1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 :)

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.