Learning the Weirstrass P function, which is relevant for elliptic functions, is related to the inverse of elliptic integrals was like seeing two worlds collide, incredible

@Alex_Deam

Жыл бұрын

Elliptic integrals historically got their name because of trying to calculate the arc length of an ellipse. If you apply the same idea to circles and do the integral, you end up with something proportional to the angle (i.e. the familiar arc length = r*theta). And an angle is just the result of inverting a trig function, and a trig function is just a periodic function with only one independent period, as you'd expect for a circle. Likewise, the Weierstrass p-function is a periodic function with two independent periods, and hence is (related to) the inverse for the arc length of an ellipse, an entity with two periods.

@leif1075

Жыл бұрын

@@Alex_Deam The two periods you refer to being the two constants of the ellipse right? The a amd b in the ellipse formula foci a d I forgot what the other is directrix or maybe the two foci I forget

@ethanjensen7967

Жыл бұрын

I LOVE the Weierstrass P function!

@ericsmith1801

Жыл бұрын

Elliptic functions are also involved in cryptography. Elliptic functions make the ecryption process more durable.

@ethanjensen7967

Жыл бұрын

@@ericsmith1801 I'm super grateful for the applied mathematicians who love the applications. You guys do incredible work and help me get paid to do what I love!

4:01 accidentally wrote u'=u, but it was said correctly as u'=-u

09:50 when you reach Michael's mastery over mathematics, letters become numbers.

@intron9

Жыл бұрын

noob detected

@PavloPravdiukov

Жыл бұрын

Since letters denote unknown numbers, I see no problem with that

A mini-series on the Weierstrass p-function would be pretty nice, if you feel like it

@lexinwonderland5741

Жыл бұрын

seconded! he's got some videos on elliptic functions/integrals but I would love a deep dive!

Hi, french student here. I would just like to explain how we are teached to solve differential equations of the type y'=f(y) (which we call "équations séparées", literally "separated equations"). We want to divide by f(y) at some point so that we can use the fact that a primitive of y'/f(y) is ln(|f(y)|). But to do that, we have to make sure f(y) is never equal to 0 on our interval. I'll detail the procedure with the example of y'=y^2. First, we see that the function y_0 defined as y_0 = 0 is a global solution (which means it's defined on R and not only an interval of R). Then, let y be a solution of the equation so that y(t0) = y0 > 0. Suppose there exists a t1 such that y(t1) 0 for all t€I if there exists t1€I such that y(t1)>0. We can show that y(t)

@MarcoMate87

Жыл бұрын

A primitive of y'/f(y) is not ln(|f(y)|), because (ln(|f(y)|))' = f'(y)y'/f(y).

@justintroyka8855

Жыл бұрын

We call them that in English too - "separable equations".

@sinuhearialdo

Жыл бұрын

Super interesting! Your English in this paragraph is excellent too, no need to apologize :)

A note that the identity commutes with any other operator would have been nice. Otherwise the factorization doesn't work.

@sharpnova2

Жыл бұрын

he did basically mention that when he talked about i ='ing i^2

@leif1075

Жыл бұрын

Why the hell is he using e^-× and not e to POSITIVE X..you can use both..

@ARVash

Жыл бұрын

What would it mean for an identity to not commute? I'm not trying to be difficult, but I'm used to identity meaning "do nothing", so it's hard to imagine the order of a non-changing change to matter.

@alexfekken7599

Жыл бұрын

@@ARVash The point Kristian is making is that A^2 - B^2 is generally not equal to (A+B)(A-B) when A and B are operators.

@JivanPal

Жыл бұрын

@@leif1075 He correctly said "u' = -u" but incorrectly/accidentally wrote "u' = u". The only valid family of solutions is u = Ae^(-x). If it were supposed to be u' = u, then _only_ Ae^x would be correct, _not_ both.

I love y' = y^2 because it blows up to infinity in finite time (calling the independent variable time). This is remarkable compared to its weaker exponential cousin y' = y. It demonstrates the limitations on existence of solutions (the solution does not exist for all times). It's treated at length for these reasons in the opening chapter of VI Arnold's ODEs book.

@pierrecurie

Жыл бұрын

Or you can place the singularity in the past, and say that it's converging to 0.

Great video Mike! I love the "warm up" exposition, it made for a very enjoyable lecture. Keep the applied math videos coming! Your explanations were also very applied for a pure math person, which I appreciate from my physics background =)

Fantastic video! I would absolutely LOVE more content on elliptic functions/integrals, I know you've done videos on them in the past but I would appreciate a deep dive on the concept

That's called the "first integral" trick. Very central in physics and it's used a lot in the action formulation which leads to conserved quantities (because the first integral is a constant) such as energy, momentum, charge, and so many other things.

@chouayabdelali3241

Жыл бұрын

yesssssssss sir. The first encounter of any physicist with the first integral trick is with the law of conservation of energy.

@element4element4

Жыл бұрын

It essentially implies that there is a one-parameter Lie group of symmetries of this differential equation, underlying the conserved quantity. Indeed very fundamental in a lot of physics.

@leif1075

Жыл бұрын

Wjat trick multiplying by y lrime..that is BS!! No one would.ever think of that unless shown it first..that's what makes it so frustrating and not cool..I don't see why anyone would do it or think of it..differentiating makes more sense or integrating..

@andrewhone3346

5 ай бұрын

​@@leif1075y prime is an integrating factor. There are systematic ways to look for them.

While not complete as an answer, it is interesting to note that when you substitute y=ax^n into the said DFQ, you arrive at n=-2 and a=6, so by mere polynomial observation, y=6x^(-2) is actually a solution...

One more thing to do with the "difficult to integrate" autonomous first-order ODE at 10:50 is to plot the ode-vector-field, this allows one to see the general shape/charater of the solutions even though its functional form is challenging to get at.

If you let yourself use a little Physics as a shortcut, you can take a shortcut to 11:06 The probem y''' = y^2 can be derived from the problem of a particle in a string, only noting that the elastic force is here quadratic in x (f = ky^2). Since f only depends on y, it's a conservative force and a potential energy can be calculated as U = ∫ f dy = ∫ ( k y^2 ) dy = k y^3/3 Moreover, since it's 1D motion, the speed v = dy/dt (I choose t, the video uses x). By conservation of energy, m (dy/dt)^2 / 2 + k y^3/3 = E0, Where E0 is system initial mechanical energy. Here in particular m = k = 1 (so the original problem is y''' = y^2)

@MattMcIrvin

Жыл бұрын

yeah, any time I see a y'' = some function of y, I'm going to think of it as a 1D classical mechanics problem, with a particle moving around in some kind of potential.

@nullpter

Жыл бұрын

That's such an interesting observation to me!

@nullpter

Жыл бұрын

Time to learn some physics

@MrTknight33

Жыл бұрын

Multiplying the equation by y' is equivalent to demonstrating the energy conservation principle ! Because force (y'') * speed (y') is an equation on power equal to 0 here.

@michaelallen1432

5 ай бұрын

​@@nullptermath is just physics without units. 😂

Elliptic function = brakes screech Great job as always

I studied this EDO at graduation and got an interesting result, that it has a unique solution for certain requirements, if you want I can translate and send you

Love this stuff. Thank you!!!!!!!

This was great. Thank you, professor!

This lecture was really incredible. Thanks!

I got to the simple solution by starting with an ansatz of y = a x^b, where a and b are contstants. This yielded the particular solution y = 6 / x^2. Then I realized that the solution can be horizontally shifted because the differential equation is autonomous: y = 6 / (x+C)^2.

15:46

It might be my physical background talking, but whenever i see a differential equation independent of x, I know there should be some conserved "integral of motion". And thus I always look for some "multiplying by y' "--like trick, that allows me to integrate it once. Pretty sure one can formulate some rigorous purely ODE rule for this.

@hectormartinpenapollastri8431

Жыл бұрын

it's the noether conserved quantity associated to the time translation symmetry. basically the energy of the equation.

@Taric25

4 ай бұрын

If you multiply by y', you may falsely come up with an incorrect solution that y equals a constant such as zero, which may not be the case.

If you have initial conditions you could just use laplace transformations. Way easier for particular applications

@armandocamacho2772

Жыл бұрын

Fourier transforms are also handy for this problem. Laplace shines when the initial conditions are zero otherwise you end up with some slightly nasty partial fraction decomposition or inversion integrals. Fourier is a little more flexible in regards to this.

Love this differential equation and its solution. Thanks a lot.

as someone who stopped doing math after basic calculus, i really enjoy getting high and then watching university level math videos like this because it might as well be magic to me. right around the time he broke out the identity operator was when i went “what the fuck” and by the time he broke out the weierstrauss p function and wrote down the cube root of 6 all my mind could think of was that meme about stop doing math they have played us all for fools? absolutely brilliant video michael 10/10 keep up the good work

@jcpoly3205

2 ай бұрын

solving differential equations is like the movie airbud but math “let me just multiply this by y’” obviously makes mathematical sense but also is very much in line with “ain’t no rule says a dog can’t play basketball”

Thank you for your math channel. You gave me so much peace during the pandemic.

Hi Michael ! Using your trick (multiplying both parts by y’) we can also solve another equation y”=-y^3, which seems to me more interesting from cybernetics, control system point of view. Solving y”=-y^3 or rather ÿ=-y^3 we will get a curve which looks pretty much like sin or cos but in fact it is an elliptic sin or cos (or combination). And while in the case of linear “oscillator” (ÿ=-k*y) the frequency of the oscillation is constant whatever the initial conditions are, in the non linear case ÿ=-y^3 the frequency is proportional to the amplitude シ

One time I looked at the case of an object falling towards earth using F=(Gm1m2)/r^2 which ended up giving me r"=A/r^2 (where A=G(m1+m2)). That turned out to be something to differentiate as well! Had to use energy concepts to help me cause I’m not too well versed in differentiation.

As a physics major I just trialled the Ansatze: y = Ax^n, solved for A=6, n=-2

@demenion3521

Жыл бұрын

same actually. and then realizing that A(x+c)^n gives the same derivatives because the equation is autonomous

@Unidentifying

Жыл бұрын

ok but, as a nutritionist, the point is the algebraic solution path here which is hard imo

@tomfan5863

Жыл бұрын

@@Unidentifying 🤣

I found the nicer solution at the end in a bit of a simpler way, though it's kind of guess-and-check. First, I guessed that the solution to y''=y^2 would be some rational function, say of degree n. Then 2 derivatives make a degree of n-2, and squaring gives a degree of 2n. Thus, in particular we must have 2n=n-2, so n=-2. Given other familiar solutions, I then guessed that the solution had to look like something of the form A/(x+B)^2. I could then solve for A by applying two derivatives and getting y''=6A/(x+B)^4. To make sure the equation y''=y^2 is satisfied, we must have A=6.

Michael beats the separation of variables within an inch of it's life.

7:34 Oh my cosh...

Yes, a bit of a surprise! Thank you Professor Penn!

1:22 I'd be curious to see that more careful version because I've never seen it.

excellent video!

Awesome solution to a seemingly simply problem

Wow I followed everything in a Michael Penn video! :)

Really cool seeing these diff. equation tricks. Here's how I did it... Found y = 6/x² manually Then asked when does 'constant over quadratic' work in general? y = n / (ax² + bx + c) Diff. once: y' = -n (2ax+b) / (ax²+bx+c)² Diff again: y" = 2n ( (2ax + b)² - a(ax² + bx + c) ) / (ax² + bx + c)³ Equate to the square of the starting function: y² = n² / (ax²+bx+c)² Divide out some common terms: 6a²x² + 6abx + 2(b²-ac) = anx² + bnx + cn This gives us some simultaneous equations, that provide the following two conclusions: 1.) numerator must be 6 times the lead coefficient: n = 6a 2.) both roots are the same b² - 4ac = 0 And so finally, we have: y = 6 / (x - r)² EDIT: Cleaned up the math a bit. Apologies. I work these problems in notepad

There is no link in the description for brilliant

There is a mistake on the lower right panel there should be written u'=-u if you are to obtain u=A exp(-x) (but what is written is u'=u..) Apart from this , this video is great thanks a lot !

@PavloPravdiukov

Жыл бұрын

I noticed that too, and he even pronounced "minus" but forgot to write it, probably because mechanically he already put a few horizontal lines.

Are there any other values of A other than 0 that yield a "standard" solution?

What should be c2 equal to give "simpler" solution?

As the second derivative of x^(-2) is 6x^(-4), the square of 6x^(-2) is 36x^(-4), which also happens to be its second derivative.

thank you

this is called "missing independent var eq". By substituting w = y' w' = dw/dt = (dy/dt)dw/dy = w dw/dy y" = y² w' = y² w dw/dy = y² w dw = y² dy ½w² = ⅓y³ + C w = sqrt(⅔y³+C) y' = sqrt(⅔y³+C) and so on....

Slick move with multiplying by y’. Should have seen that coming. Guess I need to finish my coffee lol

As a physics student at UCLA in 1968, I walked into my first Upper Division class Mechanics 101. The first thing the teacher wrote on the chalk board was the basic form of the Lagrangian energy balance differential equation. I had taken all the required math courses for a Physics student during my Lower Division two years (calculus and so forth), but it did not include how to solve differential equations. I raised my hand and when the professor called on me, I said, "That is a differential equation." (I at least knew what one looked like.) He said, "Yes, it is." I said, "The Physics math requirements did not state that I would need to solve differential equations before taking any Upper Division class, so I do not know how to solve differential equations." He looked at me and said, "You better learn fast!" I got so mad that after the class I marched over to the office of the head of the Physics Department and stated to the secretary in front of his office that somebody had better change the match requirements for Upper Division classes since it did not require classes in differential equations and the professors there all assume that the students have had such classes already. This makes Upper Division work much more difficult at UCLA. I do not know what they decided to do, but I immediately started to take differential equation courses and, with only a few more math classes needed to get a second Mathematics degree, I turned myself into a Dual Physics/Math student and 1 & 2/3 years more time at UCLA to get both degrees. (I was able to get my only career job with the US Navy as an Electronics. later Computer, Engineer two weeks after graduating. More luck.) I also was extremely lucky in that the two Differential Equations classes that I took were taught by Dr. David Sanchez, a graduate-level professor who decided to teach undergrads and turned out to be the best teacher I have ever had in anything. (Kind of like the professor in the DEAD POET'S SOCIATY but in math.) This helped me immensely, but I do not know how earlier students could have handled this differential equations solution problem without significant difficulty on top of their regular class requirements.

Damn... I remember taking partial differential equations in college (back in 1984) and had to laugh watching this. I had absolutely no idea what the hell he was saying. It's weird how much you can forget over time. Sheesh.

@MathematicalToolbox

Жыл бұрын

I'm guessing your career didn't require you to use DEs? What did you end up doing after university if you don't mind me asking?

@goodplacetostart4606

Жыл бұрын

College? I’m doing it in my ninth grade math class!!!!!

I know that this will be seen as nitpicking but since students watch these videos I feel compelled to do so. The function -1 for x0 is a NON constant smooth function with derivative equal to 0 everywhere. The trick here is that when you look for a solution of an ODE, by design, the function you get must be defined on an interval (which solves the problem: a smooth function defined on an interval whose derivative is 0 everywhere IS a constant function). To insist on this particular point when I teach, I define the solution of an ODE to be a couple (I, y) where I is an (non trivial) interval and y a differentiable function defined on I.

Great video great channel tnx

The project that got me into UC Santa Barbara was finding as many solutions to the general y^(n)=y^n so I love seeing this

What initial condition would result in A=0 ?

The simplification can be disguised with the IVP y(2)=6 ; y'(2)=-12

It is worth mentioning that this equation does not explicitly contain the independent variable x which allows a first integration by multiplying with y'. An equation of this type arises often in physics (one dimensional Poisson equation in electrostatics) y''=f(y) where y is the electric potential , -y' the electrostatic field and f(y) the potential dependent charge. By multiplying both sides by y'/2 one gets d((y')^2)= f(y)dy and y'=+-Sqrt(Integral(f(y)).

y'' = y^2 the only difficulty while solving this equation is calculating integral because we will get elliptic integral dy/sqrt(2/3y^3+C_{1})=dt

Thanks🌹

In mechanics, If we have a = dv/dt, v = ds/dt, and a = f(s) we use a = dv/dt = (dv/ds) * (ds/dt) = v * dv/ds to solve the above differential equation Now we change s to y, change t to x To solve all y'' = f(y) type of second order differential equations: y' * dy'/dy = f(y) ∫ y' dy' = ∫ f(y) dy (y')²/2 = ∫ f(y) dy (y')² = 2∫ f(y) dy

You could have skipped all of that just by looking at the function. You need to find a function y which increases its exponent when squared. Only functions doing that are of the type x^(-2n), n€N. Now derive that twice and you get y'=(-2n)x^(-2n-1), y"=(-2n)(-2n-1)x^(-2(n+1)) Take n=1, easiest case, you get y=x^(-2), y"=6x^(-4), you're off by a 6, all you need to do is work on that constant.

Which EDO book would you recommend?

@charlesmzl7727

Жыл бұрын

ODE

Just from the thumbnail & title, without yet having watched this, here's my reaction: y" = y² ? Wow! How do you begin to attack that? OK, how about a power? Like y = axᵏ ; k ≠ 0 y" = ak(k-1)xᵏ⁻⁻² ; y² = a²x²ᵏ Then we'll need to have • k-2 = 2k ; k = -2, and • ak(k-1) = a² ; a = k(k-1) = 6 So we have y = 6/x² , which works! y' = -12/x³ ; y" = 36/x⁴ = y² This may, however, not be the only solution; I have a hunch it isn't. Let's see what the prof. does for this... Fred

I liked this video but I would have appreciated a brief discussion about the general behaviour of solutions given the form of the equation before diving in to solve

i'm wondering if the nonlinear equations can be applied to any physical problem. If x is the time coordinate and y a position coordinate, then a simple initial condition of y(t=0)=0 would result in the constants of integrations being divergent. however it may be applicable to a situation where the initial velocity is zero but the position is not, describing a system in a unstable equilibrium point or something similar. don't know off the top of my head any basic physics problem with a non-linear differential equation like the ones here (i mean the ones with y^2) but it maybe could model some crazy solid state or nuclear physics type problem, or function as the Euler-Lagrange equation for some toy models for inflation or other field theory systems, maybe even self interacting systems (which i know present runaway solutions). If anyone knows of any direct physical application to these differential equations I'd very much like to know.

Deceivingly difficult? More like “Dang good information; yep!” I’ve had so much fun watching Michael’s videos over the years…

Hello there, my question is why is there a constant just on the x side after solving the integers at 2:50? (middle of the blackboard). Shouldn’t there always be a “plus constant“ part if you’re solving indefinite integers?

@richardStretcher

Жыл бұрын

If I understand your question correctly, it doesn't matter in this case. If you have "+ C" on both sides of equation, you can always move one of them to the other side and rename the resulting (C1 - C2) as + C, since this is still some constant value.

@timokapornyai3266

Жыл бұрын

@@richardStretcher thank you

7:30 : it was just a warmup! OMG 🙂

I used reduction of order to find a solution. It gives the same result that Michael got multiplying by y'. In order to find a solution, I had to make the assumption that the first constant of integration is 0. Otherwise I got a result from SageMath involving the hypergeometric function 2F1. I suppose I could have converted to series and eventually come up with it myself.

@extremovolador

Жыл бұрын

¿y''=zdz/dy...?

I'm really confused about the identity operator part

Is that shirt from the chattanooga market?!? Cool to see it out in the world 😊

I'm sorry, but what is the identity operator? I can't find anything on the internet about it. Is it what y represents or something? Like that's where y is mean't to be?

Real blackboard, real chalk. Refreshing.

I can't believe you've done this tbh The first reflex when you see a derivative being equal to a power is test functions of the form y=a*(x-b)^n. Plugging in y in the differential equation immediately gives n and a, and conditions on b. Or you can rule out quickly those solutions if it doesn't work. in this case you find 2n=n-2 (so n=-2) a^2=6a (so a=6) and no constraint on b.

I thought there was something I was missing when I solved mentally (looked at the problem from the thumbnail) since you said it was deceptively hard. After multiplying both sides by y’ the rest falls into place pretty easily.

6/x² is a very nice and easy-to-check solution :) I guess that every differential equation of the form y^(m)=y^n has a solution of the form y=c*x^n (maybe with some singular exceptions w.r.t. m and n).

1:22 things you hear in your differential equations class 😂

@MrRenanwill

Жыл бұрын

For me, this is a meme. There is no need to prove that this method is correct at our point in mathematics. Even not knowing the correct way of doing the calculations is not necessary. It should just be an exercise.

@jkid1134

Жыл бұрын

@@MrRenanwill Well, this is of course a meme, but speak for yourself. You don't know my education or my goals.

4:13 this might be my mistake, but wouldn't u=Ae^(-x) be a solution to u'= -u, not u'= u ? What am I missing here?

y(x)=0 is also a trivial solution, as well as for y'=y² and y''=y. Nice video!

Good video

I love how factoring the operators D²- I into (D+I)(D-I) just works. I need to know why that kind of magic works. I know its a difference of squares, but D is *applied* to y, not *mutiplied* to y

amazing

y'' = -c*y^2 by the way describes aerodynamic force of the oncoming air. The more particles hit the plate per second, and the faster they are going, the bigger the deceleration is. In other words, acceleration is proportional to negated square of speed.

Once again, there is trick which is "energy transformation". If we have equation: y'' = f(y) Set p = y', then dp/dx = dp/dy*dy/dx = dp*p =d(p^2)/dy so d(p^2/2) =f(y)dy p^2/2 = F(y) + C, C = p(0)^2/2 - F(y(0)). So, y' = +-sqrt(2(F(y)+C)) dy/sqrt(2(F(y)+C)) = +- dt. This integrates.

Aaaand, YOU GUESSED IT!

shorter way to solve it is by put the equation into logarithmic form and that differentiate it. Solving the 3rd derivative equation is easy. Obviously the 3d constant of the integration is 0. (because we converted the second level of DE into the 3rd level).

my wwe ass lost it when it saw "dx"

First part is simpler like this: y'' = (d/dx)y' = y' d(y')/dy=d/dy(y'^2/2), so integrating the RHS of y^2 gives y^3/3 +A, and y' = sqrt((2/3)y^3 + A)

Can somebody explain step by step what's happening at 8:06 please?

This one is quite beyond my math level lol

Why squared?

2 multiply 1/2 = 1 chain rule for a Lichtenstein

9:40 This would have been the time to multiply by 6.

ignores constant: integral of ez rational power of y includes constant: *wierstrass P function*

Nice pace. I would take pains to point out that a first order equation has a general solution with 1 arbitrary constant and a second order one, 2 arbitrary constants, so you can check immediately that your answer is reasonable.

In last example, y=0 is lost solution (dividing by y^1.5)

1:11 "We *abuse* that Leibnitz notation". I've always wondered why mathematicians get so picky about splitting the apparent fraction of dy/dx. Is it mathematically correct or not for one to do that? If it's not formally OK then, why does it work? If it's formally incorrect, why they don't change the formalities in order to make it work that way.

@anon6514

Жыл бұрын

I would also like to know this. When does treating them like fractions lead to the wrong answer?

Can you guys guide me into the detailed analysis of 1:30-1:38?

13:20 Me: "Then it's just a Bernoulli equation." Mike: "Then it's just a separable equation." Me: "How did I miss that?"

8:45 why is it from the chain rool? why 1/2?

@anon6514

Жыл бұрын

let Q = (1/2) (dy/dx)² let u = (dy/dx) So, Q = (1/2) u² dQ/du = 2(1/2)u = y' du/dx = d(dy/dx)/dx = y" the Chain rule states: (dQ/du)(du/dx) = dQ/dx i.e. y' y" = (1/2)(d/dx) (y')²

I must concur with with the comment just before mine. What *IS* the "identity operator"? That was never mentioned in the D.E. class I took, nor the Calc & Algebra, before that. It sounds like it's something really simple, something I already know, but not by that name.

(3:36) No f-ing way! I call shenanigans! You called yourself out on the "abuse" of Liebniz notation while working on y' = y², but at least that kind of abuse usually works out and it makes sense to me intuitively. But "factoring" an operator? I am in utter disbelief that that's a valid technique. It's about as credible as John Titor's claims about time travel. Citation needed! Now to watch the rest of the video…

The natural assumption that similarity between the DE's implies similarities between the solutions, gets completely exploded.

Question from a simpleton: At 3:04, why the identity operator I and not just number 1? Thanks.

@floriankubiak7313

Жыл бұрын

Probably just a formality since "d/dx" is an operator, too. Correct me if you know more.