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MATH 374 Lecture 14

MATH 374 Lecture 14. Properties of Differential Operators. 4.9: Some Properties of Differential Operators. Theorem 4.7: Let f(D) = a 0 D n + a 1 D n-1 + ... + a n-1 D + a n be a polynomial in D. Then (a) For any constant m, f(D)e mx = f(m)e mx .

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MATH 374 Lecture 14

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  1. MATH 374 Lecture 14 Properties of Differential Operators

  2. 4.9: Some Properties of Differential Operators • Theorem 4.7: Let f(D) = a0Dn + a1Dn-1 + ... + an-1D + an be a polynomial in D. Then (a) For any constant m, f(D)emx = f(m)emx. (b) For any constant a, f(D-a)[eaxy] = eaxf(D)y. 2

  3. Thm 4.7 (a) For any constant m, f(D)emx = f(m)emx. Proof of Theorem 4.7 (a) • Claim: For any integer k  1, Dk emx = mk emx. • Proof of Claim: Use induction. k = 1: D emx = m emx Induction Hypothesis: Assume the claim is true for k  1, to show it is true for k+1. Dk+1emx = D Dk emx = D(mk emx) = mk+1 emx. Therefore the claim holds! It follows that f(D) emx = (a0Dn + a1Dn-1 + ... + an-1D + an)emx = a0Dnemx + a1Dn-1emx + ... + an-1Demx + anemx = a0mnemx + a1mn-1emx + ... + an-1memx + anemx = f(m)emx 3

  4. Thm 4.7 (b) For any constant a, f(D-a)[eaxy] = eaxf(D)y. Proof of Theorem 4.7 (b) • Claim: For any integer k  1, (D-a)k[eaxy]=eaxDky. Proof of Claim: Use induction. k = 1: (D-a)[eaxy]= aeaxy + eaxDy – aeaxy = eaxDy. Induction Hypothesis: Assume the claim is true for k  1 to show it is true for k+1. (D-a)k+1[eaxy] = (D-a)(D-a)k[eaxy] = (D-a)[eaxDky] = eaxD(Dky) = eax Dk+1y Therefore the claim holds! Think of this as the function to the right of eax. 4

  5. Thm 4.7 (b) For any constant a, f(D-a)[eaxy] = eaxf(D)y. Proof of Theorem 4.7 (b) (continued) • It follows that f(D-a)[eaxy] = (a0(D-a)n + … + an)[eaxy] = a0(D-a)n[eaxy] + … + an[eaxy] = a0eaxDny + … +aneaxy = eax[a0Dny + … +any] = eaxf(D)y.  5

  6. Some Useful Results • Corollary 1: If m is a root of polynomial equation f(m) = a0mn + … + an-1m + an = 0, then f(D)emx = 0. • Proof: From Theorem 4.7 (a), f(D)emx = f(m)emx = 0·emx = 0. 6

  7. Some Useful Results • Corollary 2: For n a positive integer, m a constant, and k = 0, 1, 2, … , n-1, (D-m)n[xkemx] = 0. • Proof: Apply Theorem 4.7 (b) with y = xk and f(D) = Dn. It follows that (D-m)n[emxxk] = emxDn(xk) = 0, for k = 0, 1, 2, … , n-1.  7

  8. Example 1: Solve (D2-6D+5)y = 0. (1) • Solution: Here, f(D) = D2 – 6D + 5. It follows from Corollary 1 that if m is a root of f(m) = m2 – 6m + 5 = 0, then y = emx is a solution of (1). Since m2-6m+5 = (m-1)(m-5), roots of f(m) = 0 are m = 1 and m = 5. Hence, y1 = ex and y2 = e5x are solutions of (1). Because y1 and y2 are linearly independent (check), the general solution to (1) is: y = c1ex + c2e5x, where c1 and c2 are arbitrary constants! (See Theorem 4.4). 8

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