Square roots of elliptic operators

Consider an elliptic sesquilinear form defined on $\text{V}$ × $\text{V}$ by $\text{Ju, v] = ∫}{}_{\mathrm{\Omega }}\text{a}{}_{\mathrm{jk}}\text{?u}\text{?x}{}_{\mathrm{k}}\text{}\text{?}\text{t6v}\text{?x}{}_{\mathrm{j}}\text{+ a}{}_{\mathrm{k}}\text{?u}\text{?x}{}_{\mathrm{k}}\text{v}\text{?}\text{+ α}{}_{\mathrm{j}}\text{u}\text{}\text{?}\text{t6v}\text{?x}{}_{\mathrm{j}}\text{+ au}\text{v}\text{?}\text{dx}$, where $\text{V}$ is a closed subspace of $\text{H}{}^1\text{(Ω)}$ which contains $\text{C}{}_0{}^{\mathrm{\infty }}\text{(Ω)}$, Ω is a bounded Lipschitz domain in $\text{R}$ⁿ, $\text{a}{}_{\mathrm{jk}}\text{, a}{}_{\mathrm{k}}\text{, α}{}_{\mathrm{j}}\text{, a ? L}{}_{\mathrm{\infty }}\text{(Ω),}\text{and Re}\text{a}{}_{\mathrm{jk}}\text{ζ}{}_{\mathrm{k}}\text{ζ}{}_{\mathrm{j}}\text{? κ > 0}$ for all ζ?$\text{C}$ⁿ with ¦ζ¦ = 1. Let L be the operator with largest domain satisfying Ju, v] = (Lu, v) for all υ∈$\text{V}$. Then L + λI is a maximal accretive operator in $\text{L}{}_2\text{(Ω)}$ for λ a sufficiently large real number. It is proved that $\text{(L + λI)}{}^{\mathrm{<mtext>1</mtext><mtext>2</mtext>}}$ is a bounded operator from $\text{V}$ to $\text{L}{}_2\text{(Ω)}$ provided mild regularity of the coefficients is assumed. In addition it is shown that if the coefficients depend differentiably on a parameter t in an appropriate sense, then the corresponding square root operators also depend differentiably on t. The latter result is new even when the forms J are hermitian.