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I'm reading the proof of Proposition 8.18. in Brezis' Functional Analysis where I have a problem understanding how $u^{\prime}(0)=\alpha$ and $u^{\prime}(1)=\beta$.


Example 3 (homogeneous Neumann condition). Consider the problem $$ (21) \quad \left\{\begin{array}{l} -u^{\prime \prime}+u=f \quad \text { on } I=(0,1), \\ u^{\prime}(0)=u^{\prime}(1)=0 \end{array}\right. $$

Proposition 8.17. Given $f \in L^2(I)$ there exists a unique function $u \in H^2(I)$ satisfying (21). Furthermore, $u$ is obtained by $$ \min _{v \in H^1(I)}\left\{\frac{1}{2} \int_I\left(v^{\prime 2}+v^2\right)-\int_I f v\right\} . $$ If, in addition, $f \in C(\bar{I})$, then $u \in C^2(\bar{I})$.

Proof. If $u$ is a classical solution of (21) we have $$ (22) \quad \int_I u^{\prime} v^{\prime}+\int_I u v=\int_I f v \quad \forall v \in H^1(I) . $$

We use $H^1(I)$ as our function space: there is no point in working in $H_0^1$ as above since $u(0)$ and $u(1)$ are a priori unknown. We apply the Lax-Milgram theorem with the bilinear form $a(u, v)=\int_I u^{\prime} v^{\prime}+\int_I u v$ and the linear functional $\varphi: v \mapsto \int_I f v$. In this way we obtain a unique function $u \in H^1(I)$ satisfying (22). From (22) it follows, as above, that $u \in H^2(I)$. Using (22) once more we obtain $$ (23) \quad \int_I\left(-u^{\prime \prime}+u-f\right) v+u^{\prime}(1) v(1)-u^{\prime}(0) v(0)=0 \quad \forall v \in H^1(I) . $$

In (23) begin by choosing $v \in H_0^1$ and obtain $-u^{\prime \prime}+u=f$ a.e. Returning to (23), there remains $$ u^{\prime}(1) v(1)-u^{\prime}(0) v(0)=0 \quad \forall v \in H^1(I) . $$ Since $v(0)$ and $v(1)$ are arbitrary, we deduce that $u^{\prime}(0)=u^{\prime}(1)=0$.

Example 4 (inhomogeneous Neumann condition). Consider the problem $$ (24) \quad \left\{\begin{array}{l} -u^{\prime \prime}+u=f \quad \text { on } I=(0,1), \\ u^{\prime}(0)=\alpha, u^{\prime}(1)=\beta, \end{array}\right. $$ with $\alpha, \beta \in \mathbb{R}$ given and $f$ a given function.

Proposition 8.18. Given any $f \in L^2(I)$ and $\alpha, \beta \in \mathbb{R}$ there exists a unique function $u \in H^2(I)$ satisfying (24). Furthermore, $u$ is obtained by $$ \min _{v \in H^1(I)}\left\{\frac{1}{2} \int_I\left(v^{\prime 2}+v^2\right)-\int_I f v+\alpha v(0)-\beta v(1)\right\} . $$

If, in addition, $f \in C(\bar{I})$ then $u \in C^2(\bar{I})$.

Proof. If $u$ is a classical solution of (24) we have $$ (25) \quad \int_I u^{\prime} v^{\prime}+\int_I u v=\int_I f v-\alpha v(0)+\beta v(1) \quad \forall v \in H^1(I) . $$

We use $H^1(I)$ as our function space and we apply the Lax-Milgram theorem with the bilinear form $a(u, v)=\int_I u^{\prime} v^{\prime}+\int_I u v$ and the linear functional $$ \varphi: v \mapsto \int_I f v-\alpha v(0)+\beta v(1) . $$

This linear functional is continuous (by Theorem 8.8). Then proceed as in Example 3 to prove that $u \in H^2(I)$ and that $u^{\prime}(0)=\alpha, u^{\prime}(1)=\beta$.


My attempt If $u$ is a classical solution of (24) we have $$ -\int_I u^{\prime \prime} v + \int_I u v=\int_I f v \quad \forall v \in H^1(I) . $$

By integration by parts, $$ \begin{align*} \int_I u^{\prime \prime} v &= (u'v)(1)- (u'v)(0) - \int_I u^{\prime} v^{\prime} \\ &= \beta v(1)- \alpha v (0) - \int_I u^{\prime} v^{\prime}. \end{align*} $$

Then we have $(25)$ and in particular, $$ (26) \quad \int_I u^{\prime} v^{\prime}+\int_I u v=\int_I f v \quad \forall v \in H^1_0(I) . $$

By integration by parts again, $$ \int_I u^{\prime \prime} v = - \int_I u^{\prime} v^{\prime} \quad \forall v \in H^1_0(I) . $$

Then $$ (27) \quad \int_I (-u^{\prime \prime} + u^\prime-f) v \quad \forall v \in H^1_0(I) . $$

Then $-u^{\prime \prime} + u^\prime=f$ a.e. on $I$. It follows from $(26)$ that $u' \in H^1 (I)$ with $u''=f-u$. Then $u \in H^2 (I)$.

Could you explain how $u^{\prime}(0)=\alpha$ and $u^{\prime}(1)=\beta$?

Thank you so much for your help!

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  • $\begingroup$ It looks like you swap between $H^1$ and $H_0^1$ somewhere. Double check this, as compactly supported test functions will eliminate boundary conditions from the weak form. $\endgroup$
    – whpowell96
    Sep 12, 2023 at 16:56
  • $\begingroup$ @whpowell96 I swapped between $H^1$ and $H^1_0$ in the equality $(26)$. However, I don't know how to get $u^{\prime}(0)=\alpha$ and $u^{\prime}(1)=\beta$. Could you elaborate on this point? $\endgroup$
    – Akira
    Sep 12, 2023 at 18:31
  • $\begingroup$ If $u\in H^2$ then you can revert the partial integration that lead to (25) $\endgroup$
    – daw
    Sep 13, 2023 at 9:29
  • $\begingroup$ @daw Thank you so much for your help! I have formulated your idea as an answer posted below. $\endgroup$
    – Akira
    Oct 7, 2023 at 9:53

1 Answer 1

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If $u$ is a classical solution of (24) we have $$ -\int_I u^{\prime \prime} v + \int_I u v=\int_I f v, \quad v \in H^1(I). $$

By integration by parts (I.b.P), $$ \begin{align*} \int_I u^{\prime \prime} v &= (u'v)(1)- (u'v)(0) - \int_I u^{\prime} v^{\prime} \\ &= \beta v(1)- \alpha v (0) - \int_I u^{\prime} v^{\prime}, \quad v \in H^1(I). \end{align*} $$

Then we have $$ (25) \quad \int_I u^{\prime} v^{\prime}+\int_I u v=\int_I f v + \beta v(1)-\alpha v(0), \quad v \in H^1(I) . $$

By Lax-Milgram theorem, there is a unique $u \in H^1$ that satisfies $(25)$. In particular, $$ (26) \quad \int_I u^{\prime} v^{\prime}+\int_I u v=\int_I f v, \quad v \in H^1_0(I) . $$ and $$ (27) \quad \int_I u^{\prime} \varphi^{\prime} = \int_I \varphi(f-u), \quad \varphi \in C^1_c (I) . $$

Clearly, $(27)$ implies $u \in H^2$ with $u'' =f-u$. By I.b.P, $$ (28) \quad \int_I u^{\prime} v^{\prime} = - \int_I u^{\prime \prime} v, \quad v \in H^1_0(I) . $$

We have $(26, 28)$ implies $$ (29) \quad \int_I (-u^{\prime \prime} + u-f) v, \quad v \in H^1_0(I), $$ and thus $-u^{\prime \prime} + u=f$ a.e. on $I$. Because $u' \in H^1$, we apply I.b.P and get $$ (30) \quad \int_I u^{\prime \prime} v = (u'v) (1) - (u'v) (0) - \int_I u^{\prime} v^{\prime}, \quad v \in H^1 (I) . $$

We have $(25, 30)$ and the fact that $-u^{\prime \prime} + u=f$ a.e. on $I$ imply $$ \beta v (1) - \alpha v (0) = u' (1) v (1) - u'(0) v (0) \quad v \in H^1 (I), $$ and thus $\beta = u'(1)$ and $\alpha = u' (0)$. This completes the proof.

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