1.1 素朴に積分して解く

$\displaystyle x'(t)-x'(0)=-\int_0^t f(s) \D s.
$

$\displaystyle x(t)-x(0)-x'(0)t$ $\displaystyle =-\int_0^t\left(\int_0^r f(s) \D s\right)\D r =-\int_0^t\left(\int_s^tf(s) \D r\right)\D s$    
  $\displaystyle =-\int_0^t(t-s)f(s) \D s.$    

ゆえに ($ x(0)=0$ を用いて)

$\displaystyle x(t)=x(0)+x'(0)t-\int_0^t (t-s)f(s) \D s
=x'(0)t-\int_0^t (t-s)f(s) \D s.
$

$\displaystyle 0=x(1)=x'(0)-\int_0^1(1-s)f(s) \D s
$

であるから、

$\displaystyle x'(0)=\int_0^1(1-s)f(s) \D s.
$

ゆえに

$\displaystyle x(t)$ $\displaystyle =t\int_0^1(1-s)f(s) \D s-\int_0^t(t-s)f(s) \D s$    
  $\displaystyle =\int_0^t\left[t(1-s)-(t-s)\right]f(s) \D s +t\int_t^1(1-s)f(s)\D s$    
  $\displaystyle =\int_0^t s(1-t)f(s) \D s+\int_t^1 t(1-s)f(s) \D s.$    



桂田 祐史