u(0) = u 0, u(1) = u 1. To prove what we want we introduce a new function, where c = sup x [0,1] a(x) and ɛ 0:

Transcription

1 6. Maximum Principles Goal: gives properties of a solution of a PDE without solving it. For the two-point boundary problem we shall show that the extreme values of the solution are attained on the boundary. We also study the heat equation again. 6.1 A Two-Point Boundary Value Problem This problem has the form u (x) + a(x)u (x) = 0, x (0, 1), where a is a given function and u is given in the two endpoints, x = 0 and x = 1. The idea from calculus is to show, in a point x, that v (x) = 0 and v (x) < 0. Consider the inequality v (x) + a(x)v (x) > 0, x (0, 1) (6.1) where a is continuous. If x 0 is a local maximum of v in an interior point, then v (x 0 ) = 0 and v (x 0 ) 0, contradicting (6.1). So no local max in the interior. 1 / 19

2 Lemma 6.1 A function v C 2 ((0, 1)) C([0, 1]) satisfying (6.1), where a C([0, 1]), satisfies the following maximum principle: where V = max{v(0), v(1)}. v(x) V for all x [0, 1], Consider the two-point boundary problem with boundary condition u (x) + a(x)u (x) = 0, x (0, 1), (6.2) u(0) = u 0, u(1) = u 1. To prove what we want we introduce a new function, where c = sup x [0,1] a(x) and ɛ 0: v ɛ (x) = u(x) + ɛe (1+c)x So v ɛ (x) + a(x)v ɛ(x) = ɛ(1 + c)(1 + c + a(x))e (1+c)x, and therefore v ɛ (x) + a(x)v ɛ(x) > 0 for all ɛ > 0. 2 / 19

3 Lemma 6.1 then gives v ɛ (x) max{v ɛ (0), v ɛ (1)}. Now u(x) = v ɛ (x) ɛe (1+c)x v ɛ (x) max{v ɛ (0), v ɛ (1)} = max{u 0 + ɛ, u 1 + ɛe 1+c } The we let ɛ 0 +, and obtain u(x) max{u(0), u(1)} = max{u 0, u 1 }. Similarly, one proves a lower bound u(x) min{u 0, u 1 }. 3 / 19

4 Theorem 6.1 Suppose u C 2 ((0, 1)) C([0, 1]) is a solution of (6.2)-(6.3). Then u(x) satisfies for all x [0, 1]. min{u 0, u 1 } u(x) max{u 0, u 1 }. In this case this can also be shown directly, as the solution can be found. But the proof technique can be used more generally. 4 / 19

5 6.2 The Linear Heat Equation We use the same approach for the heat equation; derive a maximum principle. Physics: Consider a uniform rod of unit length with an initial temperature given by f (x). The temperatures at the left and right boundaries are given by u l (t) and u r (t), respectively Consider the initial-boundary value heat equation u t = u xx, x (0, 1), t (0, T ]. u(0, t) = u l (t), u(1, t) = u r (t) t (0, T ]. (6.7) u(x, 0) = f (x) x [0, 1]. We assume consistency in the sense that u l (0) = f (0) and u r (0) = f (1). We would expect the highest temperature to appear either initially or at one of the boundaries. 5 / 19

6 6.2.1 The Continuous Case Let R = {(x, t) : x [0, 1], t [0, T ]} and let v be a smooth functions satisfying the inequality v t < v xx for 0 < x < 1, 0 < t T. (6.9) Smooth means here: v cont. on R, and v t, v x, v xx cont. on the interior of R. We prove that a maximum of v occurs at the lower boundary of R is defined by B = {(x, t) : x = 0, 0 t T } {(x, t) : t = 0, 0 x 1} {(x, t) : x = 1, 0 t T }; (6.10) Suppose that (x 0, t 0 ) is a local maximum of v in the interior of R, i.e. x 0 (0, 1) and t 0 (0, T ). 6 / 19

7 Then, by the properties discussed in the previous section, we have (i) v t (x 0, t 0 ) = 0 and (ii) v xx (x 0, t 0 ) 0. This gives v t (x 0, t 0 ) v xx (x 0, t 0 ) which contradicts (6.9). A maximum cannot occur when t 0 = T either, because, if so, v t (x 0, t 0 ) 0 and v xx (x 0, t 0 ) 0, again contradicting (6.9). So: Lemma 6.2 A function v C(R) with v t, v x, v xx C((0, 1) (0, T ]) satisfying the inequality (6.9), satisfies the following maximum principle: v(x, t) V for all x [0, 1], t [0, T ], where V = sup (x,t) B v(x, t). We combine this with a regularization of u. Define for ɛ > 0 v ɛ (x, t) = u(x, t) + ɛx 2 (6.11) 7 / 19

8 Then, from the heat eq., v ɛ t = v ɛ xx 2ɛ, so v ɛ t < v ɛ xx, and the lemma gives v ɛ (x, t) V ɛ where V ɛ denotes the maximum of v ɛ on the boundary B. Therefore u(x, t) = v ɛ (x, t) ɛx 2 v ɛ (x, t) and u(x, t) V ɛ = Letting ɛ 0 + gives u(x, t) sup (f (x) + ɛx 2, u l (t), u r (t) + ɛ). (x,t) B sup (f (x), u l (t), u r (t)). (6.13) (x,t) B To find a lower bound we let w(x, t) = u(x, t) and use (6.13), and get inf of the same expression. 8 / 19

9 Theorem 6.2 Suppose u C(R), with v t, v x, v xx C((0, 1) (0, T ]), is a solution of (6.7). Then u satisfies the maximum principle inf (f (x), u l(t), u r (t)) u(x, t) sup (f (x), u l (t), u r (t)). (x,t) B (x,t) B for all (x, t) R. By continuity of u at, in particular, (0, 0) and (1, 0) we get u l (0) = f (0), u r (0) = f (1) Uniqueness and Stability A maximum principle for a linear differential equation will frequently imply a uniqueness and stability result for the solution 9 / 19

10 Let u be a solution of u t = u xx, x (0, 1), t (0, T ]. u(0, t) = u l (t), u(1, t) = u r (t) t (0, T ]. (6.14) u(x, 0) = f (x) x [0, 1]. and ū a solution of ū t = ū xx, x (0, 1), t (0, T ]. ū(0, t) = ū l (t), ū(1, t) = ū r (t) t (0, T ]. (6.14) ū(x, 0) = f (x) x [0, 1]. Define e = u ū. So e is a solution of the same heat equation but with data u l (t), u r (t) and f (x) and we apply Theorem 6.2 and get 10 / 19

11 Corollary 6.1 The problem (6.14) has at most one smooth solution. Furthermore, the solution is stable with respect to perturbations in the sense that sup (r,t) R u(x, t) ū(x, t) sup (r,t) R ( f (x) f (x), u l (t) ū l (t), u r (t) ū r (t) ) where u is the solution of (6.14) and ū is the solution of (6.15) The Explicit Finite Difference Scheme Define r = t/ x 2 and consider the explicit scheme from section 4.1: v m+1 j = rvj 1 m + (1 2r)vj m + rvj+1, m (j = 1,..., n, m 0).(6.17) Boundary conditions are and initial condition v m 0 = u l (t m ), v m n+1 = u r (t m ) (m 0) (6.18) v 0 j = f (x j ); (j = 1,..., n). (6.19) 11 / 19

12 Then one can consider the discrete rectangle R = {(x j, t m ) : x j [0, 1], t m [0, T ]} obtained from R and get a discrete maximum principle under the condition r = t/ x 2 1/2. Let V and V + be the min. resp. max of f, u l and u r over the lower boundary of the discrete rectangle. Theorem 6.3 Suppose that the grid sizes t and x satisfy (6.22), and let vj m be the numerical approximation of (6.7) generated by the scheme (6.17)-(6.19). Then V v m j V + for all grid points (x j, t m ) is the discrete rectangle R. 12 / 19

13 6.2.3 The Implicit Finite Difference Scheme Explicit schemes may be very CPU-time demanding as x is reduced. This is due to the stability condition (6.22) which forces the number of time steps to be of order O(n 2 ), where n is the number of grid points in the x-direction. Here we show that the implicit scheme satisfies the discrete maximum principle for any relevant choice of grid parameters. The implicit scheme is given by: v m+1 j v m j t = v m+1 j 1 and from boundary conditions and initial condition m+1 2vj + v m+1 j+1 x 2 (j = 1,..., n, m 0) (6.23) v m 0 = u l (t m ), v m n+1 = u r (t m ) (m 0) (6.24) v 0 j = f (x j ) (j n) (6.25) 13 / 19

14 The following maximum principle holds for all values of the grid parameters t and x. Theorem 6.4 Let v m j be the numerical approximation of (6.7) generated by the implicit scheme (6.23)-(6.25). Then V v m j V + for all grid points (x j, t m ) is the discrete rectangle R. 14 / 19

15 6.4 Harmonic Functions We studied the two-point boundary problem u = f on a bounded interval. With f = 0 this becomes u = 0 with solution the linear functions. Now we consider this equation in two dimensions. The differential equation will be studied on a bounded, connected, and open domain, Ω, in R 2. The boundary of Ω will be denoted Ω, and its clusure is denoted by Ω, so Ω = Ω Ω The Laplace operator, in two space dimensions, is defined by u = u xx + u yy = 2 u x u y 2 Definition 6.1 A function u C 2 (Ω) C( Ω) is said to be harmonic in Ω if u = 0 for all (x, y) Ω. (6.38) 15 / 19

16 Equation (6.38) is called Laplace s equation. The equation is called Poisson s equation. u = f Example 6.2 Let u(x, y) be a polynomial function of the form u(x, y) = a + bx + cy + dxy, where a, b, c, d are real coefficients. Then it is straightforward to check that u = 0. Hence, u is harmonic in any domain Ω. Example 6.3 Let r = r(x, y) = x 2 + y 2 and u(x, y) = ln(r(r, y)) Then u is continuous on R \ {(0, 0)}. We get u xx = 1 2x 2 (1 r 2 r 2 ), u yy = 1 2y2 (1 r 2 r 2 ) so u = 0 and u is Harmonic. 16 / 19

17 6.4.1 Maximum Principles for Harmonic Functions Theorem 6.7 Assume that u is harmonic in Ω. Then u satisfies the inequality M 0 u(x, y) M 1 for all (x, y) Ω where M 0 = min u(x, y) and M 1 = max u(x, y) (x,y) Ω (x,y) Ω Proof. The proof is similar to the proof of Theorem 6.2 but one uses the regularization v ɛ (x, y) = u(x, y) + ɛ(x 2 + y 2 ). Then, as u = 0, v ɛ = 4ɛ > 0, and then v ɛ cannot have a maximum of this function in the interior because then (by calculus, Hessian is negative semidefinite) v ɛ = v ɛ xx + v ɛ xx 0. So its maximum is attained on the boundary, and we let ɛ tend to zero, as before. 17 / 19

18 A function u definied on Ω is called subharmonic if ( u)(x, y) 0 for all (x, y) Ω. The proof above also works for subharmonic functions. Corollary 6.2 Assume that u is subharmonic in Ω. Then u(x, y) M 1 for all (x, y) Ω Corollary 6.3 Assume that u is harmonic in Ω. Then u(x, y) M for all (x, y) Ω where M = This follows from Theorem 6.7. max u(x, y). (x,y) Ω 18 / 19

19 An application is for the Poisson s equation u = f in Ω (6.40) u = g on Ω f is called the right-hand side and g the Dirichlet data. This problems has a solution, under certain conditions. Moreover, the solution is unique, see below. Thus, if f = 0, this means that an Harmonic function is identified with its values at the boundary of the domain. Corollary 6.8 Assume that u 1, u 2 C 2 (Ω) C( Ω) are two solutions of the problem (6.40) with the same right-hand side f and Dirichlet data g. Then u 1 = u 2. Proof. Let v = u 1 u 2. Then v = 0, so v is Harmonic. But on Ω, u 1 = u 2 = g, so v = 0 on Ω, so Corollary 6.3 shows that v = 0 on Ω. 19 / 19