An awesome system of two non-linear differential equations

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[@allykid4720]

Dividing two eq gives: xdot x + ydot y = 0, so x^2 + y^2 = C^2. Plugging it back into the first equation and taking deriv of both sides gives: x** = - C^2 x, and y** = - C^2 y, therefore: x = C sin(C^2 t + ф), y = C cos(C^2t + ф).

[@petraveryanov2572]

also got x^2 + y^2 = C^2 tight away, after u can put it in any equation and integrate

[@Chr15T]

Great. That's how a physicist typically works: using a conserved quantity. In this case, energy (or some function of it).

[@tassiedevil2200]

@Ally Kid - Nice route to constant radius - but you have a typo - the coefficient in the two Simple Harmonic DE's is C^4, matching your solutions.

[@ianfowler9340]

That's exactly what did. y ' = -x/y and take it from there. If you take his solutions, square both and add. x^2 + y^2 = a^2[cos^2(at^2 + b) + sin^2(at^2+b)] and this yields x^2 + y^2 = a^2[1] = a^2. Implicit solution but you can still solve for y. Newton always saw f '(x) = y(dot) / x(dot) where x(dot) and y(dot) were simply the horizontal and vertical components of the (tangent) velocity vector.

[@khoozu7802]

U make a big mistake: To satisfy the equations for x'=-C^2 y and y'=C^2 x U have to put x=Ccos(C^2t+Φ) And y=Csin(C^2t+Φ)

[@beanary]

Missing for me - to give an illustration what the result means - moving on a circle with constant angular velocity

[@swolf2004]

Yes. Also the angular velocity is proportional to sqrt(radius). And the physicist in me says that those are some interesting units.

[@Pengochan]

As a physicist one would immediately notice, that if x, y are cartesian coordinates for the vector r, then dr/dt is perpendicular to r since the scalar product (r,rdot)=0. So the velocity is always perpendicular to r, which will result in a circle, i.e. |r|=const. If one adds a 3rd coordinate we can describe rdot with a cross product of r x (something in the direction of z), the xdot ~ -y, ydot~x has this "crossproducty" feel to it 😉. In the solution there is a little hiccup due to the tan function, so in principle one would have to check nothing untowards is happening where cos(theta)=0. After finding rdot=0 (here the polar coordinate component, i.e. |r| of the above vector) one could put that in the derivatives for xdot, ydot at the top of the board and compare to the given equations to find thetadot=r^2

[@idjles]

And because he’s a pure mathematician he won’t tell you that that’s simple circular motion. that would require a physicist to notice

[@tomholroyd7519]

It's x**2 + y**2! How can it not be a circle. Does doing proofs while rock climbing count as applied? (Does Teaching?)

[@jursamaj]

Not sure I agree. I'm more mathematician than physicist, and I knew it was circular motion as soon as I saw the DEs.

[@andrewkepert923]

Yes, but with a caveat: although individual orbits are circular, the angular velocity varies, so it is not just describing a rotation of the whole plane. (That one is dx/dt=-y, dy/dt=x.). So maybe “simple circular motion” gives the wrong impression.

[@idjles]

@@andrewkepert923 the angular velocity in this example IS constant.

[@tassiedevil2200]

@@idjles The angular velocity (theta_dot) epends on radius of the trajectory - goes like r^2 - this is not rigid rotation of the whole plane.

[@davidwright5719]

Note this orbit is unstable. The linearized problem has a positive real eigenvalue.

[@shmuelzehavi4940]

Another approach The system of differential equations is: x ̇ = - y (x^2 + y^2 ) y ̇ = x (x^2 + y^2 ) By multiplying the first equation by 2x , the second one by 2y and adding the both we obtain: 2xx ̇ + 2yy ̇ = d/dt (x^2 + y^2) = 0 Integrating this equation we get: x^2 + y^2 = r^2 = x(0)^2 + y(0)^2 Therefore, we obtain a family of concentric circles with the center located at the point: (0, 0) and the radius, r determined by the IC. The original system of nonlinear differential equations reduces therefore to the following linear system of differential equations: x ̇ = - r^2 y y ̇ = r^2 x We define now a complex variable z by: z = x + iy and obtain the following first order linear equation: z ̇ = ir^2 z The solution is: z = rexp(i(r^2 t + φ)) , r = √(x(0)^2 + y(0)^2) , φ = atan2⁡(y(0) , x(0)) Finally, (x,y) are given by: x = r cos⁡(r^2 t + φ) , y = r sin⁡(r^2 t + φ)

[@zachariastsampasidis8880]

multiply first by x second by and then add to get zero. this implies that x^2 + y^2 is constant. so then returning that to the original pair of equations gives us the harmonic set of equations satisfied by the sine and cosine functions

[@odysseus9672]

I knew things were going to end up more complicated than necessary the minute you busted out tan(theta). Just use the fact that the position vector (x,y) = r * rhat (by definition). You then get that v = (xdot, ydot) = rdot *rhat + r*thetadot * thetahat. Your differential equation then becomes v = r^3 thetahat. From that you immediately get that rdot = 0, so r is a constant, and thetadot = r^2, so theta = r^2 * t + c. It makes a nice rotating vortex that shears with greater speed as you move outward from the origin (theta = k*t + c would be a solid rotating disk with no shear away from the origin). If you must have x and y, you have x = r*cos(r^2 *t + c) and y = r*sin(r^2*t + c), where r is to be considered as a non-variable initial condition.

[@andrewkarsten5268]

Depends on what you mean more complicated. Your method involves using vector calculus, while his only requires basic geometry and basic calculus II understanding. What is “simpler” depends on perspective.

[@Jaeghead]

5:40 You can't argue like that, just because a function is 0 at a point doesn't mean that its derivative is 0 as well. Instead we can just integrate the equation to get that r^2 is constant, which is all we need for the proof (by continuity r must also be constant, but we don't use that fact).

[@Happy_Abe]

True but we these equalities hold for all values of x and y so the solution r=0 isn’t saying that when r equals 0 it’s satisfied, it’s saying the solution would be r is identically always 0 which would in fact tell us the derivative of r is 0.

[@Jaeghead]

@@Happy_Abe But that was not what he was trying to explain, he was essentially saying "For all t: r(t)·r'(t) = 0 ⇒ r'(t) = 0 or r(t) = 0 ⇒ But if r(t) = 0 then necessarily r'(t) = 0 ⇒ We can conclude r'(t) = 0 for all t" and that logic isn't sound. He never assumed that r(t) = 0 for all t, all he knew was that either r(t) = 0 or r'(t) = 0.

[@Happy_Abe]

@@Jaeghead r(t)=0 here means it’s always 0 not that it’s 0 for some t and we need to solve for what t. The function r itself equals 0.

[@Jaeghead]

@@Happy_Abe But he never concluded that r(t) = 0 for all t (and obviously you can't as there are other solutions), so he had to explain why there couldn't be a solution with r(t) = 0 at some points and r'(t) = 0 at other points (i.e. a mixture of both cases).

[@Happy_Abe]

@@Jaeghead the equations that got him to that conclusion were for all t so this shows either r(t) is always 0 or r’(t) is always 0

[@QuantumHistorian]

With experience, some D.E. you can solve just by looking at them. This one screamed _"circular orbit!"_ at some dark recess of my mind. Checking a few key values shows that it goes counter clockwise. With that, parametrising a solution and solving for any remaining constants is easy. Of course, all problems are easy if you already know the answer lol.

[@jursamaj]

This. As soon as I saw the 2 DEs, I knew it was circles. This is so since at any given r, the DEs simplify to x•=-ky, y*=kx, which is circular motion. A moment's thought shows increasing angular velocity with increasing radius. A bit more thought yields the angular velocity going as r^2, as that's the k factor the DEs were simplified by in the previous step.

[@abrahammekonnen]

Really cool problem. The hidden geometry in the problem is very nice. Thank you for the video.

[@touitcha]

Actually this system can be solved exactly the same way even replacing "(x^2+y^2)" by an arbitrary function f(x^2+y^2). Multiplying the first equation by x and the second by y and summing will still lead to x^2+y^2=r=const and therefore f(r) is also constant.

[@SuperSilver316]

Dividing the two equations will give you dy/dx = -x/y Solving this ODE gives you the expression y^2+x^2=2D = C Plug that back into one of the original equations, take a derivative and you get second order ODE’s who’s solutions gave to be superpositions of sines and cosines. Would be cool to have some initial conditions to evaluate these constants.

[@manucitomx]

Thank you, professor. Very straightforward.

[@andrewkarsten5268]

I found the answer in a slightly different way. I used complex numbers. So take z=x+iy, then z’=x’+iy’ (yes I know, glossing over some complex analysis stuff, but just follow me here). Then, z’=x’+iy’=-y(x²+y²)+ix(x²+y²)=i(x²+y²)(x+iy)=i|z|²•z. Now, converting to polar z=re^{iθ}, z’=r’e^{iθ}+iθ’re^{iθ}, we get r’e^{iθ}+iθ’re^{iθ}=z’=i|z|²•z=ir²•re^{iθ}=ir³e^{iθ}⇒r’+iθ’r=ir³ by factoring out the exponential everywhere ⇒r’=ir³-iθ’r=ir(r²-θ’). Since r,θ∈ℝ for all t, so are their derivatives, so r’=Re(r’)=Re(ir(r²-θ’))=0 giving r=a is constant, and substituting back in gives 0=r’=ir(r²-θ’)=ia(a²-θ’). This means either a=0, so we are always at the origin, or a²-θ’=0, giving the linear formula for θ with respect to time. Plugging this back in to z=re^{iθ}, and using Euler’s equation to convert back into Cartesian coordinates, and taking the real and imaginary components to get x and y respectively, we obtain the solution provided in the video.

[@mjmasterz]

Vvv nice

[@AstroTibs]

You can tell from looking at the diffeq what the nature of the solution is. x² + y² = r² dy/dt ∝ xr² dx/dt ∝ -yr² This has the hallmarks of a rotation. Indeed the paths are circular CCW rotations, and their rates are proportional to r². If the terms were √(x² + y²) = r instead, then the rotation rate would be proportional to r, and the field would rotate like a turntable.

[@excelelmira]

-> Look at the set of equations. -> Try to plug in x=y=0. -> Works. -> Problem solved.

[@nzeches]

Actually it seemed quite obvious looking at the equations that (cos t , sin t) was a solution, because x2+y2=1. From there you can derive all solutions, and going polar seems quite an obvious path. Also note in your last step that a2x2+a2y2 = a2(x2+y2) and x2+y2=a2

[@andrejgrebenc3235]

Suggested solution is partial. It is not only the function of t, but of g(t).

[@carstenmeyer7786]

With polar coordinates *u := (x, y)^T =: r * ( cos(𝜑), sin(𝜑) )^T* and the rotation matrix *Rot_z(𝜑) := ( cos(𝜑), -sin(𝜑) // sin(𝜑), cos(𝜑) )* you can decouple the system: *[ r' ] [ x' ] [ -sin(𝜑) ] [ 0 ]* *Rot_z(𝜑) * [ ] = [ ] = r^3 * [ ] = Rot_z(𝜑) * [ ]* *[ r * 𝜑' ] [ y' ] [ cos(𝜑) ] [ r^3 ]* Multiply the system by the inverse *Rot_z(-𝜑)* from the left to obtain *r' = 0, r * 𝜑' = r^3* The first equation clearly leads to *r = R = const.* If *R* is non-zero, we may divide the second equation by *R* to get *𝜑 = R^2 * t + 𝜑_0.* Writing both into *u* yields the solution.

[@Nikolas_Davis]

This is how I solved it, although I cross-multiplied the 2 equations by cosθ and sinθ instead of using matrix multiplication to decouple them. I find this method much easier, faster and clearer than the way Michael does it here.

[@carstenmeyer7786]

@@Nikolas_Davis Thank you for the feedback! While the complete decoupling may be more systematic, I'd say Michael's way is more "basic": It does not use as many tools like *Rot_z(𝜑),* but relies on "eyeballing" which terms to eliminate to simplify the system. Of course the "eyeballing" is mathematically equivalent to the decoupling via *Rot_z(𝜑),* but I imagine the "eyeballing" may be more intuitive for some viewers.

[@Nadavot]

A nice way to way to solve this could be to define z=x+ iy This gives (dz/dt)=iz |z|^2. And (dz*/dt)=-iz* |z|^2. Together this gives d|z|^2/dt=0.

[@tassiedevil2200]

While I agree with the physics and applied math folks about the missing insight that the time derivative of the 2D vector r (= x i + y j) is orthogonal to that vector, more to the point of the exposition, once it is established that r_dot is zero, could return to the x_dot DE and the polar expression for x_dot - (top of the board) to see by substitution that theta_dot = r^2 without all the tan(theta) stuff. (Edit - just noticed @Pengochan had pointed this out earlier) Can also arrive at theta_dot = r^2 first by substituting the polar expressions into both DEs and see that the r_dot terms (and all the trig functions) can be eliminated by mult. x_dot eq by sin(theta) and y_dot eq by cos(theta) and subtracting.

[@wolfmanjacksaid]

Would love to see some DEs that result in Bessel function solutions!

[@Falanwe]

You can get the right intuition for this problem if you see (x,y) as a vector V. V.V' is obviously 0, so ||V|| must be constant

[@JoelRosenfeld]

I feel this is kinda a cheap problem. Without the x^2+y^2, it’s the ODE for a circle. This just happened take the equation for level set for the solutions and threw it in there.

[@chengkaigoh5101]

What if we divide the 2 equations to give dy/dx=-x/y and intgegrate to get x^2+y^2=constant,which can be substituted back in

[@andrejgrebenc3235]

Symple solution x'/-y=y'/x where prime is time derivative

[@PeterZaitcev]

I appreciate the will to demonetizing videos, but KZbin, however, does not - even if you turn off the monetization, it will still show ads, show a ridiculous amount of ads, and will take all revenue by himself.

[@davidseed2939]

are we allowed to guess. y=sint, x=cost

[@stephenhamer8192]

Can we use complex nos and write: z'(t) = i .|z|^2. z? We then have (|z|)' = (z.z*)' = z'.z* + z.(z*)' = z'.z* + z.(z')* (am I allowed to do this?) = i.|z|^2.z.z* + z.(i.|z|^2.z)* = i.|z|^4 - i.|z|^4 = 0; whence |z|^2 = a^2. OK, this is completely unmotivated Now write z(t) = ae^i.w(t), where w is some function of t to be determined. Differentiating, and c.f.-ing with the original DE, we have w' = a^2; whence w(t) = a^2.(t + T) Thus: z(t) = a.e^[i.a^2.(t+T)], which is Penn's result in complex form

[@adiaphoros6842]

Reminds me of the video by Dr. Barker where he derived a 3rd order differential equation whose solution set are all of the circles: (1 + (y’)^2) / y’’ = 3y’y’’ / y’’’ This seems to be a decomposed version of a more constrained version. For your next video, can you convert Dr Barker’s equation into 3 simultaneous 1st order differential equations?

[@HeyHeyder]

It seems a Hamiltonian system H=(x^2+y^2)^2/4

[@lucasdeoliveira5316]

Yeah, I'm pretty sure I've seen this system in the context of dynamic systems. Maybe in Dominic Jordan's on differential equations or Strogatz's book on chaos.

[@pwmiles56]

On the face of it, V=(k/2)(x^2+y^2) would be the potential energy term in the Hamiltonian for 2-dimensional simple harmonic motion. The kinetic energy term is T=(m/2)(xdot^2+ydot^2). But this system allows elliptical orbits. Somehow I feel that teaching applied maths from a pure maths perspective isn't the greatest approach.

[@CTJ2619]

the diff ran vids on mathmajor are awesome

[@chimetimepaprika]

At 5:30 that if/then arrow looked like a gradient nabla and I was like, "WHAAAAAAAAAA?"

[@johns.8246]

How about an initial condition like x(0)=2 y(0)=0 ?

[@The1RandomFool]

Just to let you know, I didn't put a comment in before you fixed it. I waited.

[@depressedguy9467]

Well a nice example as it is not stable

[@holyshit922]

dx/(-y(x^2+y^2))=dy/(x*(x^2+y^2))=dt/1

[@txikitofandango]

I used x = r(1-s²)/(1+s²) and y = 2rs/(1+s²) and got the same result. There's some neat cancellations this way, and you get to use the double-angle formulas for sine and cosine, as well as a nice separable diffeq to solve.

[@nikita__222]

Does it exist a real process which is described by this system of o.d.e.`s? Or - what are the possible applications of this solution to other math domains?

[@Vladimir_Pavlov]

dx(t)/dt= -y(t)*[x(t)^2 +y(t)]^n, dy(t)/dt= x(t)*[x(t)^2 +y(t)]^n => x(t)=C*sin(C^2n *t+ ф), y(t)= - C*cos(C^2n *t+ ф).

[@SlamminGraham]

"r is a function of t." "r is a constant." Okay.

[@marcelozabani8939]

Doesn't defining y and x in terms of the same r(t) and theta(t) overconstrain the set of solutions? Maybe in this case it turned out alright, but is it correct in general?

[@pwmiles56]

It is a perfectly good general method, given that x and y are different functions of r and theta. x = r cos(theta) y = r sin(theta) You need to be sure that r and theta exist for all values of x and y but that's not a problem here

[@marcelozabani8939]

@@pwmiles56 Ah, thanks! I haven't done maths in a while, sorry. The way I see it, y=xtan(theta) doesn't bind either x nor y since tangents are surjective in R, so there clearly exists some theta(t) for any given x(t) and y(t). After that, some r(t) exists to bind one of the functions (and therefore both). Would you say that's a fair assessment of why this works in general?

[@pwmiles56]

@@marcelozabani8939 Pretty much. The usual way of thinking is to check that an inverse transformation exists r = sqrt(x^2 + y^2) theta = arctan(y/x) Actually one should use the four-quadrant version of arctan, known as atan2(y, x). You don't actually use the inverse transformation, it's just a check that r and theta exist for any x, y

[@jay-oi5dj]

Just looking at the equations I saw that if x=cost and y=sint, then x^2 +y^2 = 1 and our system becomes xdot = - y and ydot = x, for which cost and sint work. Makes me think that these equations were formulated after solutions were known and that it was noticed the pythagorean trig identity could be added without changing anything.

[@unknownstoneageman81]

@8:28 Why secant² theta?

[@fabiobiffcg4980]

(x²a² + y²a²)/x² (x²a² + y²a²)/( a²cos²(theta) ) [ (x²a² + y²a²)/( a²) ] * ( 1/cos²(theta) ) [ (x²a² + y²a²)/( a²) ] * sec²(theta)

[@Happy_Abe]

Why are we using dots and not primes?

[@MarkWaller2]

It's a fairly common convention that dot means 'derivative with respect to t' (usually t = time), and prime means 'derivative with respect to x'.

[@Happy_Abe]

@@MarkWaller2 I know but I just wondered why because he usually doesn’t That’s more in physics and engineering

[@pwmiles56]

The dot was Newton's notation, introduced in his Method of Fluxions (

[@user-ts1bt7xv1y]

x•(dx/dt) = - xy(x^2 + y^2) y•(dy/dt) = xy(x^2 + y^2) x•(dx/dt) + y•(dy/dt) = 0 (2x)•(dx/dt) + (2y)•(dy/dt) = 0 d(x^2 + y^2)/dt = 0 so we have x(t)^2 + y(t)^2 = C where C is a constant so (dx/dt) = (- C)•y(t) d[(dx/dt)]/dt = (- C)•(dy/dt) d[(dx/dt)]/dt = (- C^2)•[x(t)] and dy/dt = (C)•x(t) d[(dy/dt)]/dt = C•(dx/dt) d[(dy/dt)]/dt = (- C^2)•y(t)

[@andrejgrebenc3235]

Why are you complicating. Eliminate x^2+y^2 and the solution is a circle.

[@thomasheljeschrodingertama4874]

Very poggers 💖💖🤙

[@samuelnewport4970]

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