Lecture 3: Initial and boundary value problems in 1 dimension

Transcription

1 Lecture 3: Initial and boundary value problems in 1 dimension Gantumur Tsogtgerel Assistant professor of Mathematics Math 319: Introduction to PDEs McGill University, Montréal January 10, 2011

2 Euler s finite difference method u(x + h) u(x) We have u x (x) for small h, if eg u C 1 (R) So u x = f is h something like y i+1 y i = b i+1, where y i u(ih) and b i+1 hf (ih) This can be solved as y n = y n 1 + b n = y n 2 + b n 1 + b n = = y 0 + b b n Let us rewrite the equation as with Ay = b, y 0 b y 1 y =, b = b 2, A = y n b n The dimension of KerA is 1 n (n+1) Gantumur Math 319 Lecture 3 Jan 10 2 / 10

3 Discrete Green s function So if we give y 0 = α, then the solution is unique This means we consider the equation Ãy = b where α b 1 à = and b = b (n+1) (n+1) In particular à : R n+1 R n+1 is an invertible matrix, with the inverse G := à 1 = b n Gantumur Math 319 Lecture 3 Jan 10 3 / 10

4 Discrete Green s function The rows of G correspond to how particular y i depends on b k (and α): y i = 1 α + 1 b b i + 0 b i b n The k-th column of G corresponds to the solution of the problem Ãy = e k where e k = [0,,0,1,0,0] with the 1 at k-th place Let G 0,G 1, be the columns of G Then we have In other words, y = αg 0 + b 1 G b n G n G 0 is the solution when α = 1, b 1 = = b n = 0, G 1 is the solution when α = 0, b 1 = 1, and b 2 = = b n = 0, etc To know a linear system, it is enough to know the responses of the system for a few well-chosen set of inputs Gantumur Math 319 Lecture 3 Jan 10 4 / 10

5 Discrete Green s function What if y n = β is given, and we want to find y 0,,y n 1? Then we need to integrate backwards y i = y i+1 b i+1 = y i+2 b i+1 b i+2 = = y n b i+1 b n Now the problem is Ãy = b where à = (n+1) (n+1) and and we have à 1 = b 1 b 2 b =, b n β Gantumur Math 319 Lecture 3 Jan 10 5 / 10

6 Green s function The analogue of the matrix inversion for u x = f with u(0) = α is where u(x) = α + G(x,t) = 0 G(x, t)f (t)dt, { 1 if t x 0 otherwise is called Green s function for the problem u x = f on (0, ) with the initial condition at x = 0 The function G t (x) = G(x,t) is the response to the Dirac function δ(x t) centered at t δ(x t) = 0 for x t, and δ(x t)f (x)dx = f (t) Gantumur Math 319 Lecture 3 Jan 10 6 / 10

7 Green s function For the problem u x = f on (,0) with the terminal condition at x = 0, we have 0 0 u(x) = u(0) f (t)dt = u(0) + G(x, t)f (t)dt, x so its Green s function would be { 1 if t x G(x,t) = 0 otherwise Gantumur Math 319 Lecture 3 Jan 10 7 / 10

8 Laplace operator The Laplace operator is u = u xx in 1D, u = u xx + u yy in 2D etc The two basic equations involving this operator are the Laplace equation u = 0, whose solutions are called harmonic functions, and the more general Poisson equation u = f Gantumur Math 319 Lecture 3 Jan 10 8 / 10

9 Poisson equation in 1D Consider u xx = f on the interval (0,1) subject to the initial conditions u(0) = α and u x (0) = β For α = 1, β = 0, and f = 0, the solution is u(x) = 1 For α = 0, β = 1, and f = 0, the solution is u(x) = x For α = 0, β = 0, and f (x) = δ(x t), let u(x) = G(x,t) be the solution Then in the general case we anticipate u(x) = α + βx G(x, t)f (t)dt We have G xx (x,t) = 0 unless x = t, so G(x,t) is linear in x except possibly at x = t This gives G(x,t) = 0 for x t On the other hand, we have 1 = t+ε t ε so G(x,t) = x t for x t G xx (x,t)dx = G x (t + ε) G x (t ε) = G x (t + ε), Gantumur Math 319 Lecture 3 Jan 10 9 / 10

10 Poisson equation in 1D Consider now u xx = f on the interval (0,1) subject to the boundary conditions u(0) = α and u(1) = β For α = 1, β = 0, and f = 0, the solution is u(x) = 1 x For α = 0, β = 1, and f = 0, the solution is u(x) = x For α = 0, β = 0, and f (x) = δ(x t), let u(x) = G(x,t) be the solution So we anticipate u(x) = α(1 x) + βx G(x, t)f (t)dt G(x,t) is linear in x except possibly at x = t In the vicinity of x = t: 1 = t+ε t ε G xx (x,t)dx = G x (t + ε) G x (t ε) So G(x,t) = kx for x t, and G(x,t) = (k + 1)x t for x t From G(1,t) = 0 we get k = t 1 The final result is { (t 1)x for x t G(x,t) = t(x 1) for x t Gantumur Math 319 Lecture 3 Jan / 10