Picture below is from Evans' Partial differential equations. I can't understand the red line. In my view, the speed of (2) should be $\sigma$. And I am sure it is not typo since the red line is used again later.
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1$\begingroup$ In applications, you need $y$ to be the unit vector, then $\frac{\sigma}{|y|} = \sigma$, which agrees with your intuition. However, as a function in math, in general, you can allow $y$ to be a more general vector. $\endgroup$Nicolas Bourbaki– Nicolas Bourbaki2025-10-15 07:51:03 +00:00Commented Oct 15 at 7:51
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$\begingroup$ @NicolasBourbaki The reason why I consider speed to be $\sigma$ is that the derivative of $y\cdot x -\sigma t$ is $-\sigma$, and ignore the sign. $\endgroup$Enhao Lan– Enhao Lan2025-10-15 10:27:50 +00:00Commented Oct 15 at 10:27
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2 Answers
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The speed of the wave should be the speed at which the level sets of $x\mapsto y\cdot x - \sigma t$ move. Suppose you take a level set $y\cdot x -\sigma t = c$, and vary time to produce $y\cdot x -\sigma t’ = c$. Via an explicit calculation, the distance between these parallel planes is given by
$$ d = \frac{|(c+\sigma t)-(c+\sigma t’)|}{|y|} = \frac{|\sigma|}{|y|}|t-t’|$$
To derive this formula, just calculate the distance between the points where the line parametrized by $x(s)=sy$ intersects the two planes. It follows that $\sigma/|y|$ indeed has the interpretation of speed.
Alternatively, the quantity $x\cdot y$ has units of length squared, which means $\sigma$ has units of length squared per unit time. Hence, that one needs to divide by a length scale to get the proper units for speed.
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$\begingroup$ Thank you for your clear answer. $\endgroup$Enhao Lan– Enhao Lan2025-10-16 01:03:16 +00:00Commented Oct 16 at 1:03
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Perhaps it might help to look at the one-dimensional case where the wave's displacement at position $x$ and time $t$ is given by $u(x,t) = f(ax-bt).$
If we follow what we consider to be the same point on the wave, we look at a constant value of $f,$ i.e. we have $f(ax-bt)=c$ for some constant $c.$ Assume that $f$ is locally invertible at the point we follow. Then we have $ax-bt = f^{-1}(c),$ i.e. $$ x = \frac{1}{a}\left( f^{-1}(c) + bt \right) = \frac{1}{a} f^{-1}(c) + \frac{b}{a}t . $$ From this we can directly read the velocity $v=b/a,$ i.e. the speed is $|v|=|b/a|=|b|/|a|.$
In the multi-dimensional case, $a$ is a vector containing information about the direction of the wave, but the speed is still $|b|/|a|.$
One can also use implicit differentiation to get $$ \frac{dx}{dt} = -\frac{\partial u/\partial t}{\partial u/\partial x} = -\frac{-b}{a} = \frac{b}{a}. $$
