a beautiful differential equation
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- Опубликовано: 1 окт 2024
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Answer my comment pls
Sir, when you guessed x^r might work, how did you know that that is the olONLY function that will work?
I don't like how brilliant is essentially doing the same thing that khan academy is doing except khan academy is free
I am honestly surprised he didn't do f'(x) = - f(1/x) instead. That would have given the quadratic r^2-r-1=0 and had the golden ratio he loves so much as a power. Of course it would have been a less interesting solution without the complex exponential.
It still comes out to y''=-y/x² as there are two extra factors of -1 in the mix:
f''(x) = d/dx ( -f(1/x))
= -f'(1/x) (-x⁻²). [chain]
= -x⁻² f(x) [using f'(1/x)=-f(x)]
which is the same as what Michael had. So we get the same quadratic. The difference in solution comes from the y'(1) = -y(1) step.
One small oversight Michael *does* make is that the DE doesn't assume x>0, yet the solution does. The question has two possible domains. This is easy to handle using a change of variable to -x, or modifying the working strategically, so it's not a big deal.
(unless I've messed something up - I haven't fully worked through then checked the solution.)
What do you mean bybhe '"didn't do" f'(×)= -f(1/×)? My thing WHY didn't he PLUG IN HIS GUESS I TO THE ORIGINAL EQUATION..DIDNT NSYOME ELSE.THINK LF TBIS..tou get r(x^r-1)= (1/×)^rand that gives tou r*x^2r-1=1...and from there tou can't really solve for r ..or you get that r equals x raised to a power. But r is a constant and x is a variable so it tells tounrhwee are no solutions..no real at least..
Only a modified (complex) Cauchy-Euler diffeq does what you want, namely f'(x)= i f(1/x). A solution is f(x) = A(x^\phi-i*\phi x^{\Delta}) where \Delta = 1/\phi .
Nice video. The problem was cool and the solution was pretty neat.
The proof gave me the idea for a possible video. It would be nice to make some kind of explanation on how the solution of a differential equation is an element of a vector space of dimensionality n (where n is the order of the DE). When you applied the condition y'(1) = y(1) it was like applying a constraint, reducing the dimensionality of the space by one. I think it would be cool to show this kind of concept from a geometric/linear algebraic perspective
Something that goes nicely with this topic is that every n-th order linear ODE with constant coefficients can be rewritten as an n-dimensional matrix-vector equation x'(t) = A x(t) with the solution x(t) = e^{A t} which can be analyzed using the diagonalization of A. These kinds of ODE systems are used a lot in applications like chemistry, biology, economics, etc and it can be motivation for those people to understand how all such systems can be exactly solved with a bit of linear algebra.
Not such a good place to stop -- just yet! We have found only a _candidate solution._ Meaning, that we know that every solution must have this form, or rather, that it can only be this solution. But we haven't plugged it into the original equation yet, in order to prove that it indeed is a solution.
How could that solution not satisfy the original equation?
@@whonyx6680 you always want to be careful when you achieve a solution by deriving a new differential equation from the one you want to solve. It isn't always the case that a solution to one will be a solution to the other.
@@whonyx6680 at that stage, _without checking_, it could still be the case that there is NO solution to the original equation at all. Only actually "plugging it in" will prove the existence of the solution. He has only proven the uniqueness of the solution(up to the one constant), but not its existence.
Admittedly, he has proven the harder part, and just "plugging it in" in order to compete the proof is a quite simple step left to do.
Again, my suspicion is that he skipped that easy step in order to have a shorter, more YT friendly video. But he could have at least called out that there is still one last step that "he's leaving four the viewer as homework" or something like that.
Superb! Thanks for that!
An additionnal cleanup step could be to write the final solution as C*sqrt(x)*cos(3ln(x)/2 -pi/6)
I think BPRP made a video a while back detailing the process of changing the sum of two sinusoidal into a sinusoidal with phase.
But the two sinusoids have coefficients of different magnitudes?
@@drdca8263 That isn't a problem. Try plotting the sum of two sinusoids of the form A sin(x + B), with different values of A and B, and you'll see that it remains a sinusoid.
@@chaosredefined3834 I mean of the form a sine(m x + b) though (with different values of m)
Hi Michael, i'm a physics student and since i've taken some general relativity courses, i would love you to recorver the differential forms series, maybe to get to a point where i can understand the geometrical meaning of the stoke's theorem, from a mathematician perspective. Also because i absolutely enjoyed all the chapters posted as they helped me.
at 12:52 why not just use f'(x)=f(1/x) to get f'(x) rather than fuss with the differentiation?
Wouldn't that just give f'(x)=A? Since f(1/x) = f(x) when x=1
You forgot to check that the solution satisfies the original functional equation f'(x) = f(1/x)
Also I think we need to take x >= 0 or else the solution doesn't hold.
That would be x>0. x≥0 allows x=0, hence allowing 1/0.
He checked for x=1. For continuous function, this should be same as checking for general x, up to some domain issues that might exist.
Wjat does composing the function with itself even mean? Wasn't anyone else wondering what he meant at 0:50m
@@leif1075 no. It's fairly standard maths. If g(x)=1/x, then composing it with itself is
(gºg)(x)=g(g(x))=1/g(x)=1/(1/x)=x
@@leif1075 He wasn't talking about the function f here. He meant composing the function g(x) = 1/x with itself: g°g=id. Or g=g^(-1). Or g(g(x)) = x. IOW, the g is its own inverse. This side remark isn't actually important for the video. He only tried to justify how one might get an idea to look for x^r and an _Ansatz._ I did have the very same idea myself here, but for another reason.
11:42 Why has i disppeared from B ? Shouln't B be i (c1 - c2) ?
But then, how can A be equal to a real constant times B ? A being (c1+c2),it yieds
c1+c2 = √3 i (c1-c2)
⇔
c2 = - (1-i√3)/(1+i√3) c1 = - (1-i√3)²/4 c1
or A = [1 - (1-i√3)²/4] c1
and B = i [1 + (1-i√3)²/4] c1
Both must be complex.
Where did I miss something?
OK got it!
As f(x) =y = c1 complex + c2 complex, those c's must be complex.
Provided a real constant C,
if c1=(-√3 + i)C
and c2 = (-3√3 + (1 + √3)i)C/2
then A and B are real:
A= -√3 C
B= - C
Can we prove that there are no other solutions?
I have the same question. We assumed x^r and it works, but maybe there are other solutions that could be added to his final equation.
The x^2 y'' + y = 0 equation is a second order equation, so it's solution has 2 degrees of freedom. The solution we got has two degrees of freedom, so that must be all of it.
@@DR-tx3ixyes, you define any other solution g(x)=f(x)+h(x) and you’ll find that h(x) has to be Cf(x), thus proving the uniqueness.
He did. That's actually the ONLY thing that he did prove: that there are "no other solutions." What he yet _failed _ to prove was that it is, in fact, a solution in the first place.
He skipped over that _very last hurdle, _ namely plugging his solution back into the original equation, and showing that it holds true. I guess he wants to keep his videos short, to make then YT friendly. So he skipped over that step and jumped to "good place to stop". IMO, he should have, at least, clearly articulated that he "leaves this last step as homework for the viewer", or something like that.
as long as the first derivative of the function is also differentiable, yes.
I think it would have taken me a few hours to do what Michael did in less than 16 minutes.
I did wonder if switching to polar coords about 9:00 might have helped but ...
Note: If B=0, we get the 0 function~
You forgot to multiply sin by i at 11:25
Absorbed it into the constant
You could have done a lot of simplification beforehand for f′(x), because √(x)/(2x)=1/(2√x); this means that f′(x)=1/(2√x)((A+B√3)cos(½√(3)ln(x))+(B−A√3)sin(½√(3)ln(x))). You still end up with A=B√3, though, which leads to a lot more simplification.
The solution can be further simplified by combining the two trig fxns.
sqrt(3)cos(□)+sin(□) = 2sin(□+π/3)
Bonus question: Solve f'(x) = -f(1/x). A good ole' friend pops up in the solution.
Wow, that was way more complex than it looks. I have pretty much forgotten all my DE.
Need to choose either positive branch of x or if f(x) can take complex values then negative values of x only since x=0 isn’t allowed in the initial differential equation.
You can't use negative values of x if you use positive values of x, since continuing f(x) to the complex plane leads to a discontinuity on the negative real line. If there's one branch cut, it can't be anywhere other than half the real line, because of 1/x. It can't be defined on the branch cut, because you can't take the derivative there.
Can't help feeling you drag some of these out beyond their natural lengths. "1/1 is just 1" - really?
Beauty is in the superior mathematical brain of the differential equations/analysis/abstract algebra beholder!!!
Unless A and B are complex numbers, Dr. Penn lost a factor of the imaginary unit at 11:30. This means that the answer he gives is not quite right (I think)
He implicitly defined A = c1 + c2 and B = i(c1 - c2). If you follow through defining B = c1 - c2 you'll see that you get A = i sqrt(3) B, which gives it the same value as defining B the way he did. Ultimately, the value of B would depend on imposing some sort of initial condition or boundary condition.
@@allanjmcpherson ah! that works.
14:48, I like that solution!
When you write that y=c_1x^r_1+c_2x^r_2, how are we sure that these are the only solutions? After all, this conclusion came down from the fact that you checked for a function of the form y=x^r and then used the properties of diff eqs to take the linear combination of the solution, but how are we sure that there are not other solutions that don't arise in this way?
This is a result of a theorem you can find in introductory books on ordinary differential equations (Calc 4 level in the US). The theorem essentially states that a "well-behaved" initial-value problem involving an nth-order linear ODE will always have a unique solution. Specifically, let n ∈ *N,* let I ⊆ *R* be an interval with a ∈ I, let p₀, p₁, ..., pₙ₋₁, and f: I → *R* be continuous, and let b₀, b₁, ..., bₙ₋₁ ∈ *R.* Then the equation
y⁽ⁿ⁾(x) + pₙ₋₁(x) y⁽ⁿ⁻¹⁾(x) + ... + p₁(x) y′(x) + p₀(x) y(x) = f(x)
has a unique solution y: I → *R* satisfying
y(a) = b₀, y′(a) = b₁, …, y⁽ⁿ⁻¹(a) = bₙ₋₁.
It's important to realize that this only applies to LINEAR ODEs, and then only if all the coefficients are continuous on the whole interval. Note that linear differential equations also satisfy the superposition principle, meaning the sum of any two particular solutions of a linear differential equation is another solution. So we can always form a basis for all solutions by considering different initial value problems. A simple way to do this is to set (b₀, b₁, ..., bₙ₋₁) = (1, 0, ..., 0) to get one solution, then (b₀, b₁, ..., bₙ₋₁) = (0, 1, ..., 0) for the next, etc., finally getting the nth by setting (b₀, b₁, ..., bₙ₋₁) = (0, 0, ..., 1). This gives n linearly-independent solutions which span the entire solution space. So the general solution to the differential equation will be of the form A₀ y₀(x) + A₁ y₁(x) + ... + Aₙ₋₁ yₙ₋₁(x), where the A₀, A₁, ..., Aₙ₋₁ are arbitrary real numbers and the y₀, y₁, ..., yₙ₋₁ are the particular solutions generated above.
You can check that the differential equation in this problem is ordinary and linear, has continuous coefficients over all of *R,* and has second order. So the general solution will be y = A₀ y₀ + A₁ y₁. And indeed, that's exactly what his solution for y looks like, though he calls the two real coefficients A and B.
Then, Dr. Penn adds an additional constraint that f'(1) = f(1), which comes from the functional equation at the start of the video when you substitute x = 1. This reduces the dimension of the solution space by 1, specifically by showing that A = √3 B in this case, so we end up with a single parameter in the final solution for f.
@@EebstertheGreat thanks for the exhaustive answer
11:27 - where did i go?
I’m sorry, but can anyone explain why are we sure that this is the only solution(family of solutions)?
Solving a differential equation without any ugly Integrals :-)
f(x) = 0
B)
How does one know that f is two times differentiable?
You said it has to hold for all x, but x cannot be = 0
If the solution is in the form x^r, then we get the solution in the end
But how to show, that no other solution form exists?
The equation x^2y"+y=0 is a homogeneous linear 2nd order differential equation, which would be solved rigorously in a standard DE course. One idea is to define z = yx^r (where r is a complex number satisfying r^2+r+1=0), and define w=z'x^(-2r), and then you can show that w'=0, so w is constant. You can then solve for z to get z=Ax^(2r+1)+B for constants A and B, so y=Ax^(r+1)+Bx^(-r). Therefore all solutions of the DE must have that form.
@@Notthatkindofdr thats from half to two minutes of precious content!
It should be incorporated into solution :)
Nice video!
didn't follow comp. - but go job
Brilliant can even help people like Zsa Zsa Gabor! LOL!
15:26
Fascinating ❤
Excellent problem.
You can solve for B using the original equation, i think you get B=1
How? When I use the original equation (f'(x)=f(1/x)) B cancels out. Notice that we do not have f'(x)=1/f(x)
Also notice that if B=0 we would have a solution as if f(x)=0, then both f(1/x)=0 and f'(x)=0 holds, meaning that they are the same.
@@lajont wait yeah it does nvm
Looking at the thumbnail immediately had me thinking of f(x)=√(2x). Close but not quite
That solves f'(x)=1/f(x) interestingly enough
@@insouciantFox if f were linear then that would be enough 😭. At least it's a critical component of the real solution
@@GhostyOcean I don't think linearity would suffice. The class of functions for which f(1/x)=1/f(x) is probably limited to ±x^±1 unless you want to get really devilish with abstract, non-continuous functions or the trivial y=±1.
ln x is necessary bc it converts 1/x to -x. Sin and cos are necessary to flip the sign of -x back to x.
I completely understand your reasoning. But unfortunately a √2 appears in the wrong place, right? Anyway, I didn't like the solution presented in the video at all. I mean, given an equation E, if you change it for an equation E* such that any solution of E is also a solution of E*, it doesn't mean any solution of E* is a solution of E. Am I crazy or indeed he wasn't careful about that?
I noticed that f(x)=1/x (with some constant coefficient) is a solution, how does this relate to the general form (product of square root and log) ?
That's not a solution.
If f(x) = 1/x, then f'(x) = -1/x², and f(1/x) = 1/(1/x) = x. But it is not the case that f'(x) = -1/x² = x = f(1/x) for all x.
but the solution of differential equation should not contain any constant!!
your ans contains constant B
you could get rid of b by trying values in the original equation
for example f'(2)=f(1/2) and you go from there
Not so. If there are not enough initial or other conditions, there may be constants. For example, if f'(x)=f(x) with no initial conditions, the solution is f(x)=ce^x.
You are wrong.
No? If the equation was just f'(x) = f(x), this has any solution of the form f(x) = A e^x, where A is a constant.
Your confusion probably stems from the fact that f'(1)=f(1) isn't actually a numerical boundary condition. It isn't a "starting value" like f(1)=1 would be. It is only a constraint for f in relation to f', evaluated at x=1, but it's not a fixed value. Hope that makes sense to you?