2 Two-Point Boundary Value Problems

Transcription

1 2 Two-Point Boundary Value Problems Another fundamental equation, in addition to the heat eq. and the wave eq., is Poisson s equation: n j=1 2 u x 2 j The unknown is the function u = u(x 1, x 2,..., x n ), where the x j s are spatial coordinates. In this chapter we study Poisson s eq. in one dimension (n = 1), so we get the two-point boundary value problem = f u (x) = f (x), x (, 1), u() = u(1) =. (2.1) We shall study numerical approximations and analytical properties. 1 / 21

2 2.1 Poisson s Equation in One Dimension We find an expression for the solution of (2.1), and obtain uniqueness. The fundamental theorem of calculus: u(x) = c 1 + x u (y)dy. So, using this for u (y), and assuming u is a solution of Poisson s eq. we get y y u (y) = c 2 + u (z)dz = c 2 f (z)dz so x y u(x) = c 1 + c 2 x ( f (z)dz)dy (2.5) Introduce F (y) = y f (z)dz, so by integration by parts x ( y f (z)dz)dy = x yf (y)dy = [yf (y)] x x yf (y)dy = xf (x) x yf (y)dy = x (x y)f (y)dy 2 / 21

3 So: u(x) = c 1 + c 2 x x (x y)f (y)dy and using the boundary conditions we get c 1 = and c 2 = 1 (1 y)f (y)dy. Since the constants are uniquely determined by the boundary conditions, the solution must be unique. And it is u(x) = x 1 (1 y)f (y)dy x Example 2.1 If f (x) = 1 (constant function), then u(x) = x 1 (1 y)dy x (x y)f (y)dy (2.7) (x y)dy = (1/2)x(1 x) 3 / 21

4 Introduce the Green s function: { y(1 x) if y x G(x, y) = x(1 y) if x y 1 Then we rewrite the Poisson solution from (2.7) u(x) = 1 G(x, y)f (y)dy. (2.9) The Green s function acts as an inverse, almost like Ax = b implies x = A 1 b when the matrix is invertible. Note that G is continuous, symmetric, nonnegative and piecewise linear if either x or y is fixed. Note also that the solution u is smoother than the data: u (m+2) = f (m), which follows from the Poisson eq. Theorem 2.1 For every f C([, 1]) there is a unique solution u C 2 ((, 1)) of the boundary value problem (2.1). Furthermore, the solution has the representation (2.9). 4 / 21

5 Using that G we get: Proposition 2.1 Assume that f C([, 1]) is a nonnegative function.then the corresponding solution u of (2.1) is also nonnegative. The next result is called the maximum principle and it says that the solution of the Poisson eq. is smaller than the given function f in the supremum norm: Proposition 2.2 Assume that f C([, 1]) and let u be the unique solution of (2.1). Then u (1/8) f Proof. This follows from (2.9) by a small calculation, using that 1 G(x, y)dy = (1/2)x(1 x) and the supremum of this is 1/8. 5 / 21

6 2.2 A Finite Difference Approximation Want to find approximate solution of the Poisson eq. by solving a linear algebraic system Ax = b. For a four-times continuously differentiable function g = g(x) we can use the Taylor series at x to get expressions for g(x + h) and g(x h), and show that g(x + h) 2g(x) + g(x h) h 2 = g (x) + E h (x) where the error term satisfies E h (x) M g h 2 /12 and M g = sup x g (4) (x). For instance, if g is a polynomial of degree at most 3, then the error is zero. Partition the unit interval: x j = jh (j =, 1,..., n + 1) where h = 1/(n + 1). The discrete problem is obtained by using variables v j that approximate the solution u at x j. We use the approximation for u (x) and get the linear (algebraic) equations v j 1 2v j + v j+1 h 2 = f (x j ) (j = 1,..., n), v = v n+1 =. 6 / 21

7 This may be written as Av = b where A = The right hand side column vector b is given by b = (b 1,..., b n ) (we identify n-tuples and column vectors, for typographical reasons) where b j = h 2 f (x j ) j = 1,..., n. We will later show that A is invertible. Example 2.3 With f (x) = (3x + x 2 )e x we can find the exact solution u(x) = x(1 x)e x. Figures in the book show good approximation of the discrete solution, and the error E h = seems to satisfy E h = O(h 2 ). max u(x j) v j j n+1 7 / 21

8 A is tridiagonal. How to solve tridiagonal linear systems? So more generally: α 1 γ β 2 α 2 γ.. 2. A = (2.18). β n 1 α n 1 γ n 1... β n α n and we consider Av = b. In the second-order difference matrix before we had α j = 2 and β j = γ j = 1 for each j. We use usual row-reduction: First, add β 2 /α 1 times the first row to the second row. In the resulting matrix we then get a in position (2, 1). Then we add a suitable multiple of the second row to the third row, so we get in position (3, 2). Continuing like this, adding a multiple of the ith row to the next, we end up with a row-equivalent matrix which is upper triangular. Of course, similar operations are made to the right hand side b, i.e., we work on the augmented matrix. 8 / 21

9 Then we find the solution v of Av = b by back substitution: determine v n from the last row (only one unknown), and use this value to find v n 1 from row n 1 etc. If we never get a coefficient in front of variable v k in equation k, this process can be completed. This holds if and only if A is invertible and we return to conditions that assure this. As a result (see TW) one gets the following algorithm Algorithm 2.1 δ 1 = α 1 c 1 = b 1 for k = 2, 3,..., n m k = β k /δ k 1 δ k = α k m k γ k 1 c k = b k m k c k 1 v n = c n /δ n for k = n 1, n 2,..., 1 v k = (c k γ k v k+1 )/δ k Note that the number of operations is O(n), in contrast to elimination for a full matrix which is O(n 3 ). 9 / 21

10 An n n matrix A = [a ij ] is called diagonally dominant if a ii j i a ij (i n). A slight variation is for our tridiagonal matrix A above which we call diagonally dominant if α 1 > γ 1, α k β k + γ k (k = 2, 3,..., n) where we define γ n = Lemma 2.1 Assume that the coefficient matrix A of the triangular system (2.19) is diagonal dominant and that β k for k = 2, 3,..., n. Then the variables δ k (k = 1, 2,..., n) determined by Algorithm 2.1 are well defined and nonzero. 1 / 21

11 Proposition 2.3 Assume that the coefficient matrix A of (2.19) satisfies the properties specified in Lemma 2.1 above. Then, the system has a unique solution which can be computed by Algorithm 2.1. Corollary 2.1 The system of equations defined by (2.14)-(2.15), has a unique solution that can be computed using Algorithm 2.1. The second-order difference matrix A has another important property, it is symmetric and positive definite. Recall: A symmetric n n matrix A ia called positive definite if v T Av > for all nonzero v R n. From linear algebra we know that this property holds if and only if all eigenvalues of A are positive. Another equivalent condition is that there exists an n k matrix B with linearly independent columns such that A = B T B. Every positive definite matrix A is invertible, because if Ax = O for some nonzero x, then x T Ax = x T O = ; a contradiction. 11 / 21

12 Proposition 2.4 Consider a tridiagonal system of the form (2.19) and assume that the corresponding coefficient matrix (2.18) is symmetric and positive definite. Then the system has a unique solution that can be computed by Algorithm 2.1. Proof: As just explained, A must be invertible. Moreover, the only problem that can occur in Algorithm 2.1 is that in some iteration k 1 we obtain that the entry in position (k, k) of the present matrix Ā is zero, i.e., that δ k =. But then the first k columns of Ā are linearly dependent. This is a contradiction, as A and Ā are row-equivalent, and therefore have the same rank, namely n. 12 / 21

13 2.3 Continuous and Discrete Solutions We shall show that almost all essential properties of the exact, or continuous, solution are somehow present in the approximate solution. Let L be the differential operator (Lu)(x) = u (x), f C([, 1]). So the two-point boundary value problem is: Find u C 2 ([, 1]) such that (Lu)(x) = f (x) for all x (, 1) (2.26) C 2 ([, 1]): functions that are twice continuously differentiable, and are zero at the boundaries. Let D h be the discrete functions defined at the grid points x j for j =,..., n + 1. We write v(x j ) or just v j. D h, is the subset of D h where the function is at the boundary points. Define the discrete operator which gives the finite difference approximation to the second derivative: (L h w)(x j ) = w(x j+1) 2w j + w j 1 h 2 13 / 21

14 The discrete problem: Find v C h, such that (L h v)(x j ) = f (x j ) for j = 1, 2,..., n (2.27) We use inner products (and norms) as follows. For continuous functions (as above): u, v = 1 u(x)v(x)dx. and for discrete functions u, v D h almost like the usual inner product for vectors: u, v h = h( u v + u n+1 v n+1 2 n + u j v j ) j=1 A matrix A is symmetric if A = A T, and this is equivalent to (Ax, y) = (x, Ay) for all x, y R n, where (, ) denotes the usual Euclidean inner product. 14 / 21

15 Lemma 2.2 The operator L given in (2.26) is symmetric in the sense that u, Lv = Lu, v for all u, v C 2 ((, 1)). Proof. If u, v C 2 ((, 1)), then by integration by parts and using that u() = v() = u(1) = v(1) = : Lu, v = 1 u vdx = u (x)v(x) u v dx = 1 u v dx = 1 uv dx = u, Lv By so-called summation by part we get the same property for the discrete operator: Lemma 2.3 The operator L given in (2.26) is symmetric in the sense that u, L h v h = L h u, v for all u, v D h,. 15 / 21

16 Lemma 2.4 The operators L and L h are positive definite in the following sense: (i) For any u C 2 ((, 1)) we have Lu, u and with equality only if u =. (ii) For any v D h, we have L h v, v and with equality only if v =. Proof. Follows from prev calculations: and also Lu, u = 1 (u (x)) 2 dx n L h u, u h = h 1 (v j+1 v j ) 2. j= 16 / 21

17 We showed before the uniqueness of solution of both the continuous and the discrete problem. Now we see that this also follows from the positive definiteness. Lemma 2.5 The solution u of (2.26) and the solution v of (2.27) are unique solutions of the continuous and the discrete problems, respectively. Proof. Assume Lu 1 = f and Lu 2 = f, and let e = u 1 u 2. Then Le = Lu 1 Lu 2 = f f =. So Le, e = and by positive definite property, this implies that e =, so u 1 = u 2. The discrete case is similar. 17 / 21

18 Remember that the exact solution of the Poisson solution may be written in terms of the Green s function u(x) = 1 G(x, y)f (y)dy. (2.9) For a given grid point x k = kh define a grid function G k D h, by G k (x j ) = G(x j, x k ). Using linearity of G(x, y) in x for x y, we get L h (G k )(x j ) = for j k. A computation shows that L h (G k )(x j ) = 1/h. So L h G k = (1/h)e k where e k (x j ) equal 1 if k = j, and it is otherwise. For f D h, define w D h, by n w = h f (x k )G k k=1 Then w is the unique solution of the discrete problem as n n L h w = h f (x k )(L h G k ) = f (x k )e k = f k=1 k=1 18 / 21

19 So we have v j = h n G(x j, x k )f (x k ) k=1 which is the discrete solution of (2.27). So, this is similar to the continuous solution. And since G k we get Proposition 2.5 Assume that f (x) > for all x [, 1], and let v D h, be the solution of (2.27). Then v(x j ) for all j = 1,..., n. Similarly, we also obtain a discrete maximum principle. Here we use the norm v h, = max v(x j). j=,...,n+1 Proposition 2.6 The solution v D h, of (2.27) satisfies v h, (1/8) f h, 19 / 21

20 Convergence of the discrete solution Definition 2.2 Let f C([, 1]), and let u C 2 ((, 1)) be the solution of (2.26). Then we define the discrete vector τ h, called the truncation error, by τ h (x j ) = (L h u)(x j ) f (x j ) for all j = 1,...,, n. We say that the finite difference scheme (2.27) is consistent with the differential equation (2.26) if lim τ h h, =. h Note that the error is obtained by applying u, not v, to the differential operator L h. 2 / 21

21 By using the Poisson equation, second order Taylor series and our maximum principle we can derive: Theorem 2.2 Assume that f C 2 ([, 1]) is given. Let u and v be the corresponding solutions of (2.26) and (2.27), respectively. Then u v h, f h So, we can get arbitrary small error, by choosing h small enough. 21 / 21