# Existence and Uniqueness of a PDE proved by method of continuity

Consider a nonlinear differential equation in $\mathbb{R}^d$

$$a^{ij}(x)u_{ij}(x) + F(u_x(x),u(x),x)-\lambda u(x) = 0,$$

where $u_{ij} = \frac{\partial^2 u}{\partial x_i \partial x_j}$ and Einstein summation is assumed.

Know

1. $F(\alpha,\beta,x)$ is a measurable function of $(\alpha,\beta,x)$ on $\mathbb{R}^d\times\mathbb{R}^d\times\mathbb{R}^d$ and an $f\in L^2$ exists such that we have $$F(\alpha,\beta,x)\leq K(|\alpha|+|\beta|+f(x)),$$ $$|F(\alpha_1,\beta_1,x)-F(\alpha_2,\beta_2,x)|\leq K(|\alpha_1 - \alpha_2|+|\beta_1-\beta_2|)$$ for all $\alpha,\beta,x$;
2. constant of ellipticity: $$\exists\, \kappa>0: \kappa|\xi|^2\le a^{ij}\xi^i\xi^j\le \kappa^{-1}|\xi|^2\qquad \text{for all }\xi\in \mathbb{R}^d;$$
3. uniform continuity of $a^{ij}$: $$|a^{ij}(x)-a^{ij}(x)|\le\omega(|x-y|)$$ where $\omega(\epsilon)$ is a increasing function s.t. $\omega(\epsilon)\downarrow 0$ as $\epsilon \downarrow 0$.

The goal is to prove that there exists $\lambda_0 = \lambda_0(d,\kappa,K,\omega)$ such that for all $\lambda\ge\lambda_0$ the differential equation has a unique solution $u\in W^{2,2}.$

I suspect it could be tackle with the method of continuity however the nonlinear term makes that it is difficult to build $L_t u:=(1-t)(\Delta-\lambda)u+tLu$ differential operator.

Any suggestions would be appreciated. Thank you in advance.

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