How to prove that $a(u,v)=\int_0^1 u'v' dx$ is coercive

Question

Let $u,v \in H^1_0 (0,1)$

$a(u, v)=\int_0^1 u^{\prime} v^{\prime} dx$

I did not understand how they established the inequality between the 2nd and 3rd line. Can someone explain it to me in detail?

$$ \begin{aligned} a(u, u) & =\int_0^1\left[u^{\prime}\right]^2 d x \\ & =\frac{1}{2} \int_0^1\left[u^{\prime}\right]^2 d x+\frac{1}{2} \int_0^1\left[u^{\prime}\right]^2 d x \\ & \geq \frac{1}{2} \int_0^1\left[\left(u^{\prime}\right)^2+u^2\right] d x \\ & =\frac{1}{2}\|u\|_{H^1(0,1)}^2 \end{aligned} $$

Answer 1

To follow up on my comments, the typical way you prove coercivity of this bilinear form is by using Poincare's inequality, which is stated generally in 1-D as $$\|u\|_{L^p(\Omega)}\leq C_{\Omega,p}\|u'\|_{L^p(\Omega)}\quad\forall u\in W_0^{1,p}(\Omega).$$ The subscripts on the constant $C>0$ are to indicate that $C$ depends only on $p$ and $\Omega$.

Coercitivity for $H^1_0(\Omega)$ is then typically shown like this:

$$ \begin{aligned} a(u, u) & =\int_0^1\left[u^{\prime}\right]^2 d x \\ & =\frac{1}{2} \int_0^1\left[u^{\prime}\right]^2 d x+\frac{1}{2} \int_0^1\left[u^{\prime}\right]^2 d x \\ &\geq \frac{1}{2} \int_0^1\left[u^{\prime}\right]^2 d x+\frac{1}{2C} \int_0^1\left[u\right]^2 d x \\ & \geq\frac{\theta}{2}\|u\|_{H^1(0,1)}^2, \end{aligned} $$ where $\theta = \min\{1, 1/C \}$. Notice that this inequality was reached by dividing Poincare's inequality by $C$. That is, $$\frac{1}{C}\|u\|_{L^2(0,1)}\leq \|u'\|_{L^2(0,1)}$$

The exact value of $\theta$ isn't really important as long as it is greater than zero. The conditions you have written don't contradict this as long as $C\leq 1$.

For this domain, one can actually explicitly compute the minimal value of $C$ such that the inequality is satisfied via the variational problem

$$\frac{1}{C^2} = \inf_{u\in H_0^1(0,1)}\frac{\|u'\|_{L^2(0,1)}^2}{\|u\|_{L^2(0,1)}^2}.$$

This can be solved via calculus of variations or other methods to show that $C = 1/\sqrt{\lambda_1}$, where $\lambda_1 = \pi^2$ is the first (smallest) eigenvalue of the negative Laplacian, $-\Delta$, on $(0,1)$ with homogeneous Dirichlet boundary conditions. See, for example, this post or this handout for more reading on solving this variational problem. This principle is true for any domain $\Omega$, but one can actually compute this constant easily on this domain and some other simple ones.

Answer 2

To complement whpowell96's answer: you can also prove more directly that the Poincaré constant here is less than $\frac{1}{\sqrt{2}} < 1$ without too much effort, using Cauchy-Schwarz in the following manner:

Let $u \in H^1_0(0,1)$.
Then, for $x \in [0,1]$: $$\begin{split}|u(x)|^2 = \left|\int_0^x u'(t)\mathrm{d}t\right|^2 &\leq \left(\int_0^x |u'(t)|\mathrm{d}t\right)^2\\ &\overset{\text{C-S}}{\leq} \left(\int_0^x |u'(t)|^2\mathrm{d}t\right)\left(\int_0^x 1^2\mathrm{d}t\right)\\ &\leq \|u'\|_{L^2}^2 \cdot x\end{split}$$

Therefore, by integrating: $$\|u\|_{L^2}^2 = \int_0^1 |u(x)|^2\mathrm{d}x \leq \int_0^1 \|u'\|_{L^2}^2 x \mathrm{d}x = \frac{1}{2} \|u'\|_{L^2}^2$$

Hence: $$\|u\|_{L^2} \leq \frac{1}{\sqrt{2}} \|u'\|_{L^2}$$

(As an aside, notice that only the fact that $u(0) = 0$ has been used, $u(1) = 0$ has not been used.)