Since u(x,y)=f(x^2+\frac{1}{y}) and u(x,0)=\phi(x) ,
we have u(x,0)=f(x^2+\frac{1}{\epsilon})=\phi(x) where \epsilon\rightarrow 0. This would imply that \lim_{u\rightarrow \infty}f(u)=\phi(x). Then again, that can't be right because the limit as u\rightarrow \infty should not depend on x. I'm...