Does $\lim_{t\rightarrow\infty}\sum_{n=1}^{\infty}c_n\cdot e^{(h-n^2)t}\cdot \sin(nx)$ exist?

Question

$$u(x,t)=\sum_{n=1}^{\infty}c_n\cdot e^{(h-n^2)t}\cdot \sin(nx)$$ Does $\lim_{t\rightarrow\infty}u(x,t)$ exist?

where: $c_n=\frac{4}{n^3\pi}((-1)^n-1)$. I have been told to examine when $h<1, \quad h=1,\quad h>1$. But I have no idea for this?

Answer 1

You were given a good hint. Indeed, what plays a crucial role here is the sign of the coefficient in front of $t$ in exponential, i.e. $h - n^2$.

Notice that if $h < 1$ then $h - n^2 < 0$ for every $n$ and hence $e^{(h - n^2)t} \to 0$ as $t \to \infty$. Moreover, since the series is uniformly convergen we have that $$\lim_{t \to \infty}u(x,t) = \sum_{n = 1}^{\infty}\lim_{t \to \infty}c_ne^{(h - n^2)t}\sin(nx) = 0.$$ You can procede similary in the other cases (although as you'll see you won't get the same answer).