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I am considering the following 2nd order PDE :

On a domain $R$, for some $r > 0$, \begin{equation*} \frac{1}{2}\sum_{i, j = 1}^{K}\gamma_{i}\gamma_{j}U_{x_{i}x_{j}}(x) + \sum_{i = 1}^{K}\frac{\gamma_{0}\gamma_{i}}{x_{0}}U_{x_{i}}(x) - rU(x) = -rg\,\,\,\,\,\,\,(1) \end{equation*} \begin{align} \gamma_{i} &= \gamma_{i}(a, b, x)\,\,\textrm{is Lipschitz continuous}\,\,\,\textrm{for each}\,\,i \in \{0,1,...,K\}\\ g &= g(a, b)\,\,\,\textrm{is a Lipschitz continuous function} \end{align} where, with $DU(x) = \big(U_{x_{1}}(x),...,U_{x_{K}}(x)\big)$, \begin{align*} a &= a\big(x, DU(x)\big)\\ b &= b\big(x, DU(x)\big) \end{align*}

I am considering the following iteration to show an exitence of $U$ to the above PDE:

  1. Pick any continuous functions $a_{0}(x)$ and $b_{0}(x)$ on $R$

  2. Plug them into $\gamma_{i}(a_{0}(x), b_{0}(x), x) = \gamma_{i}^{0}(x)$ and $g(a_{0}(x), b_{0}(x)) = g_{0}(x)$

  3. Solve the linear elliptic 2nd order PDE (1) on $R$ with $\gamma_{i}^{0}(x)$ and $g_{0}(x)$. Let the solution $U_{0}(x)$

  4. Let $a_{1}(x) = a(x, DU_{0}(x))$ and $b_{1}(x) = b(x, DU_{0}(x))$ be continuous functions w.r.t $x \in R$

  5. Plug them into $\gamma_{i}(a_{1}(x), b_{1}(x), x) = \gamma_{i}^{1}(x)$ and $g(a_{1}(x), b_{1}(x)) = g_{1}(x)$

  6. Solve the linear elliptic 2nd order PDE (1) on $R$ with $\gamma_{i}^{1}(x)$ and $g_{1}(x)$. Let the solution $U_{1}(x)$

  7. Iterate...

[QUESTION]

For the convergence of $\{U_{n}\}$ on $R$ to a limit,

I want to show something like a continuity property of the linear solution $U_{n}$ on the coefficients $\{\gamma^{n}_{i}, g_{n}\}_{i=0}^{K}$.

Could anyone suggest me some references for this question?

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  • $\begingroup$ This doesn’t look elliptic to me. The matrix $\gamma_i\gamma_j$ is not positive definite. $\endgroup$
    – Deane Yang
    Commented Jun 27, 2018 at 13:57
  • $\begingroup$ @DeaneYang Dear Yang, thanks for your comment. I am considering to impose any assumption that makes $\{\gamma_{i}\gamma_{j}\}_{i,j}$ a positive definite matrix so that the PDE (1) is uniformly elliptic in any compact subset of $R$. Do you think if this is reasonable for my approach? $\endgroup$
    – Philo
    Commented Jun 28, 2018 at 8:08
  • $\begingroup$ It’s never elliptic. $\endgroup$
    – Deane Yang
    Commented Jun 28, 2018 at 14:52
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    $\begingroup$ Have you checked whether the kernel of the matrix contains only $0$? $\endgroup$
    – Deane Yang
    Commented Jun 29, 2018 at 15:08
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    $\begingroup$ In that case, the PDE is indeed elliptic. Is $x_0 \ne 0$? If so, then this is a quasilinear elliptic PDE. Your approach will lead to solutions under some assumptions. You might want to consult the book by Gilbarg and Turdinger. if $x_0 = 0$ somewhere in $\R$, then more care is needed. $\endgroup$
    – Deane Yang
    Commented Jun 30, 2018 at 17:42

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