Typography of Lie groups, Lie algebras

This is a note for myself as I review my notes on finite groups, Lie groups (both the classic infinite ones and the finite simple groups of Lie type), Lie algebras, and specifically the typography for them.

First, no one agrees completely, and everyone's conventions is slightly different. What I mean by this: although everyone agrees, e.g., that the A_n family of Lie algebras refers to the same thing, some people use serif font for the "A", others use bold, some italicize, others do not, etc.

I'm inclined to follow Robert Wilson's conventions from his book The Finite Simple Groups (2009).

Families of Groups

Families of Simple Lie groups: use upright, teletype font for the family and "math italics" for the subscript if it's a variable, e.g., $\mathtt{A}_{n}$, $\mathtt{B}_{5}$, $\mathtt{C}_{m}$, $\mathtt{D}_{n^{2}}$, $\mathtt{E}_{8}$

Exceptional Finite Simple Groups of Lie Type: don't treat the formatting as special, so, e.g., the Steinberg groups would be ${}^{2}A_{n}(q^{2})$, ${}^{2}D_{n}(q^{2})$, ${}^{2}E_{6}(q^{2})$, ${}^{3}D_{4}(q^{3})$.

Sporadic simple group: these should be made upright, e.g., the Suzuki group is $\mathrm{Suz}$, the Matthieu groups look like $\mathrm{M}_{11}$, the Conway groups $\mathrm{Co}_{1}$ and $\mathrm{Co}_{2}$, and so on. BUT the exception to this rule is that the Monster group is written $\mathbb{M}$ and the Baby Monster $\mathbb{B}$.

Alternating, Cyclic, Symmetric group. These are just written as $A_{n}$, $C_{n}$, or $S_{n}$. The dihedral group, too, is $D_{n}$.

Classical Lie Groups: Here there is a double standard. For classical Lie groups over the reals or complex numbers, we write something of the form $\mathrm{GL}(n, \mathbb{F})$, $\mathrm{SL}(n, \mathbb{F})$, $\mathrm{U}(n, \mathbb{F})$, $\mathrm{SU}(n, \mathbb{F})$, $\mathrm{O}(n, \mathbb{F})$, $\mathrm{SO}(n, \mathbb{F})$, $\mathrm{Sp}(n)=\mathrm{USp}(n)$ for the compact Symplectic group, $\mathrm{Sp}(2n,\mathbb{F})$ for the generic Symplectic group.

The finite groups corresponding to these guys are written a little differently in my notes: the $n$ parameter is pulled out as a subscript, because frequently we write $q$ instead of $\mathbb{F}_{q}$ for finite fields...and then looking at $\mathrm{SL}(8,9)$ is far more confusing than $\mathrm{SL}_{8}(9)$. Thus we have $\mathrm{GL}_{n}(q)$, and so on.

Projective classical groups: the projective classical groups are prefixed by a "P", not a blackboard bold $\mathbb{P}$. E.g., $\mathrm{PSL}_{2}(7)$. At present, the projective orthogonal group wikipedia page seems to agree with this convention.

Operations

For finite groups: the Atlas of finite groups seems to have set the standard conventions for finite groups, Wilson changes them slightly. We'll find $G = N{:}H$ for the semidirect product $G = N \rtimes H = H\ltimes N$. Also $A\mathop{{}^{\textstyle .}}\nolimits B = A{\,}^{\textstyle .} B$ for a non-split extension with quotient $B$ and normal subgroup $A$, but no subgroup $B$. And $A{.}B$ is an unspecified extension.

Lie algebras. Writing notes on paper, for a given Lie group $G$, I write $\mathrm{Lie}(G)$ as its Lie algebra. (It turns out to be a functor...neat!) If I have to write Fraktur by hand, I approximate it using Pappus's caligraphy tutorial.

Huygens's Appendix to De ratiociniis in aleae ludo

Quoting from an English translation of Hugens's appendix to De ratiociniis in aleae ludo [or, On Reasoning in Games of Dice], here are the five problems given:

Problem 1. A and B play together with a pair of Dice upon this Condition, That A shall win if he throws 6, and B if he throws 7; and A is to take one Throw first, and then B two Throws together, then A to take two Throws together, and so on both of them the same, till one wins. The Question is, What Proportion their Chances bear to one another? Answ. As 10355 to 12276.

Problem 2. THREE Gamesters, A, B, and C, taking 12 Counters, 4 of which are White, and 8 black, play upon these Terms: That the first of them that shall blindfold choose a white Counter shall win; and A shall have the first Choice, B the second, and C the third; and then A to begin again, and so on in their turns. What is the Proportion of their Chances?

Problem 3. A lays with B, that out of 40 cards, i.e. 10 of each different Sort, he will draw 4, so as to have one of every Sort. And the Proportion of his Chance to that of B, is found to be as 1000 to 8139.

Problem 4. HAVING chosen 12 Counters as before, 8 black and 4 white, A lays with B that he will blindfold take 7 out of them, among which there shall be 3 black ones. Quaere [Lat., "Search out"], What is the Proportion of their Chances?

Problem 5. A and B taking 12 Pieces of Money each, play with 3 Dice on this Condition, That if the Number 11 is thrown, A shall give B one Piece, but if 14 shall be thrown, then B shall give one to A; and he shall win the Game that first gets all the Pieces of Money. And the Proportion of A's Chance to B's is found to be, as 244,140,625 to 282,429,536,481.

This is also known as the "gambler's ruin" problem. Pascal first posed this problem to Fermat (and it was communicated to Huygens via Carcavi in a letter dated September 28, 1656). To be clear: A keeps tossing the dice until the game is over.

For more about the history and context of these problems, see Anders Hald's A History of Probability and Statistics and Their Applications Before 1750 pp.74–78.

Bernoulli's Problems from Ars Conjectandi

Part III of Bernoulli's Ars Conjectandi discusses 23 problems. Here I'd like to collect them.

Problem 1. Someone with two tokens, one white and one black, hidden in an urn, offers a prize to three people, A, B, and C, on the condition that whoever draws out the white token will win the prize and if all fail to do so none will win the prize. A draws first and replaces the token, B second, C third. What are their lots?

Problem 2. Other things remaining as posited in the previous problem, if the organizer of the game of chance, wanting to deny to himself any right to the prize, tells the others to divide the prize among themselves if none of them should choose the white token, what then will be their lots?

Problem 3. Six players A, B, C, D, E, F, by order of the organizer, who favors the later players more than the earlier, attempt a game of chance. The first two, A and B, begin to play separately. The one who wins plays with the third, C, and whichever of them comes out on top competes with D, and so forth until the last player F, so that the one who emerges victorious from the last match will win the prize. Let it be supposed that each pair compete between themselves with equal lot, that is, that neither of the players has a stronger expectation of winning than the other. What are the players' lots?

Problem 4. Other things assumed to be as before, if we imagine that there is not an equal lot in any game, but that any player competing with the second player after himself has two times as many cases for winning as for losing and with the third player four times as many cases, with the fourth player eight times as many cases, and so forth—excepting only the first two players, whom we assume to compete in an equal context—then it is asked whether, in this case, all six players acquire an equal right to the offered prize, since the twice smaller expectations of the preceding proposition are compensated for the double ratio of cases.

Note that Bernoulli's notion of probability is frequentist, and the probability of an event $A$ is $\Pr(A) = N_{A}/N$ the number of outcomes $N_{A}$ which include $A$ divided by the total number of outcomes $N$. Bernoulli calls $N_{A}$ and $N$ the "number of cases".

Problem 5. A contends with B that from 40 playing cards, that is, 10 of each suit, he will choose 4 cards such that there is one of each suit. What is the ratio of their lots?

This is from Huygens's appendix to his De ratiociniis in aleae ludo, problem number three. The trick is to think in terms of "success" and "failure".

Problem 6. Taking 12 tokens, 4 white and 8 black, A contends with B that blindfolded he will choose 7 tokens from these, among which there will be 3 white ones. What is the ratio of the lots of A to the lot of B?

This is from Huygens's appendix again, problem number four. The trick is to think in terms of "success" and "failure". Bernoulli notes the ambiguity in Huygens's phrasing of the problem, and calculates both (1) the probability of obtaining exactly 3 white tokens, and (2) the probability of obtaining at least 3 white tokens.

Problem 7. Some number of players A, B, C, etc., draw cards in order and alternately from a hand of playing cards, of which one is a face card and the rest are nonface cards, on the condition that the person who draws the face card wins. Let A draw first, B second, C third, and so forth to the last, after which A proceeds to take the following card, and so forth until the end of the game. What is the ratio of their lots?

Problem 8. With other things as before, if in the hand of cards there are several face cards and if the person is judged to win who draws the first face card, then what is the ratio of lots?

Problem 9. With other things posited as before, if the players agree with each other that the one who draws more face cards will win and that if two or more players draw an equal number of face cards, they will divide the stake equally among them, while the rest who draw a smaller number of face cards will get nothing, then what is the ratio of their lots?

Problem 10.Four players A, B, C, D, having made among themselves the same agreement as in the preceding problem, play with 36 cards of which 16 are face cards and distribute to each player the cards alternately in order. It happens, however, that when 23 cards have been distributed, A has received by lot 4 face cards, B has 3, C has 2, D has 1, so that there remain 13 cards among which there are 6 face cards. The fourth player D (who is next in order to receive a card), seeing that almost all hope of his winning has vanished, wishes to sell his right to one of the others. How much should he sell it for and what are the expectations of the individual players?

Problem 11. It is proposed, with six throws of a die, to throw its six faces, each one once, so that none of the faces is repeated. What is the expectation of doing this?

Problem 12. It is proposed, with six throws of a die, to throw the six faces in order, on the first throw 1 point, on the second 2 points, on the third 3, etc. What is the expectation of achieving this?

Problem 13. Three players (A, B, C) each having written in front of himself the first six numerals, play with a die alternately, on the condition that whoever throws any number of points deletes that number from his written numbers, or, if he does not have it any more, the following player proceeds to throw. This continues until someone first deletes all his written numerals. It happens, however, that after this game has been played for a while, A still has 2 numbers before him, B has 4, and C has 3, and it is A's turn to throw. What are their lots?

Problem 14. Two players, A and B, throwing a die on a gaming tray, agree between themselves that each one will get as many throws as the number thrown on the die and that the one who throws the most points altogether will take the stake. If, however, it happens that they both get the same number of points, then they will also equally divide the stake between themselves. Afterwards, however, one of the players, B, tiring of the game, chooses instead to take a certain number of points in place of the uncertain throw of the die and tot take 12 points as his share. A assents. Which of the two players has the greater hope of winning and by how much?

Problem 15. Other things remaining as before, player B requests to be conceded as his share of points the square of the number of points on the first throw. Now what is the ratio of lots?

Problem 16.The valuation of lot in the game called Cinq et Neuf.

In France, Denmark, Sweden, Belgium, lower Germany, and neighboring regions, a kind of game is played that they call Cinq et Neuf; it is played by two people, A and B, with two dice; one of them, A, receives unending turns. These are the conditions of play: if A on the first turn throws a 3 or 11 or any pair (un doublet, ein Pasch), that is, two ones, two twos, threes, etc., then A wins. If A throws a 5 or 9, the other player, B, wins. If A throws any other number of points, namely 4, 6, 7, 8, or 10, then neither of the players wins, but the game continues until a 5 or 9 is thrown, in which case B is the winner, or until exactly the same number of points is thrown again as was thrown on the first throw, in which case A wins. The condition concerning the throwing of a 3 or 11 or of any pair does not aid A except on the first throw. On these assumptions what is the ratio of lots?

Problem 17. The valuation of the lot in a certain other kind of game of chance.

I remember once seeing here at the time of the weekly market a certain peddler who was explaining the following kind of game in the marketplace and attracting to it those passing by. There was a circular disk made quite level, mounting upwards for a little while toward the center. The border was surrounded by 32 contiguous and equal small pockets or openings that were marked into four distinct classes or series by numbers written four times in order from I up to VIII. A dice box hung perpendicularly over the middle of the disk. The one about to make a trial of fortune dropped through the cavity of the dice box four little balls to be received by an equal number of compartments in the circumference of the disk. He received as a prize what the sum of the numbers in the compartments indicated, which might be larger or smaller depending on the sum, as the prizes in the table indicate. For each throw of the balls, the player was to pay four coins. What is the player's expectation?

Points	Coins	Cases
4	120	1
5	100	16
6	30	52
7	24	128
8	18	245
9	10	416
10	6	664
11	6	976
12	6	1369
13	5	1776
14	3	2204
15	3	2560
16	3	2893
17	2	3088
18	2	3184
19	3	3088
20	3	2893
21	3	2560
22	3	2204
23	4	1776
24	4	1369
25	6	976
26	8	664
27	12	416
28	16	245
29	24	128
30	25	52
31	32	16
32	180	1

Problem 18. On the card game called in the vernacular Trijaques.

Very common among the Germans is the kind of game called Trijaques, which has an affinity to the French game Brelan. From a deck of cards 24 are taken (with the rest set aside), six from each suit, the nines, tens, jacks, queens, kings, and aces, which herafter will be referred to by their initials NTJQKA. These cards have the following priorities: in first place is the ace, followed by the king, and then the queen, jack, and ten, but higher than all these are the nines, together with the jack of clubs (which, accordingly, we will consider to be a nine, so that there are 5 nines and only 3 jacks). The preeminence of the nines, which, rather like the cards in the Spanish game Hombre called Matadors, they call robbers, murderers, or assassins, consisting in the fact that they are judged to be of any suit or number one pleases. Thus 2 nines with an ace, or 1 nine with 2 aces, make 3 aces or three-of-a-kind (un Tricon) of aces. One, 2, or 3 nines, joined with 3, 2, or 1 kings make four-of-a-kind of kings. One or two nines together with three or two cards of the same suit make four of that suit, for example, four hearts, spades, clubs, etc., the sort of combination of cards customarily called a flush (ein Fluss), which is valued beyond the number of points. The nine or the ace is worth 11 points while the other ranks of cards are each worth 10 points.

This is the way the game is played: To each of the players in order two cards are dealt. After these are privately examined, the first player is free to stake any sum of money. Another player who wishes to join the play stakes an equal amount or even adds something extra if he pleases, which the first player must match if he does not wish to lose his stake. After this is completed, each player who remains in the game is dealt two more cards face up on the table, so that part of each player's hand is visible to the other players and part is hidden. Then they begin to wager money again, alternately increasing the stake as before and challenging the others to bid or concede. In the end each player shows his hand to the others and the one who has the best hand takes the whole stake. Now four-of-a-kind is better than a flush, and a flush is better than three-of-a-kind, but 4 nines are best of all. In the other cases of three-of-a-kind and four-of-a-kind, the order follows the order of the cards, and in flushes the order follows th enumber of points. So, for example, 3 or 4 aces beats 3 or 4 kings, and a flush of 43 points beats one of 42 points. If none of the players has three- or four-of-a-kind or a flush, then the one who has the most points of the same suit takes the stake. If two hands are equal in every way, for example, if 2 players have three-of-a-kind or four-of-a-kind of the same rank, or if they each have a flush containing an equal number of points, then the one who is closer to the dealer or who received his cards first wins.

This seems like a "toy model" of Texas hold 'em.

Problem 19. Suppose any kind of game in which the organizer of the game or cashier (the banker of the game) has some advantage consisting in the fact that the number of cases in which he wins is a little larger than those in which he loses and furthermore in that the number of cases in which the banker remains in office for the following game is slightly larger than the number in which the office is transferred to another player. How great is the advantage of the banker?

Bernoulli considers a throw of dice where the number of cases in which the banker wins to the number of cases in which he loses is in ratio of p to q (greater to less). Also the number of cases in wihch the banker retains his position for the following throw to the number of cases in which the position is transferred to a fellow player to be in ratio of m to n (again of greater to less).

Problem 20. The evaluation of the lot in the card game commonly called Capriludium or Bockspiel.

Two or more players use playing cards. One competes against the others and performs the office of banker (he has the bank), after he has shuffled the cards and divided the pack into as many hands as there are players including himself. Then the individual players buy individual hands for a price laid down, leaving the last to the banker. Finally, the banker turns the hands over and reveals their bottom cards, besides which no others are considered. Once this has been done, the banker is expected to pay those whose cards are of a higher rank than his as much as each one laid out for the chance. But those to whom fell a card of a lower rank than or equal to the card of the banker suffers the loss of their stake, ceding this rather to the payment of the victorious banker, who also continues to perform his office as long as he has defeated at least one of his opponents, nor does he leave the position of banker unless he has been bested by absolutely all his fellow players.

Bernoulli considers problem 19 as a simpler "toy model" of problem 20.

Problem 21. On the game of Basset.

This game is most celebrated because of the innumerable tumults and tragedies it once excited on all sides here and there, especially in Italy and in France because of which it was proscribed from these regions not long after and prohibited under severe penalty. At the time when the play of this game flourished most greatly in the hall of the King of France, D. Salvator (Sauveur), a French mathematician and preceptor of the Dauphin, subjected the expectations of the players to calculation, and published some tables of them in the Parisian Journal des Scavans for February 1679, with a brief synopsis. From this journal, I will review those aspects of the nature and constitution of the game that seem necessary for the examination of the Tables and for the eliciting of the calculations that the author suppressed.

After the banker has taken the whole deck of playing cards and shuffled it, each of the players before him at the table lays out a card of any rank (taken from elsewhere) together with a chosen sum of money. Then the banker turns over his deck of cards, revealing the bottom card, and deals out the cards successively two at a time in order until he has exhausted the pack. When he is dealing each of the pairs of cards, the first or preceding card favors the banker and the latter or following card is advantage of the opponent. For instance, if the first card turned up is a king, then the banker takes whatever has been laid on kings. If, however, the king comes up second, the banker must give to each of the players in turn as much as they have laid on kings. Up to this point of the rules of the game, no one has any advantage over another. But in addition the following rules should be noted:

1. If both cards are of the same weight and rank as a displayed card (which they call doublets, and we will call twins), in which case the profit and loss ought to balance each other out, only the banker benefits and takes whatever is laid on the cards of that rank.

2. Any player, even in the middle of a hand, has the renewed right to lay new money on any card. When this happens it may occur that only 1, 2, 3, or all 4 of the cards of that rank remain in the pack—which will markedly affect his lot. But it should be observed that a pair of which he has seen either card is judged to be of no value with respect to the rank that has been bought; and indeed if the latter card of this sort should agree in rank, not only does it profit nothing for the player, being as it were premature [trop jeune], but it also puts an end to the round, with respect to the option of betting on another card. If, on the other hand, that rank should appear on the first card of the following pair, it has a diminished value [c'est une face] and to the banker goes a profit of only two-thirds of the stake.

3. The prior card of the first pair, moreover, since there may be some suspicion that it may have been seen on the part of the banker, is always of diminished value, and can only give the winner two-third of the prize.

4. Since, when there is only one card left of a rank for which the players contest, there is no opportunity for twin cards, in which the advantage of the banker consists it is ruled in favor of the banker that the last card of all, which ought otherwise to benefit the player, should be considered null.

Salvator's article seems to be available online, well, it's been scanned by Google Books pp. 43 et seq.

Problem 22. There is a certain kind of game in which the number of all the cases is a, the number of some of them is b, and the number of the rest is $a-b=c$. Titius buys single throws of a single die by paying single coins to Caius. However many times he throws one of the b cases, he receives m coins back from Caius, but however many times he throws one of the c cases, he receives nothing. If, however, he throws one of the c cases continuously for n times, Caius is obliged to return all of his n coins. What are the lots of Titius and Caius?

Problem 23. On the game of blind dice [blinde Wurffel].

By this name are called those six dice adopted by most of our peddlers, which are cube-shaped like ordinary dice, but appear blank, with individual dice marked with points on only one face, one die with one point, another with two, a third with three, up to the sixth, which has one of its faces marked with six points. Thus, all together, only $1+2+3+4+5+6=21$ points are found. Imposters display this sort of dice in the marketplace for cheating the populace, together with prizes for all the numbers of points from 1 up to 21 in amounts that can be seen in the adjoined table. Then a person who wishes to risk his fortune, having paid coin to the peddler, throws the dice onto the tray, and if he has thrown some numbers of points, takes the prize assigned for that number. If, however, no points fall to him, he loses his coin.

This assumed, anyone who wants to investigate the lot of these players should note the following:

1. That the number of all the cases in six dice of this sort, no different from ordinary dice, is 46,656, namely as many as the sixth power of six.

2. That the number of cases that carry absolutely no points is equal to the sixth power of five (15625) since on any die there are five blank faces of which any one can be combined with any of the five blank faces of another die, and of these combinations again any can be combined with the five blank faces of the third die, so that the number of preceding cases is always quintupled.

3. That any number of points may be produced by one, two, or more dice. If it is produced by one, on the other five dice there will be no point showing, so that, since there are five blank faces on each die, the number of cases in which that happens will be 3125 (the fifth power of 5). If the number of points is produced by two dice, on the remaining four there will be no points, so that noe, the number of cases in which this occurs will be 625 (the fourth power of 5).

4. The same number of points can be produced not only frequently by more or fewer dice, but also sometimes by the same number of dice in several ways. Thus 12 points can be produced by three ways and two ways by four dice (on three dice the points can be 1,5,6; or 2,4,6; or 3,4,5; and on four dice, the points can be 1,2,3,6; or 1,2,4,5).

Problem 24. Assuming again the situation in the previous problem, if the master of the game contracts with the player that he wishes to be bound to restore to the latter all his coins if no points fall to him on five throws in a row, what then will be the lot of each?

Method of Characteristics

First-order partial differential equations may be analyzed (and sometimes, even solved) by using the Method of Characteristics. We consider a family of curves, along which the solution to the partial differential equation is constant. (At least, intuitively.)

When we cannot solve a first-order PDE, we can still extract useful information about it using the method of characteristics. This post will just give some intuition and basic algorithms for the linear case, and generalize it to the semilinear and quasilinear cases.

Linear Partial Differential Equations

Example 1. Suppose we are working in 2-dimensions and we have a first-order partial differential equation, say the transport equation: \begin{equation} L[u] = \partial_{t}u(x,t) + c\partial_{x}u(x,t)=0. \end{equation} Here $c$ is a nonzero constant (the wave speed). How can we solve it?

We could consider a curve $x(s)$, $y(s)$ such that the total derivative of the solution to our PDE along this curve is our original partial differential equation: \begin{equation} \frac{\D}{\D s}u(x(s),y(s)) = \left.L[u]\right|_{\vec{x}=\vec{x}(s)}=0 \end{equation} This means we have several ordinary differential equations to solve \begin{equation} \frac{\D}{\D s}t(s) = 1 \end{equation} and \begin{equation} \frac{\D}{\D s}x(s) = c. \end{equation} These have solutions of the form \begin{equation} t(s) = s + k_{1},\quad\mbox{and}\quad x(s)=cs + k_{2} \end{equation} where $k_{1}$, $k_{2}$ are constants of integration.

In particular, observe that $x(s) - ct(s)$ is a constant. This means that any (sufficiently smooth) function $f(-)$ satisfies the transport equation: \begin{equation} (\partial_{t} - c\partial_{x})f(x - ct) = 0. \end{equation} This seems like some black magic, or bizarre accident. What's going on?

Example 1'. Let's try solving the transport equation with a different scheme. Let's try to change coordinates to make the transport equation of the form \begin{equation} b(\xi,\eta)\partial_{\xi}u(\xi,\eta)=0. \end{equation} How can we do this?

Step 1: Find direction. We want $\xi$ to point in the direction of $(t,x)=(1,c)$. Why? Well, we could then treat the PDE as the directional derivative $(1,c)\cdot\nabla u = 0$. We can construct the line through the origin and $(1,c)$ using the formula $t=x/c$. Supposing we didn't know how to construct a line (or it's impossible for some reason), we could determine the direction $\theta(t)$ by the differential equation \begin{equation} \frac{\D\theta(t)}{\D t} = \frac{c}{1} \end{equation} which gives the solution $\theta(t) = ct + k$ for some constant of integration $k$.

Step 2: Construct axes. We then want to use $\theta(t)$ as the formula for $x$, i.e., we want to construct lines \begin{equation} g(x,t) = x - ct \end{equation} so when we plug in $x=\theta(t)$ we have \begin{equation} g(\theta(t),t) = k \end{equation} be constant. Now we assert this is the $\xi$-axis for $k=0$.

The $\eta$-axis would be perpendicular to this, i.e., $x=-t/c$ or $t=-cx$ is the $\eta$-axis.

Step 3: Cleaning up. Along the $\eta$-axis, $\xi$ is constant. Along $\{(t,x)\in\RR^{2}\mid t=-cx+k\}$ (for some "intercept parameters" $k$) parallel to the $\eta$-axis, we expect $\xi$ to be constant. That is to say, we have \begin{equation} \xi(t,x) = t+cx \end{equation} and similar reasoning suggests $\eta$ is constant along lines $\{(t,x)\in\RR^{2}\mid t=k + x/c\}$ parallel to $\xi$, which is equivalent to asserting \begin{equation} \eta(t,x) = x - ct. \end{equation} Why? Well, given any point in the line $(t_{k},x_{k})\in\{(t,x)\in\RR^{2}\mid t=k + x/c\}$, we see $\eta(t_{k},x_{k})=k$ is constant. Similarly, for any point $(t_{k}',x_{k}')\in\{(t,x)\in\RR^{2}\mid t=-cx+k\}$, we find $\xi(t_{k}',x_{k}')=k$ is constant, too.

If we change coordinates, then we have \begin{equation} \xi=\xi(x,t)=Ax+Bt \end{equation} and \begin{equation} \eta=\eta(x,t)=Cx+Dt. \end{equation} Using the chain rule, we find \begin{equation} \begin{split} \partial_{x} &= \partial_{x}\xi(x,t)\partial_{\xi} + \partial_{x}\eta(x,t)\partial_{\eta}\\ &= A\partial_{\xi} + C\partial_{\eta} \end{split} \end{equation} and similarly \begin{equation} \begin{split} \partial_{t} &= \partial_{t}\xi(x,t)\partial_{\xi} + \partial_{t}\eta(x,t)\partial_{\eta}\\ &= B\partial_{\xi} + D\partial_{\eta}. \end{split} \end{equation} We then have \begin{equation} (B\partial_{\xi} + D\partial_{\eta})u + c(A\partial_{\xi} + C\partial_{\eta})u=0. \end{equation} Rearranging terms, we have \begin{equation} (cA + B)\partial_{\xi}u + (cC + D)\partial_{\eta}u=0. \end{equation} Erm, progress?

Yes, progress! Because we have discovered what $A$, $B$, $C$, and $D$ are! We did it in step 1. We see $B=C=1$, $A=c$, and $D=-c$. This tells us that \begin{equation} (cC + D)\partial_{\eta}u=(c-c)\partial_{\eta}u=0 \end{equation} identically.

Our partial differential equation becomes \begin{equation} \boxed{(c^{2}+1)\partial_{\xi}u = 0.} \end{equation} And we conclude that a generic solution must look like \begin{equation} u(\xi, \eta) = f(\eta) = f(x - ct) \end{equation} for sufficiently smooth $f(-)$. This is precisely the result we got before.

Example 2. Let's consider a harder PDE, with variable coefficients: \begin{equation} a(x,y)\partial_{x}u(x,y) + b(x,y)\partial_{y}u(x,y)=0. \end{equation} There are two ways to tackle this:

1. Geometrically, find curves $x(s)$ and $y(s)$ such that $\D u(x(s),y(s))/\D s$ is our original PDE; or
2. Algorithmically change variables, so we have $\eta=\eta(x,y)$ and $\xi=\xi(x,y)$ such that we choose $\eta$ to satisfy $a(x,y)\partial_{x}\eta + b(x,y)\partial_{y}\eta=0$ identically, and $\xi$ is whatever works, so our PDE becomes $(\mbox{stuff})\partial_{\xi}u=0$.

The algorithm changes slightly, because now we have variable coefficients, but it is morally the same.

Step 1. Find the direction $\theta(x)$ such that \begin{equation} \frac{\D\theta(x)}{\D x} = \frac{b(x,\theta(x))}{a(x,\theta(x))}. \end{equation} Before this was easy because $\theta(x)=(b/a)x$ for constant coefficients, which was used as the slope of the $\xi$-direction.

Step 2. Since $\theta(x)$ from step 1 is unique up to some constant of integration $k$, "invert it" to produce some function $g(x,y)$ such that \begin{equation} g(x,\theta(x))=k \end{equation} it produces the constant of integration.

Step 3. Set $\eta$ equal to this function obtained in step 2: \begin{equation} \eta(x,y) = g(x,y). \end{equation} The assertion is that $\eta$ is constant in the $(a(x,y), b(x,y))$ direction, and so any arbitrary (but sufficiently smooth) function $f(\eta)$ satisfies the original PDE.

If you look back at example 1', you'll see that $\eta=g(x,y)$, too. You'll also see the steps carried out as described.

So how do these two methods relate to each other? Well, if we parametrize $y=y(x)$ instead of $y=y(s)$ and $x=x(s)$, then we recover the algorithmic method from the geometric method. In other words, the geometric method is more general and contains (as a special case) the algorithmic method.

More general case with variable coefficients. If we have something of the form \begin{equation} a(x,y)\partial_{x}u + b(x,y)\partial_{y}u + c(x,y)u = f(x,y) \end{equation} then everything we did can carry over. The only difference is our partial differential equation becomes \begin{equation} A(\xi,\eta)\partial_{\xi}u + C(\xi,\eta)u = F(\xi,\eta) \end{equation} which is a first-order ODE. We know how to solve them!

Semilinear Equations

For a semilinear PDE of the form \begin{equation} a(x,y)\partial_{x}u + b(x,y)\partial_{y}u = f(x,y,u) \end{equation} the geometric method carries over. But we now have the system of ODEs: \begin{equation} \frac{\D x(s)}{\D s} = a(x(s),y(s)) \end{equation} \begin{equation} \frac{\D y(s)}{\D s} = b(x(s),y(s)) \end{equation} as before, and also \begin{equation} \frac{\D w(s)}{\D s} = f(x(s),y(s),w(s)). \end{equation} This new equation intuitively encodes the characteristic of the solution: $w(s) = u(x(s), y(s))$.

Our PDE along this curve is then such that \begin{equation} \frac{\D}{\D s} (u(x(s),y(s)) - w(s)) = 0. \end{equation} Care must be taken if working with initial conditions or boundary values.

Quasilinear Equations

For a quasilinear PDE of the form \begin{equation} a(x,y,u)\partial_{x}u + b(x,y,u)\partial_{y}u = f(x,y,u) \end{equation} the geometric method carries over. But we now have the system of ODEs: \begin{equation} \frac{\D x(s)}{\D s} = a(x(s),y(s),w(s)) \end{equation} \begin{equation} \frac{\D y(s)}{\D s} = b(x(s),y(s),w(s)) \end{equation} \begin{equation} \frac{\D w(s)}{\D s} = f(x(s),y(s),w(s)). \end{equation} Unlike the semilinear case, we have more complications to deal with, which I'll discuss in another post. What sort of complications? Let's look at an example.

Example 3. Consider the nonlinear transport equation (where wave-speed is the magnitude of the wave): \begin{equation} \partial_{t}u + u\partial_{x}u = 0. \end{equation} The characteristics satisfy the system of equations \begin{equation} \frac{\D t(s)}{\D s} = 1 \end{equation} \begin{equation} \frac{\D x(s)}{\D s} = w(s) \end{equation} \begin{equation} \frac{\D w(s)}{\D s} = 0. \end{equation} We can solve these equations \begin{equation} t(s) = t_{0} + s \end{equation} \begin{equation} x(s) = x_{0} + w_{0}s \end{equation} \begin{equation} w(s) = w_{0}. \end{equation} We have \begin{equation} s = t(s) - t_{0} \end{equation} and hence \begin{equation} x(s) - w_{0}(t(s) - t_{0}) = x_{0} \end{equation} or more precisely: \begin{equation} x(s) - w(s)t(s) = \mbox{constant}. \end{equation} This tells us the generic solution looks like \begin{equation} u(x,t) = f(x - ut) \end{equation} which is an implicit equation. Here $f(-)$ is determined by the initial conditions. There is no general solution beyond this.

Exercise 1. Plug in this "solution" $u(x,t) = f(x - ut)$ to the nonlinear transport equation. Is it really a solution?

References

1. Walter Strauss, Partial Differential Equations: An Introduction.
2. Julie Levandosky, First-Order Equations: Method of Characteristics. Handout, Stanford University, Math 220A, Fall 2002.

Lagrangian Description of Fluids

I'd like to just summarize the Lagrangian description of fluids quickly, because some texts have horrible typos and errors.

We begin with describing the position of fluid parcels using a function $\vec{x} = \vec{x}(\vec{x}_{0}, t)$ parametrized by position $\vec{x}_{0}$ at time $t=t_{0}$. Time evolution is described by the function \begin{equation} \vec{x}(\vec{x}_{0}, t) = \phi_{t}(\vec{x}_{0}) \end{equation} such that (locally) $\phi_{t}$ is invertible; i.e., we can write \begin{equation} \vec{x}_{0}(\vec{x}, t) = {\phi_{t}}^{-1}(\vec{x}(\vec{x}_{0},t)). \end{equation} Now we may introduce the Fluid Velocity in the Lagrangian description as \begin{equation} \vec{q} = \frac{\partial\vec{x}(\vec{x}_{0},t)}{\partial t}. \end{equation}

The Eulerian picture may be obtained by writing \begin{equation} \vec{u}(\vec{x}, t) = \vec{q}(\vec{x}_{0}(\vec{x},t), t) = \left.\frac{\partial\vec{x}(\vec{x}_{0},t)}{\partial t}\right|_{\vec{x}_{0}={\phi_{t}}^{-1}(\vec{x})} \end{equation}

Example 1. Consider a flow on the unit disc, with trajectories described by \begin{equation} \begin{pmatrix} x(t)\\ y(t) \end{pmatrix} = \begin{pmatrix}\cos(\omega t) & -\sin(\omega t)\\ \sin(\omega t) & \cos(\omega t) \end{pmatrix} \begin{pmatrix} x_{0}\\ y_{0} \end{pmatrix}. \end{equation} Observe that $x(0)=x_{0}$ and $y(0)=y_{0}$. We find we can invert this equation to determine the initial positions from the trajectory: \begin{equation} \begin{pmatrix} x_{0}\\ y_{0} \end{pmatrix} = \begin{pmatrix}\cos(\omega t) & \sin(\omega t)\\ -\sin(\omega t) & \cos(\omega t) \end{pmatrix} \begin{pmatrix} x(t)\\ y(t) \end{pmatrix} \end{equation} which permits us to write the Lagrangian parameters (initial positions) in terms of the current position of the fluid parcel. We find the time derivative of the trajectories \begin{equation} \vec{q} = \frac{\D}{\D t} \begin{pmatrix} x(t)\\ y(t) \end{pmatrix} = \omega\begin{pmatrix}-\sin(\omega t) & -\cos(\omega t)\\ \cos(\omega t) & -\sin(\omega t) \end{pmatrix} \begin{pmatrix} x_{0}\\ y_{0} \end{pmatrix}. \end{equation} Plugging in the equations relating $x_{0}$ and $y_{0}$ in terms of $x(t)$ and $y(t)$ gives us (after some simple matrix multiplication): \begin{equation} \begin{pmatrix}u(x,y,t)\\ v(x,y,t)\end{pmatrix} =\left.\frac{\D}{\D t} \begin{pmatrix} x(t)\\ y(t) \end{pmatrix}\right|_{\vec{x}_{0}=\vec{x}_{0}(\vec{x},t)} = \begin{pmatrix} -\omega y\\ \omega x \end{pmatrix}. \end{equation} This is a steady flow (since time doesn't explicit appear in the equation for $u(x,y,t)$ or $v(x,y,t)$), and just a rotation. If we want to generalize this, we could replace $\omega t$ with a dimensionless, strictly increasing function of time $f(t)$, which would usually not be steady (e.g., if $f(t) = \omega t + \alpha t^{2}$ for positive constants $\omega$, $\alpha$, then $u=-(\omega + 2\alpha t)y$ and $v=(\omega + 2\alpha t)x$).

Exercise 1. Find the density $\rho(x,y,t)$ for the flow from example 1 by solving the continuity equation, or some other way. If you do it some other way, prove your solution satisfies the equation of continuity.

Exercise 2. Is the flow from example 1 compressible or incompressible?

Example 2 (Childress, Example 2.2). Consider the one-dimensional flow with $u(x,t) = 2xt/(1 + t^{2})$. Suppose we want to find the Lagrangian description of the flow, then we can rewrite this as \begin{equation} \frac{\D x(t)}{\D t} = \frac{2x(t)\cdot t}{1 + t^{2}} \end{equation} with initial condition $x(0)=a$. This has the obvious solution \begin{equation} x(t) = a(1 + t^{2}). \end{equation} This illustrates how to translate a solution in the Euler description to find the trajectories in the Lagrangian description.

Exercise 3. Solve the Euler flow equations and continuity equation for both the pressure and density. "Someone on the internet" asserts $\rho(x,t)=x$ is a valid solution: is it? [Hint: use method of characteristics to determine density.]

References

1. C.C. Mei, Methods of Describing Fluid Motion. MIT course notes for 1.63 "Advanced Fluid Mechanics", 2001.
2. Lei Li, Math 575-Lecture 1. Duke University
3. Stephen Childress, An Introduction to Theoretical Fluid Mechanics. AMS Press, 2009.

Fluid Parcels and the Continuum Hypothesis

When working out the equations of motion (and the physics) of fluids, we tend to inherit an 18th-century perspective on the nature of matter: a fluid is a continuum, not a collection of molecules. This makes terminology rather confusing to modern students, since atomic theory is taught in primary school.

Textbook definitions explain a fluid parcel is an infinitesimal amount of fluid, presumably in a simply-connected (and, for algebraic topologists, connected) region. Older texts may use the term "fluid particle" as a synonym for fluid parcel. Another synonym found in the literature (classic and modern) is "fluid element".

The idea is that a fluid may be described as a continuous collection of fluid parcels, just as a domain may be described as a continuous collection of "infinitesimal volumes".

Puzzle 1. Why not describe a fluid parcel as a volume element or more rigorously as a volume form or density?

Well, that may all be well-and-good for a mathematical formalization describing a fluid parcel, but (a) what does it mean for the equations? And (b) when does a physical system behave like a fluid?

Continuum Hypothesis

What this means for the equations is that we can conceptually "keep zooming in" on a fluid parcel, and each fluid parcel "acts like" a physical body. So we can meaningfully describe a fluid parcel's velocity, density, pressure, temperature, etc.

Proposition 2 (Continuum Hypothesis). A fluid parcel may be described like a point-particle, and has the physical characteristics of the bulk (e.g., velocity, pressure, density, etc.).

Remark 3. I have seen the literature argue the "amount of matter" in a fluid parcel is constant (which means its volume can vary, otherwise density is constant). I'm not sure if this is standard or not? If so, I suppose we could conceptually think of the number of molecules in a parcel is fixed...but this may be "chalk and cheese".

Knudsen Number

When can we apply fluid mechanics to describe a physical body? Fortunately, we have a (dimensionless) parameter measuring how badly fluid mechanics approximates a physical body: the Knudsen number.

Definition 4. The Knudsen Number for a body is \begin{equation} \mathrm{Kn} := \lambda/L \end{equation} where L is the length scale of the body, and $\lambda$ is the mean free path of its constituent molecules.

The rule/heuristic is: we may apply fluid mechanics if \begin{equation} \mathrm{Kn}\ll 1. \end{equation} If $\mathrm{Kn}\lt0.01$, then it's perfectly fine to use fluid mechanics. For the values between $0.01\leq\mathrm{Kn}\lt0.10$, then it's debatable-but-doable...the cut-off here is rather fuzzy. Larger values demand using statistical mechanics instead of fluid mechanics.

Example 5. Air at 1013 hPa has a mean free path of approximately 68nm (see Jennings' The mean free path in air Journal of Aerosol Science 19, no.2 (1988) pp.159–166). For geophysical models, the length scale would be on the order of 8.5 kilometers (the scale atmosphere height is about 8.5km). Thus the Knudsen number for air would be about $\mathrm{Kn}\approx (68\times 10^{-9})/(8.5\times 10^{3}) = 8\times 10^{-12}\sim 10^{-11}$. Hence fluid mechanics would be a good approximation at this scale. For a more careful analysis of the mean-free path of air, see some calculations by David Pace.

Example 6. Water can be estimated to have a mean free path of $\lambda\approx 2.5\times10^{-10}\,\mathrm{m}$ (c.f., handout). For a cup of water, which holds 1 cup (approximately 236588 cubic millimeters), its scale is the cuberoot of this number: $L\approx 62\times10^{-3}\,\mathrm{m}$. Hence the Knudsen number for a cup of water is approximately $\mathrm{Kn}\approx (2.5\times 10^{-10})/(62\times 10^{-3})\approx 4\times 10^{-9}$. Fluid mechanics can describe a cup of water very well, as we expect!

Remark 7. There is a more generally setting, treating matter as continuous, which is the field of "continuum mechanics". The condition for determining appropriateness of continuum mechanics to a problem seems to be the Hill–Mandel principle, where if $F$ is the deformation gradient and $P$ is the stress tensor in equilibrium, then $\langle P : F\rangle = \langle P \rangle : \langle F \rangle$, where $A : B = \mathrm{tr}(AB^{T}) = \sum_{i,j}(A)_{ij}(B)_{ij}$ and brackets indicate microscopic averaging. For a review with emphasis on finite elements, see arXiv:1509.06621.

There are three physical length scales we can work with:

1. $L_{\text{molec}}$ Molecular scale (which is approximated by the mean free path)
2. $L_{\text{fluid}}$ Fluid parcel scale (which could be approximated by the size of a drop or droplet, which pharmacists have standardized and ordained to be 1/20 of a milliliter — approximately a sphere with radius π millimeters)
3. $L_{\text{macro}}$ Macro-scale of the phenomena (the size of the container of fluid, or some standard scale).

This is another way to heuristically reason about applicability, and the condition here becomes $L_{\text{molec}}\ll L_{\text{fluid}}$...but we usually have this for most engineering purposes (unless someone is trying to use individual water molecules to cool their quantum computer, or something) and geophysical models.

References

1. Simon J.A. Malham, Introductory fluid mechanics. (PDF) Lecture notes dated September 15, 2014.
2. Pieter Wesseling, Principles of Computational Fluid Dynamics. Springer, 2000.
   This is my only book which discusses Knudsen number.
3. Carlo Marchioro, Mario Pulvirenti, Mathematical Theory of Incompressible Nonviscous Fluids. Springer, 1994.
   Chapter 1, §1, discusses "number of molecules in a parcel", for example.

Continuity Equation

We had introduced the Euler flow equations, and saw it gave us (in $n$ dimensions) $n$ equations in $n+2$ unknowns. This requires introducing another 2 equations for us to solve the equations of motion. Today we will do half the required job, we will introduce one additional equation: the conservation of mass, or continuity equation.

Euler's Derivation

A relatively clever derivation begins by examining a fluid parcel of "small volume" $V$ and density $\rho$, so small such that $\int_{V}\rho\,\D^{n}x\approx\rho V$. Since the particles in the fluid parcel remain in the parcel over time, we have its total mass be constant: \begin{equation} \frac{\mathrm{D}(\rho V)}{\mathrm{D} t} = 0. \end{equation} Since this fluid parcel is nonempty, $\rho V\gt0$, so taking the logarithm of the mass and then taking its time derivative gives us: \begin{equation} \frac{1}{\rho}\frac{\mathrm{D}\rho}{\mathrm{D} t} + \frac{1}{V}\frac{\mathrm{D} V}{\mathrm{D} t} = 0. \end{equation} Supposing we have the fluid parcel be described as a rectangle with edges $\delta x$, $\delta y$, $\delta z$, then we see \begin{equation} V = \delta x\,\delta y\,\delta z \end{equation} hence Taylor expanding the volume as a function of time gives us \begin{equation} V + \frac{\mathrm{D} V}{\mathrm{D} t}\delta t = \left[1 + \left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}\right)\delta t\right]\delta x\,\delta y\,\delta z. \end{equation} Collecting these together give us the equation \begin{equation} \frac{\mathrm{D}\rho}{\mathrm{D} t} + \rho\left(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z}\right) = 0. \end{equation} This is the continuity equation.

Remark 1. We should observe that the divergence of the velocity vector field, as it appears in Eq (4), describes the dilation of the volume element. It is the "expansion" of the fluid at that point.

Puzzle 2. This derivation seems to use the Lagrangian description; in general, however, we would need to include a factor of the Jacobian to describe the volume of the fluid parcel. Work out this derivation. Cheaters may consult Childress.

Modern Textbook Derivation

Consider now a fixed cube in the fluid, whose side lengths are again $\delta x$, $\delta y$, $\delta z$. What amount of mass flows into the cube across the $\delta y\delta z$ face?

We find, assuming the flow moves away from the origin, \begin{equation} \begin{pmatrix}\mbox{mass which}\\ \mbox{flows in}\end{pmatrix} = \left(\rho - \frac{1}{2}\frac{\partial(\rho u)}{\partial x}\delta x\right)\delta y\,\delta z. \end{equation} Similarly, the amount flowing out \begin{equation} \begin{pmatrix}\mbox{mass which}\\ \mbox{flows out}\end{pmatrix} = \left(\rho + \frac{1}{2}\frac{\partial(\rho u)}{\partial x}\delta x\right)\delta y\,\delta z. \end{equation} We then have the net mass gained be $-\delta x\,\delta y\,\delta z\partial_{x}(\rho u)$. Doing this for the other faces, we get \begin{equation} \begin{pmatrix}\mbox{net mass}\\ \mbox{flows}\end{pmatrix} = \nabla\cdot(\rho\vec{u})\,\delta x\,\delta y\,\delta z. \end{equation} On the other hand, the mass flux varies over time as \begin{equation} \begin{pmatrix}\mbox{mass}\\ \mbox{flux}\end{pmatrix} = \frac{\partial}{\partial t}(\rho\,\delta x\,\delta y\,\delta z). \end{equation} Setting equals to equals, and simplifying, we find \begin{equation} \frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\vec{u})=0 \end{equation} which is another, equivalent, form of the continuity equation.

Exercise 3. Prove Eq (10) is equivalent to Eq (5).

References

I have profited most from Lamb's presentation, though Childress takes greater care when working through the derivation in the Lagrangian description.

1. Leonhard Euler, "Principes généraux du mouvement des fluides". Mémoires de l'académie des sciences de Berlin 11 (1757) 274–315; English translation arXiv:0802.2383
2. Horace Lamb, Hydrodynamics. Dover, sixth ed., 1932; §7 (archive)
3. Stephen Childress, An Introduction to Theoretical Fluid Mechanics. AMS Press, 2009.

Derivation of Euler Equations for Fluid Flow

Overview: the Euler [flow] equations are the equations of motion for an idealized fluid. We just derive them and conclude with a few remarks.

Derivation

The derivations of the Euler equation boil down to the same game plan: derive $F=ma$ for a fluid parcel. (A "fluid parcel" is an infinitesimal volume of fluid, more a heuristic than a rigorous notion.)

The forces acting on a fluid parcel, in a simplified idealized setting, are just the fluid's pressure. We may find it by integrating over the surface of the fluid parcel: \begin{equation} \vec{F} = -\oint_{\partial V}p(\vec{x},t)\vec{n}\,\D A \end{equation} where $\vec{n}$ is the outward-pointing normal vector on the bounding surface $\partial V$ of the fluid parcel. (The negative sign is due to the normal vector being outward-pointing.) The divergence theorem yields \begin{equation} \vec{F} = -\int_{V}(\nabla p(\vec{x},t))\,\D V. \end{equation} This is half the equation of motion.

The other half is the "mass times acceleration" term. We generically describe a fluid by its velocity vector field $\vec{u}(\vec{x}, t)$. Now we can sum over its acceleration vector for each point in the fluid parcel, weighted by its density $\rho$, to give us \begin{equation} m\vec{a} = \int_{V} \rho(\vec{x}, t)\frac{\mathrm{D}\vec{u}(\vec{x},t)}{\mathrm{D}t}\,\D V. \end{equation} Here we use Stokes' notation \begin{equation} \frac{\mathrm{D}\vec{u}(\vec{x},t)}{\mathrm{D}t} = \partial_{t}\vec{u}(\vec{x},t) + (\vec{u}(\vec{x},t)\cdot\nabla)\vec{u}(\vec{x},t) \end{equation} which is the material time derivative.

Now we set equals to equals by Newton's second Law, we get \begin{equation} \int_{V} \rho(\vec{x}, t)\frac{\mathrm{D}\vec{u}(\vec{x},t)}{\mathrm{D}t}\,\D V=-\int_{V}(\nabla p(\vec{x},t))\,\D V \end{equation} for arbitrary regions $V$. This gives us Euler's equations of motion \begin{equation} \rho(\vec{x}, t)\frac{\mathrm{D}\vec{u}(\vec{x},t)}{\mathrm{D}t} =-\nabla p(\vec{x},t). \end{equation} Usually books divide through by density $\rho$, and add an external force term (usually Earth's gravitational field $\vec{g}$) to the right-hand side.

Theorem 1. Euler's flow equations of motion for fluid is \begin{equation} \frac{\mathrm{D}\vec{u}(\vec{x},t)}{\mathrm{D}t} = \frac{-1}{\rho(\vec{x}, t)}\nabla p(\vec{x},t). \end{equation}

A few remarks are in order:

1. This is an idealization insofar as friction between fluid parcels is negligible (which is reflected by a very small viscosity).
2. Exact solutions to the Euler flow may be found in Marchioro and Pulvirenti's Mathematical Theory of Incompressible Nonviscous Fluids. One large class of solutions include potential flow of incompressible fluids, where $\vec{u}=\nabla\varphi$ the velocity is the gradient of a scalar potential. Incompressibility amounts to $\nabla\cdot\vec{u}=0$ in this case. (Note to future me: insert link to incompressibility condition when I get around to writing it)
3. Intuitively, compressible fluids are gases, incompressible fluids are liquids.
4. In $n$ dimensions, Euler equations has $n$ unknowns from the velocity vector, 1 from the pressure, and 1 from the density. But there is one equation per dimension, i.e., there are $n$ equations for $n+2$ unknowns. Consequently, Euler equations are under-determined, and we need 2 more equations to determine all unknowns. We usually add an equation for the conservation of mass (the continuity equation) and either work with incompressible fluid or compressible fluids with additional equations from thermodynamics.

References

The most elegant (and difficult) presentation of Euler's equations I have found is in Landau and Lifshitz. It's the subject of the first chapter of their Fluid Mechanics, which is just amazing (but nearly impenetrable if you don't already have a good familiarity with fluids).

1. Leonhard Euler, "Principes généraux du mouvement des fluides". Mémoires de l'académie des sciences de Berlin 11 (1757) 274–315; English translation arXiv:0802.2383
2. Landau and Lifshitz, Fluid Mechanics.
3. Demetrios Christodoulou, "The Euler Equations of Compressible Fluid Flow". Bulletin of the American Mathematical Society 44, no.4 (2007) 581–602
4. John K. Hunter, An Introduction to the Incompressible Euler Equations. Notes dated Sep 25, 2006.
5. M. Zingale, Notes on the Euler equations. Dated April 16, 2013.
6. Chorin and Marsden, A mathematical introduction to fluid mechanics. Springer, third ed., 1993.
7. C. Marchioro, M. Pulvirenti, Mathematical Theory of Incompressible Nonviscous Fluids. Springer, 1994. Discusses exact solutions to Euler flow.

Describing Fluids

There are two ways to describe fluids: the Lagrangian frame, and the Eulerian frame.

Consider some fluid, which has its initial position in some region, say $\mathcal{B}_{0}\subset\RR^{3}$ (for "body" at time 0). We take $\mathcal{B}_{0}$ to be an open subset of 3-space, though we could consider $\mathcal{B}_{0}\subset\RR^{n}$ open, for some fixed $n$.

As the fluid moves, its constituent particles also move, occupying a (potentially different) region $\mathcal{B}_{t}$ at time $t$. We could introduce a Time Evolution Operator \begin{equation} \mathcal{M}_{t}\colon\mathcal{B}_{0}\mapsto\mathcal{B}_{t} \end{equation} which takes some time $t$ and transforms the initial body to the body at the specified time.

Suppose we take $\vec{a}$ to be a point in the initial body $\vec{a}\in\mathcal{B}_{0}$. Take the position of that point to be a function \begin{equation} \vec{x} = \mathcal{X}(\vec{a}, t) \end{equation} which is such that $\vec{a} = \mathcal{X}(\vec{a}, 0)$. This is the Lagrangian Coordinate of the fluid identified by initial position $\vec{a}$.

Remark 1. We could actually choose any arbitrary parametrization of the fluid; i.e., we don't need $\vec{a}$ to be the initial position of the fluid particle. We could have chosen $\vec{a}$ to be the position at some other particular moment, or we could use some other scheme (provided the label was unique).

But in practice, it's usually the case that $\vec{a}$ is the initial position of a fluid particle. Take care when reading the literature, because someone may have a brilliant new idea which requires a particularly idiosyncratic choice for $\vec{a}$.

The Lagrangian picture resembles that in analytical mechanics: we care about the history of each particle. (It's also coincidentally the starting point for the Hamiltonian/"Canonical" formalism in fluid dynamics, just as it's the starting point for the Hamiltonian formalism in analytical mechanics.) Although mathematically and physically intuitive, elegant, and appealing...it has the disadvantage that, when we do measurements in the lab, it's at a specific point in time (not over the history of the particle).

We could take this opposite view, starting with a fixed point $\vec{x}$ inside the fluid $\vec{x}\in\mathcal{B}_{t}$. Then we take the quantities of interest as functions $f(\vec{x},t)$. This is the Eulerian description of the fluid. Convention compels us to then examine the velocity vector $\vec{u}(\vec{x}, t)$ at each point in the fluid...although there's nothing to stop us from extending $\vec{u}$ to $\RR^{n}$ in general, provided we demand $\vec{u}=0$ outside the fluid.

Relating the two pictures. The velocity vector field relates to the Lagrangian description by taking the time derivative of the Lagrange coordinates: \begin{equation} \left.\frac{\partial\mathcal{X}}{\partial t}\right|_{\vec{a}} = \vec{u}(\mathcal{X}(\vec{a}, t), t). \end{equation} This gives us a system of differential equations we'd need to solve.

Remark 2: Caution with notation. Here I must stress the warning which Abelson and friends give in The Structure and Interpretation of Classical Mechanics about ambiguity with notation. Some authors (especially from physics) write the partial derivative with respect to, say, time to mean: \begin{equation} \frac{\partial}{\partial t}f(\vec{x}(t), t) = \left.\frac{\partial f(\vec{q}, t)}{\partial t}\right|_{\vec{q}=\vec{x}(t)}. \end{equation} Undergraduates and mathematicians are understandably confounded by this choice of notation (what ever happened to the "chain rule"?). Well, that's handled by the total derivative with respect to time: \begin{equation} \frac{\D}{\D t}f(\vec{x}(t), t) = \left.\frac{\partial f(\vec{q}, t)}{\partial t} +\sum_{j}\frac{\partial f(\vec{q},t)}{\partial q_{j}}\frac{\D x_{j}(t)}{\D t}\right|_{\vec{q}=\vec{x}(t)}. \end{equation} This notation has been "grandfathered in", and there's nothing we can do to change it. I'm so, so sorry.

Also note, if we wanted to be "completely general", we would need to consider derivatives with respect to the Lagrange parameter $\vec{a}$ — the curious reader may consult with pleasure Lamb's Hydrodynamics (§§13–14).

Example 1 (Acheson). Consider a 2-dimensional fluid with $\vec{u}=(u(x,y,t), v(x,y,t))=(-\Omega y, \Omega x)$ where $\Omega\gt0$ is a constant angular speed. We have two differential equations \begin{equation*} \partial_{t}x = -\Omega y(t),\quad x(0)=a \end{equation*} and \begin{equation*} \partial_{t}y = \Omega x(t),\quad y(0)=b. \end{equation*} Then we see (by differentiating with respect to time again, in both equations) that \begin{equation*} \partial_{t}^{2}x = -\Omega^{2}x,\quad\mbox{and}\quad\partial_{t}^{2}y=-\Omega^{2}y \end{equation*} which have general solutions of the form \begin{equation} (x(t),y(t)) = (a\cos(\Omega t)-b\sin(\Omega t), b\cos(\Omega t) + a\sin(\Omega t)). \end{equation} (End of Example 1)

Example 2 (Acheson). Consider now the Rankine Vortex in two dimensions. In polar coordinates, it's defined as \begin{equation} \dot{\theta} = \begin{cases} \Omega & r\leq R\\ \Omega(R/r)^{2} & r\geq R \end{cases} \end{equation} and $\dot{r}=0$, where $\Omega\gt0$ is some fixed angular speed, $R\gt0$ is the width of the vortex. We see then that $r=r_{0}$ is constant, and \begin{equation} \theta(t) = \theta_{0} + \begin{cases} \Omega t & r\leq R\\ \Omega t (R/r)^{2} & r\geq R \end{cases} \end{equation} where $\theta_{0}$ is the initial angle at time zero. We can change coordinates to see \begin{equation} x(t) = r\cos(\theta(t)),\quad y(t)=r\sin(\theta(t)) \end{equation} which lets us determine $r$ and $\theta_{0}$ in terms of $x(0)=a$ and $y(0)=b$. So $r^{2} = a^{2} + b^{2}$ and $\theta_{0} = \operatorname{atan2}(b,a)$ (recall the definition of atan2). Also note there is a singularity when $a=b=0$, since $\theta_{0}$ is not well-defined. (End of Example 2)

Note, I lifted these example flows from D.J. Acheson's excellent Elementary Fluid Dynamics; but Acheson doesn't really discuss the Lagrangian treatment of fluids (at least, not in the first few chapters I've read).

References

For the most part, I follow Childress's An Introduction to Theoretical Fluid Mechanics, though I have independently arrived at a similar line of reasoning. It is a joy to see it elsewhere, and Dr Childress probably thought of this before I was even born.

1. Stephen Childress, An Introduction to Theoretical Fluid Mechanics. AMS Press, 2009.
2. Landau and Lifshitz, Fluid Mechanics.
3. Chorin and Marsden, A mathematical introduction to fluid mechanics. Springer, third ed., 1993.
4. D.J. Acheson, Elementary Fluid Dynamics. Oxford University Press, 2009 reprint, chapter 1.

Configuring Mathjax

Mathjax is mildly straightforward, thanks to its instructions.

The `tags: 'ams'` is responsible for automatic equation numbering.

I am also trying to use $ for inline math (TeX delimiters which I prefer).

I've also added a few custom macros, some borrowed from Springer's style classes, others just shorthand. They're implemented using mathjax's configmacros, which may turn out to be a terrible idea...

If (somehow) I botch things later on, and need to backtrack to start over with the mathjax configuration, then this is what it looks like:

<script>
MathJax = {
  loader: {
    load: ['[tex]/ams']
  },
  tex: { 
    macros: {
      RR: "\\mathbb{R}",
      D: "\\mathrm{d}",
      E: "\\mathrm{e}",
      I: "\\mathrm{i}"
    },
    tags: 'ams',
    inlineMath: [['$', '$'], ['\\(', '\\)']],
    packages: {'[+]': ['ams']},
    autoload: {color: []}
  }
};
</script>
<script type="text/javascript" id="MathJax-script" async
  src="https://cdn.jsdelivr.net/npm/mathjax@3.0.0/es5/tex-chtml.js">
</script>