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Numerical Methods for Partial Differential Equations CAAM 452 Spring 2005 Lecture 9 Instructor: Tim Warburton

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CAAM 452 Spring 2005 Today Change of plan –I will go through in detail how to solve homework 3 step by step.

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CAAM 452 Spring 2005 Homework 3 Q1) Build a finite-difference solver for Q1a) use your Cash-Karp Runge-Kutta time integrator from HW2 for time stepping Q1b) use the 4 th order central difference in space (periodic domain) Q1c) perform a stability analysis for the time-stepping (based on visual inspection of the CK R-K stability region containing the imaginary axis) Q1d) bound the spectral radius of the spatial operator Q1e) choose a dt well in the stability region Q1f) perform four runs with initial condition (use M=20,40,80,160) and compute maximum error at t=8 Q1g) estimate the accuracy order of the solution. Q1h) extra credit: perform adaptive time-stepping to keep the local truncation error from time stepping bounded by a tolerance.

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CAAM 452 Spring 2005 Q1a – Use Cash-Karp RK for Time- Stepping We will use a vector version of Cash-Karp: Where f is a vector valued function – in our case it will be a linear operator acting on a vector argument.

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CAAM 452 Spring 2005 Where the b,c coefficients are http://www.library.cornell.edu/nr/bookcpdf/c16-2.pdf Provided in cashkarp.m on the web site. Original paper at: http://portal.acm.org/citation.cfm?id=79507&coll=GUIDE&dl=GUIDE&CFID=38381912&CFTOKEN=60548034

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CAAM 452 Spring 2005 Homework 3 Q1) Build a finite-difference solver for Q1a) use your Cash-Karp Runge-Kutta time integrator from HW2 for time stepping Q1b) use the 4 th order central difference in space (periodic domain) Q1c) perform a stability analysis for the time-stepping (based on visual inspection of the CK R-K stability region containing the imaginary axis) Q1d) bound the spectral radius of the spatial operator Q1e) choose a dt well in the stability region Q1f) perform four runs with initial condition (use M=20,40,80,160) and compute maximum error at t=8 Q1g) estimate the accuracy order of the solution. Q1h) extra credit: perform adaptive time-stepping to keep the local truncation error from time stepping bounded by a tolerance.

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CAAM 452 Spring 2005 Recall: 4 th Order Central Difference Scheme The fourth order central difference derivative acts on any vector and gives the following value for each entry of a result vector: Where the increments on the indexing is done modulo M. Since we will use this operator repeatedly I made a function

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CAAM 452 Spring 2005 Code Each time step consists of evaluating the 6 intermediate vectors k1,k2,..,k6 And piecing them together. Notice – here I set up u at the start as a vector, and delta4 accepts a vector (and dx) as arguments and returns a vector. i.e. each of k1,k2,..,k6 is a vector.

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CAAM 452 Spring 2005 Alternate Code I could also have used loops Note: u is a row vector of length M so the k’s are a matrix of size 6 x M

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CAAM 452 Spring 2005 Homework 3 Q1) Build a finite-difference solver for Q1a) use your Cash-Karp Runge-Kutta time integrator from HW2 for time stepping Q1b) use the 4 th order central difference in space (periodic domain) Q1c) perform a stability analysis for the time-stepping (based on visual inspection of the CK R-K stability region containing the imaginary axis) Q1d) bound the spectral radius of the spatial operator Q1e) choose a dt well in the stability region Q1f) perform four runs with initial condition (use M=20,40,80,160) and compute maximum error at t=8 Q1g) estimate the accuracy order of the solution. Q1h) extra credit: perform adaptive time-stepping to keep the local truncation error from time stepping bounded by a tolerance.

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CAAM 452 Spring 2005 Cash-Karp The Cash-Karp Runge-Kutta integrator in our case is: Notice, we do not need the time component on the right hand side, because our finite-difference right hand side is independent of time.

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CAAM 452 Spring 2005 Linear Stability Analysis We achieve the linear stability analysis by assuming f is linear in u:

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CAAM 452 Spring 2005 cont We replace mu*dt with nu

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CAAM 452 Spring 2005 cont We wish to remove all the intermediate variables and express the scheme in a one-step form like: Where the multiplier function pi(nu) is a 6 th order polynomial in nu. It is certainly possible to find all the coefficients of this polynomial by hand but a little messy. We can take a short cut – assuming Cash-Karp is actually a 5 th order scheme as claimed we know that the first 6 terms of the polynomial multiplier must be the first 6 terms of the Taylor series for

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CAAM 452 Spring 2005 cont So we know that: Where C is to be determined. However, looking at the definition of the stages we see that there is only one contribution to the 6 th order term: So

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CAAM 452 Spring 2005 Stability Condition We need to plot the stability region, so we determine the margin of stability by finding nu such that The curve of points in the (root) complex plane with magnitude 1 can be parameterized in theta by So for each theta in [0,2pi) we seek the corresponding nu such that:

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CAAM 452 Spring 2005 Matlab roots Command The matlab roots command can be used to find roots of a polynomial. If the polynomial is say: Then construct a vector of coefficients (highest order first): And invoke: roots(a) A vector of the roots (if any) of the polynomial will be returned.

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CAAM 452 Spring 2005 Adapting RKabstab.m So I took the routine from the class web page and modified it slightly I hard coded it to use the multiplier polynomial for the Cash-Karp I changed the plotting to use a scatter plot

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CAAM 452 Spring 2005 Absolute Stability for Cash-Karp This is a classic picture. What is interesting is that it does not convincingly cover the imaginary axis!.

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CAAM 452 Spring 2005 On The Imaginary Axis Here I used Matlab’s polyval to evaluate the multiplier polynomial for nu on the imaginary axis. Conclusion – the absolute stability region is just to the left of the imaginary axis (not a big issue here)

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CAAM 452 Spring 2005 Homework 3 Q1) Build a finite-difference solver for Q1a) use your Cash-Karp Runge-Kutta time integrator from HW2 for time stepping Q1b) use the 4 th order central difference in space (periodic domain) Q1c) perform a stability analysis for the time-stepping (based on visual inspection of the CK R-K stability region containing the imaginary axis) Q1d) bound the spectral radius of the spatial operator Q1e) choose a dt well in the stability region Q1f) perform four runs with initial condition (use M=20,40,80,160) and compute maximum error at t=8 Q1g) estimate the accuracy order of the solution. Q1h) extra credit: perform adaptive time-stepping to keep the local truncation error from time stepping bounded by a tolerance.

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CAAM 452 Spring 2005 Q1d) Bound the spectral radius of the derivative operator All the eigenvalues of the 4 th order central difference are on the imaginary axis, with values (recall from Lecture 6): So the gotcha is that there is no dt such that the eigenvalues*dt are all exactly inside the stability region. However, we will estimate the largest eigenvalue and make an assumption..

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CAAM 452 Spring 2005 Q1d) continued We can estimate the largest magnitude eigenvalue directly.

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CAAM 452 Spring 2005 Homework 3 Q1) Build a finite-difference solver for Q1a) use your Cash-Karp Runge-Kutta time integrator from HW2 for time stepping Q1b) use the 4 th order central difference in space (periodic domain) Q1c) perform a stability analysis for the time-stepping (based on visual inspection of the CK R-K stability region containing the imaginary axis) Q1d) bound the spectral radius of the spatial operator Q1e) choose a dt well in the stability region Q1f) perform four runs with initial condition (use M=20,40,80,160) and compute maximum error at t=8 Q1g) estimate the accuracy order of the solution. Q1h) extra credit: perform adaptive time-stepping to keep the local truncation error from time stepping bounded by a tolerance.

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CAAM 452 Spring 2005 Q1e) Since we are only integrating for 8 periods let’s choose a reasonable maximum |nu|<=1 So for close to absolute (linear) stability Implies that we can choose: i.e. the time step restriction is: We can further reduce the right hand side of the inequality to reduce marginal growth.

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CAAM 452 Spring 2005 Homework 3 Q1) Build a finite-difference solver for Q1a) use your Cash-Karp Runge-Kutta time integrator from HW2 for time stepping Q1b) use the 4 th order central difference in space (periodic domain) Q1c) perform a stability analysis for the time-stepping (based on visual inspection of the CK R-K stability region containing the imaginary axis) Q1d) bound the spectral radius of the spatial operator Q1e) choose a dt well in the stability region Q1f) perform four runs with initial condition (use M=20,40,80,160) and compute maximum error at t=8 Q1g) estimate the accuracy order of the solution. Q1h) extra credit: perform adaptive time-stepping to keep the local truncation error from time stepping bounded by a tolerance.

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CAAM 452 Spring 2005 Q1f: Test Runs I already set up this initial condition for the web demos: I modified the testrig.m file to call the centraldifference4CKRK54.m file It looks like reasonable 4 th order convergence.. We examine the error ratios to be more certain T=4

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CAAM 452 Spring 2005 Homework 3 Q1) Build a finite-difference solver for Q1a) use your Cash-Karp Runge-Kutta time integrator from HW2 for time stepping Q1b) use the 4 th order central difference in space (periodic domain) Q1c) perform a stability analysis for the time-stepping (based on visual inspection of the CK R-K stability region containing the imaginary axis) Q1d) bound the spectral radius of the spatial operator Q1e) choose a dt well in the stability region Q1f) perform four runs with initial condition (use M=20,40,80,160) and compute maximum error at t=8 Q1g) estimate the accuracy order of the solution. Q1h) extra credit: perform adaptive time-stepping to keep the local truncation error from time stepping bounded by a tolerance.

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CAAM 452 Spring 2005 Q1g) Estimate Order of Accuracy of Solution MMax error at t=4Order of accuracy 200.44353855151728 400.128529348068531.79 800.020401880076492.66 1600.001340762635093.93 3200.000083194253744.01 Cool – asymptotically it looks like a fourth order code (in space)

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CAAM 452 Spring 2005 Homework 3 Q1) Build a finite-difference solver for Q1a) use your Cash-Karp Runge-Kutta time integrator from HW2 for time stepping Q1b) use the 4 th order central difference in space (periodic domain) Q1c) perform a stability analysis for the time-stepping (based on visual inspection of the CK R-K stability region containing the imaginary axis) Q1d) bound the spectral radius of the spatial operator Q1e) choose a dt well in the stability region Q1f) perform four runs with initial condition (use M=20,40,80,160) and compute maximum error at t=8 Q1g) estimate the accuracy order of the solution. Q1h) extra credit: perform adaptive time-stepping to keep the local truncation error from time stepping bounded by a tolerance.

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CAAM 452 Spring 2005 Q1h) Estimating the Local Truncation Error Given the k vectors we can also construct a lower order (4 th order in time) approximation to the updated solution: Then we can ball park what the time step should be:

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CAAM 452 Spring 2005 Q1h) Code Form alternative embedded fourth order approximation Estimate reasonable dt Check to see if we should reduce time step. Modify last time step to land on EndTime

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CAAM 452 Spring 2005 Q1h) Testing I started with a fairly large time step and let the code adjust itself MMax error at t=4Order of accuracy 200.44353855318971 400.128529347190141.79 800.020401904801042.66 1600.001340751907653.93 3200.000083194621834.01 These are pretty much unchanged from before.

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CAAM 452 Spring 2005 Q1h) Adaptive Time Stepping (started with dt=100*dx) Q) Why does it take more time steps for the Lower resolution runs?

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CAAM 452 Spring 2005 Note on the Cash-Karp RK Check out: http://www.math.sfu.ca/~cbm/mscthesis/cbm-mscthesis.ps.gz There is an interesting treatment of RK schemes, with the idea of parameterizing embedded RK schemes (n+1 stage, n’th order) by choosing the coefficient in the multiplier directly. Increasing this parameter a little will make sure that at least part of the imaginary axis is inside the absolute stability region.

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