The Laplace equation, cylindrically or spherically symmetric case

Transcription

1 Numerisce Metoden II, 7 4, und Übungen, 7 5 Course Notes, Summer Term 7 Some material and exercises Te Laplace equation, cylindrically or sperically symmetric case Electric and gravitational potential, waves or vibrations are just a few typical applications were u = (Laplace equation) u = q (Poisson equation) occur. In tese equations, denotes te Laplace-Operator, wic in treedimensional cartesian xyz-coordinates @y Te cylindrical coordinate system uses polar coordinates r and to locate a point in te x y plane and te coordinate z for te eigt of te point above te plane. Te Laplace operator in tis coordinate system @r + + For functions depending on r (distance from z axis) only (cylindrical symmetry), tis reduces : () Similarly, in sperical symmetry (were r measures te distance from te : () Tese differential operators (,) occur in te following examples, wic terefore in fact are just special cases of partial differential equations, reduced to ordinary differential equations by symmetry. Many features of our simple examples are also typical for real Laplace- and Poisson equations. We use examples in cylindrical symmetry ere just because one-dimensional versions of te Laplace operator () in cartesian geometry would not include first derivatives. Solutions in tat case are so simple tat tere is not muc to learn from. (On te oter and: wat s wrong wit simple examples? We sall discuss also te simple cases.)

2 Steady-state diffusion or eat conduction A cylindrically symmetric Laplace equation, cf. () governs stationary eat conduction or diffusion wit axial symmetry. An example would be a cylindrical pipe, interior radius r, exterior radius r. Te temperature = (r) in te wall of te pipe follows te equation rr + r r = for r int < r < r ext : (4) Diriclet boundary conditions specify te temperature at te interior and exterior walls, (r int ) = int (r ext ) = ext ( inner wall temperature ), ( outer wall temperature ), (Note: tere are several ways to denote derivatives of a function like (r). d (r) dr ; ; rr all stand for te same differential operator. Depending on context, one way or anoter may be more convenient to work wit.) For tis problem, te exact solution is known. We will use it to ceck te accuracy of our numerical approximations. (r) = int log(r=r ext ) + ext log(r int =r) : log(r int =r ext ) Alternatively, we migt prescribe te eat flow rate q at te inner boundary (A Diriclet condition at te rigt and a Neumann condition at te left end). In tat case, Te exact solution ere is r (r int ) = ; (r ext ) = ext : (r) = ext + r int log r r ext : For tis differential equation, Neumann boundary conditions on bot sides are no good coice. If te corresponding rates are equal, te solution is unique only up to an arbitrary additive constant. If te rates are not equal, no steady state can develop; tus, no solution exists Tis simple example demonstrates tat boundary value problems do not necessarily ave a solution. (in contrast to initial value problems, were existence of a solution is guaranteed in te vicinity of te starting point for a wide class of functions) Diriclet and Neumann conditions are also called boundary conditions of te first and second kind, respectively. Tere are tird-kind boundary

3 conditions as well (some attac to tem te names Robin or Sturm, or call tem radiation conditions), of te form c p(r) + c p (r) =. Also nonlinear boundary conditions may occur, like c p(r) 4 + c p (r) =, wic would model termal radiation according to Stefan-Boltzmann s law. Cooling of cylindrical objects To find solutions of te time-dependent eat equation in cylindrical symmetry, t = a rr + r r we set (r; t) = e at u(r), wic reduces te problem to an ordinary differential equation for u(r), u rr + r u r + u = ; < r < R (5) wic is (after multiplying by r and a transformation of variables r r) a form of Bessel s differential equation. Symmetry dictates a Neumann boundary condition u r = at te center r =. For an outer Diriclet boundary condition u = at r = R, it turns out tat nontrivial solutions exist for certain values of only (te trivial solution would be u(r) = for all r). Te solution involves J, te Bessel function of order zero of te first kind, u(r) = AJ (r) were A is an arbitrary constant (remember: te problem is a omogeneous one; any multiple of a solution is also a solution). Te outer boundary condition ten requires to be a root of J (R) = Steady one-dimensional convection-diffusion equation Let c = c(x) denote te concentration of some quantity in a flow wit velocity w and a coefficient of diffusion D. Ten Dc wc = for < x < L; c = c at x = ; c = c L at x = L;

4 models a steady-state situation were convective transport is balanced by diffusion. Tis problem as te exact solution x c(r) = c + epe L e Pe (c L c ): Here Pe is te Peclet number, Pe = wl=d. Ferzinger and Perić comment on tis problem: Because it is so simple, tis problem is often used as a test of numerical metods, including bot discretization and solution scemes. [... ] Tere are few actual flows in wic tis balance plays an important role. Normally, convection is balanced by eiter a pressure gradient or diffusion in te direction normal to te flow. [... ] Indeed, use of tis problem as a test case as probably produced more poor coices of metod tan any oter in te field. Discretization on an Equidistant Grid, Diriclet Conditions We discuss te example + r = for = (r); < r int < r < r ext (r int ) = int ; (r ext ) = ext : Te first step in te application of finite difference metods is te selection of a set of mes points in te interval (r int ; r ext ). Let us simply take n equidistant radius points r j, r j = r int + j wit mes size = r ext r int ; j = ; : : : ; n n + Boundary points are r = r int and r n+ = r ext. Next, we represent te differential equation by difference equations involving values of te unknown pressure at te mes points. A straigtforward way to do so is to replace te derivatives by appropriate difference quotients. Standard approximations are te difference formulae f (x) f (x) f(x + ) f(x ) f(x + ) f(x) + f(x ) Notation: For te value (r) of te unknown temperature at te mes point r j we write j. At mes point r j+ = r j +, it is j+, similarly j at r j. Tis way we may write j ( j+ j ) 4

5 j ( j+ j + j ) Substitution leads to difference equations, one per interior mes point. Te temperature values j at all interior mes points fulfill approximately r j j + j + r j j+ = for j = ; : : : ; n: (6) Tis is a system of n linear algebraic equations. In matrix form, A x = b wit (7) A = 6 4 ( + r ) ( r ) ( + r ) ( r ) ( + r ) ( rn ) ( + rn ) ( rn ) 7 5 ; ~x = 6 4. n ; ~ b = ( r ) int. ( + r n ) ext Note tat tis matrix is not symmetric, but diagonally dominant as long as < r. : 7 5 Exercises Exercise Solve te diriclet problem (4) for steady-state radial flow for two cases.. Strong influence of radial geometry: r int = :; r ext = ; int = ; ext = : 5

6 . moderate influence of radial geometry: To evaluate te results, plot r int = ; r ext = ; int = ; ext = :. Te error as a function of r for some values of n, say n = ; ;.. Te maximum error as a function of n for a range of n as large as computer power allows. Sow te error in a log-log-plot and estimate te asymptotic beavior as n. An important quantity is te slope of te curve. Wat we sall find out: For large n te maximum error decreases proportional to n. Tis quadratic convergence, owever, does not become apparent unless te grid size is comparable to te interior radius r int. Te memory required is proportional to n. Can you specify te memory requirements of your program more precisely? Exercise Derive a difference approximation for te convection-diffusion equation and set up te system of linear equations. Is te matrix symmetric? diagonally dominant? Try to solve te simple case D = ; L = ; w = ; c = ; c L = for various, and compare results wit te exact solution. Cange w = and try again. Evaluate your results wit similar plots as in Exercise. Exercise Te matrix (7) in our radial flow example is not symmetric. Sow tat by multiplying te i-t equation by r i a symmetric matrix results. Systems wit symmetric matrices are teoretically and numerically in many ways easier to andle tan non-symmetric ones. Implement te symmetric version and ceck wit te original version. Exercise 4 Test te sensitivity of te system (7) to small perturbations in te rigt-and side. In our MATLAB code, insert % o r i g i n a l s o l u t i o n p = A\b ; %s o l u t i o n wit p e r t u r b e d r i g t and s i d e % t r y d i f f e r e n t p e r t u r b a t i o n s e r r=r a n d ( n, ). ; %e r r =(rand (n,).5).; b = b+e r r ; p e r r = A\b ; 6

7 Te relative error in b is norm(err)/norm(b), wile te relative error in te solution is norm(p-perr)/norm(p). How muc larger is te relative error in te solution in relation to te relative error of te rigt-and side? Calculate tis quotient (a measure of te sensitivity of te linear system) for different values of n and estimate te beavior for large n. Sow a loglog-plot. In MATLAB, c = cond(a) returns te -norm condition number, te ratio of te largest singular value of A to te smallest. Te condition number of a matrix measures te sensitivity of te solution of a system of linear equations to errors in te data. It gives an indication of te accuracy of te results from matrix inversion and te linear equation solution. It also is an upper bound on te sensitivity, as calculated before. Values of cond(a) near indicate a well-conditioned matrix: errors in te data generate errors of te same magnitude in te solution. Add a plot of te condition number depending on n to your loglog-plot. Is te condition number a quite sarp or a rater pessimistic estimate on te sensitivity in tis example? Discretization on an Equidistant Grid, Mixed Boundary Conditions Just a few modifications in our first program are necessary to implement a Neumann condition at te inner radius and a Diriclet condition at te outer boundary. We are now going to solve our second example of flow in porous media, te problem p + r p = for p = p(r); r w < r < r e p (r w ) = ; (8) p(r e ) = p e : We may use te same grid and te same difference formula as before, but tere is one problem: For te first interior gridpoint r, te difference approximation is r p + p + r p = : (9) However, te boundary condition does not give us any value for p, it just specifies te derivative tere. We can exploit tis information by using a one-sided approximation of te first derivative, f (x) = f(x) + f(x + ) 7 f (x) + O( ): ()

8 In our case, we set p p = ; or p = p : () Inserting for p in Equation (9), we obtain a difference equation for te first interior gridpoint at a Neumann boundary, + p + p = r r r : () Tere is an inconsistency wit using tis approximation along wit te rest of te difference equations. Because of te one-sided approximation (), is a fist-order approximation only. At all oter points, te difference equations are second-order approximations to te original differential equation. Te first-order approximation of te Neumann condition does indeed contaminate our final results in te wole region. In general, te order of approximation of a problem is te lowest order used in any parts of te problem. Sometimes, owever, a bad approximation at just a few points will not affect te overall accuracy. To get a second-order accurate approximation at r, we can replace te second derivative in using te differential equation. In our case, p (r ) = p p p (r ) = r p (r ) = r p (r ) + O( ) from te differential equation from te boundary condition From tese equations follows p = p + r () More commonly, owever, Neumann boundaries are treated by modifying te grid. Instead of r = r w ; r = r w + ; te modified grid at a Neumann boundary as All gridpoints are given by r = r w ; r = r w + r i = r w + (i =) wit = r e r w n + = ; i = ; : : : ; n 8

9 Te point r is now outside of te region were our problem is defined. (It is called a gost point.) As a result of tis modification, te approximation p = p is of second order. However, tis sceme does not compute te pressure p wf directly, since tere is no gridpoint at r w. We may set p wf = (p + p )= + O( ). Exercises Exercise 5 Solve te mixed-type problem (8) and test te tree ways to implement Neumann boundary conditions. Data: r w = :; r e = ; p (r w ) = = ; p e = : To evaluate te results, plot te maximum error as a function of n in a log-log-plot and estimate te asymptotic beavior as n. Te difference between first-order and second-order accuracy sould become clearly apparent. Exercise 6 Solve for r w = ; and r e = te inomogeneos differential equation wit omogeneous mixed-type boundary conditions (quasi-steadystate flow), p + r p = for r w < r < r e ; p (r e ) = ; p(r w ) = : Evaluate te accuracy of your solution as in Exercise. (If you ave coded tis exercise, te modifications in te code sould be rater marginal) Exercise 7 Discretize te differential equation (5). Assume simple values for a and. Wat appens to te rigt-and side of te system? Wat solution gives MATLAB s A\b? Any ideas ow to proceed in tis case? Deriving Difference Approximations for Derivatives Consider a smoot function f(x). ( Smoot ere means tat as many derivatives of f exixt as we may need.) We will illustrate, by way of example, a metod to find approximations to derivatives. It is sometimes called te metod of undetermined coefficients. 9

10 Example: Derive an approximation for f at gridpoint x using values at gridpoints x ; x and x of an equidistant grid wit spacing. Expand f and f in Taylor series around x f = f f + f 6 f f iv + O(5 ) f = f + f + f + 6 f f iv + O(5 ) We now seek a linear combination of te tree values f ; f and f approximating f in te form f af + bf + cf : To determine te coefficients a; b and c, we insert te Taylor series and collect terms in f ; f ; : : :. We find f (a + b + c)f + ( a + c)f + ( a + c)f (a + c)f iv + O(5 ) (a + c)f + Since we ave tree degrees of freedom (tree undetermined coefficients), we can plan on setting te coefficients of f and f equal to zero and te coefficient of f equal to one. We are left wit te following system of equations. a + b + c = a + c = a + c = Solving te above system of equations gives a = ; b = ; c = : As an additional bonus, tis solution makes te coefficient of f equal to zero too. Te coefficient of f iv turns out to be =. Hence, we get te approximation f f + f = f + f iv + O( ): Tis is te well-known centered difference approximation tat we ave used already.

11 As an additional exercise, you may derive a five-point centered approximation to f. Te system of equations for five coefficients a; b; c; d and e in tis case will be a + b + c + d + e = a b + d + e = 4a + b + d + 4e = 8a b + d + 8e = 6= 6a + b + d + 6e = (Using an algebraic manipulator suc as Maple or Matematica makes solving suc systems muc easier) Te solution is a = ; b = ; c = ; d = ; e = : An alternative metod: find an interpolating polynomial p(x) for te data points (x i ; f i ). Differentiate te polynomial wit respect to x and evaluate at x. If you do tis by and, you sould use te Lagrange interpolation formula: Te polynom p(x) tat interpolates te n + pairs of values (x ; f ); (x ; f ); : : : ; (x n ; f n ) is given by p(x) = f L (x) + f L (x) + + f n L n (x) ; were L k (x) = (x x )(x x ) (x x k )(x x k+ ) (x x n ) (x k x )(x k x ) (x k x k )(x k x k+ ) (x k x n ) for eac k = ; ; : : : ; n. In Matematica, tis metod conveniently finds difference approximations for iger derivatives. Te commands data={{-,f[-]},{-,f[-]},{,f[]},{,f[]},{,f[]}}; poly=interpolatingpolynomial[data,x]; appr=d[poly,{x,}]/.x-> //Togeter

12 find an interpolating polynomial and evaluate its tird derivative at x =. Matematica gives appr = f( ) + f( ) f() + f( ) : We may find te truncation error by te command Series[appr,{,,}] wic produces te output f () () + f (5) () 4 + O() 4 : Since many tables provide difference approximations for f ; f ; : : :, it is normally not necessary to derive basic formulae. However, for irregular grids, complicated equations or some specific boundary conditions, tese metods are of great value. Non-Equidistant Grid Write f = f(x ); f = f(x); f + = f(x + + ); Ten f (x) = = r i r i ; + = r i+ r i : f (x) = (f + f ) + + (f f ) f + f f f f f ( + ) + + ( ) f iv Note tat in te second approximation, for + 6= now tere is a loworder truncation error term involving f. Equidistant grids usually provide approximations wit iger-order truncation errors. So you ave to balance te gain by finer messize against possibly iger truncation error. Linear solvers Te Tomas Algoritm for tridiagonal Systems Te so-called Tomas Algoritm is just a form of elimination for solving tridiagonal systems of linear equations. Llewellyn H. Tomas used it around 95

13 to solve elliptic partial differential equations. Te attribution to Tomas seems to be more common in some engineering disciplines tan it is in numerical analysis. Let A be a tridiagonal matrix and b te rigt-and side of a system Ax = b. A = 6 4 d e c d e c d e c n d n e n ; b = b b b. b n : 7 5 c n d n b n Transform it to upper triangular form, by Gaussian elimination. Use backsubstitution to solve te resulting equivalent system Ax = b e b e b e b A = ; b = :. 6 4 e 7 6 n 5 4 b 7 n 5 Te expressions for te entries in A and b follow from a step-by-step application of te usual elimination procedure. for i = ; : : : ; n: e i = e = e d ; b = b d ; e i d i c i e i ; b = b i c i b i i ; d i c i e i Backsubstitution: x n = b n, und f"ur i = n ; : : : ; : Tis algoritm is stable if 8 >< >: d i > ; i = ; ; : : : ; n; d > je j; x i = b i e i x i+: d i jc i j + je i j und c i 6= ; e i 6= ; i = ; : : : ; n ; d n jc n j: b n

14 Essentially tese conditions require A to be (weakly) diagonally dominant. Tey are sufficient but not necessary. 4