Download e-book for iPad: A Course in Ordinary and Partial Differential Equations by zalman rubinstein

By zalman rubinstein

ISBN-10: 1483230988

ISBN-13: 9781483230986

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Example text

3. EXISTENCE THEOREMS 45 Hence deduce that φη£χ) '—^—-f{x,y) ε <2· Ni such that ε 10nk(*) - N2. Finally show | (x + Δχ) — φ(χ) - / ( * , j D < ε. Ax Give a similar argument for Ax < 0 11. (The abstract approach to existence and uniqueness: Contraction principle) Consider a vector space B. Suppose each element fe B has a norm ||/|| as in Section 3. Furthermore, suppose that each Cauchy sequence {fn},fne B in the norm converges to an element of B.

1 ctyfk\x) = 0 for all k, 0 < k < n - 1 (by yi°\x) we mean yi(x)). It follows that the columns of the Wronskian determinant are linearly dependent, hence the determinant equals zero. On the other hand it is easy to show that PROOF: w(x) = K exp - j a&) dt\9 where A' is a constant. (35) 56 ORDINARY DIFFERENTIAL EQUATIONS Indeed Vi w'(x) y2 yn yi yn' = (n-2) v(n-2) /i yi v(n) Jl ... v(n) /2 ... ) Equation (35) follows immediately. Now if w(x0) = 0, then w(x) = 0 for x e [a, b]. Consider the function n y(x)= Σciyi(χ)> i= 1 where the ct are chosen so as to satisfy the conditions Ìciy(ik)(xo) = 0, fc = 0, 1 w - 1.

30) was satisfied for y' = 0. If, however, Eq. (30) is satisfied for y = y(x) identically in y\ this will be a singular solution. As an example take F = y*-2y'2 + 1- / = 0, 3 Fy. = 4y' - Ay' = 0. Eliminating y we have yV-l) = 0; y = ± 1 satisfies F = Fy> = 0 since Fy. = 0 for y' = 0 or y' = ± 1. On the other hand, y = 0 is not a singular solution. The Clairot equation F = y — xy' — y'2 = 0, Fy= -x-2y' =0 or y+ has a singular locus y+ x2 x2 T"T =0 T = 0· This is also a singular solution since in this case x2 ---*y'-y'2 x2 = o, - x - iy = o 52 ORDINARY DIFFERENTIAL EQUATIONS is an identity in x and y'.

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A Course in Ordinary and Partial Differential Equations by zalman rubinstein

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