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Let $I$ be the open interval $(0, 1)$. I'm trying to solve a problem in Brezis' Functional Analysis, i.e.,

Exercise 8.24

  1. Prove that for every $\varepsilon>0$ there exists a constant $C_{\varepsilon}$ such that $$ |u(1)|^2 \leq \varepsilon\left\|u^{\prime}\right\|_{L^2}^2+C_{\varepsilon}\|u\|_{L^2}^2 \quad \forall u \in H^1(I) . $$
  2. Prove that if the constant $k>0$ is sufficiently large, then for every $f \in L^2(I)$ there exists a unique $u \in H^2(I)$ satisfying $$ (1) \quad \begin{cases} -u^{\prime \prime}+k u=f \quad \text{on} \quad I, \\ u^{\prime}(0)=0 \quad \text{and} \quad u^{\prime}(1)=u(1). \end{cases} $$ What is the weak formulation of problem (1)? What is the associated minimization problem?

In below attempt of (2.), I don't need that $k$ is sufficiently large. I only need $k \ge 0$. Could you verify if my attempt contains some subtle mistakes?


Let $K := \{v \in H^1(I) : v'(0)=0, v'(1)=v(1)\}$. Then $K$ is a closed subspace of $H^1(I)$. If $u$ is a classical solution to $(1)$, then $$ \int_I [-u''v + kuv] = \int_I fv, \quad \forall v \in K, $$ which (by integration by parts) is equivalent to $$ (2) \quad u(1) v(1) + \int_I [u'v' + k uv] = \int_I fv, \quad \forall v \in K. $$

Then $(2)$ is the weak forlulation of $(1)$. We define a symmetric bilinear form $a$ on $K$ by $$ a(u, v) := u(1) v(1) + \int_I [u'v' + k uv]. $$

It follows from (1.) that $a$ is continuous. Because $k$ is non-negative, $a$ is coercive. We define $\varphi \in (H^1(I))^*$ by $$ \varphi (v) = \int_I fv, \quad \forall v \in K. $$

By Lax-Milgram theorem, $(2)$ has a unique solution $u \in K$. The associated minimization is $$ u= \operatorname{argmin}_{v \in K} \left \{ \frac{1}{2} a(v, v) - \varphi (v)\right \}. $$

Notice that $(2)$ implies $$ \int_I u'v' = -\int_I (ku-f)v, \quad \forall v \in C^\infty_c (I), $$ which implies $u \in H^2(I)$ with $u''=ku-f$.

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Your definition of $K$ is not well-defined because it involves $v'(0)$ and $v'(1)$ while we only know $v \in H^1 (I)$. The equation $(2)$ is also wrong because the wrong sign of $u(1) v(1)$, and this is where we need $k$ to be large enough. Below is a fix.


If $u$ is a classical solution to $(1)$, then $$ \int_I [-u''v + kuv] = \int_I fv, \quad \forall v \in H^1 (I), $$ which (by integration by parts) implies $$ (2) \quad -u(1) v(1) + \int_I [u'v' + k uv] = \int_I fv, \quad \forall v \in H^1 (I). $$

Then $(2)$ is the weak forlulation of $(1)$. We define a symmetric bilinear form $a$ on $H^1(I)$ by $$ a(u, v) := -u(1) v(1) + \int_I [u'v' + k uv]. $$

It follows from (1.) that $a$ is continuous and that if $k>0$ is sufficiently large then $a$ is coercive. We define $\varphi \in (H^1(I))^*$ by $$ \varphi (v) = \int_I fv, \quad \forall v \in K. $$

By Lax-Milgram theorem, $(2)$ has a unique solution $u \in H^1(I)$. The associated minimization is $$ u= \operatorname{argmin}_{v \in K} \left \{ \frac{1}{2} a(v, v) - \varphi (v)\right \}. $$

Notice that $(2)$ implies $$ \int_I u'v' = -\int_I (ku-f)v, \quad \forall v \in C^\infty_c (I), $$ which implies $u \in H^2(I)$ with $u''=ku-f$. By integration by parts, $(2)$ implies $$ (3) \quad (u'(1)-u(1))v(1) -u'(0) v(0) =0, \quad \forall v \in H^1 (I), $$ which implies $u'(1)=u(1)$ and $u'(0)=0$.

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