Numerical Methods for Partial Differential EquationsOverviewPhysical ExamplesDivide and ConquerSimplificationFurther SimplificationNecessary Information to Solve The IBVPAnswerBrief SummaryPeriodic CaseAnalytical SolutionFourier Series Representation (p4 GKO)Returning to the Advection EquationcontSlide 15Note on Fourier ModesSlide 17Add Diffusion Back InSlide 19Slide 20What Did Diffusion Do??Categorizing a Linear ODESolving the Scalar ODE NumericallyODE PrototypeODE Time Stepping TopicsReading for Next WeekNumerical Methods for Partial Differential Equations CAAM 452Spring 2005Instructor: Tim WarburtonCAAM 452 Spring 2005Overview•Our final goal is to be able to solve PDE’s of the form:•This is a conservation law with some form of dissipation (under assumptions on A)•We will discuss boundary conditions, solution domain , and suitable solution spaces for this equation later.( )( )( )( )( ), ,, , ,, , ,,uutu u x y tu x y tu x y tx y�+� =� � +�====f A gf fg gA Arrg gr rr r( )0,[ , ]x yt t T�W�CAAM 452 Spring 2005Physical Examples•These and similar equations and vector analogs are pervasive:–Fluid mechanics (Euler equations, compressible Navier-Stokes equations, magnetohydrodynamics).–Electromagnetics (Maxwell’s equations)–Heat equation–Shallow water equations–Atmospheric models–Ocean models–Bio-population models (morphogenesis, predator prey, epidemiology)–…..( )uut�+� =� � +�f A grrg gCAAM 452 Spring 2005Divide and Conquer•It is highly non-trivial to solve these equations analytically (i.e. with smarts, pen and paper).•We can forget the idea of writing down closed form solutions for the general case.•We will consider the component parts of the equations and discuss techniques to solve the reduced equations.•Some very reduced models admit exact solutions which allow us to check how well we are doing.•Finally we will put different methods together and aim for the big prize.CAAM 452 Spring 2005Simplification•Let’s choose a simple example, namely the 1D advection diffusion equation.•This PDE is first order in time and second order in space.22u u uc dt x x� � �- =� � �CAAM 452 Spring 2005Further Simplification•We can simplify even further by dropping the second order diffusion or dissipation term:•This PDE is first order in time and first order in space.•Volunteer to solve this equation analytically?.0u uct x� �- =� �CAAM 452 Spring 2005Necessary Information to Solve The IBVP•The Initial, Boundary, Value Problem represented by the PDErequires some extra information in order to to be solvable.•What do we need?.0u uct x� �- =� �CAAM 452 Spring 2005AnswerIn this case, because of the hyperbolic nature of the PDE(solution travels from right to left with increasing time), weneed to supply:a) Extent of solution domainb) What is the solution at start of the solution process: u(x,0)c) Boundary data: u(b,t)d) Final integration time.xtNeed to specify the solution at t=0As we just sawwe also need tospecify inflowdatax=a x=bCAAM 452 Spring 2005Brief Summary•There is a checklist of conditions we will need to consider to obtain a hopefully unique solution of a PDE1) The PDE (duh)2) Boundary values (also known as boundary conditions)3) Initial values (if there is a time-like variable)4) Solution domainCAAM 452 Spring 2005Periodic Case•Suppose we remove the inflow and imagine that the interval [a,b) is periodic.•Further suppose we wish to solve for the solution at some non-negative time T. •We can indicate this by the following specification:( )[)[ ]( )[)( ) ( )( )[)1) Find , such that , , 0,0given( ,0) ,, , [0, ]2) Evalute , ,ou x t x a b t Tu uct xu x u x x a bu a t u b t t Tu x T x a b" � �� �- =� �= " �= " �" �CAAM 452 Spring 2005Analytical Solution•Volunteer:•For this PDE to make sense we should discuss something about u0, what?( )[)[ ]( )[)( ) ( )( )[)1) Find , such that , , 0,0given ( ,0) ,, , [0, ]2) Evalute , ,ou x t x a b t Tu uct xu x u x x a bu a t u b t t Tu x T x a b" � �� �- =� �= " �= " �" �CAAM 452 Spring 2005Fourier Series Representation (p4 GKO)( )( ) ( )( )( )wwwwpwpwwp=�=- �-� - ��==�1 Assume that C , is 2 periodic.Then has a Fourier series representation:1ˆ 2ˆwhere the Fourier coefficients are given by1ˆ Theorem:2i xifff x f eff e( )( )p�20Finally, the series converges uniformly to xf x d xf xIn other words, we can express a sufficiently smooth function in terms of an infinitetrigonometric polynomial. The fhats are the Fourier coefficients of the polynomial.CAAM 452 Spring 2005Returning to the Advection Equation•We wills start with a specific Fourier mode as the initial condition:•We try to find a solution of the same type:( )[)[ ]( ) ( )[)1) Find 2 -periodic , such that 0,2 , 0,0 given1ˆ( ,0) = 0,22 where is a smooth 2 -periodic function of one frequency i xu x t x t Tu uct xu x f x e f xfwp pw ppp w" � �� �- =� �= " �( ) ( )1, ,ˆ2i xu x t e u twwp=CAAM 452 Spring 2005cont•Substituting in this type of solution the PDE:•Becomes an ODE:•With initial condition0u uct x� �- =� �( )1 1ˆ,ˆ ˆ2 2ˆ0ˆi x i xu u duc c e u t e i cut x t x dtdui cudtw ww wp pw� � � �� �� � � �- = - = -� � � �� �� � � �� � � �� �� - =( ) ( )ˆ,0ˆu fw w=CAAM 452 Spring 2005cont•We have Fourier transformed the PDE into an ODE.•We can solve the ODE:•And it follows that the PDE solution is:( ) ( )( ) ( ) ( )ˆ0ˆˆ, ,0ˆ ˆˆ,0ˆi ct i ctdui cuu t e u e fdtu fw www w ww w�- =�� = =��=�( ) ( )( ) ( )( ) ( )( )( )( ) ( )1: , ,ˆ21ˆ ˆsolution : , ,ˆ21ˆinitial condition: 2i xi x cti cti xansatz u x t e u tu t e f u x t e f f x ctf x e fwwwwwpw w wpwp+�=���= � = = +��=���CAAM 452 Spring 2005Note on Fourier Modes•Note that since the function should be 2pi periodic we are able to deduce:•We can also use the superposition principle for the more general case when the initial condition contains multiple Fourier modes:w��( ) ( )( )( )( ) ( )1ˆ21ˆ,2i xi x ctf x e fu x t e f f x twwwwwwwpwp=�=- �=�+=- �=� = = +��CAAM 452 Spring 2005cont•Let’s back up a minute – the crucial part was when we reduced the PDE to an ODE:•The advantage is: we know how to solve ODE’s both analytically and numerically (more about this later on).0u uct x� �- =�