Laplace Mean Value Formula
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- Опубликовано: 25 авг 2024
- Suppose that a function u equals to its average value on every ball and every sphere, what can we say about u? It turns out that u has to solve Laplace’s equation! Conversely, if u solves Laplace’s equation, then u must satisfy the above mean-value property.
In this video, I state and prove the mean-value property for Laplace’s equation (and its converse). It’s a pure delight for people who like multivariable calculus, because we’ll use a change of variables, the divergence theorem, and the polar coordinates formula. Enjoy!
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Couldn’t have found a better prof to explain a proof so complete so clearly
When I feel down and don't want to study math, I go to this guy, he makes everything sound so interesting!
A constant function is equal to its average but I'm guessing this will be nontrivial solutions?
Of course! Look at my video on Laplace’s equation!
I wish I had the opportunity to study under you. Thank you very much.
Hmm remind you Gauss’ Law anyone? This is particularly useful in electromagnetism.
Until now, I had confused about the mean value formula. This is clear after watching your lecture. thank you sir ❤
"if you love multivariable calculus you're in for a treat"...
Me: "RIP"
edit: nvm that was some pretty multivariate calculus :D
Wonderful and very clear presentation. You are an excellent professor.
Thank you!!! 😄
Great proof sir ❤
Wow! Excellent explanation! Thank you so much!
it's not easy Dr Peyam's, a lot of questions. Thanks !
Great theorem, great proof, great person in the video, as always. I understood almost everything - I should really study multivariable calculus if I want to watch videos about it.
I have two questions which don’t directly concern this video but are quite important for the understanding of videos before so I would want them answered:
1st: The proof of the Banach Fixed Point Theorem was wonderful, but there is something I obviously didn‘t quite get:
Consider the function f(x) = 1 - x ^ 2 like in the thumbnail. For the metric space we want (J, |•|) for some interval J and the normal real metric d(a, b)=| a - b |. Now, one can derive* that, for f to be a contraction, it must hold that for all a, b from J: 1 > |a+b|, which would imply that J = (-0.5, -0.5), otherwise there could be two numbers from J whose sum is greater than or equal to 1. That contradicts the theorem because there is no fixed point there, it’s at 1/sqrt(2)!
*Here’s my derivation:
d( f(a), f(b) ) = | (1 - a^2) - (1 - b^2) | = | b^2 - a^2 |
Wlog, let b = a + h => b^2 = a^2 + 2ah + h^2
q | b - a | = q | h | >= | 2ah + h^2 | = | h | | 2a + h|
q >= 2a + h = a + b
Can you spot the mistake? I am pretty confused!
2nd: Probably this question is either pretty dump or the solution super obvious, but can someone please tell me, how I would evaluate the double integral over a complex region D (it could also be in R^2) of the function |z - ζ| dζ? What I mean by that is, I want to integrate up the distances between a given point z and all points ζ in D. I only found examples for double integrals over a region of a function in terms of x and y dA.
Or, to formulate it differently, a region is given in parametric form. To evaluate a integral over that region with respect to the elements of the region, I take what I know: When you want to evaluate the winding number, which is the contour integral over the path of dζ/(ζ-z), you can say ζ = φ(t) (the parametric form of the path) => dζ = φ’(t)dt, so your integrand is φ’(t)dt/(φ(t)-z). Can one do a similar substitution with a parametric region?
Do I need to use a similar technique with the Jacobian as Peyam needed to evaluate the integral? Or even easier: What principle do I need to study to answer my question?
Thank you for sharing this amazing proof.
I have never understood it when I read Evans.
Who can understand it just by reading it???
Can you please explain that jacobian part in some lecture video....
Your ways of expression is really admirable.
There’s a video on the Jacobian
A function that satisfies Laplace's equation is called a harmonic function.
And yes, any harmonic function, averaged over a sphere, equals its value at the center of that sphere.
And by integrating spherical shells from radius 0 to r, you can show that it's also equal to the average over the ball.
Fred
Pure gold
And also thanks so much for making these awesome videos. Much appreciated.
Hello Dr.you are amaizing teacher I am interested to see you .I need to get boundary integral solution for laplace equation
Did this in todays PDE lecture. :D
Whoa, what are the odds? 😄
12:58 "Chen Lu" xDDDDD im dead
Just did that at grad class for Partial D.E.!
Cool!!! What are the chances? 😄😄😄
@@drpeyam Thats a difficult calculation Dr. Peyam. I'll be back to you when i finish some Stohastic Processes grad class 😄
Thanku very much sir👍👍
This is awesome, i can use this to check if a laplace pde solution over some boundary conditions is correct right? cause im sure solving laplaces pde's is one of dante's hell circles
I think so!
It’s a nice PDE, you should check my videos on Laplace’s and Poisson’s equations in case you haven’t!
Hlo sir, It is very helpful for me. I want to understand how z=y-x/r represent a point on a ball with center 0 and radius 1? Also, how you differentiate it?
x shifts the center to 0 and /r makes the radius 1. And this video shows how to differentiate IG
@@drpeyam sir one more question, why 'u' is independent of r?
Very good video. But I got lost at the very end. How can you assume Laplacian(U) > 0 at x = 0, without loss of generality? And why does the region of integration change from a hypersphere to a hyperball?
If the laplacian is nonzero at a point, it is either positive or negative at that point, so assume it’s positive. And for your second question, it follows from the polar coordinates formula in the video
I just watched the video again. Would you mind pointing me to which time in the video that you had phi'(r) = 1/(N*alpha(N)r^(N-1))*integral of Laplacian(U(y))dy over B(x,r)? Is there an obvious reason for that to be true that I'm missing? Thanks.
Why the jacobian is (n-1 x n-1) matrix size and not n? isnt y in R^n?
But it’s the surface measure on the sphere and the sphere is n-1 dimensional
@@drpeyam ok !! Thanks. I was thinking of it as a change of variables of a linear map( a translation to the origin) shouldnt be the same? y is in R^n but i need n-1 variables to describe the sphere so is as it lies hidden a parametrization of the sphere ? I mean whats the jacobian of T(z)=y-x/r :R^n-->R^n
@@drpeyam also does it happen you have any videos on maximum principles? Which is exactly next thing on Evans about harmonic functions? These are the only videos that make pdes understantable
I have some videos on the maximum principle
Properties of harmonic function pe videos bana do
?
Sir make videos on properties of harmonic function
Strong maximum principle ,regularity property
This is definitely intentse
If a function non constant then f (x3) is between local minimum and maximum. Let here is only one max and min. Let f(x1) = min, f (x2) = max and x3> x2> x1> 0. But f(x1) is between local minimum and maximum. We got a contradiction. So f(x)=const. That's all
There are other solutions, that’s the point!
Can you give an example? I can't imagine this.
This idea is for several variables, you are thinking in one variable.
@@franciscoabusleme9085: Yes, because in 1 variable, Laplace's equation says merely that the function's 2nd derivative = 0. And that means f is linear.
IOW, a harmonic function of a single variable, is a linear function.
And the average of a linear function at any 2 points, x=a & x=b, or averaged over the interval between them, always equals its value at the midpoint, x=½(a+b).
In 2 or more D, there are lots of more interesting harmonic functions. Ex: f(x,y) = x²-y², a "saddle" function.
Fred
I love you❤
Inhuman happiness
voice is so annoying but good content tough