Let $I$ be the open interval $(0, 1)$. Assume that $p \in C^1(\bar I)$ and $q \in$ $C(\bar I)$ such that $p(x) \geq \alpha>0$ and $q(x) \geq \alpha>0$ for all $x \in \bar I$. Here $\alpha$ is a constant.
I am trying to solve a problem in Brezis' Functional Analysis
Exercise 8.25 For $k \ge 0$ and $f \in L^2 (I)$, we consider $$ (1) \quad \begin{cases} -(p u')' + q u=f \quad \text{on} \quad I,\\ p(0)u'(0)=ku(0), u(1)=0. \end{cases} $$
Check that $(1)$ has a unique (weak) solution in $H^2 (I)$, denoted by $u_k = S_k f$.
Show that $S_k \ge 0$, i.e. $\langle S_k f, f \rangle_{L^2} \ge 0$.
Let $k_1 \ge k_2 \ge 0$. Prove that $S_{k_2} - S_{k_1} \ge 0$.
There are possibly subtle mistakes that I could not recognize in my below attempt of (6.). Could you please have a check on it?
- Let $K := \{w \in H^1(I) : w(1)=0\}$. Then $K$ is a closed subspace of $H^1(I)$. If $u$ is a classical solution to $(1)$, then $$ (2) \quad \int_I [-(p u')'w + q uw] = \int_I fw, \quad \forall w \in H^1 (I), $$ which (by integration by parts) implies $$ (3) \quad k u(0) w (0) + \int_I [p u'w' + q uw] = \int_I fw, \quad \forall w \in K. $$
We define a continuous symmetric bilinear form $a$ on $K$ by $$ a(u, w) := k u(0) w (0) + \int_I [p u'w' + q uw]. $$
We have $$ a(u, u) = k |u(0)|^2 +\int_I [p |u'| + q |u|^2] \ge \alpha \|u\|^2_{H^1}, $$ which implies $a$ is coercive. By Lax-Milgram theorem, there is a unique $u \in K$ such that $$ (4) \quad a(u, w) = \int_I fw, \quad \forall w \in K. $$
From $(3)$, we also have $$ \int_I (p u')w' = -\int_I (qu-f)w, \quad \forall w \in C_c^\infty (I), $$ which implies $(p u')'=qu-f$ and thus $pu' \in H^1 (I)$. Because $p \ge \alpha >0$ and $p' \in C (\bar I)$, we get $\frac{1}{p} \in H^1 (I)$.
Corollary 8.10 Let $I$ be a (possibly unbounded) open interval of $\mathbb R$. Let $u, v \in W^{1, p}(I)$ with $1 \leq p \leq$ $\infty$. Then $u v \in W^{1, p}(I)$ and $(u v)^{\prime}=u^{\prime} v+u v^{\prime}$. Furthermore, the formula for integration by parts holds: $$ \int_y^x u^{\prime} v=u(x) v(x)-u(y) v(y)-\int_y^x u v^{\prime} \quad \forall x, y \in \bar{I} . $$
By above corollary (in the same book), we get $u' \in H^1 (I)$ and thus $u \in H^2 (I)$. It follows from $(3)$ and $(p u')'=qu-f$ and integration by parts that $$ ku(0)w(0)-p(0)u'(0)w(0)=0, \quad \forall w \in K, $$ which implies $ku(0) = p(0) u'(0)$.
It follows from $(4)$ that $$ \langle S_k f, f \rangle_{L^2} = \int_I f (S_k f) = a (S_k f, S_k f) \ge 0 $$
Let $a_{k_i}$ be the bilinear form induced by $k_i$. By Lax-Milgram, $$ \begin{align*} u_{k_i} &= \operatorname{argmin}_{w \in K} \left \{ \frac{1}{2} a_{k_i}(w, w) - \int_I f w \right \}. \end{align*} $$ We have $$ a_{k_1}(w, w) \ge a_{k_2}(w, w), \quad w \in K, $$ which implies $$ \frac{1}{2} a_{k_1}(u_{k_1}, u_{k_1}) - \int_I f u_{k_1} \ge \frac{1}{2} a_{k_2}(u_{k_2}, u_{k_2}) - \int_I f u_{k_2}. $$ It follows from $(4)$ that $$ \frac{1}{2} a_{k_i}(u_{k_i}, u_{k_i}) - \int_I f u_{k_i} = -\frac{1}{2} \int_I f u_{k_i}, $$ which implies $\int_I f u_{k_1} \le \int_I f u_{k_2}$.