at some subsequent time t > 0. Lecture 12 MA 201, PDE (2016) 16 / 24

17 Heat conduction in a thin rod The IBVP under consideration consists of The governing equation u t = αu xx, (23) where α is the coefficient of thermal diffusivity for the specific metal. The boundary conditions for all t > 0 u(0,t) = u 0, u(l,t) = u L. (24a) (24b) The initial conditions for 0 x L u(x,0) = φ(x). (25) Lecture 12 MA 201, PDE (2016) 17 / 24

18 Heat conduction in a thin rod Due to the nonhomogeneous boundary conditions (24), the direct application of the method of separation of variables will not work. We intend to convert the given IBVP into two BVPs: one resembling the problem we solved previously the other taking care of the nonhomogeneous terms. Seek a solution in the form: u(x,t) = v(x,t)+h(x), (26) where v(x,t) is the dependent variable and h(x) is an unknown function of x alone. Lecture 12 MA 201, PDE (2016) 18 / 24

19 Heat conduction in a thin rod Now substituting (26) into (23) v t = α[v xx +h (x)] (27) subject to v(0,t)+h(0) = u 0, v(l,t)+h(l) = u L, (28a) (28b) and v(x,0)+h(x) = φ(x). (29) Now equations (27)-(29) can be conveniently split into two BVPs as follows: Lecture 12 MA 201, PDE (2016) 19 / 24

20 Heat conduction in a thin rod BVP I: BVP II: v t = αv xx, v(0,t) = 0 = v(l,t), v(x,0) = φ(x) h(x). αh (x) = 0, h(0) = u 0, h(l) = u L. The solution of BVP I is known to us from the previous problem which is v(x,t) = B n sin nπx n 2 π 2 L e α L 2 t, (30) n=1 where B n = 2 L L 0 (φ(x) h(x))sin nπx L dx, n = 1,2,3,... (31) Lecture 12 MA 201, PDE (2016) 20 / 24

21 Heat conduction in a thin rod The solution for BVP II can be easily found in two steps by integrating h (x) and h (x) : h(x) = Ax+B. Upon using the boundary conditions, we get B = u 0 and A = u L u 0 Hence h(x) = u L u 0 L x+u 0. (32) Hence the solution u(x,t) for our IBVP is given by the sum of (30) and (32). Lecture 12 MA 201, PDE (2016) 21 / 24

22 Heat conduction problem with flux and radiation conditions The previous two problems that we considered are examples where we have fixed the temperatures at the ends of the rod, and are examples of Dirichlet or fixed endpoint conditions. There are two other types of problems concerning the heat conduction in a thin rod. These types of problems contain other types of boundary conditions; the first is a Neumann or flux condition, the second is a Robin or radiation condition. Lecture 12 MA 201, PDE (2016) 22 / 24

23 Heat conduction problem with flux and radiation conditions An insulation-type flux condition at, say, x = 0 is a condition on the derivative of the form u x (0,t) = 0,t > 0; because the flux is proportional to the temperature gradient (Fourier s heat flow law states flux = αu x, where α is the conductivity) Heat flux is defined as the rate of heat transfer per unit cross-sectional area. The insulation condition requires that no heat flow across the boundary x = 0. 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 takes the form αu x (0,t)+bu(0,t) = 0,t > 0, where b is a constant. Lecture 12 MA 201, PDE (2016) 23 / 24

24 Heat conduction problem with flux and radiation conditions If b > 0, then the flux is negative which means that heat is flowing from the rod into its surroundings (radiation) if b < 0, then the flux is positive, and heat is flowing into the rod (absorption). A typical problem in heat conduction may have a combination of Dirichlet, insulation and radiation boundary conditions. Lecture 12 MA 201, PDE (2016) 24 / 24