# Equivalence of Dirichlet problems. Gilbarg & Trudinger

I do not understand the proof of theorem 11.4 in the book "Elliptic Partial Differential Equations of Second Order" by Gilbarg & Trudinger. The reason is that I do not understand the text right before the theorem (about $T$ operator and the equivalence stated in page 281.)

The problem is, I do not see why the problems $u=\sigma Tu$ and $Q_\sigma$ are equivalent.

page 281 of Gilbarg

## 1 Answer

Just plugin the definition of $T$ and $Q_\sigma$. We have $$u = \sigma Tu \iff Tu = \frac{u}{\sigma}$$ that is, by definition of $T$ iff $v := \frac{u}\sigma$ is the (unique) solution of $$a^{ij}(x,u,Du)D_{ij}v + b(x,u,Du) = 0, \quad v|_{\partial \Omega} = \phi$$ Let here $v = \frac u\sigma$, this gives $$\frac 1\sigma a^{ij}(x,u,Du) D_{ij}u + b(x,u,Du) = 0, \quad \frac{u}{\sigma}\biggr|_{\partial \Omega} = \phi$$ which is true only iff (multiply with $\sigma$) $$a^{ij}(x,u,Du)D_{ij} u + b(x,u, Du) = 0 ,\quad u|_{\partial \Omega} = \sigma\phi$$ that is iff (by definition) $0 = Q_\sigma u$