To do this we select the simplest ode solution for

Unformatted text preview: E solution for which the numerical method does not produce an exact solution. In the case of the leap frog method, this would be any cubic polynomial, and we select y(t) = t3 . Substitution into the two expressions for the local discretization error yields t3 ; (tn ; 2h)3 ; 3(t ; h)2 = 6Ch2 : n n 2h The location of the point n is irrelevant for this exact solution and we nd C = 1=3. Thus, 2 (5.1.4b) = h y000( n) n 2 (tn;2 tn ): n 3 Problems 1. Using the method of undetermined coe cients, we can make some general observations regarding the coe cients of (5.1.2). Show, for example, that k X i=0 4 i =0 (5.1.5a) if (5.1.2) is exact when y(t) = 1. Additionally, if (5.1.2) is exact when y(t) = tq , q = 1 2 : : : , show k X i=1 iq i+q k X i=0 iq;1 i = 0 q = 1 2 ::: : (5.1.5b) These are order conditions for the linear multistep method. 5.2 Newton Divided-Di erence Polynomials Speci c multistep formulas will be derived by approximating y(t) or f (t y) by interpolating polynomials and, respectively, di erentiating or integrating these polynomials. We've already seen examples where one-step methods were constructed by using interpolating polynomials with collocation. Thus, in this section, we'll review polynomial interpolation in an abstract setting that is removed from our from our primary task of solving di erential equations. The interpolation problem consists of nding a polynomial Pk (t) of degree k that interpolates a function f (t) at k + 1 distinct points t0 t1 : : : tk , i.e., f (tj ) = Pk (tj ) j = 0 1 : : : k: (5.2.1) In practice, we express the polynomial in a convenient basis for the application. The obvious basis of powers of monomial terms tk Pk (t) = a0 + a1 t + a2 t + : : : + ak = 2 k X i=0 leads to the linear algebraic system f (t0) = a0 + a1 t0 + a2 t2 + : : : + ak tk 0 0 f (t1) = a0 + a1 t1 + a2 t2 + : : : + ak tk 1 1 f (tk ) = a0 + a1tk + a2 t2 + : : : + ak tk k k 5 ai tk (5.2.2) and is rarely convenient. The Lagrange basis k Y t ; tj = Li (t) = j =1 j 6=i ti ; tj (t ;...
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