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I am dealing with a mean curvature type equation as following: $\displaystyle{\sum_{i,j=1}^{2}}(\delta_{ij}-\frac{u_{i}u_{j}}{1+|Du|^{2}})u_{ij}=(1+|Du|^{2})^{\frac{1}{2}-\frac{1}{2\alpha}}$, where $\alpha>1$ fixed. $u$ is convex and defined on the entire $R^{2}$ suppose when $|x|$ is large, $C_{1}|x|^{\alpha}\leq|Du(x)|\leq C_{2}|x|^{\alpha}$, where $C_{1}$ and $C_{2}$ are fixed positive constants. Then is there such estimate that: when $|x|$ is large $|D^{2}u|\leq C_{3}|x|^{\alpha-1}$ for some fixed constant $C_{3}$.

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You can find out the right power in the radial case, $u=f(\sqrt{x^2+y^2})$. Your equation becomes $$ \frac{\displaystyle 1+ \left(\frac{1}{r^2}-1\right)(f^\prime)^2}{\displaystyle 1+\frac{1}{r^2}(f^\prime)^2} f^{\prime\prime} + \frac{1}{r} f^{\prime} = \left(1+\frac{1}{r^2}(f^\prime)^2\right)^{\frac{1}{2}-\frac{1}{2\alpha}} $$ Now for $r$ large, the assumption $C_1<r^{-\alpha} |f^\prime| <C_2$ means, if you write $g(r)= r^{-\alpha} f^\prime $, $$ (-1 + o(1)) {r^2} f^{\prime\prime} + r^{\alpha-1} g = r^{\displaystyle \frac{(\alpha-1)^2}{\alpha}} (g^2(r)+o(1)), $$ Since $\alpha>1$, $\alpha-1 > \frac{(\alpha-1)^2}{\alpha}$, and therefore
$$ f^{\prime\prime} \approx \frac{1}{r^2}f^\prime = O(r^{\alpha-3}). $$ So it looks like the non diagonal term is all right. In the general case, I would guess (to be checked) that it would the same, the worst term coming from the diagonal, $$ \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} u, $$ removing two powers of $r$, thus $r^{\alpha-1}$ indeed.

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