In section 2.3.1, Evans PDE, it considers the solution of heat equation $u_t=\Delta u$ on $\mathbb{R}^n\times (0,\infty)$ with the following form: $$ u(x,t)=v(\frac{|x|^2}{t}) $$ Let $v(y)=w(|y|)$ and $r=|y|$. By direct calculations, one can get $$ r^{n-1}w'+\frac{1}{2}r^nw=a $$ for some constant $a$.
It says that assuming $\lim\limits_{r\rightarrow\infty}w,w'=0$, then we concluse $a=0$.
My question is how to get $a=0$ by using the assumption. I am not sure how to obtain $$ \lim\limits_{r\rightarrow\infty}r^{n-1}w'+\frac{1}{2}r^nw=0 $$ from the assumption.
I personally find the answer to this question is, although not difficult, not very straightforward though. Hope my answer is helpful to students who study Evans in the future. If anyone has any easier, more straightforward answer, please let me know as well.
I will divide the problem into three cases depending on $n$. The idea behind them are actually similar.
When $n=1$, by method of integrating factor we can observe that \begin{align} \frac{d}{dr}\bigg(e^{r^2/4}w(r)\bigg)=ae^{r^2/4} \end{align} Integrating over a finite interval $[0,r]$ gives \begin{align} w(r)=e^{-r^2/4}\left(c+\int^r_0ae^{s^2/4}ds\right) \end{align} From this we already can deduce that \begin{align} \lim_{r\to+\infty}w(r)=0 \end{align} (The part $ce^{-r^2/4}$ is trivial; the other part follows by applying L'Hopital's rule together with the Fundamental theorem of calculus.)
Thus, in my opinion, to deduce that $a=0$, what we really need is the assumption that \begin{align} \lim_{r\to+\infty}w'(r)=0 & & (1) \end{align} Indeed, we can compute that \begin{align} w'(r)=-\frac{c}{2}\frac{r}{e^{r^2/4}}+a\left[1-\frac{1}{2re^{r^2/4}}\int^r_0e^{s^2/4}ds\right] \end{align} which implies that \begin{align} \lim_{r\to+\infty}w'(r)=a \end{align} Hence, if we assume the decaying property (1), we will have $a=0$ as desired.
When $n\geq 2$, again we apply the method of integrating factor, after dividing both sides of the equation by $r^{n-1}$, and obtain \begin{align} \frac{d}{dr}\bigg(e^{r^2/4}w(r)\bigg)=\frac{ae^{r^2/4}}{r^{n-1}} \end{align} Note that now the R.H.S. is not well-defined at $r=0$. Thus if we naively integrate over $(0,r]$, the integral will be an improper one, whose convergence needs to be further analysed.
Thus, let me first integrate it over a finite interval $[\rho,r]$, where $0<\rho<r<\infty$, and obtain \begin{align} w(r)=e^{-r^2/4}\left[e^{\rho^2/4}w(\rho)+\int^{r}_{\rho}\frac{ae^{s^2}/4}{s^{n-1}}ds\right] & & (2) \end{align} We hope that when we take limit as $\rho\searrow 0$, everything remains fine. The only place that is possibly problematic is of course then the improper integral.
When $n=2$, by simple comparison test we have \begin{align} \int^{r}_{\rho}\frac{e^{s^2}/4}{s^{n-1}}ds \geq \int^{r}_{\rho}\frac{1}{s}ds =\ln\frac{r}{\rho}\to \infty \qquad\text{as }\rho\to 0 \end{align} and so, if $a\neq 0$, the R.H.S. of (2) becomes $\pm\infty$.
When $n\geq 3$, the idea is similar: \begin{align} \int^{r}_{\rho}\frac{e^{s^2}/4}{s^{n-1}}ds \geq \int^{r}_{\rho}\frac{1}{s^{n-1}}ds =\frac{1}{n-2}\left[\frac{1}{\rho^{n-2}}-\frac{1}{r^{n-2}}\right] \to \infty\qquad\text{as }\rho\to 0 \end{align} Hence things can go well only if $a=0$.
Personally I do not find the conditions $\lim_{r\to\infty}w,w'=0$ being used in the deduction, and so it indeed sounds a bit confusing for Evans to write so. Of course I may be wrong, and I welcome any correction.