II. FINITE DIFFERENCE METHOD 1 Difference formulae

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y 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 y’’(x)+(3*x−2)*y’(x)+(x^2+1)*y(x)=cos(x/2), y(−6*pi)=0, y(6*pi)=0 −20 −10 0 10 20 Figure 5: The difference method applied to the the convection-diffusion problem ¡y ′′ (x)+(3x¡2)y ′ (x)+(x 2 +1)y(x) = cos(x/2) with Dirichlet conditions y(¡6π) = 0 , y(6π) = 0. ¸ ǫz 2 1 + (p 2z1 ¡ p 2z2) 2 + . . . + ( p 2zn−2 ¡ p 2zn−1) 2 + ǫz 2 n−1 , which is > 0 for any nonzero z 2 R n−1 . Hence we have the following result. Thm. II.3.1 (Existence and uniqueness of solution of the discretized problem) The discretized version of the convection-diffusion problem above described admits a unique solution if h = (b ¡ a)/n is chosen sufficiently small with respect to the convection term p(x), as in (10). Ex.II.3B (Alternative discretizations are possible) The restriction (10) can be eliminated by discretizing in another way the term p(x)y ′ . More precisely, we set ¡ ¯yi+1 ¡ 2¯yi + ¯yi−1 h 2 ¡ ¯yi+1 ¡ 2¯yi + ¯yi−1 h 2 + p(xi) ¯yi+1 ¡ ¯yi h + p(xi) ¯yi ¡ ¯yi−1 h x 14 + q(xi)¯yi = f(xi), if p(xi) · 0 + q(xi)¯yi = f(xi), if p(xi) > 0
This scheme, which is known as upwind difference method, , may be written in the form ¡ ¯yi+1 ¡ 2¯yi + ¯yi−1 h 2 ¡ (jpij ¡ pi)¯yi+1 ¡ 2jpij¯yi + (jpij + pi)¯yi−1 2h + qi¯yi = fi The upwind scheme is less accurate than the preceding one, but it is solvable without restrictions on the choice of h. Indeed, the matrix of the corresponding linear system turns out to be with diagonal predominance. II.3.C NEWTON’S METHOD In the scalar case with f 2 C (1) , f ′ (x) 6= 0, the solution of the equation f(x) = 0 is approximated by a sequence xn starting from a (sufficiently near) point x◦. The idea, at the k¡th step, is to replace the curve f by the tangent line at xk, i.e. to solve the linear equation 0 = f(xk)+f ′ (xk)(x¡xk): xk+1 = xk ¡ [f ′ (xk)] −1 f(xk). In several dimensions suppose that f is regular with a non singular Jacobian matrix : f : R m ! R m , f 2 C 1 , J non singular. Let us take an initial point x◦ sufficiently near the true solution of the equation f(x) = 0. At the k¡th step, the idea is to replace the function y = f(x) by its Taylor’s formula at the first order, i.e. to solve the linear system 0 = f(xk)+J(xk)(x¡ xk), where J is the Jacobian: xk+1 = xk ¡ [J(xk)] −1 f(xk). Ex.II.3D (The difference method applied to a nonlinear problem) y ′′ (x) = 1 2 (1 + x + y)3 , x 2 (0, 1) y(0) = y(1) = 0 Setting f(x, y) := 1 2 (1 + x + y)3 , let’s notice that ∂f ∂y = 3 2 (1 + x + y)2 ¸ 0 Then it can be shown that such a boundary value problem admits a unique solution. Let us discretize the interval (0, 1) by the points xi = ih, with 15
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II. FINITE DIFFERENCE METHOD 1 Difference formulae

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