---
tag: R1R6024PCZ
kind: proof
name: Proof of the solutions for repeated roots
msc: [34A30]
proves: 0HRVY67CNV
uses: [0QDXMG46QW, 0A6PGTTP2A, 0H4KNNCHPY, 0GK68PZ6KH, 096V7H13H9, 0EVHJFTG9Z, 03XVP8Q4GY, 0JP0W07Y9A, 0K3MWA7DMC, 0KJVQV3E0F, 0B5Z1EC66J, 0GJM24EXP9, 0MH1WK89PP, 0EX9GFJ0B3, 0RXX2PT1MQ, 0V9A43H7WV, 0M0TKFRX2A, 0P9BNBW2W3, 0E8HNVBCSP]
sources:
  - url: https://en.wikipedia.org/wiki/Linear_differential_equation
---

## Proof

We factor the operator through $(D - r)^m$.

1. Let $D$ be differentiation on the infinitely differentiable functions $\R \to \R$. For a real polynomial $q(t) = \sum_i b_i t^i$, let $q(D) u = \sum_i b_i u^{(i)}$. The equation is $p(D) y = 0$, with $p$ the [[characteristic polynomial|0QDXMG46QW]].
2. For real polynomials $q_1, q_2$, we have $q_1(D)\left(q_2(D) u\right) = (q_1 q_2)(D) u$. Indeed, $D^i$ is linear by [[linearity of differentiation|0M0TKFRX2A]] and $D^i D^j = D^{i+j}$. Expanding both sides gives the same sum.
3. Let $s \in \R$ and $k \ge 0$. By the [[product rule|0V9A43H7WV]], the [[chain rule|0RXX2PT1MQ]] and [[the derivative of the exponential|0EX9GFJ0B3]], $D\left(x^k e^{s x}\right) = k x^{k-1} e^{s x} + s\, x^k e^{s x}$. So $(D - s)\left(x^k e^{s x}\right) = k x^{k-1} e^{s x}$, which is $0$ for $k = 0$.
4. Applying step 3 $j$ times, $(D - s)^j \left(x^k e^{s x}\right)$ is a constant multiple of $x^{k-j} e^{s x}$ for $j \le k$. It is $0$ for $j = k + 1$, and so for each $j > k$.
5. Solutions. Since $r$ has [[multiplicity|03XVP8Q4GY]] $m$, $p(t) = h(t) (t - r)^m$ for a real polynomial $h$. Let $0 \le k \le m - 1$.
6. By steps 2 and 4, $p(D)\left(x^k e^{r x}\right) = h(D)\left((D - r)^m \left(x^k e^{r x}\right)\right) = h(D)\, 0 = 0$. So $x^k e^{r x}$ solves the [[equation with constant coefficients|0A6PGTTP2A]] on $\R$.
7. Lemma. Let $s_1, \dots, s_k$ be distinct reals and $q_1, \dots, q_k$ real polynomials with $\sum_j q_j(x) e^{s_j x} = 0$ for all $x \in \R$. Then each $q_j$ is the zero polynomial. We argue by induction on $k$.
8. For $k = 1$, [[$e^{s_1 x} > 0$|0P9BNBW2W3]], so $q_1(x) = 0$ for every $x \in \R$. A nonzero polynomial [[has finitely many roots|0KJVQV3E0F]], so $q_1 = 0$.
9. Let $k \ge 2$. Multiply by $e^{-s_k x}$ and use the [[exponential of a sum|0E8HNVBCSP]]: $\sum_{j < k} q_j(x) e^{\mu_j x} + q_k(x) = 0$, where $\mu_j = s_j - s_k \ne 0$ are distinct.
10. For a polynomial $q$ and $\mu \ne 0$, step 3 gives $D\left(q(x) e^{\mu x}\right) = \left(q' + \mu q\right)(x)\, e^{\mu x}$. If $q \ne 0$ has degree $d$ and leading coefficient $a$, then $q' + \mu q$ has degree $d$ and leading coefficient $\mu a \ne 0$, by the [[derivative of a polynomial|0MH1WK89PP]].
11. Let $e$ be one more than the degree of $q_k$, or $e = 0$ if $q_k = 0$. Differentiating the relation of step 9 $e$ times kills $q_k$. By step 10, it gives $\sum_{j < k} \tilde q_j(x) e^{\mu_j x} = 0$ with $\tilde q_j \ne 0$ whenever $q_j \ne 0$.
12. By the induction hypothesis, each $\tilde q_j = 0$, so each $q_j = 0$ for $j < k$. Then $q_k(x) e^{s_k x} = 0$, and $q_k = 0$ by step 8.
13. Independence. If $\sum_{k=0}^{m-1} c_k x^k e^{r x} = 0$ on $\R$, the lemma with one exponent gives $\sum_k c_k x^k = 0$ as a polynomial. So each $c_k = 0$, and the $m$ functions are [[linearly independent|0EVHJFTG9Z]].
14. Now assume all roots of $p$ are real. We show $p(t) = \prod_{i=1}^{n} (t - c_i)$ with real $c_i$, by induction on $n$. For $n = 0$ the product is empty.
15. For $n \ge 1$, the [[fundamental theorem of algebra|0JP0W07Y9A]] gives a root $c$ of $p$, real by assumption. By the [[factor theorem|0K3MWA7DMC]] in $\R[t]$, $p = (t - c)\, p_1$ with $p_1$ real and monic of degree $n - 1$.
16. Each complex root $z$ of $p_1$ is a root of $p$, since $p(z) = (z - c) p_1(z)$. So it is real, and the induction hypothesis applies to $p_1$.
17. Let $\rho_1, \dots, \rho_s$ be the distinct values among the $c_i$, with $\rho_j$ occurring $m_j$ times. Then $p(t) = \prod_j (t - \rho_j)^{m_j}$ and $m_1 + \dots + m_s = n$. Each root $z$ of $p$ makes some factor vanish, so the roots are exactly $\rho_1, \dots, \rho_s$.
18. The multiplicity of $\rho_j$ is $m_j$. It is at least $m_j$ by step 17. Suppose $(t - \rho_j)^{m_j + 1}$ divides $p$, say $p = (t - \rho_j)^{m_j + 1} g$.
19. Since [[$\R[t]$ is an integral domain|0B5Z1EC66J]], we may [[cancel|0GJM24EXP9]] $(t - \rho_j)^{m_j}$. This gives $\prod_{i \ne j} (t - \rho_i)^{m_i} = (t - \rho_j)\, g$. At $t = \rho_j$ the left side is nonzero and the right side is $0$, a contradiction.
20. So the families for the distinct roots have $m_1 + \dots + m_s = n$ functions in total, all solutions by step 6.
21. They are linearly independent. A vanishing combination can be grouped as $\sum_j q_j(x) e^{\rho_j x} = 0$, where the coefficients of $q_j$ are the coefficients used for $\rho_j$. By the lemma each $q_j = 0$, so all coefficients vanish.
22. The space $S$ of solutions on $\R$ has dimension $n$, by [[the dimension theorem|0GK68PZ6KH]]. By [[the basis criterion|096V7H13H9]], these $n$ independent solutions form a basis, that is, a [[fundamental system|0H4KNNCHPY]].

## Idea

The operator $p(D)$ factors like the polynomial $p$, and the factor $(D - r)$ lowers the power of $x$ in $x^k e^{r x}$ by one. So $(D - r)^m$ kills $x^k e^{r x}$ for $k < m$. Independence follows because differentiating a polynomial times an exponential keeps the degree, while differentiating a polynomial alone lowers it.