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Willie Wong
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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 ( 1 + \ell \Delta t)} \leq \dfrac{1}{\ell \Delta t}$$\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

$$ \ell = C_1 + C_2 $$

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 $\dfrac{1}{\ell \Delta t ( 1 + \ell \Delta t)} \leq \dfrac{1}{\ell \Delta t}$

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

$$ \ell = C_1 + C_2 $$

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

$$ \ell = C_1 + C_2 $$

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Willie Wong
  • 39k
  • 4
  • 94
  • 176

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 $\dfrac{1}{\ell \Delta t ( 1 + \ell \Delta t)} \leq \dfrac{1}{\ell \Delta t}$

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

$$ \ell = C_1 + C_2 $$