I'm studying a paper (see citation below) on numerical analysis, and came across this estimate. I am unable to figure out what was done in the final step, and I am not certain if this was just a typo in the paper.


**Preliminary information**: $\Delta{t}$ is the time step, $n$ is the iteration count from $0$ to final iteration $(N+1)$ such that $T=N\Delta{t}$ is the final time at $n=N+1$, $\vec{u}$ is a solution vector, $\vec{f}$ is a vector of forcing functions, and $E\_k$ is the energy at iteration $n = k$. We also have various positive constants $C\_1, C\_2, C\_3, C\_a$.


**Here is the estimate in the text**:


$E\_n + \frac{C\_a\Delta{t}^2(C\_1+C\_2)}{1+(C\_1+C\_2)\Delta{t}}\|\nabla\vec{u}^{n+1}\|^2+\frac{C\_a\Delta{t}^2(C\_1+C\_2)}{3(1+(C\_1+C\_2)\Delta{t})}\|\nabla\vec{u}^n\|^2$


$\le C\_3\|\vec{f}^{n+1}\|^2\Delta{t} + (1 + (C\_1 + C\_2)\Delta{t})E\_{n-1}$


$\le e^{(C\_1+C\_2)T}E\_0 + \frac{C\_3}{C\_1+C\_2}e^{(C\_1+C\_2)T} \max\limits\_{n}\|\vec{f}^{n+1}\|^2$


Paper: *Chen, Wenbin; Gunzburger, Max; Sun, Dong; Wang, Xiaoming*, [**Efficient and long-time accurate second-order methods for the Stokes-Darcy system**](http://dx.doi.org/10.1137/120897705), SIAM J. Numer. Anal. 51, No. 5, 2563-2584 (2013). [ZBL1282.76094](https://zbmath.org/?q=an:1282.76094).


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Here is my attempt at what happened at the last step, before the final inequality:


**EDIT: Most recent attempt, just as the answer below came in:**


For clarity, take $X\_n = \frac{C\_a\Delta{t}^2(C\_1+C\_2)}{1+(C\_1+C\_2)\Delta{t}}\|\nabla\vec{u}^{n+1}\|^2+\frac{C\_a\Delta{t}^2(C\_1+C\_2)}{3(1+(C\_1+C\_2)\Delta{t})}\|\nabla\vec{u}^n\|^2$


I will use following prelimary estimates $\forall\ x,n\in\mathbb{R}\_+$. The second employs the first.
$(1+x)^n \le e^{nx}$ and
$\sum\_{k=1}^n(1+x)^k \le n(1+x)^n \le n e^{nx}$


So the estimate was
$$E\_n + X\_n \le C\_3\Delta{t}\|\vec{f}^{n+1}\|^2+(1 +\ (C\_1+C\_2)\Delta{t})E\_{n-1}$$
$$\le C\_3\Delta{t}\left(\frac{1+(C\_1+C\_2)\Delta{t}}{(C\_1+C\_2)\Delta{t}}\right)\|\vec{f}^{n+1}\|^2+(1 +\
(C\_1+C\_2)\Delta{t})E\_{n-1}$$
i.e.
$$E\_n + X\_n \le \left(1+(C\_1+C\_2)\Delta{t}\right)\left(\frac{C\_3}{C\_1+C\_2}\|\vec{f}^{n+1}\|^2+E\_{n-1}\right) \tag{1}$$
Since $X\_n \ge 0$ we also have
$$E\_n \le \left(1+(C\_1+C\_2)\Delta{t}\right)\left(\frac{C\_3}{C\_1+C\_2}\|\vec{f}^{n+1}\|^2+E\_{n-1}\right) \tag{2}$$
Using (2) in (1),
$$E\_n + X\_n \le \Bigl(1+(C\_1+C\_2)\Delta{t}\Bigr)$$
$$\cdot\left(\frac{C\_3}{C\_1+C\_2}\|\vec{f}^{n+1}\|^2
+ \Bigl(1+(C\_1+C\_2)\Delta{t}\Bigr)\left(\frac{C\_3}{C\_1+C\_2}\|\vec{f}^{n}\|^2+E\_{n-2}\right)\right)$$
Repeating the process $n-1$ times,
$$E\_n + X\_n \le \frac{C\_3}{C\_1+C\_2}\sum\_{k=1}^n \|\vec{f}^{k+1}\|^2
\Bigl(1+(C\_1+C\_2)\Delta{t}\Bigr)^k+ \Bigl(1+(C\_1+C\_2)\Delta{t}\Bigr)^n E\_0$$
i.e.
$$E\_n + X\_n \le \frac{C\_3}{C\_1+C\_2}\max\_n\|\vec{f}^{n+1}\|^2\sum\_{k=1}^n 
\Bigl(1+(C\_1+C\_2)\Delta{t}\Bigr)^k+ \Bigl(1+(C\_1+C\_2)\Delta{t}\Bigr)^n E\_0$$
Using the first preliminary estimate,
$$E\_n + X\_n \le \frac{C\_3}{C\_1+C\_2}\max\_n\|\vec{f}^{n+1}\|^2 n 
\Bigl(1+(C\_1+C\_2)\Delta{t}\Bigr)^n+ \Bigl(1+(C\_1+C\_2)\Delta{t}\Bigr)^n E\_0$$
And using the second preliminary estimate,
$$E\_n + X\_n \le \frac{nC\_3}{C\_1+C\_2}\max\_n\|\vec{f}^{n+1}\|^2 
e^{(C\_1+C\_2)T}+ e^{(C\_1+C\_2)T} E\_0$$


There is a certain $n$ multiplying the first term, which is not present in the paper.



Resolving an inequality in the final step of derivation of an *a priori* energy estimate
It looks to me like a discrete version of Gronwall's inequality. 


If you have a sequence of numbers satisfying


$$ E\_n \leq k \Delta t + (1 + \ell \Delta t) E\_{n-1} $$


You can rewrite 


$$ A\_n = (1 + \ell \Delta t)^{-n} E\_n $$


to get


$$ A\_n \leq \frac{k \Delta t}{(1 + \ell \Delta t)^n} + A\_{n-1} $$


which implies


$$ A\_n \leq A\_0 + k\Delta t \sum\_{m = 1}^n \frac{1}{(1 + \ell \Delta t)^m} $$


The sum is a geometric series bounded by $\dfrac{1}{\ell \Delta t}$ (see comment below for the computation)


So you conclude


$$ A\_n \leq A\_0 + \frac{k}{\ell} $$


This gives


$$ E\_n \leq (1 + \ell \Delta t)^n ( E\_0 + \frac{k}{\ell} ) $$


which is exactly what is claimed, using that in the above derivation you can use 


$$ k = C\_3 \max \| f^{n+1} \|^2 $$


and 


$$ \ell = C\_1 + C\_2 $$


