Re: [問題] 應該算是簡諧運動立論的基礎吧,可是課ꔠ…
※ 引述《bisconect (隨便你叫)》之銘言：
: if
: 1) f(0) = g(0)
: 2) f'(0) = g'(0)
: 3) for any t, f''(t) = h(f(t)) and g''(t) = h(g(t))
: then
: for any t, f(t) = g(t)
: 請問以上這敘述是否為真?
symbol
\int: integral
del: Dirac delta function
h<f: h(f)
f(t) - g(t)
= [ f(t) - f(0) ] - [ g(t) - g(0) ]
= \int_0^t [ f'(t') - g'(t') ] dt'
= \int_0^t dt' \int_0^t' dt" [ f''(t") - g''(t") ]
= \int_0^t dt' \int_0^t' dt" [ h(f(t")) - h(g(t")) ]
If h is linear,
then
f(t) - g(t)
= \int_0^t dt' \int_0^t' dt" h ( f(t") - g(t") )
However,
f(t) - g(t)
= \int dt" del(t-t") [ f(t") - g(t") ],
it turns to be that
[\int_0^t dt' \int_0^t' dt" h< - \int dt" del(t-t") ] ( f(t") - g(t") ) = 0
Because t is arbitrary,
f(t") - g(t") must be zero.
QED.
ps: h has to be linear. If h is nonlinear, the statement is
not necessarily true.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 70.171.59.112
※ 編輯: chungweitw 來自: 70.171.59.112 (11/15 03:06)
→
11/15 15:43, , 1F
11/15 15:43, 1F
討論串 (同標題文章)
完整討論串 (本文為第 2 之 4 篇)：