\f\fQ1. A. For any D , the integral 1 h[] ( x)dx, D x 0 Return a real number, thus it maps any vector in D into a real number, therefore is a distribution. 0, x 0 g ( x) x,0 x 1 0, x 1 B. g '[] g ( x)' ( x)dx g ( x )( x) g ' ( x)( x) dx ' ( x)( x)dx g 1 1 ( x) dx ( x 1)( x) dx 0 0 g ' ( x) 1 lim h 0 Q2. g (1 h) g (1) 1 ( x 1) h u ' '[] (t )' ' (t )dx u t' ' (t )dx sin 0 sin t' (t ) cos t' (t )dx 0 0 sin t' (t ) cos t(t ) t(t )dx sin 0 0 0 (0) t(t )dx sin 0 (t )(t )dt t(t )dx sin 0 B. For t>0, we have (sint)''+sint =0 Satisfies the homogeneous equation, thus we just need to verify that, at t=0 (u)''+u = (t) t t u ( ) u ( ) 2 u ' (0) lim 2 t 0 t sin(t / 2) 0 lim t 0 t t cos 2 t t u ( ) u ( ) 2 u ' ' (0) lim 2 t 0 t t cos 0 2 lim t 0 t and lim u ' ' (t ) 0 lim(cos 0 u (t ) lim(u' ( ) u' ( )) sin tdt 0 0 cos 1) 1 2 and the integration of delta function on the right hand side equals to 1, thus u(t) is the solution to the equation u ' ' (t ) u (t ) (t ) In the sense of distribution. Q3 ln r * 2rdr 0 r ln 0 1 d d ( r (r )) * 2rdr r dr dr 2 r ln 0 d d (r (r )) dr dr dr d d 2 ln r[r (r )] 2 (r )dr 0 dr dr 0 2 ln r[r d (r )] 2 (0) 0 dr Since ln r r 0 r 0 1/ r 3 lim r 0 limr ln r lim r 0 We have ln r * (r )2rdr 2(0) 0 2 (r ) (r ) dr 0 Q4 The general solution to the homogeneous equation is u Ax B Since the boundary condition is U(0) = 0 and du/dx(L)=0 For 0