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Rice CAAM 452 - Lecture Notes

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Numerical Methods for Partial Differential Equations CAAM 452 Spring 2005 Instructor Tim Warburton Overview Our final goal is to be able to solve PDE s of the form u f A u g t u u x y t f f u x y t x y t t0 T g g u x y t A A x y 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 CAAM 452 Spring 2005 Physical Examples u f A u g t 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 CAAM 452 Spring 2005 Divide 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 2005 Simplification Let s choose a simple example namely the 1D advection diffusion equation u u 2u c d 2 t x x This PDE is first order in time and second order in space CAAM 452 Spring 2005 Further Simplification We can simplify even further by dropping the second order diffusion or dissipation term u u c 0 t x This PDE is first order in time and first order in space Volunteer to solve this equation analytically CAAM 452 Spring 2005 Necessary Information to Solve The IBVP The Initial Boundary Value Problem represented by the PDE ut c ux 0 requires some extra information in order to to be solvable What do we need CAAM 452 Spring 2005 Answer In this case because of the hyperbolic nature of the PDE solution travels from right to left with increasing time we need to supply a Extent of solution domain b What is the solution at start of the solution process u x 0 c Boundary data u b t d Final integration time t As we just saw we also need to specify inflow data x a x b Need to specify the solution at t 0 x CAAM 452 Spring 2005 Brief Summary There is a checklist of conditions we will need to consider to obtain a hopefully unique solution of a PDE 1 The PDE duh 2 Boundary values also known as boundary conditions 3 Initial values if there is a time like variable 4 Solution domain CAAM 452 Spring 2005 Periodic 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 u x t such that x a b t 0 T u u c 0 t x given u x 0 uo x x a b u a t u b t t 0 T 2 Evalute u x T x a b CAAM 452 Spring 2005 Analytical Solution Volunteer 1 Find u x t such that x a b t 0 T u u c 0 t x given u x 0 uo x x a b u a t u b t t 0 T 2 Evalute u x T x a b For this PDE to make sense we should discuss something about u0 what CAAM 452 Spring 2005 Fourier Series Representation p4 GKO Theorem Assume that f C1 is 2 periodic Then f has a Fourier series representation 1 f x 2 f e i x where the Fourier coefficients f are given by f 1 2 2 e i x f x dx 0 Finally the series converges uniformly to f x In other words we can express a sufficiently smooth function in terms of an infinite trigonometric polynomial The fhats are the Fourier coefficients of the polynomial CAAM 452 Spring 2005 Returning to the Advection Equation We wills start with a specific Fourier mode as the initial condition 1 Find 2 periodic u x t such that x 0 2 t 0 T u u c 0 t x given 1 i x u x 0 f x e f x 0 2 2 where f is a smooth 2 periodic function of one frequency We try to find a solution of the same type u x t 1 i x e u t 2 CAAM 452 Spring 2005 cont Substituting in this type of solution the PDE u u c 0 t x Becomes an ODE u u 1 i x 1 i x du c c e u t e i cu dt t x t x 2 2 du i cu 0 dt With initial condition u 0 f CAAM 452 Spring 2005 cont We have Fourier transformed the PDE into an ODE We can solve the ODE du i cu 0 dt u 0 f u t ei ct u 0 ei ct f And it follows that the PDE solution is 1 i x e u t 2 solution u t ei ct f ansatz u x t 1 i x initial condition f x e f 2 u x t 1 i x ct e f f x ct 2 CAAM 452 Spring 2005 Note 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 1 i x f x e f 2 1 i x ct u x t e f f x t 2 CAAM 452 Spring 2005 cont Let s back up a minute the crucial part was when we reduced the PDE to an ODE u u c 0 t x du i cu 0 dt The advantage is we know how to solve ODE s both analytically and numerically more about this later on CAAM 452 Spring 2005 Add Diffusion Back In So we have a good handle on the advection equation let s reintroduce the diffusion term u u 2u c d 2 t x x Again let s assume 2 pi periodicity and assume the same ansatz This time u u u c d 2 t x x 2 1 i x u e u t 2 du i cu d 2u dt du ic d 2 u dt CAAM 452 Spring 2005 cont Again we can solve this trivial ODE du i c d 2 u dt u u 0 e u u 0 e ic d 2 t i ct x d 2t e CAAM 452 Spring 2005 cont The solution tells a story u u 0 e i ct x d 2t e The original profile travels in …

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