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- 1. How to Solve a Partial Diﬀerential Equation on a Surface Tom Ranner University of Warwick T.Ranner@warwick.ac.uk http://go.warwick.ac.uk/tranner University of Warwick, Graduate Seminar, 3rd November 2010
- 2. Where do surface partial diﬀerential equations come from? Partial diﬀerential equations on surfaces occur naturally in many diﬀerent applications for example: ﬂuid dynamics, materials science, cell biology, mathematical imaging, several others. . .
- 3. Example Surfaces – Cell Biology Figure: An endoplasmic reticulum (ER)
- 4. Example Surfaces – Pattern Formation Invited Article R253 Figure 1. Examples of self-organization in biology. Clockwise from top-left: feather bud patterning, somite formation, jaguar coat markings, digit patterning. The ﬁrst is reprinted from Widelitz et al 2000, β-catenin in epithelial morphogenesis: conversion of part of avian foot scales into feather buds with a mutated β-catenin Dev. Biol. 219 98–114, with permission from Elsevier. The second is courtesy of the Pourqui´ Laboratory, Stowers Institute for Medical Research, and e the remainder are taken from the public image reference library at http://www.morgueﬁle.com/. (a) Activator (b) Inhibitor 80 80 8 15 Taken from R E Baker, E A Gaﬀney and P K Maini, Partial diﬀerential equations for self-organisation in cellular 60 60 Distance (y) Distance (y) and development biology. Nonlinearity 21 (2008) R251–R290. 40 10 40 4 20 20
- 5. Example Surfaces – Dealloyed surface The surface can be the interface of two materials in an alloy. 9738 C. Eilks, C.M. Elliott / Journal of Computational Physics 227 (2008) 9727–9741 Fig. 3. Simulation on a large square, t ¼ 0:04; t ¼ 0:1 and t ¼ 0:2. Taken from C The geometric motion ofNumerical simulation of dealloying by surface nonvanishing right hand evolving surface plane. Eilks, C M Elliott, the surface has little inﬂuence, except for providing for a dissolution via the side for the conservation law of gold on the surface, since gold from the bulk is accumulating on the surface. While the concrete ﬁnite element method. Journal of Computational Physics 227 random distribution in the bulk, the lengthscales of the appearance of the structure obviously depends on the particular (2008) 9727–9741. structure depend only on the particular values of the parameters in the equation. After the phases have separated, etching still continues in the areas with a small concentration of gold, while the motion is negligible in regions covered by gold yielding a maze like structure of the surface. The origins of this shape can still be explained by the initial phase separation which ﬁxed the gold covered regions that proceeded to move into the bulk. So at this stage the simulation does not necessarily show the mechanisms for the emergence of a nanoporous structure. By undercutting the gold rich portion of the surface, the area of the surface that is not covered by gold increases. In the last stages of this simulation new components of the gold rich phase emerge at the bottom of the surface. Additionally the inter- face separating gold rich and gold poor phases shows no effect of coarsening as for the planar Cahn-Hilliard equation, but instead becomes more complicated. These two effects can be seen as signs that the model shows increasing formation of morphological complexity. We explore them in more detail in the following examples. Note however that due to self-inter- sections the surface is not embedded at later stages, as can be seen in Fig. 4, where the cross sections along the plane parallel
- 6. 1D Heat Equation The one dimensional heat equation is given by ut = uxx on (0, 1) × (0, T ) u(0, t) = u(1, t) = 0 for t ∈ (0, T ) u(x, t) = u0 (x) for x ∈ (0, 1). Joseph Fourier solved this in 1822 using separation of variables. The idea is to write u(x, t) = X (x)T (t) and derive a system of decoupled odes for X and T , which can then be solved simply.
- 7. 2D Heat Equations In two dimensions the heat equation on a square Ω = (0, 1) × (0, 1) becomes ut = ∆u := uxx + uyy in Ω × (0, T ) u = 0 on ∂Ω × [0, T ] u(x, 0) = u0 (x) on Ω × {0}. This can be solved using Fourier analysis. We rewrite u(x, t) = uk (t)e ik·x , ˆ k∈Z2 and derive a system of odes of uk . ˆ
- 8. What is the surface Heat Equation? For a d-dimensional hypersurface Γ we would like to write down something like ut = ∆u but u is only deﬁne on Γ. So instead we have ut = ∆Γ u, where ∆Γ is the Laplace-Beltrami operator.
- 9. Surface gradients For a function η : Γ → R we deﬁne the surface gradient Γη of η by Γη = η − (ν · η) where ν is the outward pointing normal on Γ and η is the gradient in ambient coordinates.
- 10. Surface gradients For a function η : Γ → R we deﬁne the surface gradient Γη of η by Γη = η − (ν · η) where ν is the outward pointing normal on Γ and η is the gradient in ambient coordinates. The Laplace-Beltrami operator ∆Γ is given by the surface divergence of the surface gradient ∆Γ η = Γ · Γ η.
- 11. Surface Heat Equation The surface heat equation on a closed surface Γ is given by ut = ∆Γ u on Γ × [0, T ] u(x, 0) = u(x) on Γ.
- 12. Surface Heat Equation The surface heat equation on a closed surface Γ is given by ut = ∆Γ u on Γ × [0, T ] u(x, 0) = u(x) on Γ. How do we solve this equation? separation of variables Fourier analysis parametrisation approximation. . .
- 13. Approximation of the 1D Heat Equation – using ﬁnitediﬀerences Let’s ﬁrst go back to the 1D problem is demonstrate some possible methods. Take the 1D heat equation ut = uxx , and let’s approximate the derivatives by ﬁnite diﬀerences. We divide (0, 1) into N intervals of length ∆x, (xj , xj+1 ) for j = 0, . . . N. Our approximate solution u h solves for j = 1, . . . , N − 1 and t ∈ (0, T ) h u h (xj−1 , t) − 2u h (xj , t) + u h (xj+1 , t) ut (xj , t) = . ∆x 2
- 14. Approximation of the 1D Heat Equation – using ﬁnitediﬀerences From, for j = 1, . . . , N − 1 and t ∈ (0, T ) h u h (xj−1 , t) − 2u h (xj , t) + u h (xj+1 , t) ut (xj , t) = , ∆x 2 we wish to discretize in time. We approximate the time derivative on the left hand side in the same way, but there is a choice on the right-hand side to evaluate at t = tk or tk+1 . We choose for numerical reasons the Backwards Euler method t = tk+1 . So we have a linear system u h (xj , tk+1 ) − u h (xj , tk ) u h (xj−1 , tk+1 ) − 2u h (xj , tk+1 ) + u h (xj+1 , tk+1 = ∆t ∆x 2
- 15. Approximation of the 1D Heat Equation – using ﬁnitediﬀerences The solve strategy is then 1. Initialise u h (xj , 0) = u0 (xj ) for j = 0, . . . , N 2. For k = 0, 1, 2, . . . solve the linear system ∆t h u h (xj , tk+1 ) − u (xj−1 , tk+1 ) − 2u h (xj , tk+1 ) + u h (xj+1 , tk+1 ) ∆x 2 = u h (xj , tk ) for j = 1, . . . , N − 1, and u h (xj , tk+1 ) = 0 for j = 0, N.
- 16. Approximation of the 1D Heat Equation – using ﬁnitediﬀerences This was implemented in MATLAB with the following result:
- 17. Approximation of the 2D Heat Equation – using ﬁnitediﬀerences The same method can be implemented for the 2D problem, with the following result:
- 18. Can ﬁnite diﬀerences work on a surface? This method works best on a regular grid, which is almost always impossible on a surface. One must parameterise the surface ﬁrst! Projections or embeddings can be used to solve a the surface pde using this type of method. An example of this type of method is the closest point method. Instead, we can try to approximate the solution rather than the problem.
- 19. Approximation of the 1D Heat Equation – using ﬁniteelements Again we split the domain (0, 1) into N intervals of length h. We deﬁne the space Vh of ﬁnite element functions to be Vh = {ηh ∈ C0 (0, 1) : ηh |(xj ,xj+1 ) is aﬃne linear for each j}. We would like to solve h h ut = uxx so that u h (·, t) ∈ Vh for all t, but the ﬁnite element functions don’t have two derivatives in space!
- 20. Approximation of the 1D Heat Equation – using ﬁniteelements The ﬁnite element functions do have a ﬁrst derivative almost everywhere, so we put the equations in integral form to remove one of the derivatives.
- 21. Approximation of the 1D Heat Equation – using ﬁniteelements The ﬁnite element functions do have a ﬁrst derivative almost everywhere, so we put the equations in integral form to remove one of the derivatives. We multiply by a test function φ and integrate with respect to x 1 1 ut φ = uxx φ, 0 0 we then integrate by parts using the boundary condition u(0) = u(1) = 0 to get 1 1 ut φ + ux φ x = 0 0 0 Now all the terms in the above equation exist for all 1 u, φ ∈ Vh ⊆ H0 (0, 1). This is called the weak form of the heat equation.
- 22. Approximation of the 1D Heat Equation – using ﬁniteelements We wish to ﬁnd u h (·, t) ∈ Vh such that for all time t 1 1 h h ut φ + ux φx = 0. 0 0 We would like this to be true for all φ ∈ Vh , but this is equivalent to being true for all basis functions φj ∈ Vh .
- 23. Approximation of the 1D Heat Equation – using ﬁniteelements We wish to ﬁnd u h (·, t) ∈ Vh such that for all time t 1 1 h h ut φ + ux φx = 0. 0 0 We would like this to be true for all φ ∈ Vh , but this is equivalent to being true for all basis functions φj ∈ Vh . We can ﬁnd a basis for Vh by setting φj (xi ) = δij . Our problem is to ﬁnd u h (·, t) ∈ Vh for t such that 1 1 h h ut φ j + ux (φj )x = 0 for j = 1, . . . , N. 0 0 Notice that the boundary condition is automatically satisﬁed if u h ∈ Vh .
- 24. Approximation of the 1D Heat Equation – using ﬁniteelements We can decompose u h in terms of the basis function φj to get N u h (x, t) = Ui (t)φj (x). i=0 The equations become 1 1 Ui,t φi φj + Ui (φi )x (φj )x = 0, 0 i 0 i If we write U(t) = (U0 (t), . . . , UN (t)) and deﬁne the mass matrix M and stiﬀness matrix S by 1 1 Mij = φi φj Sij (φi )x (φj )x , 0 0 we can write a matrix system MUt = SU.
- 25. Approximation of the 1D Heat Equation – using ﬁniteelements We discretize in time using backwards Euler again to get the following solve strategy: 1. Initialise U 0 as Uj0 = u0 (xj ). 2. For each k = 0, . . ., solve the linear system (M + ∆tS)U k+1 = MU k .
- 26. Approximation of the 1D Heat Equation – using ﬁniteelements This method was implemented in MATLAB with the following result:
- 27. Approximation of the 2D Heat Equation – using ﬁniteelements This method can also be used for the 2D problem:
- 28. tinuously; the method should also be robust enough to tolerate diﬀerent resolutions and boundaries; for database indexing, each class index should be small for storageCan we use ﬁnite elements for surface partial diﬀerential and easy to compute. Conformal mapping has many nice properties to make it suitable for classiﬁcation problems. Conformal mapping only depends on the Riemann metric and is indepen-equations? dent of triangulation. Conformal mapping is continuously dependent of Riemann metric, so it works well for diﬀerent resolutions. Conformal invariants can be repre- sented as a complex matrix, which can be easily stored and compared. We propose to use conformal structures to classify general surfaces. For each conformally equivalent class, we can deﬁne canonical parametrization for the purpose of comparison. Geometry matching can be formulated to ﬁnd an isometry between two surfaces. We can use what is called a surface ﬁnite element method, in By computing conformal parametrization, the isometry can be obtained easily. For which we approximate the domain Γ also. surfaces with close metrics, conformal parametrization can also give the best geometric matching results. Fig. 1. Surface & mesh with 20000 faces 1.1. Preliminaries. In this section, we give a brief summary of concepts and Justiﬁcation of this method including well posedness, stability and convergence can be found in G. Dziuk and C. M. notations. Elliott, Surface ﬁnite elements for parabolic equations. J.realization |K| is homeomorphic Let K be a simplicial complex whose topological Comput. Math. 25(4), (2007) 385–407. Image taken from Xianfeng Gucompact 2-dimensional manifold. SupposeConformal Structures of surfaces. Communications in to a and Shing-Tung Yau Computing there is a piecewise linear embedding Information and Systems. 2(2) (2002), 121–146. (1) F : |K| → R3 . The pair (K, F ) is called a triangular mesh and we denote it as M . The q-cells of K
- 29. Solving the surface heat equation – using surface ﬁniteelements We deﬁne Γh to be a polyhedral approximation of Γ (made of triangles) with vertices xi . Vh is the space of piecewise linear functions on Γh with basis φj given by φj (xi ) = δij We look to solve the weak form of the surface heat equation on Γh : h h ut φ + Γh u · Γh φ = 0. Γh Γh
- 30. Solving the surface heat equation – using surface ﬁniteelements We deﬁne the surface mass matrix M and surface stiﬀness matrix S by Mij = φi φj Sij = Γ h φi · Γ h φj Γh Γh Using the same notation for U as before we have MUt + SU = 0.
- 31. Solving the surface heat equation – using surface ﬁniteelements We discretize using backwards Euler to get the solve strategy 1. Initialise U 0 by Uj0 = u0 (xj ) 2. For k = 0, 1, . . ., solve the linear system (M + ∆tS)U k+1 = MU k .
- 32. Solving the surface heat equation – using surface ﬁniteelements This method was implemented using the ALBERTA ﬁnite element toolbox on S 2 to get:
- 33. Optimal Partition Problem Given a surface Γ, a positive integer m and parameter ε > 0, we wish to minimised the following energy functional for a vector-valued function u = {uj } ∈ (H 1 (Γ))m : 2 E ε (u) = (| Γ u| + 2Fε (u)), Γ where Fε is in interaction potential of the form m 1 Fε (u) = 2 ui2 + uj2 . ε i=1 j<i Based on Quang Du and Fangda Lin, Numerical approximations of a norm-preserving gradinet ﬂow and applications to an optimal partition problem Nonlinearity 22 (2009) 67–83.
- 34. Optimal Partition Problem – Gradient Decsent If we minimised E ε subject to uj L2 (Γ) = 1 for each j we have the following norm preserving gradient decent equations: uj,t − ∆Γ uj = λj (t)uj − Fε,uj (u) on Γ × R+ |uj |2 = 1 for j = 1, . . . , m. Γ along with the initial condition u(0, x) = g (x) ∈ H 1 (Γ, Ξ), with gj L2 (Γ) = 1. Here Ξ is the singular subset in Rm m Ξ = y ∈ Rm : yj2 yk = 0, yk ≥ 0 for 1 ≤ k ≤ m . 2 j=1 k<j
- 35. Optimal Partition Problem – Partition? Since the vector-valued function u takes values on in Ξ, the above system is equivalent to the following problem For a given surface Γ, partition Γ into m region Γj such that the sum j λj , with λj the ﬁrst eigenvalue of ∆Γ over Γj with a Dirichlet boundary condition, minimised.
- 36. Optimal Partition Problem – How to solve? To progress from step tn to tn+1 , given u n we calculate u n+1 by 1. Set u n+1 as the solution the heat equation at t = tn+1 ˆ ut = ∆Γ u for x ∈ Γ, tn < t < tn+1 u(tn , x) = u n (x). 2. use (Gauss-Seidel) solver of decoupled ODEs: sequentially for j = 1. . . . , m, 2uj uj,t = − 2 (˜in+1 )2 + u (ˆin+1 )2 for x ∈ Γ, tn < t < tn+1 u ε i<j i>j uj (tn , x) = ujn+1 (x) ˆ Set u n+1 as the solution u(tn+1 , x) of this system at t = tn+1 , ˜ 3. normalisation: we set u n+1 via ujn+1 ˜ ujn+1 = for j = 1, . . . , m, ujn+1 ˜ L2 (Γ)
- 37. Optimal Partition Problem – Numerical Results This equation was discretize using a surface ﬁnite element method and solved here for m = 6 on S 2 .
- 38. Summary Methods such as Fourier series and separation of variables don’t work on general surface so we must approximate solutions. Finite diﬀerence methods work by approximating the diﬀerential operator by diﬀerence quotients, but only work best on rectangular domains. Finite element methods approximate weak solutions by dividing up the domain into ﬁnitely many subdomains and can be extended to surface problems. A surface ﬁnite element method can be used to solve many diﬀerent types of problems on surfaces.

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