Let \(\vartheta_2\) be as defined in the proof of (0x68c9184d) . Then there exists a numerical constant \(C>0\) such that
\[ \lvert \Delta \vartheta_1(z(x,y))\rvert^2 \le C \frac{\beta^2(1 + \alpha_-^2)^2}{\delta^4 R_2^4}. \]
Proof
We write \(\lesssim\) if \(\le\) holds up to a positive numerical constant.
By the product rule for the divergence , we obtain
\[ \lvert \Delta(\vartheta_1(z))\rvert^2 = \lvert \div (\vartheta_1'(z) \nabla z)\rvert^2 \le \lvert \vartheta_1''(z)\rvert^2 \lvert \nabla z\rvert^4 + \lvert v_1'(z)\rvert^2 \lvert \Delta z\rvert^2. \]With bounds on \(\vartheta_1'\), \(\vartheta_1''\) (see (0x68edf29b) ), \(R_2< 1\), \(\delta< 1\), \(\beta>1\), and estimates on \(\nabla z\) and \(\Delta z\) (see (0x68ecae27) ), we obtain
\[ \lvert \Delta v_1(z)\rvert^2 \lesssim \frac{1+\alpha_-^4}{\delta^4 R_2^4}+\frac{\beta^2}{\delta^4 R_2^2} \lesssim \frac{\beta^2(1 + \alpha_-^2)^2}{\delta^4 R_2^4}. \]