Accurate Gradient Computation for Interface Problems

Transcription

1 Accurate Gradient Computation for Interface Problems Nnenna Anoruo, Kevin Luna, William Reese, Justin Rivera Faculty Advsior: Zhilin Li North Carolina State University February 11, 016

2 Outline Problem : Solve ODE/PDE Interface Problems and Derivatives Piecewise nonconstant coecient problems Polar Coordinates (D) Spherical Coordinates (3D) Achieve second order accurate solutions and derivatives using FD method Approach Worked on nonconstant coecent interface problems and their derivatives Solved using Finite Dierence Method/IIM Worked on axis symmetric spherical and polar coordinate Work in progress Used the Immersed Interface Method 1D Stefan Problem 1D Moving Boundary Problem

3 Acheivements Derivatives on boundaries for regular problems Piecewise constant coecient 1D general problem Polar coordinates (D) Spherical coordinates (3D)

4 An Example of an Interface Problem Rubber band example u = Cδ(x α), 0 < x < 1, 0 < α < 1 u(0) = u(1) = 0 Consider an elastic material with two xed end points. The solution is continuous, but the derivative is not

5 Interfaces What is an interface? An interface is essentially a point (1D), a curve (D), or a surface (3D), that connects dierent states of a system Some examples include : ice and water, a string with dierent materials on it, the heating of an embedded metal block in another material with a dierent heat conductivity, etc. Simple model for Interface Problem (β(x)u x ) x = f(x) + vδ(x α), 0 < x < 1, 0 < α < 1 Where f(x) C (0, 1), and β(x) C 1 (0, α) C 1 (α, 1) β(x) = { β(x) 0 < x < α β(x) + α < x < 1 (1) Discontinous β(x) β(x) x

6 Mathematical Formulation of Rubber Band Problem in general Two equivalent ways to formulate the problem: On a single domain: (βu ) = Cδ(x α) 0 < x < 1 u(0) = u(1) = 0 Equivalently, splitting the domain so that the point source is avoided (βu x ) x = 0 x (0, α); (βu x ) x = 0 x (α, 1) [u] = 0, [βu x ] = C Note here [u] notation actually implies the following: [u] x=α = lim u(x) lim u(x) x α + x α

7 Immersed Interface Method (βu x ) x σu = f(x) + Cδ (x α), 0 < x < 1, 0 < α < 1 Peskin's method approximates the delta function approximation is 1st order accurate Immersed Interface Method reformulates the problem USING the jump condition (βu x ) x = f(x), x (0, α) (α, 1)

8 Finite dierence scheme at an irregular grid point We begin with the general FD stencil, γ 1 U j 1 + γ U j + γ 3 U j+1 = f j To nd the scheme at the irregular points, we want to minimize the local truncation error at α T j = γ 1 U j 1 + γ U j + γ 3 U j+1 f(x j ) C j We use method of undetermined coecients to solve for γ 1, γ, γ 3,and C j such that the local truncation error is minimized in magnitude by using the interface relations u + = w + u, u + x = β β + u x + v β +, u+ xx = β β + u xx

9 Finite dierence scheme at an irregular grid point T j = γ 1 u j 1 + γ u j + γ 3 u j+1 f(x j ) C j u(x j 1 ) = u + u x (x j 1 α) + (x j 1 α) u xx + O(h 3 ) u(x j ) = u + u x (x j α) + (x j α) u xx + O(h 3 ) u(x j+1 ) = u + + u + x (x j+1 α) + (x j+1 α) u + xx + O(h 3 ) = u + w + β β + u x (x j+1 α) + β (x j+1 α) β + u xx + O(h 3 )

10 FD at the irregular grid point, lll Taylor Expansion + interface relations + undetermined coecient method T j = γ j 1 u(x j 1 ) + γ j u(x j ) + γ j+1 u(x j+1 ) f j C j = γ j 1 (u + u x (x j 1 α) + (x j 1 α) u xx) + O(h 3 )) +γ j (u + u x (x j α) + (x j α) u β xx) + γ j+1 β + C(x j+1 α) +γ j+1 (u + w + β β + u x (x j+1 α) + β (x j 1 α) β + u xx) + O(h 3 )) = u (γ j 1 + γ j + γ j+1 ) + u x (γ j 1 α) + γ j (x j α) β +γ j+1 β + (x j+1 α)) + u (x j 1 α) xx(γ j 1 β (x j+1 α) γ j+1 β + ) f j C j x j α) + γ j +

11 Set-up equation The linear system for the coecients: γ j 1 + γ j + γ j+1 = 0 γ j 1 (x j 1 α) + γ j (x j α) + γ j+1 β γ j 1 (x j 1 α) + γ j (x j α) β + (x j+1 α) = 0 β (x j+1 α) + γ j+1 β + = β The correction term is ( ) C j = γ j+1 w + C β β + (x j+1 α)

12 Derivatives at Interface (Two Sided) To approximate u (x) we use the IIM on the approximated solutions of u(x) which were given by the IIM We collect the terms by u and its derivatives and set u equal to 1 and the other linear equations equal to 0 as follows: γ 1 + γ + γ 3 = 0 γ 1 (x j 1 α) + γ (x j α) + γ 3 β γ 1 (x j 1 α) and the correction term + γ (x j α) C j = γ 1 (w + β + (x j+1 α) = 1 β (x j+1 α) + γ 3 β + = 0 c β + (x j+1 α)). When this system is solved for γ 1, γ, and γ 3 we are left with the sided stencil for u (x) at an interface.

13 1-D Interface w/ Nonconstant Coecients In general 1D Interface problems where x (0, α) (α, 1) are expressed as: (β x u) x σu = f For our purposes we explored functions that have nite jumps at the interface α In all our examples we dened β to be: { 1 + cos(x) β(x) = if x α 1 + x if x > α and σ to be: σ(x) = { (1 + x) if x α ln( + x) if x > α Note: The appropriate Dirichlet boundary condition was chosen in each example according to the actual solution

14 1-D Interface w/ Non-constant Coecients Consider the following example: (10) ( ( cos(x) + 1)) sin(10x) 10 sin(x) cos(10x) (1 + x) cos(10x) if x < α f(x) = (30)( 30(x + 1) cos(30x) x sin(30x)) ln( + x) cos(30x) if x < α. The exact solution to the example can be shown as follows: { sin(10x) if x α u(x) = cos(30x) if x > α ()

15 Solution and Error Plot Still considering the same example: This gure shows the computed solution with n = 180 grid divisions. The actual solution has discontinuity that is connected by a plotter. This gure displays the error dierence of the estimated solution and actual solution at each grid point Error (difference) Plot

16 Error Plots on Log-log Scale Grid renement analysis of approximated solution with 15 renements starting at n = 0. Note the slope of the line is.1 implying that it is second order accurate. Grid renement analysis of the right derivative at the interface using one and two sided IIM methods. The slope of the lines are 1.08 and respectively. 1 These new results can be seen on Dr.Li's website: zhilin/

17 Polar coordinates We considered axis symmetric cases of interface problems in polar coordinates 1 r (rβu r) r = f(r) We generate a uniform grid r i = R 1 + i r, i = 0, 1,..., m, r = R R 1 m Standard dierence scheme for polar ( r i 1 + r i+ 1 1 r i 1 U i 1 r i ) U i + r i+ 1 U i+1 ( r) = f(r i )

18 Polar coordinates When the origin is in the boundary there is a singularity We use a staggered grid to address this issue r i = (i 1 ) r, r = R m 1, i = 1,...m. When i = 1 we have to use U 1 + U ( r) + 1 r 1 U r = f(r 1) For all other points on the grid we use the standard dierence scheme We can apply IIM at an interface if needed

19 Polar Coordinates with Axis Symmetry, Ex1 1 r (rβu r) r = f(r), 0 < r < 1, α = 3 { β = 1 r α β = β + = 100 r > α { f (sin(r)+r cos(r)) = f(r) = r r α f + = 9 cos(9r) r 81 sin(9r) r > α

20 Polar Coordinates with Axis Symmetry, Ex1 ctd. Exact Solution u(r) = { u = cos(r) u + = sin(9r) r α r > α Surface of Approximation

21 Solution and Error Plot Still considering the same example: This gure shows the computed solution with n = 160 grid divisions. The actual solution has discontinuity that is connected by a plotter. This gure displays the error dierence of the estimated solution and actual solution at each grid point

22 Error Plots on Log-log Scale Grid renement analysis of approximated solution with 8 renements starting at n = 0. Note the slope of the line is.1 implying that it is second order accurate. Grid renement analysis of the right derivative at the interface using one and two sided IIM methods. The slopes of the lines are 0.96 and 1.94 respectively.

23 Polar Coordinates with Axis Symmetry, Ex 1 r (rβu r) r = f(r), 0 < r < 1, α = 3 { β = 1 r α β = β + = 100 r > α f(r) = { f = sin(r)+r cos(r) r f + = r ( sin(r))+sin(r) r cos(r) r 3 r α r > α

24 Polar Coordinates with Axis Symmetry, Ex ctd. Exact Solution u(r) = Surface of Approximation { u = cos(r) u + = sin(r) r r α r > α

25 Solution and Error Plot Still considering the same example: This gure shows the computed solution with n = 160 grid divisions. The actual solution has discontinuity that is connected by a plotter Solution and Approximation plot for β (r u r ) r = rf(r) Approximation Exact Solution This gure displays the error dierence of the estimated solution and actual solution at each grid point Error (difference) Plot u(r) 0.6 U-u(x) r x

26 Error Plots on Log-log Scale Grid renement analysis of approximated solution with 8 renements starting at n = 0. Note the slope of the line is 1.99 implying that it is second order accurate Solution Error Plot on a log-log Scale for the IIM Immersed Interface Method Grid renement analysis of the right derivative at the interface using one and two sided IIM methods. The slopes of the lines are 0.9 and.06 respectively Two-Sided Method(IIM) One-sided method loglog Plot of U x + (α) Error log 10 ( U-u ) log 10 ( U x - (α)-ux - (α) ) log 10 (h) log 10 (h)

27 3D Spherical coordinates with Radial Symmetry We considered axis symmetric cases of interface problem in spherical coordinates 1 r (r βu r ) r = f(r) We generate the staggered grid r i = (i 1 ) r, r = R m 1, i = 1,...m. The standard dierence scheme for spherical coordinates (r i 1 ) U i 1 ( (r i 1 ) + (r i+ 1 ) ) U i + (r i+ 1 ) U i+1 ( r) = r f(r i )

28 Spherical coordinates For r i=1 the dierence scheme is ( ) r 1 U r + U1 + U r 1 ( r) U 0 4 = f(r 1) For all other points on the grid we use the standard dierence scheme We can apply IIM at an interface if needed

29 3D Spherical coordinates with Radial Symmetry, Ex1 1 r (r βu r ) r = f(r), 0 < r < 1, α = 3 { β = 1 r α β = β + = 100 r > α { f = 3r(4 sin(r 3 ) + 3r 3 cos(r 3 )) r α f(r) = f + = 6 cos(r ) 4r sin(r ) r > α

30 3D Spherical coordinates with Radial Symmetry, Ex1 ctd. Exact Solution u(r) = { u = cos(r 3 ) u + = sin(r ) r α r > α Surface of Approximation

31 Solution and Error Plot Still considering the same example: This gure shows the computed solution with n = 160 grid divisions. The actual solution has discontinuity that is connected by a plotter. This gure displays the error dierence of the estimated solution and actual solution at each grid point

32 Error Plots on Log-log Scale Grid renement analysis of approximated solution with 8 renements starting at n = 0. Note the slope of the line is 1.9 implying that it is second order accurate. Grid renement analysis of the right derivative at the interface using one and two sided IIM methods. The slopes of the lines are 0.99 and 1.99 respectively.

33 Work in Progress Stefan problem ( xed and moving boundaries) Parabolic interface problems Moving interface problems

34 Acknowledgements We would like to thank NCSU Department of Mathematics for hosting the REU program Faculty Advisor Dr. Li The NSF for grant DMS The NSA for grant H