Lecture 1. Finite difference and finite element methods. Partial differential equations (PDEs) Solving the heat equation numerically

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1 Finite difference and finite element methods Lecture 1

2 Scope of the course Analysis and implementation of numerical methods for pricing options. Models: Black-Scholes, stochastic volatility, exponential Lévy. Options: European, American, Asian, barrier, compound... In this course: Focus on deterministic (PDE based) methods Finite difference methods (FDM) Finite element methods (FEM) This course will be complemented by the course Monte Carlo methods in autumn 2009.

3 Organization of the course 14 lectures (2 hours) + 13 exercise classes (1 hour). No lectures on April, 14. Testat: 70% of solved homework assignments (theoretical exercises + MATLAB programming). Examination: On Tuesday, May 26, Written, closed-book examination includes theoretical and MATLAB programming problems. Examination takes place on ETH-workstations running MATLAB under LINUX. Own computer will NOT be required.

4 Outline Partial differential equations (PDEs)

5 Definitions and notation Let α = (α 1,..., α d ) N d 0 be a multiindex. Set α = d i=1 α i. For u : R, x = (x 1,..., x d ) R d define Let k N 0. Then D α u(x) := α u(x) x α 1 1 xα d d = α 1 x 1 α d x d u. D k u(x) := {D α u(x) : α = k} is the set of all partial derivatives of order k. If k = 1, we regard the elements of D 1 u(x) =: Du(x) as being arranged in a vector Du = ( x1 u,..., xd u).

6 If k = 2, we regard the elements of D 2 u(x) as being arranged in a matrix x1 x1 u x1 xd u D 2 u =.... xd x1 u xd xd u In the following: write xi x j instead of xi xj. Hence, the Laplacian u of u can be written as u := d xi x i u = tr(d 2 u). i=1 A partial differential equation (PDE) is an equation involving an unknown function of two or more variables and certain of its derivatives.

7 Let R d be open, x = (x 1,..., x d ) and N k 1. Definition An expression of the form F (D k u(x), D k 1 u(x),..., Du(x), u(x), x) = 0, x is called a k-th order PDE, where F : R dk R dk 1 R d R R is given and is the unknown. u : R

8 Let a ij (x), b i (x), c(x) and f(x) be given functions. For a linear 2nd order PDE in d + 1 variables, F has the form F (D 2 u, Du, u, x) = d d a ij (x) xi x j u+ b i (x) xi u+c(x)u f(x). i,j=0 Assume that the matrix A(x) := {a ij (x)} d i,j=0 is symmetric with real eigenvalues λ 0 (x) λ 1 (x) λ d (x). i=0 Definition Let S = {0,..., d}. At x R d+1, the PDE is called (i) elliptic λ i (x) 0, i sign(λ 0 (x)) =... = sign(λ d (x)) (ii) parabolic!j S : λ j (x) = 0 rank(a(x), b(x)) = d + 1 (iii) hyperbolic λ i (x) 0, i!j S : signλ j (x) signλ k (x), k S \ {j} The PDE is called elliptic, parabolic, hyperbolic on, if it is elliptic, parabolic, hyperbolic x.

9 Examples t u u = f(t, x) is parabolic (set x 0 = t). The Poisson equation u = f(x) is elliptic. The wave equation tt u u = f(t, x) is hyperbolic (set x 0 = t) The Black Scholes equation for the value of a European option v(t, s) t v 1 2 σ2 s 2 ss v rs s v + rv = 0 with σ, r 0 is parabolic at (t, s) (0, T ) (0, R) and degenerates to a ordinary differential equation as s 0. Note: PDEs can have infinitely many solutions. To obtain a unique solution, we have to pose boundary conditions.

10 Outline Partial differential equations (PDEs) Partial differential equations (PDEs)

11 The PDE Partial differential equations (PDEs) Let = (a, b) R be a open interval and let J := (0, T ), T > 0. Find u : J R such that Remark t u xx u = f(t, x) in J u = 0 on J u(0, ) = u 0 in (i) The equation u(0, ) = u 0 in is the initial condition. (ii) The equation u = 0 on J is the boundary condition. Here it is of Dirichlet type and homogeneous. oal: approximate u(t, x).

12 Discretization of the domain Computational domain J is replaced by discrete grid: {(t m, x i )}, i = 0,..., N + 1, m = 0,..., M, where x i are space grid points with space step size h and t m are the time levels with time step size k: x i = a + ih, h = b a N + 1, t m = mk, k = T M. We represent the exact solution u(t, x) by its values on the grid: u(t, x) {u m i = u(t m, x i )}, i = 0,..., N +1, m = 0,..., M. The goal is to approximate the values {u m i }. Values of the solution between grid points are then found by some interpolation.

13 Difference Quotients (= Finite Differences) We want to approximate the derivatives of u using only its values on the grid. First, let us consider a function f(x) of one variable. Assume that f C 2. Then, using Taylor s formula, f (x) = f(x + h) f(x) h + h 2 f (ξ), ξ [x, x + h]. If f i = f(x i ) are the values of f on the grid {x i }, we obtain f (x i ) = f i+1 f i h Similarly, for f C 4 + O(h) =: ( δ + x f) i + O(h). f (x i ) = f i+1 2f i + f i 1 h 2 + O(h 2 ) =: ( δ xx f ) i + O(h2 ).

14 FD scheme Partial differential equations (PDEs) Let θ [0, 1]. We replace the PDE t u xx u = f by the set of algebraic equations E m i = θf m+1 i + (1 θ)f m i i = 1,..., N, m = 0,..., M 1 u 0 i = u 0 (x i ) i = 1,..., N u m k = 0 k {0, N + 1}, m = 0,..., M, where E m i Ei m := k 1( u m+1 i u m i is the finite difference operator = um+1 i k u m i [θ um+1 i+1 2um+1 ) [ θ(δxx u) m+1 i + (1 θ)(δ xx u) m i i + u m+1 i 1 h 2 + (1 θ) um i+1 2um i + u m ] i 1 h 2. ]

15 FD scheme in matrix form Introduce the column vectors u m = (u m 1,..., um N ), E m = (E m 1,..., E m N ), f m = (f m 1,..., f m N ) and the tridiagonal N N matrix = h 2 tridiag( 1, 2, 1). Then the FD-scheme E m = θf m+1 + (1 θ)f m becomes, in matrix form: iven u 0 = (u 0 (x 1 ),..., u 0 (x N )) R N, for m = 0,..., M 1 find u m+1 R N such that ( I } + {{ θk } =:B or ) u m+1 + ( I + (1 θ)k }{{} =: C ) u m = k[θf m+1 + (1 θ)f m ], }{{} =:F m Bu m+1 = Cu m + kf m, m = 0,..., M 1.

16 Variational formulation We do not require the PDE to be hold pointwise. Take a smooth test function v C0 () satisfying v(a) = v(b) = 0. Multiply the PDE with v and integrate by parts: t uv dx xx uv dx = fv dx d dt uv dx [ x u(t, x)v(x)] x=b x=a + }{{} x u x v dx = fv dx =0 The variational or weak formulation of the heat equation reads: Find u such that u(0) = u 0 and such that v C0 () d u(t, x)v(x)dx + u (t, x)v (x)dx = f(t, x)v(x)dx. dt

17 alerkin discretization Let V N be a finite (N) dimensional subspace of H 1 0 (). The idea is to approximate u(t, x) by an element u N (t, x) V N, for each t J. Find u N (t, x) V N such that u N (0, x) = u 0,N (x) and such that v N V N d dt u N (t, x)v N (x)dx+ Let {b j } N j=1 be a basis of V N. Then u N (t, x) = N j=1 u N,j(t)b j (x), where u N(t, x)v N (x)dx = u N (t) = (u N,1 (t), u N,2 (t),..., u N,N (t)) is a vector of unknown functions. Similarly, v N V N v N (x) = N i=1 v N,ib i (x). f(t, x)v N (x)dx.

18 Hence (we skip the argument x in ) d u N (t)v N + u N dt (t)v N = f(t)v N, v N V N d ( )( ) u N,j (t)b j v N,i b i dt j i ( ) ( ) + u N,j (t)b j v N,i b i = f(t) v N,i b i j i i [ ] v N,i u N,j b j b i + u N,j b jb i f(t)b i = 0 i j [ ] v N M u N (t) + Au N (t) f N (t) = 0, v N R N M u N (t) + Au N (t) = f N (t).

19 Semi discrete scheme Thus, d dt (u N (t), v N ) + a ( ) u N (t), v N = (f(t), vn ), is equivalent to the ODE M u N (t) + Au N (t) = f N (t), v N V N where M (mass matrix) and A (stiffness matrix) are N N matrices with M ij = b j (x)b i (x)dx, A ij = b j (x)b i (x)dx. Similarly, f N (t) R N with entries f N,i (t) = f(t, x)b i (x)dx.

20 Fully discrete scheme We discretize in time. Write u m N := u N(t m ), f m N := f N(t m ), where the time levels t m, m = 0,..., M are as before. Proceeding exactly as in the FDM, the fully discrete scheme reads: iven u 0 N = (u 0(x i )) N i=1 RN, for m = 0,..., M 1 find R N such that u m+1 N Mk 1( u m+1 N um N ) ( ) +A θu m+1 N +(1 θ)um N = θf m+1 N +(1 θ)f m N.

21 Choice of V N Setting B := M + kθa, C := M k(1 θ)a and F m m+1 N := θf + (1 θ)f m this can be written as N N Bu m+1 N = Cu m N + kf m N, m = 0,..., M 1. It remains to chose a space V N. Probably the simplest choice: V N is the space of piecewise linear, continuous functions. Let T := {a = x 0 < x 1 < < x N+1 = b} = {K i } N+1 i=1 be an equidistant mesh on with K i := (x i 1, x i ). (x i as before). Set V N = S 1 0 := { u C 0 0 () : u K i is affin linear on K i T }. Note: dimv N = N.

22 g replacements A basis {b i } N i=1 of V N is given by the so-called hat-functions b i : [a, b] R 0, b i (x) = max{0, 1 h 1 x x i }, i = 1,..., N. b 1 (x) b 2 (x) b k (x) b N (x) a = x 0 x 1 x 2 x k 1 x k x k+1 x N 1 x N x N+1 = b With this basis, we find for the mass- and stiffness matrix M = h/6 tridiag(1, 4, 1), A = h 1 tridiag( 1, 2, 1).

23 For both FDM and FEM, we have to solve M systems of N linear equations of the form Bu m+1 = Cu m + kf m, m = 0,..., M 1. where F m = θf m+1 + (1 θ)f m and FDM FEM u m vector of u m i u(t m, x i ) coeff. vector of u N (t m, x) B I + kθ M + kθa C I k(1 θ) M k(1 θ)a f m = h 2 tridiag( 1, 2, 1) A = h 1 tridiag( 1, 2, 1) f(t m, x i ) f m = i f(t m, x)b i (x)dx

24 Matlab coding (FDM) function error = heateq_fdm(a,b,t,n,theta) h = (b-a)/(n+1); k = h; e = ones(n,1); I = speye(n); = h^(-2)*spdiags([-e 2*e -e],-1:1,n,n); B = I+k*theta*; C = I-k*(1-theta)* x = [a+h:h:b-h] ; u0 = x.*sin(pi*x); f = -(1-pi^2)*x.*sin(pi*x)-2*pi*cos(pi*x); u = zeros(n,t/k+1); u(:,1) = u0; for j = 1:T/k F = k*f*(theta*exp(-j*k)+(1-theta)*exp(-(j-1)*k)); u(:,j+1) = B\(C*u(:,j)+F); err(j) = norm(u(:,j+1)-exp(-k*j)*u0); end error = sqrt(h)*max(err)

25 Example Partial differential equations (PDEs) Let = (0, 1), T = 1, and u(x, t) = e t x sin(πx). We measure the discrete L (0, T ; L 2 ())-error defined by e := sup h 1 2 ɛ m l2, m ɛ m 2 l 2 := i u(t m, x i ) u m i 2. For θ = 0.5 and k = h, we obtain, in terms of the mesh width h, convergence both of FDM and FEM of second order, i.e., e = O(h s ), s = 2.

26 FDM FEM 10 5 e 10 6 eplacements s = h

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