Large and small time PDE solution

Question

I have the following solution for a PDE

$$ u(x,t)=(2x+4t-10)+2e^{\frac{-1}{2}t}+\sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)e^{\frac{-1}{4}n^2\pi^2t}}{n^3\pi^3(n^2\pi^2-2)} $$

I want to find the first two terms in the asymptotic expansion of u(x, t) as $t\rightarrow 0$ and $t\rightarrow \infty$(fixed $x$). Any hints or suggestions will be appreciated.

Answer 1

The sum $\Sigma(\cdots)$ is an asymptotic series as $t \to \infty$, so the first two terms of the expansion of $u(x,t)$ as $t \to \infty$ are simply the two terms which do not decrease exponentially, namely

$$ u(x,t) \sim \underbrace{4t}_\text{first} + \underbrace{2x-10}_\text{second} + \cdots $$ as $t \to \infty$.

Now, the first two terms of the expansion as $t \to 0$ should be what you get when you replace $e^{-n^2\pi^2t/4}$ and $e^{-t/2}$ by the first two terms of their Taylor series. That is,

$$ e^{-n^2\pi^2t/4} \approx 1 - \frac{n^2\pi^2}{4}t \qquad \text{and} \qquad e^{-t/2} \approx 1 - \frac{1}{2}t, $$

so we should have

$$ \sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)e^{-n^2\pi^2t/4}}{n^3\pi^3(n^2\pi^2-2)} \approx \sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n^3\pi^3(n^2\pi^2-2)} - \frac{1}{4} t \sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n\pi(n^2\pi^2-2)}. $$

We expect this to work since the new series we get after multiplying the summand by $n^2$ still converges, but as far as I know this does not constitute a proof. Anyway, we should therefore have

$$ u(x,t) \sim 2x - 10 + 2 - t + \sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n^3\pi^3(n^2\pi^2-2)} - \frac{1}{4} t \sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n\pi(n^2\pi^2-2)} + \cdots $$

or, after grouping terms,

$$ u(x,t) \sim \underbrace{2x - 8 + \sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n^3\pi^3(n^2\pi^2-2)}}_\text{first} - \underbrace{\left(1 + \frac{1}{4}\sum_{n=1}^{\infty} \frac{(-1)^n cos(\frac{n\pi}{2}x)}{n\pi(n^2\pi^2-2)}\right) t}_\text{second} + \cdots $$ as $t \to 0$.

Proving this could be tricky since the estimate $e^{-n^2\pi^2t/4} \approx 1 - n^2\pi^2t/4$ doesn't hold for large $n$. One can probably proceed by truncating the sum after $\lfloor 1/t \rfloor$ terms and showing that the error accrued isn't too large.