What guarantees that ODEs don't blow up in finite time? Picard-Lindelöf just guarantees a unique solution, right? Does it guarantee, that the solution exists for all times? I thought of Picard-Lindelöf because you mentioned Lipschitz at the end... Now I am wondering, if there is an analog to Lipschitz continuity for weak or distributional derivatives. Your videos always get me excited and thinking about more stuff, after they are over :)
@VeryEvilPettingZoo I think that when you create the weak formulation, the resultant inner product form must be coercive and the forcing function must be weakly Lipschitz. But it has been 30 years since grad school.
Having the singularity is cool, it makes the solution an analytic continuation. Imagine that a car is one unit of space in front of you and starts driving directly away from you at a speed proportional to the square of the distance between you. After only 2 units of time the car will have circumnavigated the universe and will be arriving one unit behind you, coming up with the same speed that it left with.
What I like about this video is: It puts the question upside down. I.e.: You try a solution, and then find out what conditions it works under. Normally You represent the conditions first, and then derive the calculation. Normally in practical work it is the other way round. Now -1/x-c = 1/c-x Depending on what it is: It could be interpreted as a fuelload that limits the speed of a craft. As the fuel is used the speed increases. Just to take an example off the top of my head.
Well, you divided by y in order to solve the diff eq. But in order to do that, you must assume that y=/= 0. That is why the solution is lost. This will also happen with the separation of variables
@@dackid2831, I understand it, I did not mean that I am in 2 grade and don’t know that we can’t divide by 0. I meant: “why didn’t you consider 2 cases: 1) y=0, so we can’t divide 2) y is not 0, so we can divide”
So I actually don't know if you will get a solution apart from the constant function y=0. It may be possible you do get something, but then you'd be dealing with a non-linear differential equation which may get messy. It might be worth seeing if Laplace transforms can help you get a solution with y=0
it has weird implication on the maclaurin series of 1/(1-x). i mean, if you square the series, does it equal to the derivative of the series? or this is the reason we have this lipschitz definition of continuity.
Place the variables on one side of "=" to eulerize it with ideation, and then change "y" to "g", and, if it doesn't assume presence and completeness, then it's fairly real bad with distinction.
It is essentially a stronger form of uniform continuity. For a uniformly continuous function, delta can be chosen arbitrarily smaller than epsilon, whereas Lipschitz continuity requires that such a delta is no more than a certain factor L smaller than epsilon. For differentiable functions, Lipschitz continuity just means the derivative is bounded. See the wikipedia page en.wikipedia.org/wiki/Lipschitz_continuity.
Dr. Peyam, are you still a professor at UCLA? Would love to see a video about the research you are working on and collabing with. Also, could I just use the Euler-Lagrange equations for B.C.?
Although the bahaviour looks bizarre - it has to to with the non-compact nature of the real line. If you compactify the real line by adjoining a point at infinity (for example by stereographically projecting a circle to the real line) the equation becomes very "tame". The solution with initial condition y(0) = 1 reaches the point at infinity for finite x* - but y=infinity becomes just a regular point for the d.e. For x > x* the solution projects on the negative real line and approaches y=0 from the left.
But y^2 IS Lipschitz continuous! This does NOT guarantee a Solution on all of R, but only on a small enough interval. But this small intervall will NEVER contain the blow-up!
@@drpeyam You are right, I have to specify the interval. But y^2 is Lipschitz continuous on ANY compact interval, so I could take y in [-1000, 1000] with a Lipschitz constant of e.g. 3000. So does not in any affect my argument.
cool Eq in a square: BTW can u solve (*) y' =-ay⋅(1-y)^2, i.e., WTF y(x) s.t solves (*) for a>0? I Think LambertW function is involved. maybe W(0,x) or W(-1,x).
C=infinity in the limit; what i love about math is that sometines you can just force infinity into a solution and it sometimes makes sense. If y=1/infinity that's (in the limit) zero, and zero happens to be valid
Actually I'd prefer another version of explanation of Lipschitz continuity: Imagine a double-sided 2D cone sliding on the graph of the function with a constant slope. We say that a function is Lipschitz if all points of the graph lie inside the cone *locally*.
If I remember well an extremely useful condition which guarantees that the maximal solution of an initial value problem y'(t) = f(t, y(t)) is indeed global is to require f to have a linear growth in y uniformly with respect to t on the whole interval of t. In fact f being locally lipschitz in y uniformly with respect to t is a special case of this.
That's really annoying :( it seems like every power accept 1 and 0 wouldn't be Lipschitz continuous. Does it mean we would have problems with, for example, using Tylor series to express the function on the right side, every partial sum of which isn't Lipschitz continuous even if the actual function is?
Yes, and apparently this is the solution that works with the y(0) = 0 initial condition. Would it be legal to say the y = 0 solution fits in with the general solution y = -1/(x+c) if we let c = ∞?
@@desmondhutchinson6095 Thinking about it, that's probably correct. The graph of y = -1/(x+c) always has two asymptotes, one being y = 0 and the other x = -c. As c gets bigger, the x = -c asymptote approaches -∞, while the graph itself moves closer to the y = 0 asymptote.
when pi m was doing it at the first place y’/y^2 = 1 , this instantly split the case into y=0 and y is not 0 and he only did the case which y is not 0. So your ans is correct and is come from the other case.
It is not the non Lipschitzness that makes the solution blow up. One can easily alter this to make it Lipschitz. Just say that y'=f(y) where f(y)=y^2 on [-2,2] and f(y)=4 on (-infinity,-2] union [2,infinity). With the initial condition y(0)=1, we have the same solution to the differential equation on [-2,2]. But note that f is clearly Lipschitz as it is continuously differentialble on [-2,2] and constant outside of [-2,2]. So it is not true that if f(y) is globally Lipschitz that y does not blow up in finite time. One have a problem with my above function f, but bump functions can be used to make my counter example C^infinity. So it is not even true that f(y) globally Lipschitz and C^infinity implies that y does not blow up in finite time.
Lipschitz continuous: th-cam.com/video/iXy2_9V4pZU/w-d-xo.html
f'(x) = f(f(x)): th-cam.com/video/cXpFDlIIczg/w-d-xo.html
Equation with a twist: th-cam.com/video/Gtb9kx5WQXA/w-d-xo.html
ODE existence uniqueness theorem: th-cam.com/video/CeB2qyUEdaQ/w-d-xo.html
Love to see a more formal explanation of Lipschitz continuity!
Lipschitz continuous: th-cam.com/video/iXy2_9V4pZU/w-d-xo.html
Dr Peyam Thanks!
3:20 I see my best friend
Btw, I gave the same exact problem on my calc 2 exam lol
Your students must have loved you 😂
What guarantees that ODEs don't blow up in finite time? Picard-Lindelöf just guarantees a unique solution, right? Does it guarantee, that the solution exists for all times? I thought of Picard-Lindelöf because you mentioned Lipschitz at the end...
Now I am wondering, if there is an analog to Lipschitz continuity for weak or distributional derivatives. Your videos always get me excited and thinking about more stuff, after they are over :)
@VeryEvilPettingZoo I think that when you create the weak formulation, the resultant inner product form must be coercive and the forcing function must be weakly Lipschitz. But it has been 30 years since grad school.
Btw, on what interval is x^2 not lipschitz continuous? Or we don’t specific an interval for it.
The interval is important, it’s not Lipschitz on R or any unbounded interval
@@drpeyam what is the particular theorem from differential equations? is there a version that depends instead on uniform continuity?
No need to use differentials, just use the reverse chain rule (or u-sub if that's what you wanna call it).
y= -1/(x+c)
Having the singularity is cool, it makes the solution an analytic continuation.
Imagine that a car is one unit of space in front of you and starts driving directly away from you at a speed proportional to the square of the distance between you. After only 2 units of time the car will have circumnavigated the universe and will be arriving one unit behind you, coming up with the same speed that it left with.
"But not on this channel" hahaha
Love that line
What I like about this video is: It puts the question upside down. I.e.:
You try a solution, and then find out what conditions it works under.
Normally You represent the conditions first, and then derive the calculation. Normally in practical work it is the other way round.
Now -1/x-c = 1/c-x
Depending on what it is: It could be interpreted as a fuelload that limits the speed of a craft. As the fuel is used the speed increases. Just to take an example off the top of my head.
Uncut version is always more fun!
Why have we lost the case when y=0? It is solution as well. Or is it just because we can’t exponentiate 0?
Well, you divided by y in order to solve the diff eq. But in order to do that, you must assume that y=/= 0. That is why the solution is lost.
This will also happen with the separation of variables
@@dackid2831, I understand it, I did not mean that I am in 2 grade and don’t know that we can’t divide by 0. I meant: “why didn’t you consider 2 cases: 1) y=0, so we can’t divide
2) y is not 0, so we can divide”
So I actually don't know if you will get a solution apart from the constant function y=0.
It may be possible you do get something, but then you'd be dealing with a non-linear differential equation which may get messy.
It might be worth seeing if Laplace transforms can help you get a solution with y=0
it has weird implication on the maclaurin series of 1/(1-x).
i mean, if you square the series, does it equal to the derivative of the series?
or this is the reason we have this lipschitz definition of continuity.
Place the variables on one side of "=" to eulerize it with ideation, and then change "y" to "g", and, if it doesn't assume presence and completeness, then it's fairly real bad with distinction.
Thanks Dr. Peyam.
Is lipschitz continuity similar to uniform continuity?
It is essentially a stronger form of uniform continuity. For a uniformly continuous function, delta can be chosen arbitrarily smaller than epsilon, whereas Lipschitz continuity requires that such a delta is no more than a certain factor L smaller than epsilon. For differentiable functions, Lipschitz continuity just means the derivative is bounded. See the wikipedia page en.wikipedia.org/wiki/Lipschitz_continuity.
@@FromTheMountain oh wow, that's really interesting! Thanks!
Dr. Peyam, are you still a professor at UCLA? Would love to see a video about the research you are working on and collabing with.
Also, could I just use the Euler-Lagrange equations for B.C.?
No, I moved from UCI to ASU! I’m not really collaborating with anyone right now, and I’ll talk about my research once I published my paper
Although the bahaviour looks bizarre - it has to to with the non-compact nature of the real line. If you compactify the real line by adjoining a point at infinity (for example by stereographically projecting a circle to the real line) the equation becomes very "tame". The solution with initial condition y(0) = 1 reaches the point at infinity for finite x* - but y=infinity becomes just a regular point for the d.e. For x > x* the solution projects on the negative real line and approaches y=0 from the left.
And then C Señor...hahaha.
Si señor Peyam, bien hecho (Yes Mr. Peyam, well done)
But y^2 IS Lipschitz continuous! This does NOT guarantee a Solution on all of R, but only on a small enough interval. But this small intervall will NEVER contain the blow-up!
You can’t say y^2 is lipschitz continuous, you have to specify the interval. As you said, that is the main problem here
@@drpeyam You are right, I have to specify the interval. But y^2 is Lipschitz continuous on ANY compact interval, so I could take y in [-1000, 1000] with a Lipschitz constant of e.g. 3000. So does not in any affect my argument.
Is Lipschitz continuity related in any way to system stability?
cool Eq in a square:
BTW can u solve (*) y' =-ay⋅(1-y)^2, i.e., WTF y(x) s.t solves (*) for a>0?
I Think LambertW function is involved. maybe W(0,x) or W(-1,x).
C=infinity in the limit; what i love about math is that sometines you can just force infinity into a solution and it sometimes makes sense. If y=1/infinity that's (in the limit) zero, and zero happens to be valid
It’s an autonomous DE right?
Actually I'd prefer another version of explanation of Lipschitz continuity: Imagine a double-sided 2D cone sliding on the graph of the function with a constant slope. We say that a function is Lipschitz if all points of the graph lie inside the cone *locally*.
Doc, I wanna know more on Lipschitz continuity
See description
If I remember well an extremely useful condition which guarantees that the maximal solution of an initial value problem y'(t) = f(t, y(t)) is indeed global is to require f to have a linear growth in y uniformly with respect to t on the whole interval of t.
In fact f being locally lipschitz in y uniformly with respect to t is a special case of this.
Wouldn't have guessed that's the reason. Interesting
Are you iranian peyam?
Bale
@@drpeyam oh my gosh!
واقعا باعث افتخاره👌🌹
@@drpeyam you make very interesting and useful videos 👍
I like them
Keep it up...
رائع
Next do f'(x)=f(f(x))
Already done, see description
@@drpeyam omg I forgot about that video
Watching this instead of doing my calculus homework!!
That's really annoying :(
it seems like every power accept 1 and 0 wouldn't be Lipschitz continuous.
Does it mean we would have problems with, for example, using Tylor series to express the function on the right side, every partial sum of which isn't Lipschitz continuous even if the actual function is?
y=0 is a solution
Yes, and apparently this is the solution that works with the y(0) = 0 initial condition. Would it be legal to say the y = 0 solution fits in with the general solution y = -1/(x+c) if we let c = ∞?
@@zanti4132 Good question, I’m not sure
@@desmondhutchinson6095 Thinking about it, that's probably correct. The graph of y = -1/(x+c) always has two asymptotes, one being y = 0 and the other x = -c. As c gets bigger, the x = -c asymptote approaches -∞, while the graph itself moves closer to the y = 0 asymptote.
when pi m was doing it at the first place y’/y^2 = 1 , this instantly split the case into y=0 and y is not 0 and he only did the case which y is not 0. So your ans is correct and is come from the other case.
It is not the non Lipschitzness that makes the solution blow up. One can easily alter this to make it Lipschitz. Just say that y'=f(y) where f(y)=y^2 on [-2,2] and f(y)=4 on (-infinity,-2] union [2,infinity). With the initial condition y(0)=1, we have the same solution to the differential equation on [-2,2]. But note that f is clearly Lipschitz as it is continuously differentialble on [-2,2] and constant outside of [-2,2]. So it is not true that if f(y) is globally Lipschitz that y does not blow up in finite time.
One have a problem with my above function f, but bump functions can be used to make my counter example C^infinity. So it is not even true that f(y) globally Lipschitz and C^infinity implies that y does not blow up in finite time.
2:15 "si señor!!"....... A moment later..... wait a minute xD
f(x)=-1/(x+c)
Let f prime for respect equal f for respect squared, then the solution will be presented by Dr P.
Very interesting
y'(x) = y(x^2) next? (:
Then you'd have the whole trifecta:
y'(x) = (y(x))^2
y'(x) = y^2(x)
y'(x) = y(x^2)
I like that! If by y^2 you mean y(y(x)) then I’ve done that already!
@@drpeyam yes, I just wrote it that way, so the three different equations look more pleasing/uniform
Lipshitz... nearly spat out my cereal. lol sorry.
I literally did this two days ago
Well... yes but actually no.
Hello!
Man, loved your explanation, but the "Dangerous" is exaggerated