[微分方程] 判斷解的行為

Question

問兩題 QQ
1. 成大考古題

y\' = (y-1)(y-2)(y-3) , y(0) = 4

試判斷 y(t) 的行為在 t -> oo, 不能求出 y(t) 的解去看
這個好像是用定理 ?
只要在適當條件下有 y\'(0) > 0 那 y\'(t) 就會大於 0 for all t 是嗎 ?
不過我書上翻不到類似的定理 QQ

2. 書上後面的習題, 原來是一個 x(t) , y(t) 的非線性聯立微分方程, (0,0) 是一個 critical point, 然後做 x = rcosθ , y = rsinθ後變成

r\' = r^3
θ\' = 1

證明 (0,0) 是一個不穩定的 critical point

第一題我想出來了, 用唯一定理, 先假設若存在 t_0 使得 y'(t_0) < 0, 那由 IVT => 存在 t* 使 y'(t*) = 0, 設 y(t*) = y*,
用它當初始條件去解, 由唯一定理知此解必是 y(t) = y*, 故
y'(t) = 0 for all t (=>

Answer 1

Question 1.Lemma. y(t) > 3.5 for all t > 0.Proof. Suppose otherwise, and lett* = inf{t > 0: y(t) ≤ 3.5} > 0.Clearly y(t*) = 3.5, and for all t in (0,t*) we have y(t) > 3.5 and furthermore yʹ(t) = (y(t)-1)(y(t)-2)(y(t)-3) > 1.875. y is continuous on [0,t*] and differentiable on (0,t*), so by the Mean Value Theorem, there exists c in (0,t*) such thatyʹ(c) = (y(t*) - y(0))/t* = -0.5/t*= ≤ 0,which leads to a contradiction. ∎Claim. y(t) is strictly increasing for all t, hence y(t) → ∞ as t → ∞.Proof. By the lemma, yʹ(t) = (y(t)-1)(y(t)-2)(y(t)-3) > 1.875 for all t.The rest follows.Question 2.Claim. (0,0) is not a stable critical point of this system.Proof. Say (x(t),y(t)) is a solution of the ODE with x(0)2+y(0)2 = r0 > 0. Transforming the solution to polar coordinates (r(t),θ(t)), we find that r satisfies the initial value problem
rʹ(t) = r3,
r(0) = r0 > 0,

and hence r(t) = (r0-2 - 2t)-1/2. Clearly r(t) blows up in finite time and hence escapes any neighborhood of r0. Hence (0,0) is not stable.  ∎