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I have a simple question on elliptic pdes, actually I can not understand clearly from definitions. Thats why I want to try think on an example. Let us have an elliptic pde $$-A \Delta u(x)+Bu(x)=F(x)$$ where $A,B$ are constants and $x\in\mathbb{R}^3$. Thus, I can define an elliptic operator $L$, $$Lu=(-A \Delta+B)u=F(x).$$ Associated bilinear form to $L$ is given $b:H \times H\rightarrow\mathbb{R}$ for $u,v\in H$$$b\left(u,v\right)=\int \limits_{\mathbb{R}^3}A\sum\limits_{i=1}^{3}\frac{\partial u(x)}{\partial x_{i}}\frac{\partial v(x)}{\partial x_{i}}dx+B\int \limits_{\mathbb{R}^3}u(x)v(x)dx.$$ Then I am trying to check Lax-Milgram's Lemma,$$\begin{eqnarray} \left|b\left(u,v\right)\right|&=&\left|\int \limits_{\mathbb{R}^3}A\sum\limits_{i=1}^{3}\frac{\partial u(x)}{\partial x_{i}}\frac{\partial v(x)}{\partial x_{i}}dx+B\int \limits_{\mathbb{R}^3}u(x)v(x)dx\right|\\&\leq&\int \limits_{\mathbb{R}^3}A\left|\sum\limits_{i=1}^{3}\frac{\partial u(x)}{\partial x_{i}}\frac{\partial v(x)}{\partial x_{i}}dx\right|+B\int \limits_{\mathbb{R}^3}\left|u(x)v(x)dx\right|\end{eqnarray}$$ As you see, I am failed in the first step and can not see even how to apply Cauchy-Schwarz here. Coerciness is then much worse then continuity. Can you please help me on this example?

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We can assume $A=1$ so that your equation is $$ -\Delta u +bu= f $$ with $b=A^{-1}B$ and $f=A^{-1}F$. Your bilinear form, which I'll call $B$, then satisfies $$ |B(u,v)|=\left| \int_{\mathbb{R}^3} \nabla u \cdot \nabla v dx +b \int_{\mathbb{R}^3} uvdx \right| \leq \| \nabla u\| _{L^2} \|\nabla v \|_{L^2} +|b| \| u\|_{L^2}\| v\|_{L^2}\\ \leq \| u\|_{H^1}\| v\| _{H^1} +|b| \| u\|_{H^1}\| v\|_{H^1} $$ where the first inequality is Cauchy-Schwarz applied to each summand, and the second follows from the definition of the $H^1$ norm.

For coercivity, I can only prove it when $b> 0$ (when it is obvious since in this case $B(u,u)=\| \nabla u\|_{L^2}^2 +b\| u\|_{L^2}^2$).

Edit: A variant of $f_k(x)=|x|^{-\frac{3}{2}-\frac{1}{k}}$ shows that when $b\leq0$ the bilinear form is not coercive.

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  • $\begingroup$ thank you very much for your response. If I do not bother you, I want to ask the step with cauchy-schwarz. After applying C-S you switch to the $L^2$ norms and you write $|\int \limits_{\mathbb{R}^3}\left|\sum\limits_{i=1}^{3}\frac{\partial u(x)}{\partial x_{i}}\frac{\partial v(x)}{\partial x_{i}}dx\right|\leq\left\|\nabla u\right\|_{L^2} \left\|\nabla v\right\|_{L^2}.$ Could you please clear this step a bit more? $\endgroup$ – pcepkin Jan 28 '13 at 0:00
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    $\begingroup$ @pcepkin: It's a vector valued version of C-S: let $f_i=\partial_iu$, $g_i=\partial_i v$, $a_i=\| f_i\|_{L^2}$, $b_i=\| g_i\|_{L^2}$ then we have $$\left|\int\sum_i f_ig_i\right|\leq \sum_i a_ib_i \leq \left( \sum_i a_i^2 \right)^{1/2}\left( \sum_I b_i^2 \right)^{1/2}$$ (the first inequality is the usual C-S applied to each summand and the second is C-S in $\mathbb{R}^n$) and you can check that this is the inequality that you want. $\endgroup$ – Jose27 Jan 28 '13 at 0:10
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I assume you want $H = H^1(\Bbb R^3)$ here.

Note that \begin{align} |b(u,v)| & \leq \max \{|A|, |B|\} \left| \int_{\Bbb R^3} \sum_{i = 1}^3 \frac{\partial u}{\partial x_i} \frac{\partial v}{\partial x_i} ~dx + \int_{\Bbb R^3} uv ~dx \right| \\ & \leq \max \{|A|, |B|\} |(u,v)_{H^1(\Bbb R^3)}| \\ & \leq \max \{|A|, |B|\} \|u\|_{H^1(\Bbb R^3)} \|v\|_{H^1(\Bbb R^3)}. \end{align}

On the other hand, \begin{align} b(u,u) & = A \int_{\Bbb R^3} |\nabla u|^2 ~dx + B \int_{\Bbb R^3} u^2 ~dx \\ & \geq \min\{A, B\} \|u\|_{H^1(\Bbb R^3)}^2. \end{align} Note that we need that $B/A > 0$ in order for the above to prove coercivity of $b(u,v)$. (Jose27's answer takes a nicer approach by dividing through by $A$ at the start)

From the above we see that $b(u,v)$ satisfies the hypotheses of the Lax-Milgram lemma.

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  • $\begingroup$ Why $H=H^2$? As far as I know (and that's not much), those equations are usually stated in terms of $H^1$-space. $\endgroup$ – Giuseppe Negro Jan 27 '13 at 21:43
  • $\begingroup$ @GiuseppeNegro: You're right, I was thinking about how we want 2 detivatives for the Laplacian to make sense, but the bilinear form just needs 1 derivative. I've edited my answer accordingly. $\endgroup$ – Henry T. Horton Jan 27 '13 at 21:57
  • $\begingroup$ Ok, thank you for the feedback. The other question I have regards Lax-Milgram's lemma. Shouldn't we check that $$b(u,u)\ge C\lVert u\rVert_{H^1}^2?$$ That is, without the absolute value. $\endgroup$ – Giuseppe Negro Jan 27 '13 at 22:08
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    $\begingroup$ I don't understand how $A$ and $B$ got squared in the lower estimate for $|b(u,u)|$. $\endgroup$ – user53153 Jan 27 '13 at 22:27
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    $\begingroup$ @GiuseppeNegro Since $b$ is continuous and $b(u,u)$ is real, the inequality $|b(u,u)|\ge C\|u\|^2$ implies that either $b(u,u)\ge C\|u\|^2$ or $-b(u,u)\ge C\|u\|^2$, so L-M applies to one or the other. So the lemma can be stated with $|b|$. $\endgroup$ – user53153 Jan 27 '13 at 22:31

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