Bourbaki

Proposition 10

Tag 08D9, permanent. Cite it as [Bourbaki, Tag 08D9].

Functions of a Real Variable, chapter IV, § 2. LINEAR DIFFERENTIAL EQUATIONS

Let g and h be two relatively prime polynomials such that f=ghf = gh. The subspace of solutions of (42) is the direct sum of the subspaces of solutions of the two equations

g(D)x=0,h(D)x=0. g(D)x = 0, \qquad h(D)x = 0.

Now, by the Bezout identity (Alg., VII. 2, th. 1) there exist two polynomials p(D)p(D) and q(D)q(D) such that p(D)g(D)+q(D)h(D)=1p(D)g(D) + q(D)h(D) = 1. For every solution xx of (42) one can write x=y+zx = y + z, where y=p(D)g(D)xy = p(D)g(D)x and z=q(D)h(D)xz = q(D)h(D)x; and then h(D)y=p(D)(f(D)x)=0h(D)y = p(D)(f(D)x) = 0 and g(D)z=q(D)(f(D)x)=0g(D)z = q(D)(f(D)x) = 0. On the other hand, if both g(D)x=0g(D)x = 0 and h(D)x=0h(D)x = 0, one deduces that

x=p(D)(g(D)x)+q(D)(h(D)x)=0, x = p(D)(g(D)x) + q(D)(h(D)x) = 0,

which completes the proof.

With the preceding notation one can write

f(D)=∏j=1q(D−rj)nj f(D) = \prod_{j=1}^q (D - r_j)^{n_j}

and prop. 10, by induction on qq, shows that the subspace of solutions of (42) is the direct sum of the subspaces of solutions of the qq equations

(D−rj)njx=0(1≤j≤q).(43) (D - r_j)^{n_j} x = 0 \qquad (1 \leq j \leq q). \tag{43}

Now, for every complex number rr one has

D(ertx)=ert(D+r)x(44) D(e^{rt} x) = e^{rt}(D + r)x \tag{44}

so (43) is equivalent to

Dnj(e−rjtx)=0 D^{n_j}(e^{-r_j t} x) = 0

and thus has as solutions the functions erjtpj(t)e^{r_j t} p_j(t), where pjp_j runs through the set of polynomials of degree ≤nj−1\leq n_j - 1; thus one recovers prop. 9 of IV, p. 194.

Assuming that the homogeneous equation (41) has been solved (that is, assuming that the characteristic roots have been found), one knows that the method of variation of parameters allows one to find the solutions of the inhomogeneous equation

Dnx−a1Dn−1x−…−an−1Dx−anx=b(t)(45) D^n x - a_1 D^{n-1} x - \ldots - a_{n-1} Dx - a_n x = b(t) \tag{45}

where b(t)b(t) is an arbitrary regulated function (IV, p. 182); we note that if b(t)b(t) is indefinitely differentiable on an interval J then all the integrals of (45) are indefinitely differentiable on J. In the particular case b(t)=eαtp(t)b(t) = e^{\alpha t} p(t), where pp is a polynomial (with complex coefficients) and α\alpha is an arbitrary complex number, one obtains an integral of (45) more simply by the following method. Put x=eαtyx = e^{\alpha t} y; the equation

f(D)x=eαtp(t) f(D)x = e^{\alpha t} p(t)

can, by (44), be written

f(α+D)y=p(t) f(\alpha + D)y = p(t)

or, by Taylor’s formula applied to the polynomial f(D)f(D),

f(n)(α)n!Dny+f(n−1)(α)(n−1)!Dn−1y+⋯+f′(α)1!Dy+f(α)y=p(t).(46) \frac{f^{(n)}(\alpha)}{n!} D^n y + \frac{f^{(n-1)}(\alpha)}{(n-1)!} D^{n-1} y + \cdots + \frac{f'(\alpha)}{1!} Dy + f(\alpha) y = p(t). \tag{46}

Let mm be the degree of the polynomial p(t)=∑k=0mλktm−kp(t) = \sum_{k=0}^m \lambda_k t^{m-k}; if f(α)≠0f(\alpha) \neq 0 (that is, if α\alpha is not a characteristic root), there exists one and only one polynomial u(t)=∑k=0mcktm−ku(t) = \sum_{k=0}^m c_k t^{m-k} of degree mm which is a solution of (46), for the coefficients ckc_k are determined by the system of linear equations

f(α)ck+(m−k+11)f′(α)ck−1+(m−k+22)f′′(α)ck−2+⋯+(mk)f(k)(α)c0=λk(0≤k≤m) f(\alpha)c_k + \binom{m-k+1}{1} f'(\alpha)c_{k-1} + \binom{m-k+2}{2} f''(\alpha)c_{k-2} + \cdots + \binom{m}{k} f^{(k)}(\alpha)c_0 = \lambda_k \qquad (0 \leq k \leq m)

which clearly admits one and only one solution. If, on the other hand, α\alpha is a characteristic root, and hh is its multiplicity, the preceding calculation shows that there is one and only one polynomial of degree mm such that every solution of Dhy=v(t)D^h y = v(t) is an integral; in other words, every polynomial solution of (46) is then of degree m+hm + h ("resonance").

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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