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TVD (total variation diminishing) finite difference methods that produce non-oscillatory solutions are based on the total variation. In LeVeque's book the total variation of a function $q(x)$ is defined as (p. 109)

$TV(q)=\sup\sum_{j=1}^N\left|q(\xi_i)-q(\xi_{i-1})\right|$

where "the supremum is taken over all subdivisions of the real line $-\infty=\xi_0<\xi_1<\ldots < \xi_N=\infty$". I do not understand that last part. Assuming a specific value for $N$ and a function $q(x)$, we can carry out the summation. I don't understand where the supremum comes into it or why it is necessary.

Brian Zatapatique
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1 Answers1

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The supremum in the definition is necessary to capture the maximum variation in the function.

Here's an example to illustrate what happens if we ignore the supremum and take any fixed partition of the real line:

Consider the function $\sin(x)$, and let the subdivision of the real line be the zeros of $\sin(x)$. Then its variation over that subdivision is zero, even though its total variation is infinite because it does not have compact support. We could choose a similar example with compact support by replacing $\sin(x)$ with the zero function sufficiently far away from the origin and the point would be the same.

The basic idea of total variation is to capture the unsigned, total vertical distance traveled by the function. Oscillations like the sine function will travel a nonzero vertical distance, and will travel a greater vertical distance over a fixed interval as their frequency increases. To capture that vertical distance accurately for all functions, you need the supremum to be able to adapt the subdivision to the function. Then you can detect the degree to which your solution "oscillates", which makes it a useful property to track for numerical methods for solving hyperbolic partial differential equations.

Geoff Oxberry
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  • Ok, so if I understand you correctly, the supremum is necessary to accurately represent the TV of the function for arbitrary $\xi_i$. But couldn't this be avoided by taking $\lim_{N\rightarrow\infty}$ of the summation rather than $\sup$? – Brian Zatapatique Jun 20 '13 at 07:35
  • No. Suppose I instead had a function equal to zero everywhere but $[-\pi, \pi]$, and equal to $\sin(x)$ on $[-\pi, \pi]$. I could cluster an infinite number of points outside of $[-\pi, \pi]$, and zero points within $[-\pi, \pi]$, and come up with the wrong result that the total variation of this function is zero. Having an infinite number of points says nothing about the distribution of those points on the real line. – Geoff Oxberry Jun 20 '13 at 07:55
  • Fair point. So assuming the points were distributed smartly, it appears the $\sup$ could be replaced by $\lim_{N\rightarrow\infty}$. – Brian Zatapatique Jun 21 '13 at 07:17
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    The whole point of the supremum is to achieve what you describe as a smartly distributed set of points partitioning the real line. Even taking a sequence of equipartitions of the real line in the limit as $N$ goes to infinity will not necessarily yield a sequence whose limit is the supremum. Your intuition would work if we were talking about the Riemann integral -- some undergraduate textbooks define the Riemann integral over an equipartition of the real line as its width goes to zero (i.e., $N \rightarrow \infty$). However, for total variation, you can construct pathological counterexamples. – Geoff Oxberry Jun 21 '13 at 08:50