boundary value problem $-u''+u=\delta_y$ where $u'(0)=u'(1)=0$ 
Consider the boundary value problem (BVP) $$-u''+u=\delta_y\;\text{on}\; I=(0,1)$$ $$u'(0)=u'(1)=0,$$where $y\in I$, $\delta_y:H^1(I)\to\mathbb{R}$ defined by $\delta_y(v)=v(y)$. For all $y\in I$ find a weak solution $u\in H^1(I)$ of the BVP. 

This means, I have to consider $$\int_0^1{u'(t)v'(t)+u(t)v(t)dt}=v(y)$$ for all $v\in H^1((0, 1))$ and maybe it's useful to consider $(0,y)$ and $(y,1)$ and $\int_0^y{u'(t)v'(t)+u(t)v(t)dt}=v(y)$ and $\int_y^1{u'(t)v'(t)+u(t)v(t)dt}=v(y)$, but I'm not sure. How do I find $u$? 
 A: The thing to notice is that $-u''+u=0$ for $x < y$ and for $x > y$. So you have to piece together eigenfunctions. On the left, you want $u'(0)=0$ and on the right you want $u'(1)=0$. Then solutions of these problems are
$$
                  u(x)=A\cosh(x),\;\;\;\;\;\;\; 0 \le x < y  \\
                  u(x)=B\cosh(x-1), \;\;\;\; y < x \le 1.
$$
Then you have to piece these together at $y$ in such a way that the function is continuous in $x$ at $x=y$, and the derivative has a jump discontinuity of $-1$ at $x=y$. That way $-u''$ gives you a delta function. There are two conditions
$$
                A\cosh(y)-B\cosh(y-1) = 0 \\
                B\sinh(y-1)-A\sinh(y) = -1.
$$
That's a simple 2x2 system for $A$, $B$, and the determinant of the coefficient matrix is a Wronskian, which does not vanish. So everything is good. I'll let you determine the values of $A$ and $B$.
A: You can work with the ODE directly to get a "formal" solution and then check that it satisfies the weak form of the equation. One way to do this is to use the Laplace transform; the definition of the Dirac delta and of the Laplace transform tells you that the Laplace transform of $\delta_y$ is $e^{-ys}$. The Laplace transform of the left side is $-s^2U+su(0)+u'(0)+U=(-s^2+1)U+su(0)$. So you have
$$U=\frac{su(0)-e^{-ys}}{s^2-1}.$$
You can use partial fraction decomposition and the inverse Laplace transform to solve this for $u$, keeping $u(0)$ as a free parameter. We already used $u'(0)=0$, so you then need to choose $u(0)$ to make $u'(1)=0$. You should then check that the solution satisfies the weak equation (since the strong equation does not make pointwise sense at $y$).
