Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 107 / 3 / b

For a compactly supported variation $v+t\varphi$, differentiation of the area functional at $t=0$ gives the first variationAfter integration by parts, this is the minimal surface equation for a graphFor $v_R(x)=R^{-1}v(Rx)$ one has $D{v}_R(x)=Dv(Rx)$. The equation is invariant under this scaling, so $v_R$ solves it on $B_1$ for every $R>0$.

Differentiate the equation for $v_R$ with respect to $x_k$. The derivative $w=D_k{v}_R$ is a weak solution ofwhereThe eigenvalue in the direction of $p$ is $(1+|p{|}^2{)}^{-3/2}$ and every orthogonal eigenvalue is $(1+|p{|}^2{)}^{-1/2}$. Thus a uniform bound on $|Dv|$ makes this a uniformly elliptic operator with constants independent of $R$.

Suppose $|Dv|\le M$. The coefficient matrices in part (ii) then have uniform ellipticity constants depending only on $M$. Applying the Harnack inequality for uniformly elliptic divergence-form equations to the nonnegative solutions $M+D_k{v}_R$ and $M-D_k{v}_R$ gives a scale-independent oscillation contractionScaling back,For fixed $r$, iterate this estimate with $R=4^{m}r$ and use the global bound $|D_{k}v|\le M$ to obtain ${\mathrm{osc}}_{B_r}{D}_{k}v\le 2M{\theta }^m\to 0$. Every partial derivative is therefore constant, so $v(x)=a\cdot x+b$ is an affine function. This is a bounded-gradient Bernstein theorem for entire minimal graphs.