I. Recurrence formula

For operators A_n+1(t) defined in a Banach space X, the recurrence formula is introduced by Y. Iwata ^[1]:

(1)  A_n+1 = (∂_t + A₁) A_n,      n = 1, 2, 3 ….

It shows a relation of operators or a relation of infinitesimal generators in X. In case of n = 1, this formula has the same form as the Miura transform, so that the same form as the Riccati’s differential equation. Essentially speaking, if non-trivial u satisfies the n-th order differential equation

(2)  ∂_tⁿ u = A_n u,

then the solution u satisfies one-order higher equation

(3)  ∂_tⁿ⁺¹ u = A_n+1 u

at the same time, where for a given infinitesimal generator A_n, another infinitesimal generator A_n+1 is determined by Eq. (1) automatically. Here u and ∂_tⁿ u are assumed to commute for any natural number n, then A_n is formally represented by

(4)  A_n = u^-1 ∂_tⁿ u

for a certain non-trivial solution u. This representation generally works for representing the infinitesimal generator, as

(5)  A_n u = u^-1 (∂_tⁿ u) u = u^-1 u (∂_tⁿ u) = ∂_tⁿ u

is true. Note that, once the representation (4) is accepted, An should be treated as consisting of the two components u^-1 and ∂_tⁿ that are generally dependent on t.

2. Solution

2.1. Formal Argument

By solving Eq. (1), a sequence of operators A₁, A₂, … A_k … is obtained. The general term of a sequence is represented by

(6)  A_k = (∂_t + A₁) A_k-1 = (∂_t + A₁)^k-1 A₁,

where the operator "∂_t + A₁" is taking from the left-hand side, and ∂_t and A_k depending generally on t are not necessarily assumed to commute. This shows a sequence of formal solutions, and the interplay between ∂_t and A_k leads to the disappearance of term as seen in following concrete examples. It means that a set of infinitesimal generators exists at the least, and therefore the existence of a set of differential equations sharing one common solution follows. Consequently, beginning with the first order equation ∂_t u = A₁ u, the concept of general form of n-th order differential equation

(7)  ∂_tⁿu = (∂_t + A₁)^n-1 A₁u

is obtained.

2.2. Proof

The proof is tersely shown. If the representation (2) is true, the concept of connecting different equations by Eq. (1) is readily proved to be true. For a certain u satisfying the differential equations up to (n-1)th order, by substituting Eq.(4) to Eq. (1), the right-hand side of Eq. (1) is calculated as

(8)  (∂_t + A₁) A_n = (∂_t + A₁) u^-1 ∂_tⁿ u
                         = ∂_t (u^-1 ∂_tⁿ u) + A₁ u^-1 ∂_tⁿ u
                         = - u^-2(∂_t u) ∂_tⁿ u + u^-1 ∂_tⁿ⁺¹ u + u^-2 (∂^t u) ∂_tⁿ u
                         = u^-1 ∂_tⁿ⁺¹ u

which is expected to be equal to An+1. Note again that the commutation assumption is applied.

3. Examples

3.1. Ordinary Differential Equation (finite-dimensional X)

Since X is assumed to be only a Banach spaces, ordinary differential equations in the usual sense can be included by taking as finite-dimensional Euclidean space. For simplicity, let us take one-dimensional case. let X be one-dimensional Euclidean space R and let A₁ = α(t): R → R be a continuous function. To avoid mistakes and misunderstandings, the coefficient α is denoted by α(t) if α is assumed to be t-dependent. The differential equation is

(9)  ∂_t u = α(t) u,

and its general solution is represented by the indefinite integral

(10)  u(t) =C e^∫α(t)dt

for a certain constant C ∈ R, where the indefinite integral can be replaced with the definite integral with its range from any constant t₀ to t. Using the recurrence formula, the infinitesimal generator

(11)  A₂ = (∂_t + α(t)) α(t) = ∂_t α(t) + α(t)²

and the related 2nd order equation

(12)  ∂_t²u = (∂_t α(t)) u + α(t)² u

follows. Let us examine the validity of recurrence formula by substituting Eq.(10) to Eq.(12). The left-hand side of Eq.(12) becomes

(13)  ∂_t²u = ∂_t²(C e^∫α(t)dt )
               = ∂_t (C α(t) e^∫α(t)dt )
               = (∂_t α(t)) C e^∫α(t)dt + α(t)² C e^∫α(t)dt

which is equal to the right hand. Let us further examine by taking n = 3,

(14)  A₃ = (∂_t + α(t)) (∂_t α(t) + α(t)²) = ∂_t²α(t) + 3 α(t) ∂_t α(t) + α(t)³

and

(15)  ∂_t³u = (∂_t²α(t)) u + 3 α(t) ∂_t α(t) u + α(t)³u

follows. Let us examine the validity of recurrence formula by substituting Eq.(10) to Eq.(15). The left-hand side of Eq.(15) becomes

(16)  ∂_t³u = ∂_t³(C e^∫α(t)dt )
               = ∂_t {(∂_t α(t)) C e^∫α(t)dt + α(t)² C e^∫α(t)dt}
               = (∂_t²α(t)) C e^∫α(t)dt + (∂_t α(t)) α(t) C e^∫α(t)dt +2 α(t) (∂_t α(t)) C e^∫α(t)dt + α(t)³ C e^∫α(t)dt
               = (∂_t²α(t)) C e^∫α(t)dt +3 α(t) (∂_t α(t)) C e^∫α(t)dt + α(t)³ C e^∫α(t)dt

which is equal to the right hand. Essentially the same discussion is valid to n > 3.

Next let us take cases in which α(t) is independent of t (i.e. α(t) = 0). In those cases, the differential equation is

(17)  ∂_t u = α u,

and its general solution is represented by the indefinite integral

(18)  u(t) =C e^tα

for a certain constant C ∈ R. Using the recurrence formula, the infinitesimal generator

(19)  A₂ = (∂_t + α) α = α²

and the related 2nd order equation

(20)  ∂_t²u = α² u

follows, and therefore

(21) A_n = αⁿ with ∂_tⁿu = α_n u

follow. In case of n = 2, let us examine the validity of recurrence formula by substituting Eq.(18) to Eq.(20). The left-hand side of Eq.(20) becomes

(22)  ∂_t²u = ∂_t²(C e^tα )
               = ∂t (C α e^tα )
               = α² C e^tα

which is equal to the right hand. Essentially the same discussion is valid to n > 2.

3.2. Partial Differential Equation (infinite-dimensional X)

Let k(t): R →R be a continuous function. Let us take one-dimensional partial differential equation (A₁ = k(t) ∂_x) in infinite dimensional Hilbert space X = L²(-∞, ∞). The differential equation is

(23)  ∂_t u = k(t) ∂_x u

and its solution is represented by the indefinite integral

(24)  u(t,x) = C exp[∫k(t)dt + x]

for a certain constant C ∈ R, where the indefinite integral can be replaced with the definite integral with its range from any constant t₀ to t. Using the recurrence formula, the infinitesimal generator

(25)  A₂ = (∂_t + k(t) ∂_x) k(t) ∂_x = (∂_t k(t)) ∂_x + k(t)² ∂_x²

and the related 2nd order equation is formally written by

(26)  ∂_t²u = (∂_t k(t)) ∂_x u + k(t)² ∂_x²u.

Let us examine the validity of recurrence formula by substituting Eq.(24) to Eq.(26). The left-hand side of Eq.(26) becomes

(27)  ∂_t²u = ∂_t²(C exp[∫k(t)dt + x])
               = ∂_t(C k(t) exp[∫k(t) dt + x] )
               = (∂_t k(t)) C exp[∫k(t)dt + x] ) + k(t)² C exp[∫k(t)dt + x]

which is equal to the right hand, where

(28)  ∂_x u = C exp[∫k(t)dt + x]

is valid. Let us further examine by taking n = 3,

(29)  A₃ = (∂_t + k(t) ∂_x) ((∂_t k(t)) ∂_x + k(t)² ∂_x²)
             = (∂_t² k(t)) ∂_x + 2 k(t) (∂_t k(t)) ∂_x² + k(t) (∂_t k(t)) ∂_x² + k(t)³ ∂_x³
             = (∂_t² k(t)) ∂_x + 3 k(t) (∂_t k(t)) ∂_x² + k(t)³ ∂_x³

and

(30)  ∂_t³u = (∂_t² k(t)) ∂_x u +3 k(t) (∂_t k(t)) ∂_x²u + k(t)³ ∂_x³u

follows. Let us examine the validity of recurrence formula by substituting Eq.(24) to Eq.(3). The left-hand side of Eq.(30) becomes

(31)  ∂_t³u = ∂_t³(C exp[∫k(t)dt + x])
               = ∂_t {(∂_t k(t)) C exp[∫k(t)dt + x] ) + k(t)² C exp[∫k(t)dt + x] }
               =(∂_t²k(t)) C exp[∫k(t)dt + x] )+k(t)(∂_t k(t)) C exp[∫k(t)dt + x]
                 + 2k(t)(∂_t k(t)) C exp[∫k(t)dt + x] + k(t)³ C exp[∫k(t)dt + x]
              =(∂_t²k(t)) C exp[∫k(t)dt + x] ) + 3k(t)(∂_t k(t)) C exp[∫k(t)dt + x] + k(t)³ C exp[∫k(t)dt + x]

which is equal to the right hand. Essentially the same discussion is valid to n > 3.

Next let us take cases in which k(t) is independent of t (i.e. k(t) = k). In those cases, the differential equation is

(32)  ∂_t u = k ∂_x u,

and its general solution is represented by the indefinite integral

(33)  u(t,x) = C exp[kt + x]

for a certain constant C ∈ R. It has a form of d’Alembert’s solution. Using the recurrence formula, the infinitesimal generator

(34)  A₂ = (∂_t + k∂_x) k∂_x = k² ∂_x²

and the related 2nd order equation

(35)  ∂_t²u = k² ∂_x² u

follows, and therefore

(36)  A_n = kⁿ ∂_xⁿ with ∂_tⁿu = k_n ∂_xⁿ u

follow. In case of n = 2, let us examine the validity of recurrence formula by substituting Eq.(33) to Eq.(35). The left-hand side of Eq.(35) becomes

(37)  ∂_t²u = ∂_t²(C exp[kt + x])
               = ∂_t (C k exp[kt + x])
               = k² C exp[kt + x]

which is equal to the right hand. Essentially the same discussion is valid to n > 2.

Abstractly speaking, let U be a solution of 1st order equation ∂_t U = k ∂_x U. For U,

(38)  ∂_t ∂_x = k^-1 ∂_t (U^-1 (∂t U))
                = -k^-1 U^-2 (∂_t U)² + k^-1 U^-1 (∂_t²U)

If U = C exp[kt + x], then

(39)  ∂_t ∂_x = -k^-1 exp[kt + x]^-2 k² exp[kt + x]² + k^-1 exp[kt + x]^-1 (k² exp[kt + x])
                = -k^-1k² + k^-1 k²
                = 0.

It shows the possible disappearance of ∂_t ∂_x term not in Eq.(25) but in Eq.(34), although it is not explicitly seen in the abstract and formal discussion.