I am stuck on the following problem that says:

If all the zeroes of the polynomial $b_ny^n+b_{n-1}y^{n-1}+ \ldots..+b_1y+b_0$ have negative real parts and if $g(t)$ is any solution to the ordinary differential equation $b_n\frac{d^ng}{dt^n}+b_{n-1}\frac{d^{n-1}g}{dt^{n-1}}+ \ldots..+b_1\frac{dg}{dt}+b_0g=0$,then $\lim_{t \to \infty}g(t)=?$

Thanks in advance for your time.


There is a theorem that goes as follows: let $D$ be the differentiation operator, and given a (monic) polynomial $p(x) = x^{n} + \ldots + b_0,$ define $p(D) = D^{n} + \ldots + b_0 I$, where $D^{j}$ is the $j$-th derivative and $I$ is the identity map. If $p(x) = (x-\lambda_1)^{n_1}\ldots (x-\lambda_k)^{n_k},$ then all solutions $g$ to the differential equation $p(D)(g) = 0$ are given as a linear combination (over $\mathbb{C}$) of the functions $e^{\lambda_1 x}, \ldots, x^{n_1 - 1}e^{\lambda_1 x}, \ldots, e^{\lambda_k x}, \ldots, x^{n_k -1}e^{\lambda_k x}.$ This finishes your question quickly. If you'd like a proof, I can post it in the morning.

This is a relatively long proof, but here it is.

Proposition 1: Let $p(x) = (x-\lambda)^n.$ Then $g$ is a solution to $p(D)(g) = 0$ iff $g$ is a linear combination of $e^{\lambda x}, \ldots, x^{n-1}e^{\lambda x}.$

Proof: Let $g(x) = e^{\lambda x} h(x).$ Then note $(D-\lambda)(e^{\lambda x}h(x)) = (e^{\lambda x} h(x))^{\prime} - \lambda e^{\lambda x} h(x) = e^{\lambda x}h^{\prime}(x).$ Then we can consider a composition of $(D-\lambda)$ with itself $n$ times. There are a couple of properties here to verify; namely, given two polynomials $p_1$ and $p_2,$ we have $(p_1 + p_2)(D) = p_1(D) + p_2(D),$ and $(p_1p_2)(D) = p_1(D)\circ p_2(D) = p_2(D) \circ p_1(D).$

Continuing, $p(D)(g) = 0 \iff (D-\lambda)^{n} h(x) = 0 \iff e^{\lambda x}h^{(n)} (x) = 0.$ That is, $h^{(n)} (x) = 0.$ I trust that you have seen the proof that $h$ must then be a polynomial of degree $\le n-1,$ so we are done.

More generally, we have

Theorem: Let $p(x) = (x-\lambda_1)^{n_1} \ldots (x-\lambda_k)^{n_k}.$ Then $p(D)(g) = 0$ iff $g$ is a linear combination of $x^{i} e^{\lambda_j x},$ where $0 \le i \le n_j - 1.$

There are a number of ways to prove this, but here is one that has its roots in linear algebra.

Proposition 2: Let $p$ be as above, and let $p_i (x) = (x-\lambda_i)^{n_i},$ so that $p = p_1\ldots p_k.$ Let $T: V\to V$ be a linear operator, where $V$ is a vector space (over $\mathbb{C}$ here). Then $\ker(p(T)) \subseteq \bigoplus \ker (p_i(T))$ (in fact equality holds, but we don't need that here).

Proof: Define $q_i = \frac{p}{p_i}$ for each $1 \le i \le k.$ Note then that the $q_i$ are relatively prime, whence there exist polynomials $r_1, \ldots, r_k$ such that $r_1 q_1 + \ldots + r_k q_k = 1.$ Let $g_i = (r_i q_i)(E)(g).$ Then $g = g_1 + \ldots + g_k.$ Moreover, $p_i(E)(g_i) = (p_i(E)\circ (r_iq_i)(E))(g) = (p_i r_i q_i)(E)(g) = (r_i p_i q_i)(E)(g) = r_i(E)(g)\circ p(E)(g) = 0$ since $g \in (\ker p(E)),$ so $g_i \in \ker(p_i(E)),$ and we are done.

The other direction in the theorem is easy to check :)

  • $\begingroup$ thanks a lot . If possible,you can also post a proof in the morning. $\endgroup$ – learner May 5 '13 at 5:54
  • $\begingroup$ @learner I posted the proof yesterday $\endgroup$ – cats May 6 '13 at 7:37
  • $\begingroup$ thank you so much for the detailed clarification. +1 from me. $\endgroup$ – learner May 6 '13 at 12:46

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