# Brezis' proposition 8.18: how to prove that $u^{\prime}(0)=\alpha$ and $u^{\prime}(1)=\beta$?

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!

• 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. Sep 12, 2023 at 16:56
• @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? Sep 12, 2023 at 18:31
• If $u\in H^2$ then you can revert the partial integration that lead to (25)
– daw
Sep 13, 2023 at 9:29
• @daw Thank you so much for your help! I have formulated your idea as an answer posted below. Oct 7, 2023 at 9:53

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.