# Exact solution $\int_0^1 u'v'=v(1/2)$

This question concerns a variational form of the Laplace equation with homogeneous Dirichlet boundary conditions: $$-u''=f \text{ on } [0,1], u(0)=u(1)=0.$$

Let $$V=H^1_0(\Omega), \Omega=[0,1]$$ and $$f\in H^{-1}(\Omega)$$ . Solve

$$u\in H_0^1(\Omega) \text{ such that } \int_{0}^1 u'v'=f_i(v) \text{ for all } v\in H_0^1(\Omega)$$ for

i ) $$f_1(v)=\int_0^1 v(x)dx, v\in H_0^1(\Omega)$$

ii) $$f_2(v)=v(\frac12), v\in H_0^1(\Omega)$$

For the first one I integrated by parts to get $$-\int_0^1 u''v=\int_0^1 v$$ so $$u''=-1$$ and $$u(x)=-x^2/2+x/2$$ using the boundary conditions. How can I tackle the second one?

Integrating by parts suggests that you want to find $$u\in H_0^1(\Omega)$$ such that $$-\int_0^1 u''v = v(1/2).$$ Intuitively, this suggests that $$u$$ must be some $$H_0^1$$ function whose second distributional derivative is $$u''(x) = -\delta(x-\frac{1}{2})$$, where $$\delta$$ denotes the Dirac delta distribution. It remains to find such a function. One function whose distributional derivative is $$-\delta(x-\frac{1}{2})$$ is $$u_1' = -1_{[\frac{1}{2},1]}(x),$$ the indicator function of the interval $$[\frac{1}{2},1]$$ (with a minus sign). Integrating this function gives $$u_1(x) = -\int_0^x 1_{[\frac{1}{2},1]} = \begin{cases} C_1 & 0\leq x\leq\frac{1}{2},\\ -x+\frac{1}{2} + C_1 & \frac{1}{2}\leq x \leq 1. \end{cases}$$ This is almost great, but unfortunately this function fails to satisfy the boundary conditions $$u_1(0) = u_1(1) = 0$$ for any choice of $$C_1$$. However, this can be fixed, as there is another function whose distributional derivative is $$-\delta(x-\frac{1}{2})$$, namely $$u_2'(x) = 1_{[0,\frac{1}{2}]}(x).$$ Integrating this function gives $$u_2(x) = \int_0^x 1_{[0,\frac{1}{2}]} = \begin{cases} x + C_2 & 0\leq x\leq\frac{1}{2},\\ \frac{1}{2} + C_2 & \frac{1}{2}\leq x \leq 1. \end{cases}$$ $$u_2$$ has the same problem as $$u_1$$ in terms of satisfying the boundary conditions. However, a linear combination of them might eliminate this problem. Since $$u_1$$ and $$u_2$$ both solve $$-\int_0^1 u''v = v(1/2),$$ any convex combination $$u = au_1 + bu_2$$ with $$a+b = 1$$ will yield another solution to the variational equation. To keep things simple, let us take a simple average: $$u = \frac{1}{2}(u_1 + u_2)$$. Then $$u(x) = \begin{cases} \frac{1}{2}(x + C_1 + C_2) & 0 \leq x \leq \frac{1}{2},\\ \frac{1}{2}(-x + 1 + C_1 + C_2) & \frac{1}{2}\leq x \leq 1. \end{cases}$$ To satisfy the boundary conditions we take $$C_1 + C_2 = 0$$; then $$u(x) = \begin{cases} \frac{1}{2}x & 0 \leq x \leq \frac{1}{2},\\ \frac{1}{2}(1-x) & \frac{1}{2}\leq x \leq 1. \end{cases}$$ This is an $$H_0^1(\Omega)$$ solution to the variational equation, as you can now readily check.
• +1, nice answer. Just a comment, these functions are $H^1(\Omega)$ by construction. Indeed, they are absolutely continuous functions (as integrals of $L^1$ functions) with $L^2$ a.e. derivatives. – operatorerror Nov 7 '19 at 21:06
Here is another approach: First, we test the weak formulation with $$v \in H_0^1((0,1/2))$$, i.e., $$v = 0$$ on $$[1/2,1)$$. Then, $$\int_0^{1/2} u' \, v' \, \mathrm{d}x = 0.$$ Since $$v'$$ is an arbitrary function with zero mean on $$(0,1/2)$$, we get that $$u'$$ must be constant on $$(0,1/2)$$, i.e., $$u$$ is affine on $$(0,1/2)$$. Similarly, we can check that $$u$$ is affine on $$(1/2,1)$$.
Since $$u$$ is continuous on $$[0,1]$$ and has zero boundary conditions, this already implies $$u(x) = c \, (1/2 - |x - 1/2|)$$ for some constant $$c \in \mathbb R$$ and it remains to find $$c$$. This can be achieved by using a single test function $$v$$ with $$v(1/2) \ne 0$$ or by using $$v = u$$ as a test function.