$U=U(x)$ satisfies $(U^{2})_{x}-\nu U_{xx}=f$ where $x\in R$. $U$ is positive, $C^{2}$ solution which is non-decreasing.
Now author wrote that since $U$ is monotone, the limits $U(-\infty):=\lim_{n\to -\infty}U(x) $ and $U(\infty):=\lim_{n\to \infty}U(x) $ exist which is confusing to me. I am not sure how to prove this however i know that monotonic function has a limit at positive or negative infinity which is either real number, $\infty$ or $-\infty$. But still not able to prove this.
After this integration of $(U^{2})_{x}-\nu U_{xx}=f$ from $-\infty$ to $\infty$,
we get $$U^{2}(\infty)-U^{2}(-\infty)-\nu (U_{x}(\infty)-U_{x}(-\infty))=\int_{-\infty}^{\infty}f(x)dx$$
now i need to show $U_{x}(\infty)$ and $U_{x}(-\infty)$ are zero.which also i am not able to prove. I have seen this question Proving that $\lim\limits_{x\to\infty}f'(x) = 0$ when $\lim\limits_{x\to\infty}f(x)$ and $\lim\limits_{x\to\infty}f'(x)$ exist
but how can we say $\lim_{x\to \infty}U(x)$ and $\lim_{x\to \infty}U'(x)$ exist. I know $U(x)$ is two times continuously differentiable in $R$. That means limit has to exist through out $R$. But since we are considering $R$ which is not extended real number set, how can we be sure that limit will exists at infinity?
Thanks for your precious time.
