Summer Lectures 1. Boundary Value Problems: Introduction

Transcription

1 Boundary Value Problems of Second Order ODEs and Their Approximation Professor Lin, Ping University of Dundee and USTB Summer Lectures 1. Boundary Value Problems: Introduction A general second order ode an be written as u = F (x, u, u ) (1) where F is a given function. In order to explore the theoretical properties of second order equations we shall focus on linear problems. These take the general form u = a(x)u + b(x)u + f(x). () Recall that the general solution of such an equation is made up of (1) the complementary solution: i.e., the general solution of the homogeneous equation 1 u a(x)u b(x)u =. This will have two linearly independent solutions, u 1 (x) and u (x) say, from which we construct the general solution u(x) = Au 1 (x) + Bu (x). (3) () a particular solution: denote this by P (x) (there is no general procedure for determining P ). The general solution of the inhomogeneous equation () is then u(x) = Au 1 (x) + Bu (x) + P (x). (4) This leaves a solution containing two arbitrary constants that have to be found using supplementary information (boundary conditions). Example 1 Find the general solution of u + 3u 4u = 1. Hence find the solution satisfying the boundary conditions ( bcs) u() = 1, u(1) = 3. Equivalent Formulations It is sometimes convenient to recast the ode () in dierent, but equivalent, forms by either reorganization of terms or by changes of variable. One of the goals in reformulation is to obtain an equation with no first derivative term. This can be achieved in several ways. Suppose we multiply both sides of () by a function p(x) to give p(x)u p(x)a(x)u p(x)b(x)u = p(x)f(x). (5) Now, by the rule for dierentiating a product, d dx pu = pu + p u 1 An equation in a variable u, say, is homogeneous if replacing u by cu, where c is a constant, leaves the equation unchanged. so that (5) becomes (pu ) [pa + p ]u pbu = pf and the term in u can be made to vanish by choosing p so that and we obtain where p = ap p(x) = e R a(x)dx (p(x)u ) q(x)u = g(x), (6) p(x) = e R adx, q(x) = p(x)b(x), g(x) = p(x)f(x). An alternative form that is sometimes easier to work with involves making a change of dependent variable. We seek a function M(x) so that the change of variable u = M(x)w (7) leads to an ode for w that does not contain w. We first compute u = Mw + M w, u = Mw + M w + M w which, when substituted into (), gives an equation for w where the coeicient of w is M am which, when set to zero, gives M am = M(x) = e 1 R a(x)dx. (Note: M(x) = p(x) 1/.) Then, on dividing by M (the coeicient of w ) we obtain w r(x)w = h(x), () where r(x) = (Mb M + am )/M and h(x) = f(x)/m. It can be shown (See Exercise 1) that r(x) = b(x) 1 a (x) a (x). (9) Two Point Boundary Value Problems We define a Second Order Two Point Boundary Value Problem (or bvp for short) to consist of a second order dierential equation ((), or (6) or ()), defined on an interval < x < L, say, two boundary conditions (bcs), one at x = and one at x = L. The bcs are, generally, relationships between u and its first derivative. In the vast majority of applications the relationships are linear and we have three types. At x = these take the forms (with similar types possible at x = L)

2 Summer Lectures: Boundary Value Problems: Introduction Page 1. Dirichlet: the value of u is specified, e.g., u() = 3. Neumann: the value of u is specified, e.g., u () = 1. Robin: a linear combination of u and u is specified, e.g., u() + u () =. The questions we shall be interested in are Does a given pt bvp have a solution? Is the solution unique? What can we say about the magnitude of the solution without having to solve the bvp? How can we compute a numerical approximation to the solution? Will the numerical solution converge? Example Solve the bvps defined by u a u = sin πx, (1) u + a u = 1, (11) on < x < 1 when subject to the bcs u() = 1, u(1) =. Is there a unique solution for each real value of a? a) Equation (1) has general solution u(x) = Ae ax + Be ax 1 π sin πx, a. + a The bcs lead to the equations A + B = 1 and Ae a + Be a = (note how these equations are made simpler by the fact that the particular solution satisfies homogeneous bcs) giving A = (1 + 1 e a )/ sinh a, which lead to u(x) = sinh a(1 x) sinh ax sinh a B = (1 + 1 ea )/ sinh a 1 π sin πx + a so we have a unique solution for every real value of a. It is easily shown (taking the limit a is one possibility) that there is a unique solution u(x) = (1 3x) 1 sin πx, (a = ) π so there is a unique solution for all real a. b) For equation (11), the general solution is u(x) = A sin ax + B cos ax + 1 a (a ) (1) We now have to apply the bcs: u() = 1 gives B + 1 a = 1 while u(1) = leads to A sin a + B cos a + 1 a = and so, provided that a and sin a, we have A = ( + 1/a + B cos a)/ sin a giving (after some reorganization) u(x) = (1 1/a ) sin a(1 x) ( + 1/a ) sin ax + 1 sin a a. However, this breaks down when a = nπ (n =, ±1, ±,... ). The case a = gives u(x) = 1 x(x 1) + (1 3x), (a = ). The situation can be compared to that in Linear Algebra. Suppose that C is an n n matrix and consider the solution of (C ai)x = b. (13) If a is not an eigenvalue of C, then the matrix C ai is non singular and we have a unique solution x = (C a I) 1 b. However, if a is an eigenvalue of C, then it will also be an eigenvalue of C T. Suppose that v is a corresponding eigenvector of C T, so C T v = av v T C = av T. Multiplying both sides of (13) by v T leads to v T (C ai)x = v T b = v T b. The equations will therefore be inconsistent unless it happens that b is such that v T b =. That is, we have either no solution (b is not orthogonal to the null eigenvector of C ai) otherwise there are an infinite number of solutions. This suggests that the special values of a in the bvp governed by (11) are related to eigenvalues so we look at these problems next. Eigenvalue Problems We shall suppose that our bvp has been transformed into the form (6) and consider the eigenvalue problem (pu ) + q(x)u = λr(x)u, < x < 1, u() = u(1) =. (14) We have chosen Dirichlet bcs for definiteness, in general one might have α u() β u () =, α 1 u(1) β 1 u (1) =, the important feature is that they must be homogeneous. Problems of the type (14) are known as Sturm Liouville Problems.

3 Summer Lectures: Boundary Value Problems: Introduction Page 1.3 Every value of λ for which bvp (14) has a nontrivial solution (i.e., one that is not zero for all x [, 1]) is called an eigenvalue and the corresponding solution is called an eigenfunction. We have written the ode in the form shown, rather than (pu ) q(x)u = λr(x)u in order that the eigenvalues be positive for p(x) >, q(x) >, r(x) >, < x < 1; (15) we shall always assume these conditions to hold (with the possible exception of that on q). Our aim is to discover as many properties as possible regarding the eigenvalues and eigenfunctions of (14). The following example, which is closely related to Example is quite typical. Example 3 Determine the eigenvalues and eigenvectors of the problem with bcs u() = u(1) =. u = λu, < x < 1 The form of the general solution of this ode will depend on whether λ <, λ = or λ >. We consider each case in turn. Suppose firstly that λ <. We write λ = µ then the ode has general solution u(x) = Ae µx + Be µx. The first bc implies that A + B = so u(x) = A(e µx e µx ) and the second bc now gives A(e µ e µ ) = A sinh µ = which implies that A =. Therefore the only solution when λ < is u(x), the trivial solution. Suppose, next, that λ =. Then u = has general solution u(x) = A+Bx and the bcs then give A = B =. This is again, the trivial solution so λ = cannot be an eigenvalue. Suppose, therefore, that λ >. This time we write λ = ω then we have the general solution u(x) = A sin ωx + B cos ωx. The bc u() = 1 implies that B = while the bc u(1) = gives A sin ω =. Thus, either A = (which again gives the trivial solution so we ignore this possibility) or sin ω =, i.e., ω = jπ, j =, ±1, ±,... leading to the infinite sequence of eigenvalues λ j = j π, j = 1,,... with corresponding eigenfunctions φ j = sin jπx. The next theorem shows that many of the properties of this example carry over more generally. Theorem 1.1 If p(x) > and q(x) for x (, 1), then the homogeneous bvp (pu ) + q(x)u =, < x < 1, u() = u(1) =. has only the zero solution u(x) =. (16) Proof. See lectures. Note: The bcs could be changed to any homogeneous type (except u () = u (1) = ) without aecting the conclusion. Theorem 1. The eigenvalues of (pu ) + q(x)u = λr(x)u, < x < 1, u() = u(1) =. are real, simple and positive. (17) Proof. See lectures. In view of this theorem, we can put the eigenvalues of the system in order: It can be proven that < λ 1 < λ < λ 3 <. (1) λ k as k. Theorem 1.3 If φ k and φ j are eigenfunctions corresponding to λ k and λ j, respectively (k j), then 1 r(x)φ k (x)φ j (x)dx =. (19) When r(x) > we can define an inner product on continuous functions on [, 1] by u, v = 1 r(x)ū(x)v(x)dx, (where ū(x) denotes the complex conjugate of a complex function u) and then (19) states that the eigenfunctions corresponding to distinct eigenvalues are orthogonal to each other. Defining a norm via u = u, u 1/, the eigenfunctions are often normalized so that φ j = 1. Proof. See lectures. A Sturm Liouville problem therefore generates a sequence of orthogonal functions {φ j (x)} whose span defines a vector space which we call V: V = span{φ 1, φ,... }, () meaning that any function v V can be expanded in terms of these functions: v = a j φ j (x) j=1 the diiculty being that there are an infinite number of terms in the sum (i.e., V is infinite dimensional).

4 Summer Lectures: Boundary Value Problems: Introduction Page 1.4 Example 4 Find the eigenfunctions and eigenvalues of the bvp u + 4u = λu, We can rewrite the ode as u () = u(π) =. u = (λ 4)u < x < π and Theorem 1. tells us that the eigenvalues (λ 4) of this problem are real and positive. Then, since λ > 4, we write λ = 4 + ω which leads to the general solution u(x) = A sin ωx + B cos ωx. The bc u () = gives A = while u(π) = gives B cos ωπ =. Since B cannot be zero (it would lead to the trivial solution), we must have cos ωπ =. This can only occur if ω is an odd multiple of 1 : ω = 1 (j 1), λ = 4 + (j 1 ) for j = 1,,..., and the corresponding eigenfunctions are φ j = cos(j 1 )x. The weight function is r(x) = 1 so φ j = π cos (j 1 )xdx = 1 π so the normalized eigenfunctions are φ j = π cos(j 1 )x. Notation We introduce some operator notation that will be useful later on. For an ode we define a dierential operator on an open interval < x < 1 by, for example, L u(x) (p(x)u (x)) + q(x)u(x), < x < 1. To incorporate the bcs, we define an operator on a closed interval, x 1, say. For instance with a The form of the general solution of this ode will depend on whether λ 4 <, λ 4 = or λ 4 >. As in the previous exercise, λ 4 leads to trivial solutions u (this has to be checked in each case since the bcs can have subtle influences). Dirichlet bc at x = and a Neumann condition at x = 1, we might define u(), at x = L u(x) = L u(x), < x < 1 u (1), at x = 1. An operator with a bar above it therefore incorporates the bcs. A Sturm Liouville eigenproblem can now be expressed quite succinctly: L u = λru, where, at x = Ru(x) = r(x)u(x), < x < 1, at x = 1, and a two point bvp can be written as L u = F, where a, at x = F (x) = f(x), < x < 1 b, at x = 1, to represent (pu ) + qu = f, < x < 1, u() = a, u (1) = b. Thus F contains the data of the problem: the boundary values and the right hand side of the ode. Also, the bvp L u λru = F will have a unique solution provided λ is not an eigenvalue of L v = λrv (v is the corresponding eigenfunction). If λ is an eigenvalue, then the equation will only have a solution for certain right hand sides F : it will need to be orthogonal to the eigenfunction v corresponding to λ, i.e., v, F =. In this case there will be an infinite number of solutions u = p + αv, where p is a particular solution, i.e., L p λrp = F and α is an arbitrary constant. Exercise 1 Express r(x) in (9) in terms of a(x) and b(x). Exercise Write each of the equations (1) u + u = 1, () u + 3u 4u = x,

5 Summer Lectures: Boundary Value Problems: Introduction Page 1.5 (3) Legendre s equation: (1 x )u xu + ku = (4) Bessel s equation: 3 x u + xu + (x n )u = in the forms (6) and (). Exercise 3 Consider the bvp u 4π u = f(x), < x < 1 with bcs u() = u(1) =. Show that it has no solution in the case f(x) = sin πx and an infinite number of solutions when f(x) = 1. Explain your results. [Hint: when f(x) = sin(πx), show that the general solution is u(x) = 1 4π x cos πx+a sin πx+b cos πx.] Exercise 4 Solve the bvp u + a u = sin πx, < x < 1 u() = 1, u(1) = for all a R. What are the solutions in the cases a = ±π? (c.f. Example.) Exercise 5 Repeat Example for the bcs u() = 1, u (1) =. Exercise 6 Check that the eigenfunctions in Example 4 are orthogonal (the weight function is r(x) = 1). Exercise 7 Repeat Example 4 when the u() = u(π) =. Exercise For what value(s) of a does the bvp u 9u = x a, < x < π, u() =, u(π) = have a solution? bcs are Exercise 9 Show that λ = ω is an eigenvalue of the bvp u = λu, < x < 1, u() =, u(1) = u (1) provided that ω is a root of the equation tan ω = ω. By considering the graphs of tan ω and ω, show that there are an infinite number of eigenvalues λ n and that λ n (n + 1 )π as n. Exercise 1 Derive the dierential equation satisfied by u when u + a u = 1 (u d u/dx ) is subject to the change of variable s = x (the ode should be written in terms of d u/ds using the chain rule). What about a more general change of scale s = α+lx? Use such a change of variables to deduce the exceptional values of a when solving the bvp u + a u =, with u() = 1 and u(l) =. < x < L 3 Solutions of Bessel s equation give rise to the polar coordinate equivalents of sin and cos functions.

6 Summer Lectures. Green s Functions Introduction 5 1 (x s) 5 (x s) 5 3 (x s) It is known from linear algebra that a square matrix A is nonsingular (and therefore invertible) if none of its eigenvalues is zero. Also, the solution of a system of linear equations Ax = b is given by x = Xb, where X is the square matrix X = A 1. We can find X by solving AX = I, so the jth column of X satisfies e j being the jth unit vector: Ax j = e j, (1) e j = [,,...,,1,,..., ] T the only nonzero entry being in the jth column. We can think of x j as a response to a point load in the jth position. We want to pursue a similar line with second order two point bvps. We know from Theorem 1.1 that the eigenvalues of a Sturm Liouville problem are positive (and therefore non zero) when the coeicients are all positive in the sense of (1.13) (we shall only be concerned with problems with r(x) 1). Thus, the operator L in the bvp L u = F () is invertible. What might its inverse look like? We shall consider this for the case of homogeneous bcs: F (x) = [; f(x); ] (see the Notation section of Handout 1). We have to find solutions to () when F corresponds to a point load. The critical component of such an F is f(x), the right hand side of the ode. One might think that a possible choice could have f(x) to be zero everywhere except at one point, s say, where f(s) = 1. However the net strength of such a load would be R 1 f(x)dx = and it is not suiciently strong. What we want is a function f that is strongly localized i.e., zero except for close to x = s but yet has R 1 f(x)dx = 1. It is not possible to write down an explicit formula for such a function but there are several ways in which a suitable candidate can be defined as the result of a limiting process. 4 We denote the functions 1, and 3: 1(x) = lim ε ( 1 ε for x < ε otherwise. 1 (x) = lim e (x/ε) ε πε ε 1 3(x) = lim ε π x + ε. It can be shown that R j(x s)dx = 1 for any real s and, as can be seen from Figure 1, each is strongly localized Fig. 1: The localized load functions when s =.7. (each function is drawn for three value of ε). Whichever limiting process we choose, we end up with a generalized function called the Dirac delta function, δ(x) which is defined to have the properties (1) δ(x) = for all x, () R δ(x)dx = 1. (3) R δ(x)f(x)dx = f() for any continuous function f. The limits of integration can be varied here so long as the interval of integration contains x =. It follows that Z δ(x s)f(x)dx = f(s). (3) Thus, in order to compute the inverse of L, we solve the bvp () with f(x) = δ(x s) for any s (, 1). The solution will clearly depend on both x and s and we denote it by G(x, s) called the Green s function of L ; it is the function equivalent of the matrix X in the linear algebra analogy. Example 1 Find the Green s function for the problem We have to solve u = f(x), < x < 1, u() =, u (1) =. G = δ(x s), < x < 1, G() =, G (1) =. Since δ(x s) = except when x = s, we break the problem into four parts, one in each of the intervals (, s), (s, s +) and (s, 1), where s and s + are points just to the left and right of x = s, respectively. These are then glued together to form a continuous function this is the fourth part of the process. Left x [, s): Here we solve G = to give G = Ax+B and then apply the left bc: G = at x =. Hence G(x, s) = Ax, < x < s, (4) where A is an arbitrary constant. 4 See Introductory Applications of Partial Dierential Equations, G.L. Lamb Jr., Wiley Interscience, 1995

7 Summer Lectures: Green s Functions Page. Right x (s, 1]: Again G = and, using the right bc: G = at x = 1, we deduce that G(x, s) = C, s < x < 1, (5) where C is an arbitrary constant. Continuity x = s: The solutions we have computed on [, s) and (s, 1] should be continuous at x = s: As = C. (6) Centre x (s, s +): We now integrate the ode G = δ(x s) from x = s ɛ to x = s + ɛ and use the property that This gives Z Z s+ɛ s ɛ δ(x s)dx = 1. G dx = Z s+ɛ s ɛ δ(x s)dx G (s +) + G (s ) = 1. (7) (The last result is obtained by letting ɛ ) We now use (4 7) to determine the constants A and C. Since G = A for < x < s and G = for s < x < 1, and equation (7) gives G (s ) = A, G (s +) = A = 1. This, together with (6),leads to ( x for < x s G(x, s) = s for s < x < 1. This is shown in Figure. The solid line highlights the particular function G(x,.7) s 1.5 Fig. : The Green s function for Example 1. x How can we use G to solve L u = F? The justification is a little messy (it involves multiplying the ode by the Green s function and integrating by parts twice) but the end product is very elegant..5 1 Example Use the Green s function of the previous example to obtain the solution of u = f(x), < x < 1, u() =, u (1) =. We multiply both sides by G(x, s) and integrate by parts twice (when dierentiating G, we always mean dierentiation with respect to x, s is regarded as a parameter). [ u G] 1 x= + u Gdx = u G dx = Gfdx Gfdx. (Now use G = at x = and u = at x = 1) [ug ] 1 x= u G dx = ug dx = Gfdx Gfdx. (Now use u = at x = and G = at x = 1) But, G (x, s) = δ(x s) so ug dx = u(x)g (x, s)dx = u(s) (see the property (3) of δ). Hence u(s) = G(x, s)f(x)dx or, interchanging the x and s variables, u(x) = Gfdx G(s, x)f(s)ds. () Note that G(x, s) = G(s, x) in this example (this occurs whenever the original ode does not have a u term see page 11.4.) Example 3 Use the Green s function to solve the bvp of Example when f(x) = e x. We have u(x) = = = Z x Z x G(s, x)e s ds G(s, x)e s ds + se s ds + x x xe s ds G(s, x)e s ds = (xe x e x + 1) + x(e e x ) = 1 + xe e x Example 4 Determine the Green s function for the bvp u + u = f(x), < x < 1 u() u () =, u(1) =.

8 Summer Lectures: Green s Functions Page.3 So G satisfies the ode G + G = δ(x s) for x (, 1) together with the bcs G() G () = and G(1) = and we split the solution process into four parts as before. Left : G + G = so G = Ae x + Be x for which the bc G() G () = implies that B =. hence G(x, s) = Ae x, < x < s. (9) Right : G = Ce x + De x for which the bc G(1) = implies that Ce + De 1 = D = e C. Thus, G(x, s) = C(e x e x+ ) = Ce(e x 1 e x+1 ) = a sinh(1 x), s < x < 1, (1) where it is convenient to define a new arbitrary constant a = Ce. Continuity G(s, s) = Ae s = a sinh(1 s) (11) Centre: integrating the ode from x = s ɛ to x = s + ɛ and using the property R δ(x s)dx = 1 gives Z s+ɛ s ɛ ( G + G)dx = Z s+ɛ s ɛ δ(x s)dx G (s +) + G (s ) = 1. (1) (The last result is obtained by letting ɛ and noting that lim ɛ Z s+ɛ s ɛ Gdx = since G is bounded on this interval and the length of the interval tends to zero.) We see from (9) and (1), respectively, that and so (1) leads to We finally have G(x, s) = a = e s 1, G (s ) = Ae s G (s +) = a cosh(1 s) ( e x 1 sinh(1 s) A = e 1 sinh(1 s). for < x < s e s 1 sinh(1 x) for s < x < 1 (13) which is again seen to be symmetric; it is shown below in Figure 3. Symmetry and Self Adjointness Returning to the linear algebra analogy: an n n matrix A is symmetric if its entries satisfy a i,j = a j,i which carries over to the bvp case as G(x, s) = G(s, x) symmetry of the Green s function. However, it is impractical to judge whether or not a bvp operator L is symmetric by having to calculate its Green s function. We therefore look for G(x,s) x Fig. 3: The Green s function G(x, s) for Example 4 is shown for the 5 values s = :. :.. a direct way of assessing symmetry. If we have an inner product x, y = x T y then, for a symmetric matrix A, Thus, x, Ay = x T Ay = (A T x) T y = (Ax) T y = Ax, y. x, Ay = Ax, y (14) for a symmetric matrix A. i.e., for a symmetric matrix we can move A from one side of the inner product to the other. This is the key to generalizing the idea of symmetry to bvps where the property is known as self adjointness. We consider now a dierential operator of the form L u(x) (p(x)u (x)) + q(x)u(x), < x < 1 (see the section on Notation in Handout 1) and we defer the specification of bcs for the moment. Then, with the usual L (, 1) inner product u, v = we find, by integration by parts, v, L u = v( (pu ) + qu)dx = [ vpu ] 1 x= + u(x)v(x)dx, (pu v + quv)dx = [ vpu ] 1 x= + [v pu] 1 x= + = [p(uv u v)] 1 x= + L v, u. ( u(pv ) + quv)dx The operator L will be self adjoint if the bcs are such as to make [p(uv u v)] 1 x= =. Operators for which v, L u = L v, u + boundary terms are said to be formally self adjoint. Thus, provided that p() and p(1) 5 the bcs must be such that uv = u v at both x =, 1. 5 If one or neither of these is true (for example, p(x) = x, 1 x or x(1 x)), the operator is said to be singular which should not be confused with the use of the term in linear algebra, it has quite a dierent meaning here.

9 Summer Lectures: Green s Functions Page.4 If we take the most general homogeneous bcs, then an operator defined on [, 1] would be >< a u() b u (), at x = L u(x) = L u, < x < 1 a 1u(1) + b 1u (1), at x = 1 and, since we are discussing the self adjointness of L, both u and v must satisfy the same bcs at each end. These bcs are homogeneous so, at x =, for instance, a u() b u () = & a v() b v () =. If b = it follows immediately that u()v () u ()v() = while, when b, we have u () u() = v () v() = a b and» u u()v () u ()v () () = u()v() u() v () =. v() Thus, in order to obtain a self adjoint operator, it is not suicient to consider the operator L on the interior of the interval, the bcs are also crucial and so it is L that has the property of self adjointness. We obtain a formally self adjoint operator L by adopting either of the reformulated odes (1.6) or (1.). The reason is that in each integration by parts the leading term in the integrand undergoes a change of sign so that uv ends up as u v after two integrations by parts. If there were a first derivative term present, only one integration by parts would be needed and so the term would eectively change sign. This is illustrated in Exercise. Inhomogeous bcs To solve a bvp L u = F with non homogeneous bcs such as u() = 3 and u(1) + u (1) = 1, we first find any function φ(x) that satisfies these conditions (in this case φ(x) = 3 4x/3 in general v might need to be a quadratic polynomial). We then write v(x) = u(x) φ(x) so that v satisfies homogeneous bcs and 3 3 L v = L u L φ = F 4 L φ 1 5 = 4 f(x) L φ and we solve this problem for v (from which u = v + φ). Exercise 3 Solve the bvp ( u = f(x), < x < 1 u () = 1, u(1) + u (1) = by using the appropriate Green s function. 3 5 Exercise 1 Determine the Green s functions for the bvps ( u + u = f(x), < x < 1 u() =, u(1) = ( (xu ) + 16 u = f(x), < x < 1 x u () =, u(1) = (Hint: in the second problem, seek solutions of the form u(x) = Ax n.) Exercise If L u(x) = u + u, show that u, L v = L u, v + boundary terms, where L u(x) = u u (this is known as the adjoint of L ). Find the adjoint of the operator L u = u a(x)u b(x)u.

10 Summer Lectures 3. Comparison Principles for BVPs The results we describe are from a family of techniques that generally come under the title of Maximum Principles. We shall refer to the theorems as Comparison Principles because this best captures the nature of the two special forms that we describe. We consider the family of bvps of the form p(x)u + q(x)u + r(x)u = f, < x < 1, (1) u() = α, u(1) = β, and we shall assume throughout this handout that p(x) > and r(x). We shall also use the shorthand notation L u = F, () where >< u(), at x = L u(x) = L u, < x < 1 u(1), at x = 1, L u(x) = p(x)u + q(x)u + r(x)u and >< α, at x = F (x) = f(x), < x < 1 β, at x = 1. The first theorem proves that positive loads lead to positive solutions. Theorem 3.1 If r(x) > for x (, 1), then L u implies that u. (Both inequalities hold for x 1.) By L u we mean that we solve a bvp L u = F, where F each of its components is non negative. Proof. Suppose, to the contrary, that there is a point where u is negative on [, 1]. There must be a point x = s, say, s (, 1) where u achieves a negative minimum 6 : This means that u(s) <, u (s) = and u (s). L u(s) = p(s)u (s) + r(s)u(s) < which contradicts the assertion that L u(s). Thus, there can be no point s where u achieves a negative minimum and so u(x), x [, 1]. Example 1 We deduce from the theorem that the bvp u xu (x) x u = ex, < x < 1 u() =, u(1) = has a positive solution even though we cannot write down an expression for the solution. 6 The negative minimum cannot occur at either endpoint because L u implies that u() and u(1). (3) (4) Corollary 3. L u L v implies that u v Proof. Because L is a linear transformation 7 we have L (u v) = L u L v and so, by Theorem 1.1, u v. The corollary compares two functions hence the name Comparison Principle. Theorem 1.1 also falls within this category as it compares u with the zero function. Corollary 3.3 L u = F has a unique solution. These results can be useful for homogeneous odes as the next example shows. Example Show that the solution of the bvp u xu (x) x u =, < x < 1 u() = 1, u(1) = satisfies 1 u(x). i.e., u achieves its max and min values on the boundary of the interval [, 1]. Let L u = u xu (x) x u and consider (1) v(x) = 1 (the smallest boundary value). A short calculation shows that >< 1 >< 1 L v(x) = 1 = L u(x) 1+x 1 so L u L v from which we conclude that u(x) v(x) from Cor. 3.. () With v(x) = (the largest boundary value) we find >< >< 1 L v(x) = = L u(x) 1+x so L u L v from which we conclude that u(x) v(x) from Cor. 3.. The next result is useful for finding out the size of the solution in terms of the magnitude of the load f. Theorem 3.4 Suppose that there is a non negative function v (called a comparison function) such that L v(x) 1 for x (, 1).Then, if u is the solution of the bvp L u(x) = f(x), < x < 1 u() = u(1) =, it follows that Mv(x) u(x) Mv(x), for x [, 1], where M = max x [,1] f(x). 7 L (u + v) = L u + L v and L (cu) = cl u for any constant c and any (twice dierentiable) functions u and v.

11 Summer Lectures: Comparison Principles for BVPs Page 3. Proof. The upper bound follows from the sequence of inequalities L u(x) = f(x) M ML v(x) = L (Mv(x)) for < x < 1 and, at the boundaries: L u() = u() Mv() and u(1) Mv(1). Hence L (Mv u) so that Mv u by Corollary 1.. The left inequality follows in an exactly similar way starting with L u = f M. There is often a great deal of flexibility in the choice of comparison function v we shall focus on relatively simple situations where a constant comparison function suices. This is illustrated in the next example. Example 3 Determine upper and lower bounds on the solution of the ode of Example 1 when subject to the bcs u() = u(1) =. We consider the use of a constant function v(x) = c: L (c) = c 1 + x 1 c, x 1 so that we may choose v(x) =. Since f(x) = e x e we have M = e and e u(x) e, x 1. We can immediately refine this because the data are all positive (Theorem 1): L u and so u(x) e, x 1. (The actual maximum value of the solution is approximately.1... so our upper bound is not very precise.) Inhomogeneous bcs When considering a bvp L u = F, we need an extension of Theorem 3.4: Theorem 3.5 Suppose L u implies that u and that there is a comparison function v such that L v 1. Then, if u is the solution of the bvp L u = F it follows that Mv(x) u(x) Mv(x), for x [, 1] provided that M F M. Example 4 Find upper and lower bounds on the solution of the bvp u xu (x) x u = ex, < x < 1 Now, with v(x) = c, >< c L v(x) = c u() = 1, u(1) =. c 1+x >< 1, at x = 1, < x < 1 1, at x = 1, if c 1 and c/ 1, i.e., we may choose c = (the smallest constant satisfying both inequalities). Also, M = e since 1 F e and we find that e u(x) e. If we were to use the result of Exercise we would find u(x) e. This type of problem inhomogeneous bcs and inhomogeneous ode can also be tackled by writing as the sum of two problems that we have solved earlier. For example, >< α >< α >< L u(x) = f(x) = + f(x) β β Hence, if we define the functions w(x) and z(x) as solutions of the bvps >< >< α L w(x) = f(x), L z(x) = β then L u = L w + L z, which implies that u = w + z (provided L u implies u so that there is a unique solution Theorem 1.3). Upper and lower bounds on w and z may be found using theorems above and these can be combined to give upper and lower bounds on u. General bcs When incorporating Robin bcs into L it is imperative that we use the outward derivatives at end points if we wish to prove positivity results. That is, we should define >< a u() b u (), at x = L u(x) = L u(x), < x < 1. (5) a 1u(1) + b 1u (1), at x = 1 Theorem 3.6 Suppose L is defined by (5) then L u implies u provided a, b (with not both zero) and a 1, b 1 (with not both zero). (If a = b = then no bc is applied at x =.) Proof. This follows the proof of Theorem 3.1; we have to prove that L u precludes the possibility of a negative minimum for u(x) at x = or x = 1. (a) a = b = : then no bc is applied at x = and is disallowed by the statement of the Theorem. (b) a >, b = : then the bc gives u() which clearly precludes a negative minimum. (c) a >, b > : We suppose that u has a negative minimum at x =. The bc implies that a u() b u () so that u () a b u() < from which it follows that u is decreasing at x =, contradicting the supposition that it has a minimum. Thus u().

12 Summer Lectures: Comparison Principles for BVPs Page 3.3 (d) a =, b > : so that u (). If we have strict inequality; u () <, then an argument similar to case (c) can be used to preclude a negative minimum at x = while, if u () =, we have a stationary point at x = and the argument used for a negative minimum at x = s (, 1) can be extended to this case. Similar arguments can be used at x = 1 (see Exercise 9). Exercise 1 Complete the proof of Theorem 1.4. Exercise Amend the proof of Theorem 1.4 to prove that Dv(x) u(x) Cv(x) when f is bounded by D f(x) C, where C and D are positive constants. Exercise 3 Determine upper and lower bounds on the solutions of the bvps defined by the odes a) u + u = 1 3x b) 1 1 u + u = 1 3x c) u + 1u = 1 3x when subject to the bcs u() = u(1) =. How are these bounds changed when the bcs are u() = 3 and u(1) =? Exercise 4 Solve the bvp u + 1u =, < x < 1, u() = 1, u() = to illustrate that the solution of a homogeneous ode does not lie between its boundary values when these have the same sign. i.e., it is not true here that 1 u. Exercise 5 Consider the bvp u + (1 x )u = 4x 1, 1 < x < 1 u ( 1) = u( 1) + 3, u(1) =. Exercise 7 Show that a constant function v(x) = c fails as a comparison function for the bvps u = x (13 x ), 1 < x < 1 u( 1) =, u(1) =. u + x u = x (13 x ), 1 < x < 1 u( 1) =, u(1) =. Use a comparison function v(x) = 1 (1 x ) to find upper and lower bounds for the solution in each case. Verify that u = 1 x 4 is the solution of the second bvp. Sketch (or draw in Scientific NoteBook) both the bounds for the solution and the exact solution. What do you observe? Exercise Use the bvp u a u =, < x < 1 u() =, u(1) = 1. to illustrate the fact that that odes of the type (1) do not satisfy a comparison principle when r(x) <. i.e., prove that u(x) does not satisfy u(x) 1. Exercise 9 Complete the proof of Theorem 1.6 by proving that u cannot have a negative minimum at x = 1. Exercise 1 Determine constants a, b and c so that v(x) = a + bx + cx satisfies L v 1 when >< v (), if x =, L v(x) = v (x) + xv(x), if < x < 1, v(1), if x = 1, and verify that v(x) for x [, 1]. [a = 5/, b = 1, c = 1/]. Show that it is possible to find suitable forms for L and F so that the bvp may be written as L u = F in such a way that it can be proven that L u implies u. Pay particular attention to the bc at x = 1. Explain why a constant comparison function v cannot be used in this example. Show that v(x) = 4 1 x satisfies all the requirements of a comparison function and hence find upper and lower bounds on the solution. Exercise 6 Consider the bvp u + x u = 4x 1, 1 < x < 1 u( 1) = 3, u(1) =. Show that it is possible to find suitable forms for L and F so that the bvp may be written as L u = F in such a way that it can be proven that L u implies u. Pay particular attention to proving that a negative minimum cannot occur at x =. (Use continuity of u(x) to deduce that, if u() <, then u(x) must also be negative in a suiciently small interval ( ε, ε) enclosing the origin. Show that this contradicts the inequality L u.)

13 Summer Lectures 4. Finite Dierence Approximations Let M be a positive integer and define a grid size h = 1/M. For finite dierence approximations of bvps we replace the interval x 1 by a grid of points x m = mh, m =, 1,..., M, i..e., x, x 1,..., x M at which the exact solution of the bvp has values u(x ), u(x 1),..., u(x M ). Instead of solving a dierential equation over the interval (, 1), we replace this by an (approximate) set of M + 1 algebraic equations that are solved to give the M + 1 numbers U, U 1,..., U M. If the algebraic equations are suitably designed, then we shall have U u(x ), U 1 u(x 1),..., U M u(x M ), i.e., approximations to the solution of the bvp at the grid points. We also hope that as more grid points are used (by increasing M), the level of accuracy achieved is also increased. The key step is the replacement of the dierential equation and bcs by algebraic equations and for this we shall require Taylor series expansions. Notation In this and subsequent chapters, we shall use u m, u m, u m as shorthand notation for u(x m), u (x m) and u (x m), respectively. Taylor Expansions and Approximations of Derivatives For any smooth function u we have u(x + h) = u(x) + hu (x) + 1 h u (x) + (1) where the dots indicate terms we consider to be negligibly small. By choosing x to be a generic point on the grid, x = x m, and rearranging, we have u (x m) = 1 h [u(xm+1) u(xm)] 1 hu (x m) +. () This can also be written as u m = 1 h [um+1 um] 1 hu m +. Thus, when h is small, the remainder terms 1 hu m + may be regarded as being small and we have the approximation u m 1 [um+1 um] h which gives a means of estimating the value of u at x = x m by an algebraic expression that uses values of u at the grid points x = x m and x = x m 1. This is known as the forward dierence approximation of u m; we shall use the symbol to denote a forward dierence, as defined by u m = u m+1 u m (3) Fig. 1: The slope, u m, of the tangent at A is approximated by the slope of AB for a forward dierence approximation, by the slope of AC for a backward dierence approximation and by the slope of BC for a central dierence approximation. so u m h 1 u m. In a similar fashion, starting with the Taylor series u(x h) = u(x) hu (x) + 1 h u (x) + (4) with x = x m and rearranging, we have or u (x m) = 1 h [u(xm) u(xm 1)] + 1 hu (x m) + (5) u m = 1 h [um um 1] 1 hu m +. (6) This is known as the backward dierence approximation of u m; we shall use the symbol (nabla) to denote a backward dierence, as defined by u m = u m u m 1 (7) so u m h 1 u m. Finally, if we subtract (4) from (1) we obtain, u(x + h) u(x h) = hu (x) h3 u (x) + which leads to u m = 1 h [um+1 um 1] 1 6 h u (x) + at x = x m. This is known as the second order central dierence approximation of u m and is represented in terms of the second order central dierence defined by u m = 1 [um+1 um 1]. () Thus, if we allow use of the three grid values u m 1, u m and u m+1 there are three ways of approximating u m: the forward and backward approximations (h 1 u n and h 1 u m) have remainder terms proportional to h and so are called first order approximations while the central We shall use the term smooth to indicate that all the derivatives present in our Taylor expansions exist and are continuous.

14 Summer Lectures: Finite Dierence Approximations Page 4. aprroximation h 1 u m has a remainder term proportional to h and is therefore a second order approximation. All three are illustrated in Fig. 1. Approximating Second Derivatives. Adding (4) to (1) we obtain, u(x + h) + u(x h) = u(x) + h u (x) h4 u (4) (x) + which leads to u m = h [u m+1 u m + u m 1] 1 1 h u (x) + at x = x m. Defining the second order dierence operator δ by δ u m = u m+1 u m + u m 1 (9) then we have the approximation u m h δ u m with a remainder term proportional to h. This is the only finite dierence approximation of u that makes use of just the three grid points x mmx m±1. Table 1: Dierence operators, their definitions and Taylor expansions. Second Order Central Dierence Operator δ δ u m u m+1 u m + u m 1 = h u m h4 u (4) m + O(h 6 ) (a) Forward Dierence Operator u m u m+1 u m = hu m + 1 h u m + O(h 3 ) (b) Backward Dierence Operator u m u m u m 1 = hu m 1 h u m + O(h 3 ) (c) Central Dierence Operator u m 1 [u m+1 u m 1 ] = hu m h3 u m + O(h 5 ) (d)

15 Summer Lectures 5. Approximation of BVPs Introduction We begin by considering the two-point boundary value problem (bvp) u (x) + r(x)u(x) = f(x), < x < 1, (1) u() = a, u(1) = b, where u = d u/dx and r(x), f(x) are given continuous functions on the interval [, 1]. bcs of a more general nature will be described in subsequent sections. In order to solve this problem numerically the dierential equation is approximated in such a way as to generate a system of linear algebraic equations. These may then be solved by standard techniques from numerical linear algebra. Choosing a positive integer M, a grid is defined on the interval [, 1] by X h {x m = mh, m =, 1,..., M}, where h = 1/M is the grid size. The dierential equation is then evaluated at the mth point of the grid to give The relationships u (x m) + r(x m)u(x m) = f(x m). () h δ u m = u m + O(h ) δ u m = u m+1 u m + u m 1 are given in Table 13.1 (recall that u m u(x m)). Then, u m = h (u m+1 u m + u m 1) + O(h ) and using these in conjunction with (), we obtain h δ u m + O(h ) + r mu m = f m. (3) The order terms, representing the remainder terms in the Taylor expansions, have been retained, so this equation is satisfied identically by the exact solution, u, of the dierential equation. When the order term is neglected, equation (3) will no longer be satisfied by u, but by some grid function U, say, which it is hoped will be close to u. This process leads to one possible finite dierence replacement of (1), given by U = a, h δ U m + r mu m = f m, m = 1 : M 1 U M = b. Written explicitly, 9 = ; (4) 1 [Um 1 Um + Um+1] + rmum = fm (5) h for m = 1,,..., M 1. Taken together with the boundary conditions U = a, U M = b, there are M +1 linear algebraic equations with which to determine the M + 1 grid values of U. Local Accuracy The local accuracy of this method 9, denoted by R, is defined to be the residual (or defect) when the exact solution of the dierential equation is substituted into the dierence equation. Hence, R m = h δ u m + r mu m f m, (6) where u is the exact solution of (1). Since f = u + r u, the result (a) of Table 13.1 leads to R m = u m h u (4) m + O(h 4 ) + r mu m ` u m + r mu m = 1 1 h u (4) m + O(h 4 ). (7) Strictly speaking the boundary conditions should also be taken into account when estimating the local accuracy, but since U = u = a and U M = u M = b, it follows that R = R M =. Hence, since R = O(h ), we say that the method (4) is consistent with (1) and has a consistency error of order two. Matrix Vector Formulation It is convenient, both from the point of view of implementation (and some aspects of the analysis) to rewrite the dierence equations in matrix vector form. Setting m = 1 in (5) we find 1 [U U1 + U] + r1u1 = f1 h then using the bc U = a we have «h + r1 U 1 1 h U = f1 + a h. For 1 < m < (M 1) we have 1 «h Um 1 + h + rm U m 1 Um+1 = fm h and, for m = M 1, 1 h UM + h + rm 1 «U M 1 = f M 1 + b h, where we have used the bc U M = b. Let U R M 1 denote the vector U = [U 1, U,..., U M 1] T, which contains the unknown grid values of U in their natural order, then scheme (5) may then be expressed in the matrix form AU = F, () 9 This is often called the local truncation error; we do not use this term since the definition of local accuracy for bvps conflicts with the definitions of lmms and rk methods in earlier chapters.

16 Summer Lectures: Approximation of BVPs Page 5. where A = 6 4 a 1,1 M M a, M a M,M M M a M 1,M 1 and a m,m = M + r m. We can express the system more neatly by defining T as the (M 1) (M 1), symmetric, tridiagonal matrix T = 1 h (9), (1) 7 5 which represents the approximation of the second derivative. Then A = T + D, D = diag(r 1, r,..., r M 1) is a diagonal matrix and F R M 1 is defined by f 1 + a/h 3 f 7 F = 6 4. f M f M 1 + b/h The algebraic system () will have a unique solution provided A is nonsingular; the following lemma establishes a somewhat stronger result. Lemma 14.1 The matrix A of equation () is symmetric and positive definite if r(x). Proof. The symmetry is obvious so we need to prove that U T AU > for all non zero vectors U R M 1. The condition r(x) ensures that the entries in D are non negative so that 7 5 U T AU = U T (T + D)U = U T T U + U T DU U T T U and the result will follow from the positive definiteness of T. To prove this, let L denote the (M 1) M matrix L = h Example 1 Use the method (4) to solve the dierential equation u (x) + a sec (ax)u(x) = 4a cos (ax) (11) on the interval < x < 1 with a = 5π/1 and the boundary conditions u() = u(1) = 1. The numerical solution with M = is shown by the dots in Figure 1. Also shown for comparison is the exact solution u = 1 4 tan(ax) + cos (ax). Solution Fig. 1: The exact solution (solid line) and the numerical solution (dots) for Example 1 when M =. x A direct computation reveals that LL T = T and so, defining V = L T U, U T T U = V T V. Clearly, V T V = if, and only if, V = but in this case, solving L T U = leads to U =. The matrix L is a representation of the dierence operator h 1 in that LU m = (U m+1 U m)/h. Likewise, L T represents h 1.

17 Summer Lectures: Approximation of BVPs Page 5.3 Further Notation We introduce some further operator notation in order to bring out the structure of finite dierence methods and their relationships to bvps. The bvp (1) is written as L u = F, where >< u(), at x = L u(x) = L u, < x < 1 (1) u(1), at x = 1, L u(x) = u + r(x)u and >< α, at x = F (x) = f(x), < x < 1. (13) β, at x = 1 For its approximation, we define L h to be a finite dierence operator defined by (see (4)) L h U m = h δ U m + r mu m = f m for m = 1 : M 1. The overall set of algebraic equations, including bcs are denoted by L h U = F, (14) where >< U, at m = L h U m = L h U m, m = 1 : M 1 (15) U M, at m = M, and >< α, at m = F m = f m, m = 1 : M 1. (16) β, at m = M. The local accuracy, measured at all grid points in [, 1], is then defined by R = L h u F (17) which, in view of (14), implies that R = L h u L U = L h (u U) and so L h e = R (1) where e = u U is the ge. Thus, to obtain a small ge one needs not only a small local error R but also that solutions of the finite dierence equations L h e = R with small rhs should have small solutions. We shall prove this using comparison principles in a later chapter. Exercise 1 Show that the finite dierence method U = 1, h δ U m + x mu m = x m, m = 1 : M 1 U M =. is a second order approximation of the bvp u (x) + x u(x) = x, < x < 1, u() = 1, u(1) = 9 = by using Taylor expansions to show that the local accuracy R m = h δ u m + x mu m x m satisfies R = O(h ). When M = 4, write the finite dierence equations in matrix vector form AU = F. Exercise Write down a finite dierence method to solve the bvp u (x) = x, < x < 1, u() = 1, u(1) = and check its order of accuracy. When M = 4, write the finite dierence equations in matrix vector form AU = F. Exercise 3 Determine the order of accuracy of the finite dierence method h δ U m + h 1 U m =, m = 1 : M 1, U =, U M = for solving the bvp u + u =, < x < 1, u = 1, u(1) = Exercise 4 Write the finite dierence approximation in Exercise 1 at x m in the form a mu m 1 + b mu m c mu m+1 = d m for m = 1 : M 1. Give expressions for a m, b m, c m and d m. Validation and Numerical Assessment of Accuracy Whenever a computer program has been written to implement a given method it is good practice to test it on a problem with a known exact solution (as we have done in Example 1). One has to guard against the possibility of choosing a problem which is special in the sense that the method may perform particularly well because of the structure of the exact solution. This might arise, for example, when the chosen test problem has an exact solution that is a polynomial of low degree. In such cases the local accuracy, which depends on derivatives of the exact solution, might be exactly zero and this would lead to a situation where u and U would coincide at grid points. This feature ;

18 Summer Lectures: Approximation of BVPs Page 5.4 is often used to devise problems that provide a preliminary validation of a computer code. The results shown in Figure 1 are reassuring in that the numerical solution, shown by the dots, lies close to the exact solution, shown as a solid line. However, it would be premature to conclude from this evidence alone that the method is satisfactory or that the code is error free. It is necessary to compute several solutions with an ever increasing number of grid points and to verify that U in each case gets closer to u. We shall, for the present, measure the degree of closeness using the maximum norm of u U, defined by u U h, max um Um. (19) m M The quantity u U will be referred to as the global error and is regarded as a grid function despite the fact that u is a continuous function. If u U h, as h we say that the method is convergent. During the testing phase the numerical solution is computed with a sequence of grids with M taking the values M (1) < M () < M (3) < (and the corresponding grid sizes h (1) > h () > h (3) > ) in the expectation that u U h, should get progressively smaller as M is increased (certainly when the number of grid points is suiciently large). Sample results for Example 1 are shown in Table 1. These suggest that the global error is reduced by a factor of about four whenever the grid size h is halved, that is u U h, h. Table 1: The maximum global error as a function of h for Example 1 M h u U h, from which log c may be eliminated to give log e(h (1) )/e(h () ) q. (1) log (h (1) /h () ) Typically, h (1) = h and h () = h/ and the expression simplifies to log (e(h)/e(h/)) q. log () For example, using h =.315 and the data in Table 1 it is found that q 1.9. An alternative to computing numerical estimates of q is to produce a log log 1 plot of e(h) versus h which, according to (), should give a straight line with slope q. Figure shows such a plot for Example 1. The maximum global errors for M = 4,, 16,..., 51 are shown by dots and a triangle whose hypotenuse has slope two has been superimposed. The dots lie on a line of slope when h is suiciently small which illustrates the fact that the assumption e(h) = ch q is an asymptotic estimate. Maximum error Fig. : Log log plot of maximum error versus h for Example 1. h In other problems the variation of global error with h may not be so obviously deduced. We proceed in these cases under the assumption that u U h, h q where q, when positive, is known as the rate of convergence of the method. Numerical results such as those shown in Table 1 may be used to estimate the value of q; this estimate is called the experimental order of convergence (EOC). Let e(h) denote the computed value of u U h, then, by assumption, there is a constant c such that e(h) ch q when h is suiciently small. Taking logarithms, log e(h) q log h + log c. () Thus, from computations with grid sizes h (1) and h (), we have the values log e(h (1) ) q log h (1) + log c, log e(h () ) q log h () + log c 1 In a log log plot, we plot log 1 e(h) vs. log 1 h but the scales shown on the axes are those for e(h) and h.

19 Summer Lectures 6. Boundary Conditions In the previous chapter we described the approximation of Dirichlet Problems: bvps subject to Dirichlet bcs. We now turn our attention to more general bcs, as described on page 1.. The numerical treatment of these boundary conditions is illustrated with reference to the dierential equation u (x) + r(x)u(x) = f(x), < x < 1, (1) approximated at x = x m by the second order finite dierence scheme h δ U m + r mu m = f m. () Thus, at grid points x = x m (m = 1 : M 1), «h U m 1 + h + rm U m h U m+1 = f m. Robin bcs This type of boundary condition subsumes the Dirichlet and Neumann cases. We shall assume that the Dirichlet condition u() = a holds at x = and αu(1) + βu (1) = b (3) at x = 1 for given values of the constants a, b, α, β. When β = this reduces to a Dirichlet condition and when α = we obtain a bc of Neumann type. Since Dirichlet bcs have been covered in the previous chapter we shall suppose that β and describe two common approximations of this boundary condition. Method 1: First order method The bc (3) is applied at x = 1, i.e., at the grid point x m with m = M. Since this bc is at the right end of our domain, we have to use a backward approximation (see equation (13.5)) at x = x M (= 1): u M = h 1 u M + O(h) = h 1 (u M u M 1) + 1 hu M + O(h ). (4) Neglecting the last two terms on the right, (3) leads to the numerical bc αu M + βh 1 U M = b, (5) that is, αu M + β [U M U M 1] /h = b. (6) Defining U = [U 1, U,..., U M ] T, we have AU = F, where with a i,i = + h r i (i < M), a M,M = 1 + hα/β, and F = [f 1, f,..., f M 1, F M ] T, () with F M = b/(βh). Equation (6) representing the boundary condition has been divided by βh so as to preserve the symmetry of A. Method : A second order method 11 As we shall see in Example 1 the method described above converges at a first order rate: it has a global error proportional to h. We can trace the reason for its first order accuracy to the O(h) term on the right of equation (4). To increase the order of accuracy we note, from the ode (1) that u (1) = r(1)u(1) f(1) and so, incorporating this in (4) we have u M = h 1 (u M u M 1) + 1 h(rm um fm ) + O(h ). (9) Thus, the bc αu(1) + βu (1) = b leads to αu M + β`h 1 `U M U M h(rm UM fm ) = b, i.e., `α + βh UM hβrm βh 1 U M 1 = b + 1 hβfm (1) To check that this equation is correctly scaled, it has to reduce to U M = b in the case α = 1 and β =. The local accuracy of this numerical bc is R M = `α + βh hβrm um βh 1 u M 1 b 1 hβfm = 1 6 βh u M + O(h 3 ). When (1) is combined with the finite dierence equations at internal grid points we have AU = F, with A and F defined as in (7) and () except that a M,M = 1 + hα/β + 1 h r M, F M = b/(hβ) + 1 fm, which are small perturbations of the original equations. Method 3: Staggered Grid Method Omitted Example 1 Use the methods described in this section to solve the dierential equation (14.1) on the interval < x 1 with the boundary conditions u() = 1 and au(1) + u (1) = (5/4 + 3)a and a = 5π/1. A is the M M matrix A = 1 h 6 4 a 1,1 1 1 a, a M 1,M a M,M 3, 7 5 (7) The boundary conditions have been chosen so that the exact solution of this problem coincides with that of Example 5.1. A log log plot of the maximum errors versus h is shown in in Figure 1. Method 1 is seen to be conver-gent at a first order rate (commensurate with the order of approximation of the boundary condition) while the other two methods converge at a second order rate. Moreover, Method 3 appears to give errors that are approximately times smaller than for Method. 11 The method we describe leads to exactly the same boundary equation as the Fictitious Point Method.