Let $h \in C[0,1]$ be a function. Show that the boundary value problem
$$u''(x) = h(x)\mbox{ on } [0,1], u(0) = 0, u(1)=0$$
has a unique solution given by
$$u(x)=\int_0^1 g(x,\xi)h(\xi)d\xi$$
with $g(x,\xi)=\cases{\xi(x-1) & if $0\le \xi\le x \le 1$\\ x(\xi-1) & if $0\le x \le \xi \le 1$}$
To show that this is a solution, one can use substitution into the BVP above.
But $$u''(x)=\int_0^1 \frac{\partial^2 g}{\partial x^2}(x,\xi)h(\xi)d\xi$$ and $$\frac{\partial^2 g}{\partial x^2}(x,\xi)=0$$
So $u''(x) =$ constant.
I doubt this is the way it should be. I'm really puzzled. How does one approach this problem?
Differentiating piecewise, we get $$ \frac{\partial g}{\partial x} = \left\{ \begin{matrix} \xi && 0 \le \xi < x \\ \xi-1 && x < \xi \le 1 \end{matrix} \right. $$
Thus $$ u'(x) = \int_0^x \xi h(\xi) d\xi + \int_x^1 (\xi-1)h(\xi)d\xi $$
Therefore $$ u''(x) = xh(x) - (x-1)h(x) = h(x) $$
The boundary conditions are also satisfied since $g(0,\xi)=g(1,\xi)=0$