A bump that slides
Tonight's idea
Sand lies in a pile on a conveyor belt. At the start, its height at position \(x\) is \(f(x)\). The belt runs right at a steady speed \(v\). What is the height at position \(x\) at time \(t\)?
A grain that started at \(x_0\) is now at \(x = x_0 + vt\). So the sand you see at \(x\) started at \(x - vt\), and it brought its height along.
The height at position \(x\) and time \(t\) is \(f(x - vt)\): the shape \(f\), unchanged, moved a distance \(vt\). With \(v \gt 0\) it moves right. To move it left, write \(f(x + vt)\).
Plot a grain's path with \(x\) across and \(t\) up. It is the straight line \(x - vt = x_0\), called a characteristic. The height is carried along it, unchanged.
The belt obeys a rule about rates: at each place, the height changes at \(-v\) times the slope of the pile there. That is a first-order partial differential equation. Following characteristics solves it.
This is a model: a steady belt, and sand that never slumps or spills.
Watch one: three snapshots of one bump
Take \(f(x) = \dfrac{1}{1 + x^2}\) and \(v = 1\). The height at position \(x\) and time \(t\) is \(f(x - t) = \dfrac{1}{1 + (x - t)^2}\).
Peak. The fraction is biggest when \((x - t)^2 = 0\), so the peak is at \(x = t\): at 0, then 1, then 2. Its height is 1 every time. Subtracting \(t\) moves the bump right, because \(x\) now has to be bigger to give the same height.
Width. The height is \(\frac{1}{2}\) when \((x - t)^2 = 1\), so at \(x = t - 1\) and \(x = t + 1\). At half height the bump is 2 wide at every time. Nothing about the shape changes.
Stand still. At \(x = 2\) you see \(\frac{1}{1 + 4} = \frac{1}{5}\) at \(t = 0\), then \(\frac{1}{1 + 1} = \frac{1}{2}\) at \(t = 1\), then 1 at \(t = 2\) as the peak passes. After that the heights come back down in reverse order: \(\frac{1}{2}\) at \(t = 3\) and \(\frac{1}{5}\) at \(t = 4\).
Your turn
- Keep \(f(x) = \frac{1}{1 + x^2}\). A bump has height \(f(x - 3t)\). (a) How fast is it moving, and which way? (b) Where is its peak at \(t = 2\)? (c) What is the height at \(x = 5\) at \(t = 2\)? (d) At which two times is the height at \(x = 9\) equal to \(\frac{1}{2}\)?
- (a) Write a formula for the same bump moving left at speed 2, starting with its peak at \(x = 0\). (b) Where is its peak at \(t = 3\)? (c) Another bump has height \(f(x + 2t - 9)\). Where is its peak at \(t = 0\), and which way does it move? (d) When and where does its peak meet the peak of \(f(x - t)\)?
- A belt runs right at 1.5 m each second. At \(t = 0\) the peak of a pile is at \(x = 2\) m. (a) Where is the peak after 3 s? (b) On spare paper, draw axes with \(x\) across (0 to 12) and \(t\) up (0 to 6). Draw the characteristics of the grains that start at \(x_0 = 0, 1, 2\) and \(3\). What do you notice about the four lines? (c) Which starting point \(x_0\) has its characteristic through \(x = 8\), \(t = 4\)? What does that tell you about the height there? (d) The belt ends at \(x = 11\) m. When does the peak fall off?
- Press a very long string into the same shape \(f\) and let go from rest. In the ideal-string model with wave speed 1, its height at time \(t\) is \(w = \frac{1}{2}f(x - t) + \frac{1}{2}f(x + t)\). (a) Show that \(w = f(x)\) when \(t = 0\). (b) At \(t = 10\), find \(w\) at \(x = 10\), \(x = -10\) and \(x = 0\). Describe the string. (c) Show that the height at \(x = 0\) is \(\frac{1}{1 + t^2}\). When is it \(\frac{1}{10}\)? (d) At \(t = 0.5\), find \(w\) at \(x = 0\) and at \(x = 0.5\). Has the bump split yet?
Stretch: when the tall parts go faster
In shallow water, the taller parts of a wave travel faster. Here is a toy model: each point of a shape moves right at a speed equal to its own height \(h\), so a point that starts at \(x_0\) is at \(x_0 + ht\) at time \(t\). Start with a tent: height \(1 - x\) for \(0 \le x \le 1\), \(1 + x\) for \(-1 \le x \le 0\), and 0 elsewhere. (a) Where are the peak and the two feet at \(t = 0.5\)? (b) The sides stay straight. How steep are the front and the back at \(t = 0.5\)? (c) Write the characteristics of the points that start at \(x_0 = 0\), \(0.5\) and \(1\). Where do they meet, and what does the front look like then? (d) What heights does the rule give at \(x = 1.5\) when \(t = 2\)? Why is that a problem?
Answers
- (a) Here \(vt = 3t\), so \(v = 3\): right at speed 3. (b) The peak is where \(x - 3t = 0\), so at \(x = 6\). (c) \(f(5 - 6) = f(-1) = \frac{1}{2}\). (d) The height is \(\frac{1}{2}\) when \(9 - 3t = 1\) or \(9 - 3t = -1\), so at \(t = \frac{8}{3}\) and \(t = \frac{10}{3}\). The peak passes \(x = 9\) halfway between, at \(t = 3\).
- (a) \(f(x + 2t) = \dfrac{1}{1 + (x + 2t)^2}\). At \(t = 0\) this is \(f(x)\), with its peak at 0. (b) Where \(x + 6 = 0\), so at \(x = -6\). (c) The peak is where \(x + 2t - 9 = 0\), so at \(x = 9 - 2t\). It starts at \(x = 9\) and moves left at speed 2. (d) The peak of \(f(x - t)\) is at \(x = t\). They meet when \(t = 9 - 2t\), so at \(t = 3\) and \(x = 3\).
- (a) It moves \(1.5 \times 3 = 4.5\) m, so it is at \(x = 6.5\) m. (b) The lines are \(x = x_0 + 1.5t\). Each starts on the \(x\)-axis at \(x_0\) and moves 1.5 across for each 1 up, so at \(t = 6\) they reach \(x = 9, 10, 11\) and \(12\). They are straight and parallel, and they never cross: the grains keep their order and their spacing. (c) \(x_0 = 8 - 1.5 \times 4 = 2\). That is where the peak started, so at \(t = 4\) the peak is at \(x = 8\). (d) \(2 + 1.5t = 11\) gives \(t = 6\) s.
- (a) At \(t = 0\) both terms are \(\frac{1}{2}f(x)\), and two halves of \(f(x)\) make \(f(x)\). (b) At \(x = 10\): \(\frac{1}{2}f(0) + \frac{1}{2}f(20) = \frac{1}{2}\left(1 + \frac{1}{401}\right) = \frac{201}{401} \approx 0.501\). At \(x = -10\) the two terms swap, so it is the same. At \(x = 0\): \(\frac{1}{2}f(-10) + \frac{1}{2}f(10) = \frac{1}{101} \approx 0.0099\). The string has two bumps, each about half as tall as the first one, near \(x = 10\) and \(x = -10\), with almost nothing between. (c) At \(x = 0\), \(w = \frac{1}{2}f(-t) + \frac{1}{2}f(t)\). Because of the square, \(f(-t) = f(t)\), so \(w = f(t) = \frac{1}{1 + t^2}\). It is \(\frac{1}{10}\) when \(t^2 = 9\), so at \(t = 3\) (taking \(t \ge 0\)). (d) At \(x = 0\): \(w = f(0.5) = \frac{1}{1.25} = 0.8\). At \(x = 0.5\): \(\frac{1}{2}f(0) + \frac{1}{2}f(1) = 0.5 + 0.25 = 0.75\). The middle is still the highest point, so there is still one hump. The two halves overlap at first. For this \(f\), a dip appears in the middle only after \(t = \frac{1}{\sqrt 3} \approx 0.58\). This formula for a string is named after d'Alembert.
Stretch. (a) The peak has height 1, so it is at \(0 + 1 \times 0.5 = 0.5\). The feet have height 0, so they stay at \(-1\) and \(1\). (b) The front drops 1 between \(x = 0.5\) and \(x = 1\), so its slope is \(-2\). The back rises 1 between \(x = -1\) and \(x = 0.5\), so its slope is \(\frac{2}{3}\). At the start they were \(-1\) and \(1\): the front has steepened and the back has flattened. (c) The heights are 1, 0.5 and 0, so the lines are \(x = t\), \(x = 0.5 + 0.5t\) and \(x = 1\). All three pass through \(x = 1\) at \(t = 1\). So does every other point of the front, since \(x_0 + (1 - x_0)t = 1\) when \(t = 1\). At that moment the front is a vertical cliff. These characteristics are still straight, but they are no longer parallel, so they can cross. (d) Three heights. From the front, \(x_0 + 2(1 - x_0) = 1.5\) gives \(x_0 = 0.5\) and height \(\frac{1}{2}\). From the back, \(x_0 + 2(1 + x_0) = 1.5\) gives \(x_0 = -\frac{1}{6}\) and height \(\frac{5}{6}\). And the ground at \(x_0 = 1.5\) has height 0 and has not moved. A shape has only one height at each place, so the toy model has broken down at \(t = 1\). A real wave running up a beach steepens at the front in a similar way, and then it breaks. What happens next needs more physics than this sheet has.