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MIT 3.016 Fall 2005 � W.C Carter Lecture 21 c134 Nov. 14 2005 : Lecture 21: Higher-Order Ordinary Differential Equations Reading: Kreyszig Sections: 2.1 (pp:54–70) , 2.2 (pp:72–75) , 2.3 (pp:76–80) § § §Higher-Order Equations: Background For first-order ordinary differential equations (ODEs), F (y�(x), y(x), x), one value y(xo) was needed to specify a particular solution. For second-order equations, two independent values are needed. This is illustrated in the following forward-differencing example. Mathematica r� Example: Lecture-21 A Second-Order Forward Differencing Example Recall the example in Lecture 19 of a first-order differencing scheme: at each iteration the function grew proportionally to its current size. In the limit of very small forward differences, the scheme converged to exponential growth. Now consider a situation in which function’s current rate of growth increases propor- tionally to two terms: its current rate of growth and its size. Change in Value’s Rate of Change + α (the Value) + β (Value’s Rate of Change) = 0 To Calculate a forward differencing scheme for this case, let Δ be the forward- differencing increment. � Fi+2−Fi+1 Fi+2−Fi+1 �� Fi+1 − Fi �Δ − Δ + αFi + β = 0 Δ Δ and then solve for the “next increment” Fi+2 if Fi+1 and Fi are known. Forward Difference FormulaeMIT 3.016 Fall 2005 � W.C Carter Lecture 21 c135 qcqckLinear Differential Equations; Superposition in the Homogeneous Case . . . . . . . . . . . . . . . A linear differential equation is one for which the function and its derivatives are each linear— that is they appear in distinct terms and only to the first power. In the case of a homogeneous linear differential equation, the solutions are superposable. In other words, sums of solutions and their multiples are also solutions. Therefore, a linear heterogeneous ordinary differential equation can be written as a product of general functions of the dependent variable and the derivatives for the n-order linear case: 0 = f0(x) + f1(x) dy dx + f2(x) d2y dx2 + ··· + fn(x) dny dxn = (f0(x), f1(x), f2(x), . . . , fn(x)) · �1, dy dx , d2y dx2 , . . . , dny dxn � (21-1) = �f (x) · D�ny The homogeneous nth-order linear ordinary differential equation is defined by f0(x) = 0 in Eq. 21-1: 0 = f1(x) dy dx + f2(x) d2y dx2 + ··· + fn(x) dny dxn = (0, f1(x), f2(x), ··· , fn(x)) · �1, dy dx , d2y dx2 , . . . , dny dxn � (21-2) = fhom� (x) · D�ny Equation 21-1 can always be multiplied by 1/fn(x) to generate the general form: 0 = F0(x) + F1(x) dy dx + F2(x) d2y dx2 + ··· + dny dxn = (F0(x), F1(x), F2(x), . . . , 1)) ·(1, dy dx , d2y dx2 , . . . , dny dxn ) (21-3) = �F (x) · D�ny For the second-order linear ODE, the heterogeneous form can always be written as: d2y dy+ p(x) + q(x)y = r(x) (21-4)dx2 dx and the homogeneous second-order linear ODE is: d2y dy+ p(x) + q(x)y = 0 (21-5)dx2 dx qcqckBasis Solutions for the homogeneous second-order linear ODE . . . . . . . . . . . . . . . . . . . . . . . Because two values must be specified for each solution to a second order equation—the solution can be broken into two basic parts, each deriving from a different constant. These two independent solutions form a basis pair for any other solution to the homogeneous second-order linear ODE that derives from any other pair of specified values.MIT 3.016 Fall 2005 � W.C Carter Lecture 21 c136 The idea is the following: suppose the solution to Eq. 21-5 is found the particular case of specified parameters y(x = x0) = A0 and y(x = x1) = A1, the solution y(x; A0, A1) can be written as the sum of solutions to two other problems. y(x; A0, A1) = y(x, A0, 0) + y(x, 0, A1) = y1(x) + y2(x) (21-6) where y(x0, A0, 0) = A0 and y(x1, A0, 0) = 0 (21-7) y(x0, 0, A1) = 0 and y( x1, 0, A1) = A1 from these two solutions, any others can be generated. The two arbitrary integration constants can be included in the definition of the general solution: y(x) = C1y1(x) + C1y2(x) (21-8)= (C1, C2) · (y1, y2) Second Order ODEs with Constant Coefficients The most simple case—but one that results from models of many physical phenomena—is that functions in the homogeneous second-order linear ODE (Eq. 21-5) are constants: d2y dy a + b + cy = 0 (21-9)dx2 dxMIT 3.016 Fall 2005 � W.C Carter Lecture 21 c 137 If two independent solutions can be obtained, then any solution can be formed from this basis pair. Surmising solutions seems a sensible strategy, certainly for shrewd solution seekers. Suppose the solution is of the form y(x) = exp(λx) and put it into Eq. 21-9: (aλ2 + bλ + c)e λx = 0 (21-10) which has solutions when and only when the quadratic equation aλ2 + λx + c = 0 has solutions for λ. Because two solutions are needed and because the quadratic equation yields two solutions: −b + √b2 − 4ac λ+ = 2a (21-11) −b −√b2 − 4ac λ = − 2a or by removing the redundant coefficient by diving through by a: λ+ = −β + � ( β )2 − γ 2 2 (21-12) λ = −β � ( β )2 − γ− 2 − 2 where β ≡ b/a and γ ≡ c/a. Therefore, any solution to Eq. 21-9 can be written as λ−x y(x) = C+e λ+x + C−e (21-13) Mathematica r� Example: Lecture-21 Solutions to dx2 d2y + β dy + γy = 0 dx Because the fundamental solution depend on only two parameters β and γ, the be-havior of all solutions can be visualized in the γ-β plane. 1. Insert y(x) = exp(λx) into the ODE y�� + βy� + γy = 0 and solve for a condition on λ that solutions exist (assuming real coefficients γ and β). 2. Plot the condition that the roots are complex in the γ-β plane. This is the region of parameter space that gives oscillatory solutions (because exp(r + ıθ) = exp(r)(cos(x) + ı sin(x))) 3. Plot the conditions that the λ are real—these are the monotonically growing (λ > 0) or shrinking (λ < 0) solutions 4. Plot the conditions that the real part is negative, this is the damped oscillatory region. 5. Plot the conditions that the real part is positive, this is the unbounded growth region. 6. Use the Mathematica r� function Reduce to find the three regions: (λ+ > 0, λ− > 0)—monotonically growing solutions, (λ+ > 0, λ− < 0)—one growing and one decaying solution, (λ+ < 0, λ− < 0)—monotonically decaying solutions.c� 138γγ =β24βRoots are Complex ConjugatesPositive


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