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

Question

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?

Answer 1

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.

Answer 2

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.