1 Introduction to MATLAB

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1 L3 - December 015 Solving PDEs numerically (Reports due Thursday Dec 3rd, carolinemuller13@gmail.com) In this project, we will see various methods for solving Partial Differential Equations (PDEs) using numerical tools. The first part is a quick introduction to MATLAB. We will see how to define functions using matrix notations, and how to plot them as contours or surfaces. The second part aims at solving the one-dimensional advection equation using finite differences. We will look at the stability of various schemes, and introduce the famous CFL condition (Courant-Friedrichs-Lewy). Finally in the last part, we will introduce the notions of numerical dispersion or diffusion, which can be present in certain numerical schemes. 1 Introduction to MATLAB 1.1 Defining a 1D function We will start by defining a vector x, and a function of x. Using TP11intro.m, set subsection number = 1.1 which will execute the first paragraph of the code. This defines a vector x, the function f(x) = x and plots f(x). Note that in the definition of f(x), the power is preceded with a dot : f(x) = x. which indicates that the operation should be done component by component (i.e. x. = x. x = [x 1, x,, x n] in opposition to matrix multiplication ). Uncomment the last line and compare M. with M. Using the File pop-up menu, you may save the figure in Matlab format (*.fig) or use Save as to save it in various file formats (*.eps, *.png). 1. Defining a D function We now define a two-dimensional function of x and y. Set subsection number = 1. and plot f(x, y) as a surface z = f(x, y), and using contours. Note that in MATLAB, both surf and contour take the transpose of f as input (denoted f. ). Change the function to cos(πx). sin(πy) + by uncommenting the corresponding line. The tic/toc commands return the time elapsed between the tic and the toc. So here Matlab should return in the main command window the time elapsed to compute the for loops. In the next section, we will compare this time to the one obtained when you define the function in matrix notation instead of loops. 1

2 1.3 Defining a D function using matrix notations MATLAB tends to be quite slow when loops are used, as was done in the previous section for instance (there were two for loops). Therefore, an efficient way to perform D calculations is to use matrix notations. Set subsection number = 1.3 and define F (Y, X) using meshgrid. The MATLAB function meshgrid takes vectors x and y, and returns matrices X and Y. Matlab should return the time elapsed betweent the tic and toc commands, in other words the time elapsed to define the function in matrix form. How does it compare to the loops of the previous section? 1.4 Numerical approximation of derivative Our goal is to solve differential equations numerically. In order to do so, we need to approximate the derivatives of functions. One method is called finite difference and is based on Taylor s expansion : f(x + x) = f(x) + f (x) x + f (x) From (1) it is easy to show that f (x) = x + f (3) (x) 3! f(x + x) f(x) x x + + f (n) (x) x n + O( x n+1 ). (1) n! + O( x) () and f f(x + x) f(x x) (x) = + O( x ). (3) x The first approximation, forward difference, is of order 1, while the second two-sided difference is of order hence more accurate in general. Set subsection number = 1.4 and compare the first approximation () with the second (3). Numerical instability and the CFL ( Courant-Friedrichs-Lewy ) condition In this section, we will use finite differences to solve numerically the 1D advection equation with initial condition φ(x, t = 0) = φ 0 (x)..1 Exact solution t + = 0, (4) x It is easily verified that any function of the form φ(x, t) = f(x t) is a solution of (4), so that the solution of the above system is φ(x, t) = φ 0 (x t). Open TP1PDE.m and set subsection number =.1 to plot the exact solution at t = 0, and 4 when φ 0 (x) = e x. You can see that the initial condition is advected in time, and is centered around x = t at time t.

3 . Numerical solution 1: forward difference We now use forward difference () in both x and t to solve the above equation. We denote φ n j = φ(x j, t n ), where x j = j x (j = 0, ±1, ±, ±J) and t n = n t (n = 0, 1,, N) are our grid points in space and time. The forward difference () applied to φ(x j, t n ) yields t φn+1 j φ n j t and x φn j+1 φn j. (5) x Using forward difference (5), show that equation (4) yields the following time evolution for φ n at the next time step : j = (1 + γ)φ n j γφ n j+1, (6) where γ = t x. (7) For the numerical computation, it is convenient to write (6) in matrix form: J J+1 J = 1 + γ γ γ γ γ γ γ γ φ n J φ n J, where on the last row we have assumed x-periodicity of the solution so that φ n J+1 = φn J. Set subsection number =. and plot the numerical solution. Do you recover the exact solution? Why not? This is not an easy question. In order to answer this question, let s look at the domain of dependence of the solution, exact versus numerical. Remember that the exact solution at (x j, t n ) is a function of the initial condition at (x j t n, 0), in other words it is a function of the solution at earlier times and smaller x (see figure below). But from (6), the numerical solution at (x j, t n ) depends on the solution at (x j+n, t n N ), i.e. at smaller times but larger x. So changing the initial condition at (x j t n, 0) changes the exact solution, but does not impact the numerical solution at all We can not expect this numerical scheme to be accurate! In fact it is unstable.!me$ dx$ dt$ n$ ϕ jn $=$f(ϕ j n+1,$ϕ j+1 n+1 )$=$f(ϕ j,$ϕ j+1,$ϕ j+ )$ $ n+1$ j$ j+1$ Exact$solu!on$ϕ(x j,t n )$=$ ϕ 0 (x j +t n )$depends$on$earlier$!mes$and$smaller$x$ Domain$of$dependence$of$the$numerical$ solu!on$using$forward$difference$:$earlier$!mes$and$larger$x$ In order to stabilize the code, we need to make sure that the domain of dependence of the numerical solution contains the domain of dependence of the exact solution. So we need to make sure that the numerical solution depends on smaller t and x, which can be obtained with backward differentiation in space. x$ 3

4 .3 Numerical solution : backward difference in x We still use forward difference in t but now we use backward difference in x, i.e. instead of (5) we use: In that case, it is easily shown that (6) becomes: x φn j φn j 1 + O( x). (8) x j = (1 γ)φ n j + γφ n j 1, (9) where as before γ = t/ x. Looking again at the domain of dependence of the numerical solution (see figure below), it contains the exact solution if t/ x 1 γ 1. We therefore expect the numerical solution to be stable if γ 1, and unstable if γ > 1. $$$$Exact$solu!on$$!me$ dx$ dt$ n$ n+1$ ϕ jn $=$f(ϕ j n+1,$ϕ j+1 n+1 )$=$f(ϕ j,$ϕ j+1,$ϕ j+ )$ $ j+1$ j$ ϕ(x j,t n )$=$ϕ 0 (x j +t n )$ Domain$of$dependence$of$the$numerical$ solu!on$using$backward$difference$in$x$ As before, for the numerical computation it is convenient to write (9) in matrix form: J J+1 J = 1 γ 0 0 γ γ 1 γ γ 1 γ γ 1 γ x$ where on the first row we have assumed x-periodicity of the solution so that φ n J 1 = φn J. Set subsection number =.3 and plot the numerical solution for various functions of γ. Check the stability of the numerical solution for γ and > 1. Change the maximum value of x (spatial domain) by uncommenting the corresponding line. Since we use periodic boundary conditions, the numerical solution reenters the domain on the left when the maximum x is reached. If one wants to compute the numerical solution at large times, the spatial domain needs to be large as well in order to avoid periodic reentry..4 Numerical solution for arbitrary c: the CFL condition The stability criterion discussed earlier, γ 1, is called the CFL condition. More generally, for an arbitrary advection speed c: t + c = 0, (10) x equation (9) becomes: j = (1 cγ)φ n j + cγφ n j 1. (11) 4 φ n J φ n J,

5 A similar argument based on the domain of dependence of the numerical solution yields that the condition for stability is c x/ t, i.e. the numerical speed x/ t must be faster than the physical speed c. Another interpretation of the CFL condition is that a grid cell must not be completely emptied from one time step to the next. In other words, the solution travelling at speed c must not travel more than one x per t. Set subsection number =.4 and plot the numerical solution for various functions of x, t, and c. Investigate the stability of the numerical solution for various values of c, t and x, and check that the CFL condition has to be satisfied. 3 Numerical diffusion and dispersion When solving numerically a PDE, the truncation error in the Taylor series can lead to diffusive errors, or dispersive errors. We will illustrate both here. 3.1 Exact solution with discontinuity Open TP13Error.m and set subsection number = 3.1 to plot the solution to (4). Then set subsection number = 3. to plot the solution with a step function as initial condition. Note the slight difference between Matrix time evolution and the one used in TP1PDE.m since the step function is not periodic in x, so one can not use the periodic boundary condition φ n J+1 = φn J. Instead we use the one-sided boundary condition φ n J+1 = φn J. As before, the exact solution is simply the initial condition advected at x = t. We will now see what the numerical solution looks like when the initial condition is not smooth. 3. Numerical diffusion 3..1 Some background on diffusion An advection-diffusion equation is of the form: t + c x = D φ x, (1) where D > 0 is the diffusion coefficient. The last term diffuses the solution, to see this let s see what (1) does to a Fourier mode φ(x, t) = e ikx v(t) : (1) v t = ( ick Dk )v φ(x, t) = e ik(x ct) e k Dt. The first term is the usual travelling wave solution, and the second damping term tends to diffuse the solution particularly at small scales (large k). 3.. Numerical diffusion - modified equation Set subsection number = 3.3 to plot the numerical solution looks like when the initial condition is a step function. What do you observe? The truncation error in the Taylor expansion introduces numerical diffusion, which explains the diffusion of the numerical solution that you observe with time. This can be understood by looking at the so-called modified equation, which is obtained by Taylor expansion of the numerical scheme keeping the first error 5

6 term (see Appendix 6 for a detailed derivation of the modified equation). For instance for the numerical scheme (3.3.), the modified equation is: ( t + c x = c x 1 c t ) φ x x. (13) Note that numerical diffusion D = c x/ (1 c t/ x) is positive since the CFL condition for stability is x/ t c. Set subsection number = 3.4 and plot the numerical solution for various c, x, t. How does the numerical solution depend on the diffusivity? From the modified equation, the diffusion should be 0 when c t/ x = 1. Do you recover this result numerically? 3.3 Numerical dispersion Some background on dispersion A dispersive equation is of the form: t + c x = D 3 φ x 3, (14) where D is the dispersion coefficient. The last term disperses waves, in the sense that different wavenumbers are advected at different speeds. In order to illustrate this, let s see what (14) does to a Fourier mode φ(x, t) = e ikx v(t) : (1) v t = ( ick Dik3 )v φ(x, t) = e ik(x ct), with the modified wave speed c = c + Dk. Therefore, waves with different wavenumbers travel at different speeds (faster or slower depending on the sign of D) Numerical dispersion An example of dispersive scheme is the Lax-Wendroff scheme. This scheme is similar to the one used so far (3.3.), except that second order terms are kept, yielding a third order error (dispersive): j = (1 c γ )φ n j + cγ (cγ 1)φn j+1 + cγ (cγ + 1)φn j 1. It can be shown from the modified equation that the Lax-Wendroff scheme has a dispersion coefficient D = c/6 x (1 c γ ). Set subsection number = 3.5 and plot the numerical solution for various c, x, t. what happens to the sharp discontinuity of the initial condition? How does it depend on c, x, t? Do small scale waves travel faster or slower than the discontinuity? Is this what you would expect from the equations? Is this still true if you change the initial condition from step to square by uncommenting the corresponding lines? From the modified equation, the dispersion should be 0 when c t/ x = 1. Do you recover this result numerically? 6

7 4 Application: Numerical D Advection of a passive tracer In this section, we will use a full D numerical code to investigate the impact of different discretizations on the advection of tracers. The code does not use the finite difference methods discussed in and 3. Instead it uses a method called finite volume. It is very similar to finite differences, the only difference being that the equations are written for the spatial averages over the small grid volumes x x + x (see 5 for a more detailed discussion of finite volume methods). Three options are given for the spatial discretization: up1 (lowest order hence lowest accuracy), up3 (intermediate order) and up5 (highest order). For the time discretization, there are also three choices: Heun, RK (similar order as Heun), RK3 (highest order). (RK stands for Runge Kutta methods, which are popular methods of time discretization, see 5 for a discussion of other common time stepping methods). Open Adv D.m, and pick the discretizations up3 in space and Heun in time (default values). Run the code, which advects a patch of tracer. The tracer is advected by a circular flow, and the simulation performs a full π circular advection. At the end of the simulations, the code should return the difference between the initial state and the final state. Does the square patch get advected exactly? i.e. does the patch go back to its initial position and shape at the end of the simulation? Why or why not? Reduce the order of the spatial discretization to up1, and then increase the order by choosing up5. Does the answer to the previous question change? It can be shown that up3 and up5 have numerical dispersion. This can be seen by saturating the colorbar to show smaller anomalies. The matlab command to change the colorbar range is caxis, for instance to [ ]. Uncomment the corresponding line in Adv D.m (last line) and run the code again. Do you see the numerical dispersion of up3 and up5? You can change the square patch to a circular patch. Uncomment the corresponding lines in the code, do the above results still hold? You can save a movie of the simulation by setting param.movie to 1. This will save the file ANIMATION.avi, make sure that you rename this file before running the code again, otherwise your movie will be overwritten. Also keep an eye on your simulation length if you make a movie, since matlab movies are quite large. 5 Suggested further topics: Some modern numerical tools 5.1 Multistep methods In state of the art models, various tricks have been used to increase the order of the discretization in time, for instance multistep methods. Multistep methods use information from several previous steps to calculate the next value. For instance for the differential equation dy dt the general form for an explicit two-step method is: = f(y, t), (15) y n+1 = A y n + B y n 1 + C t f(y n, t n ) + D t f(y n 1, t n 1 ). A popular scheme is the second-order Adams-Bashforth method which uses A=1, B=0, C=3/ and D=-1/. Note that if B or D are not zero, you need two past time steps n 1 & n to deduce the next time step n. Let s derive the coefficients A, B, C, D of the second-order Adams-Bashforth method: 7

8 Let s write (15) as y n+1 = y n + t n+1 t n f(y, t). Show that the second-order Adams-Bashforth method is obtained by approximating f(y, t) in the integrand by its linear interpolating polynomial between t n 1 and t n. 5. Intermediate time steps We note that these is another approach than multistep methods to increase the order of the discretization in time: the use of intermediate time steps. This is the case of the popular Runge Kutta method, which uses intermediate times between t n and t n+1 to deduce y n+1. For instance the second-order Runge-Kutta yields: 5.3 Finite volume k 1 = tf(y n, t n ) (16) ( k = tf y n + k 1, t n + t ) (17) y n+1 = y n + k (18) A powerful method for solving PDEs, particularly those containing a divergence term, is the finite volume method. Finite volume refers to the small volume surrounding each grid point (in our one-dimensional case x j x/, x j + x/). The equations are written for the spatial averages over these small grid volumes φ n j = 1 x xj+ x/ x j x/ φ(x, t)dx. Using the divergence theorem, divergence terms are written as surface integrals of fluxes in and out the small grid cell volume. Let us for instance consider the following equation: t + F x = 0. (19) This equation is similar to the one-dimensional advection equation that we studied so far, but the flux of φ is now given by the divergence of an arbitrary function F (in our case F = cφ). It is standard to use a slightly different discretization, where the advected quantities φ n j are computed at (t n, x j ) as before, but where fluxes Fj±1/ n are now computed at interfaces (t n, x j ± x/). Rewrite equation (19) for φ j (t) by integrating (19) in space (between x j 1/ and x j+1/ ), and show that: d φ j dt + 1 [ ] F x j+ 1 F j 1 = 0. This equation is exact for the volume averages, i.e. no approximations have been made during its derivation. 6 Appendix: Modified equation In order to estimate the leading order numerical error in the numerical scheme (3.3.), we Taylor expand up to order : j = (1 cγ)φ n j + cγφ n j 1 (0) φ n j + t t + t ( φ t = (1 cγ)φn j + cγ φ n j x x + x ) φ x (1) t t + cγ x x = t φ x φ + cγ t x. () 8

9 Using the leading order equation t + c x = 0 φ t = c φ x, and remembering that γ = t/ x, We recover equation (13). 7 References t + c x = ( c t + c x ) φ x. (3) Courant R., Friedrichs K. and Lewy H. (1967) On the partial difference equations of mathematical physics, IBM Journal. Durran D. (010) Numerical methods for fluid dynamics: with applications in geophysics, Springer, textbook 516pp. Iserles A. (008) A first course in the numerical analysis of differential equations, Cambridge University Press, textbook 480 pp. 9