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Published byJackeline Crain Modified about 1 year ago

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Number of grid points per wavelength

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u_t + u_x = 0 u(x, 0) = sin (Lπx) 0

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Method1: Upwind(1st order) Method2: Lax-Wendroff(2nd order) Method3: Traditional 4th order Method4: Compact schemes (4th order)

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Upwind L=2 h error grid points per wavelength 1/ / / / / / /

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Upwind L=4 h error grid points per wavelength 1/ / / / / / /

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Lax-Wendroff L h error grid points per wavelength 2 1/ / / /

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Traditional 4th order L h error grid points per wavelength 2 1/ / / /

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Compact schemes L h error grid points per wavelength 2 1/ / / /

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High Order Schemes for Resolving Waves: Number of Points per Wavelength

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First derivative f(x)=sin(kx) TEk=c*(Δx)^(p-1)*k^p p is the order of numerical scheme TEk= c*(1/N)^(p-1)*k^p, Δx=1/N If k changes to m, and we want TE to be unchanged. TEk=TEm (1/N)^(p-1)*k^p= (1/a*N)^(p-1)*m^p, a=(m/k)^(p/(p-1)).

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If p>>0, then a~m/k. So, if m=2k, then the number of grid points is also doubled since a~2. But if p is small, say p=2,then when m=2k, the number of grid points should be multiplied by 4 to insure that TE is unchanged.

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Higher derivative (1/N)^(p-q)*k^p=(1/a*N)^(p-q)*m^p, a=(m/k)^(p/(p-q)).

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1 st order scheme Scheme N error IC UW sin(x) UW sin(2x) L-F sin(x) L-F sin(2x) a=2^2=4

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2 nd order scheme Scheme N error IC FD sin(x) FD sin(2x) L-W sin(x) L-W sin(2x) a=2^(3/2)=2.83

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4 th order scheme Scheme N error IC FD sin(x) FD sin(2x) FDC sin(x) FDC sin(2x) a=2^(5/4)=2.34

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6 th order, a=2^(7/6)= th order, a=2^(9/8)=2.18

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Upwind: L changes from 2 to 4, under the same error, 1/h multiplied by 4. grid points per wavelength multiplied by 2. L-W: L=2, 4, 8, 16, under the same error. 1/h multiplied by 2.83 grid points per wavelength multiplied by 1.42

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Traditional 4th order L=2, 4, 8, 16, under the same error. 1/h multiplied by 2.37 grid points per wavelength multiplied by 1.19 Compact schemes L=2, 4, 8, 16, under the same error. 1/h multiplied by 2.37 grid points per wavelength multiplied by 1.19

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