= Solution
Let $u=v+\psi$, so $v$ has zero <Dirichlet boundary condition>. Using the supplied form of the <minimal surface equation for a graph>, set
$$
Q_\psi(v)=\widetilde Q(v+\psi)-\widetilde Q(\psi),\qquad
f_\psi=\widetilde Q(\psi)-\Delta\psi.
$$
For $\|\psi\|_{C^{2,\alpha}}\leq\beta\leq1$ and $\|v\|,\|w\|\leq1$, the difference bound implies
$$
\|Q_\psi(v)\|_{C^{0,\alpha}}\leq C\|v\|_{C^{2,\alpha}}^2+2C\beta,
$$
and
$$
\|Q_\psi(v)-Q_\psi(w)\|_{C^{0,\alpha}}
\leq\bigl[C(\|v\|_{C^{2,\alpha}}+\|w\|_{C^{2,\alpha}})+2C\beta\bigr]\|v-w\|_{C^{2,\alpha}}.
$$
The shifted arguments stay in the permitted radius-two ball, and $\|f_\psi\|_{C^{0,\alpha}}\leq C\beta^2+C_n\beta$. The <Laplacian> with zero <Dirichlet boundary condition> has trivial <null space> by the <maximum principle for harmonic functions>. Choose $\beta>0$ small enough that $2C\beta\leq\delta$ and $C\beta^2+C_n\beta\leq\varepsilon_0$, for the constants in part (c) with $L=\Delta$. Its <small-data existence for a nonlinear elliptic Dirichlet problem> supplies $v$, and $\boxed{u=v+\psi}$ solves the <minimal surface equation for a graph> with the required boundary values.
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