In this lecture we shall learn how to solve the inhomogeneous heat equation. u t α 2 u xx = h(x, t)

Transcription

1 MODULE 5: HEAT EQUATION 2 Lecture 5 Time-Dependent BC In this lecture we shall learn how to solve the inhomogeneous heat equation u t α 2 u xx = h(x, t) with time-dependent BC. To begin with, let us consider the following IBVP problem with time-dependent BC: PDE: u t = α 2 u xx x L, < t <, (1) BC: u(, t) = a(t) u(l, t) = b(t), < t <, (2) IC: u(x, ) = f(x). (3) In the previous lecture, we had discussed the solution of this problem in the case where a(t) and b(t) are constant functions (independent of t) and f(x) is a suitable given function. Notice that the function w(x, t) defined by [ ] b(t) a(t) w(x, t) = x + a(t) L satisfies the BC (2). However, w(x, t) will not satisfy the PDE (1) unless a(t) and b(t) are constant. In fact, [ b w t α 2 (t) a ] (t) w xx = x + a (t). L We now attempt to find a solution for the problem (1)-(3) of the form where v(x, t) satisfies the following problem Further, w(x, t) + v(x, t), v t α 2 v xx = u t α 2 u xx (w t α 2 w xx ) = (w t α 2 w xx ) = [b (t) a (t)]x/l a (t). v(, t) = u(, t) w(, t) = a(t) a(t) =, v(l, t) = b(t) b(t) =.

2 MODULE 5: HEAT EQUATION 21 Thus, the function v(x, t) must satisfy the following related problem with homogeneous BC, but inhomogeneous PDE: PDE: v t α 2 v xx = [b (t) a (t)]x/l a (t), x L, < t <, (4) BC: v(, t) =, v(l, t) =, < t <, (5) IC: v(x, ) = u(x, ) w(x, ) = f(x) [a() b()]x/l a(). (6) Note: When a(t) and b(t) are constants, the PDE is homogeneous. But, in this case, v(x, t) satisfies nonhomogeneous PDE. The problem (4)-(6) is a special case of the following general problem: PDE: v t α 2 v xx = h(x, t) x L, < t <, (7) BC: v(, t) =, v(l, t) =, < t < (8) IC: v(x, ) = g(x). (9) The solution procedure to the above problem was given by the French mathematician and physicist Jean-Marie-Constant Duhamel ( ). The method is known as Duhamel s principle. Suppose u 1 and u 2 are solutions of the following problems: (P 1 :) PDE: (u 1 ) t α 2 (u 1 ) xx = BC: u 1 (, t) =, u 1 (L, t) = IC: u 1 (x, ) = g(x) (P 2 :) PDE: (u 2 ) t α 2 (u 2 ) xx = h(x, t) BC: u 2 (, t) =, u 2 (L, t) = IC: u 2 (x, ) = (1) It is easy to check that v(x, t) = u 1 (x, t) + u 2 (x, t) solves (7)-(9). The solution u 1 to the problem (P 1) is known (cf. Lecture 4 in Module 5). It remains only to solve the problem (P2) for u 2. The above observation has led to the following (cf. [1]). THEOREM 1. A solution to problem (1)-(3) is given by w(x, t) + u 1 (x, t) + u 2 (x, t), where [ ] b(t) a(t) w(x, t) = x + a(t) L is the particular solution of the BC and u 1 (x, t) solves (P1) with g(x) = f(x) w(x, ) and u 2 (x, t) solves (P2) with h(x, t) = (w t α 2 w xx ) = [b (t) a (t)]x/l a (t).

3 MODULE 5: HEAT EQUATION 22 1 Duhamel s principle The basic idea of Duhamel s principle is to transfer the source term h(x, t) to initial condition of related problems. This is done in the following manner. The function defined by v(x, t; s)ds is a solution of (7)-(9) provided v(x, t; s) is a solution of the problem PDE: v t = α 2 v xx, x L, < t <, (11) BC: v(, t; s) =, v(l, t; s) =, < t <, (12) IC: v(x, s; s) = h(x, s). (13) Note that both PDE and BC are homogeneous. We use translation in time v(x, t s; s)ds to obtain an IC at t =, instead of t = s. Rewriting (11)-(13) in terms of v, we now reduce the problem to the following associated problem with IC at t = : PDE: v t = α 2 v xx x L, < t <, (14) BC: v(, t; s) =, v(l, t; s) =, < t < (15) IC: v(x, ; s) = h(x, s). (16) To illustrate the procedure let us consider the following example: EXAMPLE 2. Solve PDE: u t α 2 u xx = t sin(x) x π, < t <, (17) BC: u(, t) =, u(π, t) =, < t <, (18) IC: u(x, ) =. (19) Solution. Here h(x, t) = t sin(x). We solve the related problem: PDE: v t = α 2 v xx, x π, < t <, (2) BC: v(, t; s) =, v(π, t; s) =, < t <, (21) IC: v(x, ; s) = h(x, s) = s sin(x). (22)

4 MODULE 5: HEAT EQUATION 23 Treating s a constant, we easily obtain v(x, t; s) = se α2t sin(x). Note that v(x, t s; s)ds = = e α2t sin(x) which satisfies (17)-(19). se α2s ds = se α2 (t s) sin(x)ds [ ] (α 2 ) 1 t + (α 2 ) 2 (e α2t 1) sin(x), THEOREM 3. (Duhamel s principle, [1]) Let h(x, t) be a twice continuously differentiable function in x L, t. Assume that, for each s, the IBVP PDE: v t = α 2 v xx x L, < t <, (23) BC: v(, t; s) =, v(l, t; s) =, < t <, (24) IC: v(x, ; s) = h(x, s). (25) has a solution v(x, t; s), where v(x, t; s), v t (x, t; s) and v xx (x, t; s) are continuous (in all three variables). Then the unique solution of the problem PDE: u t α 2 u xx = h(x, t) x L, < t <, (26) BC: u(, t) =, u(l, t) =, < t <, (27) IC: u(x, ) =. (28) is given by v(x, t; s)ds. (29) Proof. Note that the function u(x, t) defined by v(x, t; s)ds satisfies the IC u(x, ) = and the BC u(, t) = u(l, t) =. Observe that v(x, t; s) satisfies the BC (24). Now, with g(t, s) = v(x, t; s), where x fixed, we have u t (x, t) = v(x, t; t) + = h(x, t) + Apply Leibniz s rule to obtain v t (x, t; s)ds α 2 v xx (x, t; s)ds. u t (x, t) = h(x, t) + α 2 u xx (x, t). By the hypothese on v(x, t; s), it follows that u(x, t) is in C 2. For the uniqueness, see Theorem 4 (of Lecture 2 of Module 5).

5 MODULE 5: HEAT EQUATION 24 REMARK 4. The solution u in (29) may be written as where v solves (14)-(16). EXAMPLE 5. Solve the IBVP: v(x, t s; s)ds u t α 2 u xx = t[sin(2πx) + 2x] x 1, < t <, u(, t) = 1, u(1, t) = t 2, < t <, u(x, ) = 1 + sin(πx) x. Solution. The function that satisfies the BC is w(x, t) = (t 2 1)x + 1. Then w(x, t)+v(x, t), where v(x, t) solves the related problem with homogeneous BC: v t kv xx = u t α 2 u xx (w t α 2 w xx ) = t sin(2πx) v(, t) = u(, t) w(, t) = v(1, t) = u(1, t) w(1, t) = v(x, ) = u(x, ) w(x, ) = sin(πx). Now, v = u 1 + u 2, where u 1 and u 2, respectively, solves (u 1 ) t α 2 (u 1 ) xx = (u 2 ) t α 2 (u 2 ) xx = t sin(2πx) (a) u 1 (, t) = u 1 (1, t) = (b) u 2 (, t) = u 2 (1, t) = u 1 (x, ) = sin(πx) u 2 (x, ) =. We know that u 1 (x, t) = e π2 α 2t sin(πx). The function u 2 is found via Duhamel s principle. The solution u 2 is given by u 2 (x, t) = where v solves the problem v(x, t s; s)ds, v t = α 2 v xx v(, t; s) = v(l, t; s) = v(x, ; s) = s sin(2πx).

6 MODULE 5: HEAT EQUATION 25 We know that v(x, t; s) = se 4π2 α 2t sin(2πx). Thus, u 2 (x, t) = s e 4π2 α 2 (t s) sin(2πx)ds = e 4π2 α 2t sin(2πx) s e 4π2 α 2s ds [ = (4π 2 α 2 ) 2 4π 2 α 2 t + e 4π2 α 2t ] 1 sin(2πx). The solution is then given by w(x, t) + u 1 (x, t) + u 2 (x, t). REMARK 6. Duhamel s principle is also applicable to problems with PDE u t α 2 u xx = h(x, t) and homogeneous BC of the forms: u x (, t) = u(l, t) = ; u(, t) = u x (L, t) = ; u x (, t) = u x (L, t) =.. Practice Problems 1. Solve the following IBVP: u t = α 2 u xx + cos(3t), < x < 1, t >, u x (, t) =, u x (1, t) = 1, t >, u(x, ) = cos(πx) 1 2 x2 x, < x < Solve the following IBVP: u t = 4u xx + e t sin(x/2) sin(t), < x < π, t >, u(, t) = cos(t), u(π, t) =, t >, u(x, ) = 1, < x < π.