Numerical Methods for Partial Differential EquationsRecall: Convergence Conditions for LMM Time-stepping MethodsConsistency of Finite Difference OperatorExample: Euler-Forward + Right DifferenceQuick NoteSecond Example (Leap Frog in Time and 4th Order Central in Space)contDefinition: Method Order AccuracyScheme NotationStabilitycont (norm notation)Well PosednessConsequences of Well PosednessSlide 14ConvergenceSlide 16Lax-Richtmyer Equivalence TheoremAccuracySolution NormsApproach 1) Compare Solution With Interpolated Numerical SolutionSlide 21Slide 22Approach 2) Compare Solution and Numerical Solution at NodesSlide 24cont (sketch of convergence)Sketch contcont (final result)Summary of Convergence Test for Finite Difference SchemesBoundary ConditionsExample Right-DifferenceSystemInterim (summation by parts) ResultEnergy Method for Semi-discrete Difference Approximation of the Upwind Finite Difference MethodSlide 34Slide 352nd Order Central + Boundary ConditionsSlide 37Interim (summation by parts) Result 2CorollaryExample 2: Energy MethodHigher Order + Boundary ConditionsClass DiscussionNext LectureNumerical Methods for Partial Differential Equations CAAM 452Spring 2005Lecture 8Instructor: Tim WarburtonCAAM 452 Spring 2005Recall: Convergence Conditions for LMM Time-stepping Methodsi. Establish that a unique solution to the ODE exists via Picard’s theorem (http://mathworld.wolfram.com/PicardsExistenceTheorem.html) ii. For time stepping Dahlquist’s Equivalence Theorem tells us that a linear multistep time-stepping formula is convergent if and only if it is consistent and stableiii. We can easily verify consistency by using Taylor expansions for the local truncation error.iv. We check stability conditions by finding roots of the stability polynomial.v. A global error analysis tells us that if the right hand side function is sufficiently smooth (p times continuously differentiable), and the LMM is stable with local truncation error then the error at a fixed time converges as ( )pO dt( )1pO dt+CAAM 452 Spring 2005Consistency of Finite Difference OperatorDefinition: A finite difference operator is consistent if it converges towards the continuous operator of the PDE as both dt0 and dx0CAAM 452 Spring 2005Example: Euler-Forward + Right Difference•The finite difference method is:•We define the local truncation error as the operator which maps the actual solution of the PDE to the correction required to make it satisfy the scheme at each time step:1n nnm mmu uc udtd++-=( ) ( )( )( )( )( )( )( ) ( )12 2*22 2*2, ,,1, ,2!, ,2!nm n m nm m nm n m nm n m nu x t u x tT u c u x tdtu dt udt x t x tdt t tc u dx udx x t x tdx x xO dt O dxd++-= -� �� �= +� �� ��� �� �- +� �� ��= +CAAM 452 Spring 2005Quick Note•Notice that in the definition of the LTE for the finite difference scheme we have not multiplied through by dt (since that would bias the LTE with respect to dt)•In this example the scheme is said to be first order method accurate in both time and space.( ) ( )( )( ) ( )1, ,,nm n m nm m nu x t u x tT u c u x tdtO dt O dxd++-= -= +CAAM 452 Spring 2005Second Example(Leap Frog in Time and 4th Order Central in Space)•Here we use the fourth order central differencing in space and Leap Frog in time:•The truncation error in this case is:•Thus we declare the method accuracy to be 2nd order in time and 4th order in space.1 1 24 012 6n nn nm mm mu u dxc u c udtd d d d+ -+ -� �-= = -� �� �( ) ( )( )( ) ( )( ) ( )1 142 3 4 5*52*34, ,,248, ,6 5!nm n m nm m nm n m nOu x t u x tT u c u x tdtdt u dx ux t c x ttO dt dxxd+ --= -� � � �� - �= +� � � �� ���= +CAAM 452 Spring 2005cont•In this case there is a discrepancy between the magnitude of the time stepping error and the spatial error. •Using this scheme may require a smaller time step than dx to ensure that the truncation errors for each part are of similar size.CAAM 452 Spring 2005Definition: Method Order Accuracy•If the local truncation error satisfies:•Then the method s order accurate in time and r’th order accurate in space.•Again – if there is a discrepancy between r and s then it may be wise to consider reducing dt (if s<r) or dx (if r<s) possibly significantly more than the CFL condition suggests.( ) ( ) as , 0n s rmT u O dt O dx dt dx= + �CAAM 452 Spring 2005Scheme Notation •For brevity we will denote the linear finite difference schemes:•where the coefficients may depend on dt,dx as:•then the scheme reads:10j p j qn j n jj m j mj j ru c ua b= =+- += =-=� �1,0j p j qn n j n jdt dx m j m j mj j rP u u c ua b= =+- += =-= -� �,0ndt dx mP u =CAAM 452 Spring 2005StabilityDefinition:A finite difference scheme for a first-order PDE isstable if there is an integer J and positive numbers dt0 and dx0 such that for any positive time T, there is a constant CT such that: ,0ndt dx mP u =1 12 20 0 00 0 for 0 ,0 and 0M J Mn jm T mm j mdx u C dx undt T dx dx dt dt- -= = =�� � < � < �� ��i.e. for a scheme to be stable it must not increase the “solution energy” beyond some “energy” injected at thestart of the time stepping.CAAM 452 Spring 2005cont (norm notation)•We define a discrete Euclidean norm on the discrete solution as:•Then the stability condition is:•Or equivalently:10: or sometimes denoted by dxm Mm mmhu dx u uu= -=� �=� ���rr20Jn jTdx dxju C u=��r r*0Jn jTdx dxju C u=��r rCAAM 452 Spring 2005Well PosednessDefinition: The initial value problem for a first-order PDE is wellposed if the following holds for all initial data u(x,0)for some choice of norm (say with integration overthe interval in x ) where the constant C(t) isindependent of the solution.( ) ( ) ( ), ,0u x t C t u x�2LCAAM 452 Spring 2005Consequences of Well Posedness•If a first order PDE is well posed, it satisfies an analog of the numerical stability we have been seeking. •There are two important consequences:–two initial conditions which are almost everywhere identical will generate two solutions which are almost everywhere identical.–two solutions which start close together will remain close together almost everywhere.( ) ( ) ( ), ,0u x t C t u x�*0Jn jTdx dxju C u=��r rCAAM 452 Spring 2005cont•The following PDE’s are well-posed22322 22u uc sut xu usut xu u uc dut t x xu u ucut t x x� �= +� �� �= +� �� � �- = +� �� �� � �-