Re: [微積] ODE
※ 引述《oblivion87 (oblivion87)》之銘言：
: 毫無頭緒，只想出這個幾乎是亂猜的方法，請問這樣判斷可以嗎，還是要硬級數解呢，有
: 什麼比較嚴謹的證明方法嗎？
: https://i.imgur.com/3Cyv9f7.jpg
你的題目應該是：
f"+f=0, (x+1)g"+xg=0 for x>0, both f and g are not trivial,
show that there is a zero of f between any consecutive zeroes of g.
pf):
Take two consecutive zeroes of g, say a and b, where a < b.
suppose that there is no zero of f in (a,b),
then f remains its sign on (a,b), say positive.
WLOG, we may assume that g is positive on (a,b) as well.
f"g-fg" = -fg+fgx/(x+1) = -fg/(x+1)
Integrate the previous equation on (a,b), we get
f'(b)g(b)-f(b)g'(b)-f'(a)g(a)+f(a)g'(a) = -∫_a^b fg/(x+1) dx
=> -f(b)g'(b)+f(a)g'(a) = -∫_a^b fg/(x+1) dx
, where RHS < 0.
By definition, g'(a)≧0 and g'(b)≦0.
They cannot be 0, since g'(a) = 0 implies g = 0.
So g'(a) > 0 and g'(b) < 0, hence LHS≧0, a contradiction.
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※ 編輯: Vulpix (61.230.70.207), 12/29/2018 12:13:13
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