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Math2221 Higher Theory and Applications of Differential Equations

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Math2221 Higher Theory and Applications of Differential Equations Math2221 Higher Theory and Applications of Differential Equations School of Maths and Stats, UNSW Session 2, 2016 Version: January 20, 2017 Part I Linear ODEs Introduction In first year, you studied second-order linear ODEs with constant coefficients. We will see that the techniques you learned can be extended to handle higher-order linear ODEs with variable coefficients. A key idea, used repeatedly throughout the remainder of the course, is linear superposition. Outline Linear differential operators Differential operators with constant coefficients Wronskians and linear independence Methods for inhomogeneous equations Solution via power series Singular ODEs Bessel and Legendre equations Linear differential operators In linear algebra, you have seen the advantages arising from the compact notation Ax = b for a system of linear equations. A similarly compact notation is of great value when dealing with a linear differential equation Lu = f: Here, L is an operator (or transformation) that acts on a function u to create a new function Lu. Notation Given coefficients a0(x), a1(x),..., am(x) we define the linear differential operator L of order m, m j Lu(x) = aj(x)D u(x) j=0 (1) X m m-1 = amD u + am-1D u + ··· + a0u; where Dju = dju=dxj (with D0u = u). We refer to am as the leading coefficient of L. For simplicity, we assume that each aj(x) is a smooth function of x. The ODE Lu = f is said to be singular with respect to an interval [a; b] if the leading coefficient am(x) vanishes for any x 2 [a; b]. (Also say L is singular on [a; b].) Linearity An operator of the form (1) is indeed linear: for any constants c1 and c2 and any (m-times differentiable) functions u1 and u2, L(c1u1 + c2u2) = c1Lu1 + c2Lu2: Hence, the set of solutions to the homogeneous equation Lu = 0 forms a vector space. Example Lu = (x - 3)u000 -(1 + cos x)u0 + 6u is a linear differential operator of order 3, with leading coefficient x - 3. Thus, L is singular on [1; 4], but not singular on [0; 2]. Example N(u) = u00 + u2u0 - u is a nonlinear differential operator of order 2. Linear initial-value problem Consider a general mth-order linear differential operator m j Lu = aj(x)D u: j=0 X Given f(x) and m initial values ν0, ν1,..., νm-1 we seek u = u(x) satisfying Lu = f on [a; b], (2) with 0 (m-1) u(a) = ν0; u (a) = ν1; : : : ; u (a) = νm-1: (3) Theorem Assume that the ODE Lu = f is not singular with respect to [a; b], and that f is continuous on [a; b]. Then the IVP (2) and (3) has a unique solution. Homogeneous problem Theorem Assume that the linear, mth-order differential operator L is not singular on [a; b]. Then the set of all solutions to the homogeneous equation Lu = 0 on [a; b] is a vector space of dimension m. Proof. Let V = f u : Lu = 0 on [a; b] g and define the linear m transformation Θ : V ! R by Θu = [u(a); u0(a); : : : ; u(m-1)(a)]T : Uniqueness of solutions means that Θ is one-one, and existence means that Θ is onto. Hence, Θ is an isomorphism, and therefore the vector space V has dimension m. General solution If fu1; u2; : : : ; umg is any basis for the solution space of Lu = 0, then every solution can be written in a unique way as u(x) = c1u1(x) + c2u2(x) + ··· + cmum(x) for a 6 x 6 b. (4) We refer to (4) as the general solution of the homogeneous equation Lu = 0 on [a; b]. Linear superposition refers to this technique of constructing a new solution out of a linear combination of old ones. Of course, this trick works only because L is linear. Example The general solution to u00 - u0 - 2u = 0 is -x 2x u(x) = c1e + c2e : Inhomogeneous problem Consider the inhomogeneous equation Lu = f on [a; b], and fix a particular solution uP. For any solution u, the difference u - uP is a solution of the homogeneous equation because L(u - uP) = Lu - LuP = f - f = 0 on [a; b]. Hence, u(x)- uP(x) = c1u1(x) + ··· + cmum(x) for some constants c1,..., cm, and so u(x) = uP(x) + c1u1(x) + ··· + cmum(x); a 6 x 6 b; uH(x) is the general solution| of the inhomogeneous{z equation} Lu = f. Reduction of order If we know (somehow) one solution u = u1(x) to a second-order, linear, homogeneous ODE u00 + p(x)u0 + q(x)u = 0; then, to find a second (linearly independent) solution, substitute u = v(x)u1(x) into the ODE and rearrange to obtain 00 0 00 0 0 (u1 + pu1 + qu1)v + u1v + (2u1 + pu1)v = 0: =0 This is just| a first{z-order,} linear ODE for the derivative of the unknown factor v(x): put w = v0 then 0 0 u1w + (2u1 + pu1)w = 0: Reduction of order (continued) Writing the ODE for w in the standard form 0 0 -1 w + (2u1u1 + p)w = 0; we seek an integrating factor 0 -1 2 I(x) = exp (2u1u1 + p) dx = u1 exp( p dx), so that d R R (Iw) = Iw0 + I0w = Iw0 + (2u0 u-1 + p)w = 0: dx 1 1 Then Iw = C for some constant C, and so C v = dx: I(x) Z Example 00 0 3x For the ODE u - 6u + 9u = 0, take u1 = e and find v. Differential operators with constant coefficients If L has constant coefficients, then the problem of solving Lu = 0 reduces to that of factorizing the polynomial having the same coefficients. Some complications occur if the polynomial has any repeated roots. Characteristic polynomial Suppose that aj is constant for 0 6 j 6 m, with am 6= 0. We define the associated polynomial of degree m, m j m m-1 p(z) = ajz = amz + am-1z + ··· + a1z + a0; j=0 X so that, formally, L = p(D). Since Djeλx = λjeλx we have λx m m-1 λx λx p(D)e = amλ +am-1λ +···+a1λ+a0 e = p(λ)e ; and so p(D)eλx = 0 () p(λ) = 0: Factorization By the fundamental theorem of algebra, k1 k2 kr p(z) = am(z - λ1) (z - λ2) ··· (z - λr) where λ1, λ2,..., λr are the distinct roots of p, with corresponding multiplicities k1, k2,..., kr satisfying k1 + k2 + ··· + kr = m: Lemma j λx j-1 λx (D - λ)x e = jx e for j > 0. Lemma (D - λ)kxjeλx = 0 for j = 0, 1,..., k - 1. Proof An elementary calculation gives (D - λ)xjeλx = jxj-1eλx + xjλeλx - λxjeλx = jxj-1eλx; as claimed, and then (D - λ)2xjeλx = (D - λ)jxj-1eλx = j(j - 1)xj-2eλx; . (D - λ)jxjeλx = j!eλx; (D - λ)j+1xjeλx = j!(D - λ)eλx = 0; k j λx so (D - λ) x e = 0 for all k > j + 1, that is, for j 6 k - 1. Basic solutions Lemma If (z - λ)k is a factor of p(z) then the function u(x) = xjeλx is a solution of Lu = 0 for 0 6 j 6 k - 1. Proof. Write p(z) = (z - λ)kq(z), so that q(z) is a polynomial of degree m - k. It follows that p(D) = (D - λ)kq(D) = q(D)(D - λ)k and so for 0 6 j 6 k - 1, p(D)xjeλx = q(D)(D - λ)kxjeλx = q(D)0 = 0: General solution Theorem For the constant-coefficient case, the general solution of the homogeneous equation Lu = 0 is r kq-1 l λqx u(x) = cqlx e ; q=1 l=0 X X where the cql are arbitrary constants. k Since (z - λq) q is a factor of p(z), r kq-1 r kl-1 l λqx Lu = cqlLx e = cql × 0 = 0: q=1 l=0 q=1 l=0 X X X X Linear independence is shown in the Technical Proofs. Distinct real roots Example From the factorization D4 - 2D3 - 11D2 + 12D =( D + 3)D(D - 1)(D - 4) we see that the general solution of u0000 - 2u000 - 11u00 + 12u0 = 0 is -3x x 4x u = c1e + c2 + c3e + c4e : Repeated real root Example From the factorization D4 + 6D3 + 9D2 - 4D - 12 =( D - 1)(D + 2)2(D + 3) we see that the general solution of u0000 + 6u000 + 9u00 - 4u0 - 12u = 0 is x -2x -2x -3x u = c1e + c2e + c3xe + c4e : Complex root Example From the factorization D3 - 7D2 + 17D - 15 = (D2 - 4D + 5)(D - 3) =( D - 2 - i)(D - 2 + i)(D - 3) we see that the general solution of u000 - 7u00 + 17u0 - 15u = 0 is (2+i)x (2-i)x 3x u(x) = c1e + c2e + c3e 2x 2x 3x = c4e cos x + c5e sin x + c3e : Simple oscillator Second-order ODEs arise naturally in classical mechanics. Consider a particle of mass m that moves along the x-axis with velocity v = x_ = dx=dt under the influence of an external applied force = f(t); a frictional resistance force = -r(v)v; a restoring force = -k(x)x: Newton's second law, d2x m x¨ = m = f(t)- r(v)v - k(x)x; dt2 leads to a second-order differential equation m x¨ + r(x_) x_ + k(x)x = f(t): Simplest case is when r(v) = r0 > 0 and k(x) = k0 > 0 are constant, giving a linear ODE with constant coefficients: m x¨ + r0 x_ + k0x = f(t): Typically interested in the case when the applied force is periodic with frequency !; for example, f(t) = F sin !t: The general solution x(t) = xH(t) + xP(t).
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