To give you an idea: Given a function f find a function u such that u'' = f. This is not too difficult by integrating. But surely if u'' = f then u'' * v = f*v for all functions v which are "good enough" (i.e. smooth e.t.c.) but then integral(f*v over [a,b]) = integral(u''*v over [a,b]) = integral(u'*v over [a,b]) - u'(b)*v'(b)+u'(a)*v'(a).
Now you can show that (for some function spaces) you can also find solutions to the problem ` integral(f*v over [a,b]) = integral(u''*v over [a,b]) = integral(u'*v over [a,b]) - u'(b)*v'(b)+u'(a)*v'(a)`. But they are not necessarily two times differenti…