Reductions in Differential Equations

    Reduction is the most crucial and effective step toward solving a DE. For reduction, algebraic methods are most effective.
  1. Riccati Equation. The substitution v = uN/u reduces the problem of solving [Bir, p.34, (8)] to the integration of a first-order quadratic DE and a quadrature (see [Bir, p.44, l.9-l.23].

  2. Strategies for reduction.
    1. Eliminating variables: Differentiation [Inc, p.63, (20.1)(20.2)]].
    2. Reducing the order of a DE: (i). Integration (Choose the easier integral from two options by integrating by parts, see [Inc, p.67, l.15 & l.17]): When a particular solution u1 of a DE L(u)=0 of order n is known, we may reduce the DE to a DE L1(v)=0 of order n-1 by letting u = u1V, where VN= v [Inc, p.103]. (ii). The DE [Inc, p.63, (20.1)] may be nonlinear, but the DE in [Inc, p.63, l.-6] is linear. (iii). Integrating factors [Inc, p.94, l.8].
    3. Reducing the order of a linear system: (i). Suppose some solutions are known [Har, p.49, chap.4, '3]. (i)N. Suppose some first integrals are known: Use the independence of first integrals to introduce new coordinates [Pon, p.185, (D)]. Solve p1 and then project the vortex line of pdq-Hdt from the extended phase space {(p,q,t)} onto the phase space {(p,q)}[Arn2, p.243, l.!13-l.!9].
    4. Reducing degree: (i). Factorization [Inc, p.60, Example 18.2]. (ii). Homogeneous in x & y: Substitute y = vx [Inc, '23].(iii). Homogeneous in y and its derivatives: Divide throughout by y m [Inc, p.87, l.8].
    5. Reducing dimension by one-parameter group of symmetries [Arn1, p.79, Problem 1]. Even if [Arn1, p.79, Theorem] looks quite algebraic, we can not say that its interpretation is complete unless its geometric meaning is fully revealed [Arn1, p.78, Lemma].

  3. Duhamel's principle:
    1. This method reduces an inhomogeneous DE to a succession of homogeneous DE's.
      Inhomogeneous ODE: Variation of parameters [Bir, p.48, Theorem 8].
      Homogeneous wave equation [Joh, p.129, (1.14)]
      Inhomogeneous wave equation [Joh, p.136, (1.40)]
      Inhomogeneous n-dim hyperbolic equation with constant coefficients [Joh, p.144, l.9].
    2. Reduction of mixed problems for the diffusion equation:
      1. {time-dependent [Sne, p.284, (14)] {time-independent [θ(x, 0, tN)=0, θ(0, t, tN)=g(tN)]}.

      2. {inhomogeneous, time-dependent [Sne, p.279, (A)]} {inhomogeneous, time-independent [Sne, p.279, (B)]} (see [Sne, p.279, Theorem 1 & its heuristic proof])
        {homogeneous, time-independent [Sne, p.280, (B1, B2, B3)]} (see [Sne, p.280, (4)]).

  4. We use reflection to reduce the mixed problem [Sne, p.283, (8)] to the initial value problem [Sne, p.285, (6)]. The mixed problem in [Joh, p.220] becomes [Sne, p.283, (8)] if we extend f from 0<x<L to x>0 first.

  5. Applications.
    1. The construction of a local Lie subgroup by its corresponding subalgebra [Po3, p.417, l.!4-l.!3].

  6. Reduction avoids redundant work.
        If we reduce the problem we are dealing with to one that we have already solved, then it is unnecessary to prove the existence of a solution using the same procedure.
    Example. [Cou, vol. 1, p.126, l.13].

  7. Reduction to a similar problem that we have solved before provides a means of solving the original problem through parallelism.
    Example 1. [Cou, vol. 1, p.124, l.6]
        Similar procedures: Compare the method in [Cou, vol. 1, p.128, l.6-l.14] with that in [Cou, vol. 1, p.23, chap.I, ' 3.1].
    Remark 1. The interpretation must be consistent.
        K(x,x) in [Cou, vol. 1, p.24, l.2] Jn(j ,j ) in [Cou, p.124, (28)]. jn in [Cou, p.124, l.-7] can be considered a vector (xi)(1 i qn) with the basis (wi)(1 i qn).
    Remark 2. The result obtained in quadratic form must be translated back to the integral form.
        After we find (xi)(1 i qn) [Cou, p.124, l.17] satisfying [Cou, vol. 1, p.124, (29)] we must translate it back to the original problem and interpret (xi)(1 i qn) as jn satisfying [Cou, vol. 1, p.124, (30)].
    Example 2.
        The proof of [Cou, vol. 1, p.133, l.- 4-p.134, l.2] is similar to that of [Cou, vol. 1, p.34, l.1-l.2]. The latter proof is based on [Cou, vol. 1, p.32, l.1-l.3] and the former proof is based on [Cou, vol. 1, p.133, l.-15-l.-10]. [Cou, vol. 1, p.34, l.1-l.2] is intuitive while [Cou, vol. 1, p.133, l.-4-p.134, l.2] is not. Consequently, for a complicated case, reduction is an indispensable step for clarification.

  8. Reducing a repeated integration to a single integration [Cou2, vol. 2, p.78, l.-12].

  9. When proving that two equations are equivalent, we use the simpler equation to prove the complicated one.
        [Guo, p.318, l.-1] and [Guo, p.319, (3)] are equivalent. Since the former equation is simpler than the latter one, we use the former one as the basis for cancellation in proving the latter equation. The simpler the pattern of cancellation is, the more easily we can recognize the pattern. Consequently, we begin with expressing dy/dz in terms of the basis dPk,m/dV.

  10. For a complicated calculation, it is important for the author to specify his strategy. To prove [Guo, p.314, (5)], Guo tells his readers to eliminate Mk, -m. Then we have the problem about which multiple to choose for each of the two equations in order to facilitate calculation. This is because Guo only says that there is a cancellation available in the middle of the calculation, but he fails to specify the method for cancellation. If he would have said, "The term Mk, -m will be cancelled if we substitute [Guo, p.313, (1)] and [Guo, p.314, (4)] into the r.h.s. of [Guo, p.314, (5)].", it would save his readers a lot of time.

  11. Suppose we are given a formula that has many terms on the left-hand side but few terms on the right-hand side. Then there are many cancellations involved in this formula. We want to know how many calculations we have saved each time we use this formula. This reduction rate must be calculated systematically. Cancellations involving plus and minus terms are classified as (+/-) cancellations. Cancellations due to the common factors of the numerator and denominator are classified as (/) cancellations.
    Example 1. (a+u)(b-c)+(b+u)(c-a)+(c+u)(a-b) = 0.
        There are 12 terms on the left-hand side. However, these (+/-)-terms cancel each other, so there are no terms left on the right-hand side.
    Example 2 [Wea1, p.125, (5)(i)].
    Step 1. 4x12 = [a2(a+v)2(a-b)(a-c)]/[(a-b)2(a-c)2a(a+u)(a+v)] = a(a+v)/[(a-b)(a-c)(a+u)].
    Consider the second equality. (/) cancellations reduce the number of terms in the numerator from 16 to 2 and reduce the number of terms in the denominator from 64 to 8. (/) cancellations reduce the number of (/) operations from 14 to 6. A similar argument applies for 4y12 and 4z12. In total, Step 1 reduces the number of (+/-) operations from 3(16+64)=240 to 3(2+8)=30 and the number of (/) operations from 42 to 18.
    Step 2. Consider x12+y12+z12. To make x12, y12 and z12 have the same denominator requires 13 ()-operations.
    (b-c)a(a+v)(b+u)(c+u)+(c-a)b(b+v)(a+u)(c+u)+(a-b)c(c+v)(a+u)(b+u)=(a-b)(a-c)(b-c)u(u-v). The factorization on the right-hand side requires 4 ()-operations. Cancellations reduce the number of terms in the numerator from 48 to 2 and the number of terms in the denominator from 64 to 8. (/) cancellations reduce the number of (/) operations from 13 to 7.

  12. A new idea may provide a more direct and effective solution than the machine of formalism.
        If we interpret "mean life" as "the expected time until death", then we can easily derive the formula in [Cot, p.11, l.12]. This formula can also be derived from the machine of formalism through a more complicated method [Lin76, p.178, l.-11].

  13. If En[f] = O(n-(k+a)), where k is a nonnegative integer and 0<a<1, then f has a continuous k-th derivative belonging to La [Zyg, vol.1, p.120, Remark (a)]. One may ask what advantages we have by using this test rather than directly using the definition of La. I would say it divides the task into countable small tasks. That is, this device should theoretically reduce the complexity of the entire task.

  14. Merging two into one
        In the proof of [Sak, p.290, (1.14)(c)], Saks divides his proof into two similar parts [Sak, p.290, l.-9-l.-2; l.-2-l.-1] by considering the quotient pn/p'n in [Sak, p.290, (1.16)]. Rudin merges the two parts into one by considering the difference qM - pN in [Ru2, p.322, (5)] and then factoring out the common factor of the two terms.