[proofplan]
We decompose $B[u, u]$ into a principal (second-order) part $I_{\text{principal}}$ and a lower-order part $I_{\text{lower}}$. Uniform ellipticity bounds $I_{\text{principal}}$ from below by $\theta \|\nabla u\|_{L^2(U)}^2$. The lower-order terms contain at most one gradient factor paired with $|u|$, so Young's inequality with a small parameter $\varepsilon$ absorbs their gradient contribution into $I_{\text{principal}}$ at the cost of a penalty proportional to $\|u\|_{L^2(U)}^2$. Choosing $\varepsilon = \theta/2$ and rewriting $\|\nabla u\|_{L^2}^2$ in terms of $\|u\|_{H^1_0}^2$ produces the Garding inequality.
[/proofplan]
[step:Decompose $B[u, u]$ into principal and lower-order parts]
Write the bilinear form evaluated on the diagonal:
\begin{align*}
B[u, u] = \underbrace{\int_U \sum_{i,j=1}^n a_{ij}\, \partial_{x_i} u\, \partial_{x_j} u \, d\mathcal{L}^n}_{=:\, I_{\text{principal}}} + \underbrace{\int_U \sum_{i=1}^n b_i\, \partial_{x_i} u\, u \, d\mathcal{L}^n + \int_U c\, u^2 \, d\mathcal{L}^n}_{=:\, I_{\text{lower}}}.
\end{align*}
Since $I_{\text{lower}}$ may be negative:
\begin{align*}
B[u, u] = I_{\text{principal}} + I_{\text{lower}} \ge I_{\text{principal}} - |I_{\text{lower}}|.
\end{align*}
[/step]
[step:Bound $I_{\text{principal}}$ from below using uniform ellipticity]
By the uniform ellipticity hypothesis, for $\mathcal{L}^n$-a.e. $x \in U$ we substitute $\xi = \nabla u(x) \in \mathbb{R}^n$:
\begin{align*}
\sum_{i,j=1}^n a_{ij}(x)\, \partial_{x_i} u(x)\, \partial_{x_j} u(x) \ge \theta\, |\nabla u(x)|^2.
\end{align*}
Integrating over $U$:
\begin{align*}
I_{\text{principal}} \ge \theta \|\nabla u\|_{L^2(U)}^2.
\end{align*}
[/step]
[step:Bound $|I_{\text{lower}}|$ from above via Young's inequality with parameter $\varepsilon$]
We bound each piece separately using the triangle inequality.
**Drift terms.** For each $i \in \{1, \dots, n\}$, apply Young's inequality $ab \le \varepsilon a^2 + \frac{1}{4\varepsilon} b^2$ with $a = |\partial_{x_i} u(x)|$ and $b = |b_i(x)|\,|u(x)|$:
\begin{align*}
\int_U |b_i|\,|\partial_{x_i} u|\,|u| \, d\mathcal{L}^n \le \varepsilon \|\partial_{x_i} u\|_{L^2(U)}^2 + \frac{\|b_i\|_{L^\infty(U)}^2}{4\varepsilon} \|u\|_{L^2(U)}^2.
\end{align*}
Summing over $i = 1, \dots, n$ and replacing $\varepsilon$ by $\varepsilon/n$:
\begin{align*}
\left|\int_U \sum_{i=1}^n b_i\, \partial_{x_i} u\, u \, d\mathcal{L}^n\right| \le \varepsilon \|\nabla u\|_{L^2(U)}^2 + \frac{n \max_i \|b_i\|_{L^\infty(U)}^2}{4\varepsilon} \|u\|_{L^2(U)}^2.
\end{align*}
**Reaction term.**
\begin{align*}
\left|\int_U c\, u^2 \, d\mathcal{L}^n\right| \le \|c\|_{L^\infty(U)} \|u\|_{L^2(U)}^2.
\end{align*}
Combining, for every $\varepsilon > 0$:
\begin{align*}
|I_{\text{lower}}| \le \varepsilon \|\nabla u\|_{L^2(U)}^2 + C_\varepsilon \|u\|_{L^2(U)}^2,
\end{align*}
where $C_\varepsilon := \frac{n \max_i \|b_i\|_{L^\infty(U)}^2}{4\varepsilon} + \|c\|_{L^\infty(U)}$.
[guided]
The lower-order terms threaten to destroy the positivity gained from ellipticity. The drift terms $\int b_i\, \partial_{x_i} u\, u$ are dangerous because they contain one gradient factor — they can grow with $\|\nabla u\|_{L^2}^2$. How do we neutralize them?
The standard $\varepsilon$-Young trick: for any $\varepsilon > 0$ and non-negative reals $a, b$, the inequality $ab \le \varepsilon a^2 + \frac{1}{4\varepsilon}b^2$ lets us trade a large gradient term for a controllable $L^2$ penalty. We apply this pointwise with $a = |\partial_{x_i} u(x)|$ and $b = |b_i(x)||u(x)|$:
\begin{align*}
|b_i(x)|\,|\partial_{x_i} u(x)|\,|u(x)| \le \varepsilon |\partial_{x_i} u(x)|^2 + \frac{|b_i(x)|^2}{4\varepsilon} |u(x)|^2.
\end{align*}
Integrating over $U$ and bounding $|b_i(x)|$ by $\|b_i\|_{L^\infty(U)}$:
\begin{align*}
\int_U |b_i|\,|\partial_{x_i} u|\,|u| \, d\mathcal{L}^n \le \varepsilon \|\partial_{x_i} u\|_{L^2(U)}^2 + \frac{\|b_i\|_{L^\infty(U)}^2}{4\varepsilon} \|u\|_{L^2(U)}^2.
\end{align*}
Summing over $i = 1, \dots, n$ gives $\varepsilon$ times $\sum_i \|\partial_{x_i} u\|_{L^2}^2 = \|\nabla u\|_{L^2}^2$ on the gradient side, but with the $\varepsilon$ multiplied by $n$. We compensate by replacing $\varepsilon$ with $\varepsilon/n$ from the start:
\begin{align*}
\sum_{i=1}^n \int_U |b_i|\,|\partial_{x_i} u|\,|u| \, d\mathcal{L}^n \le \varepsilon \|\nabla u\|_{L^2(U)}^2 + \frac{n \max_i \|b_i\|_{L^\infty(U)}^2}{4\varepsilon} \|u\|_{L^2(U)}^2.
\end{align*}
The reaction term $\int_U c\, u^2$ contains no gradient and is bounded by $\|c\|_{L^\infty(U)} \|u\|_{L^2(U)}^2$. Combining:
\begin{align*}
|I_{\text{lower}}| \le \varepsilon \|\nabla u\|_{L^2(U)}^2 + C_\varepsilon \|u\|_{L^2(U)}^2,
\end{align*}
where $C_\varepsilon = \frac{n \max_i \|b_i\|_{L^\infty(U)}^2}{4\varepsilon} + \|c\|_{L^\infty(U)}$.
[/guided]
[/step]
[step:Choose $\varepsilon = \theta/2$ and absorb the gradient term]
Substituting the bounds from the previous steps:
\begin{align*}
B[u, u] \ge I_{\text{principal}} - |I_{\text{lower}}| \ge \theta \|\nabla u\|_{L^2(U)}^2 - \varepsilon \|\nabla u\|_{L^2(U)}^2 - C_\varepsilon \|u\|_{L^2(U)}^2 = (\theta - \varepsilon)\|\nabla u\|_{L^2(U)}^2 - C_\varepsilon \|u\|_{L^2(U)}^2.
\end{align*}
Set $\varepsilon = \theta/2$. Then $\theta - \varepsilon = \theta/2$, giving:
\begin{align*}
B[u, u] \ge \frac{\theta}{2}\|\nabla u\|_{L^2(U)}^2 - \gamma_0 \|u\|_{L^2(U)}^2,
\end{align*}
where $\gamma_0 := C_{\theta/2} = \frac{n \max_i \|b_i\|_{L^\infty(U)}^2}{2\theta} + \|c\|_{L^\infty(U)}$.
[/step]
[step:Rewrite in terms of the $H^1_0$ norm to obtain the Garding inequality]
Recall that $\|u\|_{H^1_0(U)}^2 = \|\nabla u\|_{L^2(U)}^2 + \|u\|_{L^2(U)}^2$ for the [Sobolev space](/pages/1018) $H^1_0(U)$. Add and subtract $\frac{\theta}{2}\|u\|_{L^2(U)}^2$:
\begin{align*}
B[u, u] &\ge \frac{\theta}{2}\|\nabla u\|_{L^2(U)}^2 - \gamma_0 \|u\|_{L^2(U)}^2 \\
&= \frac{\theta}{2}\bigl(\|\nabla u\|_{L^2(U)}^2 + \|u\|_{L^2(U)}^2\bigr) - \gamma_0 \|u\|_{L^2(U)}^2 - \frac{\theta}{2}\|u\|_{L^2(U)}^2 \\
&= \frac{\theta}{2}\|u\|_{H^1_0(U)}^2 - \left(\gamma_0 + \frac{\theta}{2}\right)\|u\|_{L^2(U)}^2.
\end{align*}
Setting $\beta = \theta/2 > 0$ and $\gamma = \gamma_0 + \theta/2 \ge 0$ yields:
\begin{align*}
B[u, u] \ge \beta \|u\|_{H^1_0(U)}^2 - \gamma \|u\|_{L^2(U)}^2.
\end{align*}
[guided]
Why does the correction term $-\gamma \|u\|_{L^2(U)}^2$ appear? It is the price of absorbing the gradient contribution of the lower-order terms. The $\varepsilon$-Young trick lets us make the gradient coefficient of the bad terms as small as we like, but the $L^2$ penalty $C_\varepsilon$ blows up as $\varepsilon \to 0$. Choosing $\varepsilon = \theta/2$ balances the two: we retain half of the ellipticity constant on the gradient term, and all the damage is absorbed into the $L^2$ correction.
To convert from a $\|\nabla u\|_{L^2}^2$ bound to an $\|u\|_{H^1_0}^2$ bound, we use the identity $\|u\|_{H^1_0}^2 = \|\nabla u\|_{L^2}^2 + \|u\|_{L^2}^2$ for the [Sobolev space](/pages/1018) $H^1_0(U)$:
\begin{align*}
B[u, u] &\ge \frac{\theta}{2}\|\nabla u\|_{L^2(U)}^2 - \gamma_0 \|u\|_{L^2(U)}^2 \\
&= \frac{\theta}{2}\bigl(\|\nabla u\|_{L^2(U)}^2 + \|u\|_{L^2(U)}^2\bigr) - \gamma_0 \|u\|_{L^2(U)}^2 - \frac{\theta}{2}\|u\|_{L^2(U)}^2 \\
&= \frac{\theta}{2}\|u\|_{H^1_0(U)}^2 - \left(\gamma_0 + \frac{\theta}{2}\right)\|u\|_{L^2(U)}^2.
\end{align*}
With $\beta = \theta/2$ and $\gamma = \gamma_0 + \theta/2$, this is the Garding inequality.
[/guided]
[/step]
