PDE 101: Separation of Variables! ...or how I learned to stop worrying and solve Laplace's equation

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  • Опубліковано 20 січ 2025

КОМЕНТАРІ • 76

  • @Chandra-hw9hv
    @Chandra-hw9hv Рік тому +7

    I couldn't understand the whole Fourier thing for the longest time in my college, huge thanks to you for this gem of a video!

  • @YassFuentes
    @YassFuentes 2 роки тому +27

    Masterclass. How much I do love your videos. They are so much enjoyable! Thank you for your effort and time ❤️

  • @sodaangel7375
    @sodaangel7375 Рік тому +2

    An amazing video coming to save me two days before my final. This lecture is Superb!!

  • @xue7444
    @xue7444 2 роки тому +4

    Best title, my PDE class starts in 2 weeks, this video is such good timing woo-hoo

  • @mostafaatalla71
    @mostafaatalla71 2 роки тому +3

    You are fantastic, Steve! Thanks so much for all of your videos. I love it!

  • @Tyokok
    @Tyokok Місяць тому +1

    Hey Steve or anyone who can advice, would you please explain at 25:15 why G(y) doesn't have coefficient, such as Cn? (or it doesn't matter cuz eventually F(x)G(y) combines coefficients?) Thank you very much!

  • @Unidentifying
    @Unidentifying 2 роки тому +2

    27:55 doesnt such a function [G(y)] also depend on x in reality ? But then you can't do separation of variables. Not so clear to me how this then can be done, also related to the assumption. Or am i misunderstanding something, is it purely mathematical separation and in the end it works with real phenomena?

    • @Ant0ine1
      @Ant0ine1 2 роки тому +3

      The separation is mathematic. On it's own,when you solve -Gyy/G= Lambda, Lambda could be a function of x and constant in regards to y, but Lambda is also solution of Fxx/F=Lambda, which means that if Lambda is either a function of y or x or both, it cannot be solution, because -Gyy/G=Fxx/F=Lambda is valid for all x and y, so it's constant.
      The assumption at the beginning that u can be separated into F(x)G(y) is a very very strong assumption. There are many cases where it can't work, but also a lot of cases where it does: heat equation, wave equation, Laplace's Equation, Helmholtz' Equation and biharmonic equation to give a few examples.
      It's beautiful to see how elegant Physics are !

    • @Unidentifying
      @Unidentifying 2 роки тому +1

      @@Ant0ine1 wow yea, thank you sir for some elaboration.

  • @teodorbabic8007
    @teodorbabic8007 2 роки тому

    Great video and extremely useful for anyone in younger scientific community! Greetings from Serbia :D

  • @noahr9539
    @noahr9539 Рік тому +1

    This is incredibly smart...
    Well explained 👍

  • @harrydireen1061
    @harrydireen1061 Місяць тому

    Great explanation of a complex subject!

  • @pmcate2
    @pmcate2 2 роки тому +1

    @35:40 isn’t this incorrect? Once we have the infinite sum of the basis functions, these form a general solution to the original differential equation, whether the original function was separable or not.

    • @basics5427
      @basics5427 2 роки тому

      He explained what it really means in what follows. You can often try to write the solution of a PDEs with the technique of separation of variables. Sometimes you can find a general form of a possible solution. But very few times (only for problems in domains with regular shapes and very simple boundary conditions) you can find the solution that satisfy both the equation in the domain and the boundary conditions on the boundary and thus the whole differential problem

  • @fernando28795
    @fernando28795 Рік тому

    Brilliant, mate. Lovely. Preparing for the finals. Regards from Chile

  • @EndG00gle
    @EndG00gle 2 місяці тому +1

    All I want is one video without just skipping all the math.

  • @finshinggun
    @finshinggun 2 роки тому

    @44:22 Can you link the video you mentioned, please?

  • @ShinjiCarlos
    @ShinjiCarlos 2 роки тому

    At about 15:00, I would say that if you take the derivative in relation to, let's say x, to both sides, the derivative of the right hand side should be 0 by definition, just like the partial derivative of a function of y regarding to x should be. In the other hand, the derivative of a function of x regarding to x should be zero only in the case that this function is a constant.

  • @martinsanchez-hw4fi
    @martinsanchez-hw4fi Рік тому +2

    This confuses me. I understand that superposition says that if we have solutions to the same problem, their sum is also a solution (any linear combination), but here we are saying solutions to different problems add up to the solution of another new solution that contains all non zero boundaries. The most direct example I can think of is when two adjacent boundaries do not coincide in their extremes (for example f=0 in u(0,y) and f=c in u(x,0)

  • @sdpenning
    @sdpenning 3 місяці тому

    Very usefull, thank you. I wonder though how it would work if you considered the surface as another boundary. Like a thin square of copper foil where the edges and the top and bottom surfaces are in contact with air.

  • @muthukamalan.m6316
    @muthukamalan.m6316 2 роки тому +1

    thanks for phenomenal works and make public to know more

  • @WesleyDevlin
    @WesleyDevlin 9 місяців тому +12

    UA-cam is pushing the limits on advertisements. It's sad.

  • @lgl_137noname6
    @lgl_137noname6 2 роки тому

    One can only hopre that eventually there will a Brunton book integrating Brunton's view and treatment of PDE's
    Thank You.

  • @LucasVieira-ob6fx
    @LucasVieira-ob6fx Рік тому +3

    I told myself I would only watch Steve Brunton's favorite lectures... Turns out I'm watching all of them

  • @ShinjiCarlos
    @ShinjiCarlos 2 роки тому

    I have not checked, but I think that we could also extract the solutions components at the boundary as done, but alongside the functions sinh. It only remains to check that those functions form a complete set of eigentunctions. But since each of them belong to a unique $\lambda$, they do form a complete set of eigentunctions as well and could properly be used on its own right, in an equivalent Fourier trick.

  • @saadmansakib1549
    @saadmansakib1549 2 роки тому

    Thank you so much for the video, Sir !

  • @gracie1397
    @gracie1397 Місяць тому

    i just passed my electrostatics exam entirely thanks to this video

  • @indrasismitra6484
    @indrasismitra6484 2 роки тому

    This is just for a little information. Recently, while working on a problem involving heat transfer in cylindrical coordinates, I have found (from other published literature) that for some Boundary Value Problems (i.e., steady-state problems) the multiplicative, separation of variables i.e., u=F(x)*G(y) is insufficient. Another alternative that can be used is u = F(x)*G(y) + H(x) + I(y), although both the multiplicative (F(x)*G(y)) and the additive (H(x) + I(y)) parts have to be separately substituted into the original pde to find two different sets of odes consisting of different separation constants.

  • @Gullinnova
    @Gullinnova 2 роки тому

    It just so happens im reading the end of certainty right now and all that you teach is helping to illuminate the pages. I feel like i have a really good handle on his explanations thanks to you. I still dont understand what he means by "deterministic chaos" it just seems like a contradiction on its face.

  • @FelipeCondo
    @FelipeCondo 2 роки тому +2

    thanks for the video, really useful. One question. Why the solution G(y) does not have a coefficient like the F(x) function A_n and B_n?
    Thanks

    • @sebastianrada4107
      @sebastianrada4107 8 місяців тому

      You can have A_n sin + B_n cos
      The coefficient for the cos is zero due to the initial condition that requires f(0)=0
      The coefficient of the sin was not indicated but it isn't important as this coefficient would be later multiplied by A_n (the coefficient of sinh) forming a single coefficient that is found later with Fourier.

  • @nntutorials5817
    @nntutorials5817 2 місяці тому

    This is amazing. Assuming i want to solve this similar equation for two different types of boundary conditions. How do i solve the equation Along the x-axis if the BC is a No-slip BC, while along the y-axis, the BC is a periodic BC?

  • @hamidrezaalavi3036
    @hamidrezaalavi3036 2 роки тому

    at 6:56, boundary values must be consistent at corners, it is possible only if all fn is zero at corners. In this way, we can not solve problems with constant values i.e. fn =Cn.

    • @basics5427
      @basics5427 2 роки тому

      You're right in the world of strong solutions of PDEs. This could not be true in the world of weak solutions of PDEs or integral solutions of PDEs, and the numerical world of (maybe discontinuous) Finite Elements or Finite Volume methods (or Spectral Elements methods, ...)

    • @hamidrezaalavi3036
      @hamidrezaalavi3036 2 роки тому

      @@basics5427 Thank you very much.

  • @lorhancosta622
    @lorhancosta622 2 роки тому

    Honest to goodness this content is guuuhhhd!!!

  • @rogelv8td
    @rogelv8td 2 роки тому

    Thank you, very good lecture.
    Love your videos

  • @lightnlies
    @lightnlies 2 роки тому

    is there a way to check if a PDE can be solved through separation of variables a priori?

  • @stephanoseghale2370
    @stephanoseghale2370 4 місяці тому

    this is art!

  • @evanparshall1323
    @evanparshall1323 2 роки тому

    You say that because the PDE is linear then you can use superposition so that the solution is a linear combination of all the solutions. But how do you know there is not a solution that is not satisfied by the separation of variables assumption? If that is the case then the basis of solutions would be under-defined.

  • @kl-wastikc8909
    @kl-wastikc8909 Рік тому

    Hello Steve. Where do I find a similar board you're using? Or anyone who knows

  • @jamesmosher6912
    @jamesmosher6912 2 роки тому +1

    I would love to get your perspective on “similar” homogenous PDEs on irregular domains; something like a the unit square but exclude the upper right quadrant (an “L” shape domain).
    I love the beauty of the simple domains, circles, squares, rectangles, etc. but I’ve struggled to find or see a method for solving simple PDEs on irregular domains outside of numerical methods.

    • @basics5427
      @basics5427 2 роки тому +2

      Numerical is the answer for domains with general shapes and general boundary conditions.
      Anyway, if you're an analytical-addicted, for Laplace equations you could rely on conformal transformations to transform a domain with "complex" shape (you could do for example with L shape domains) to a simpler one, but it's likely to be a pain in the a*s, as well.

  • @ermi6880
    @ermi6880 Місяць тому

    I dont understand how you chose the boundaries for solving after finding the separate differential equations

  • @ShinjiCarlos
    @ShinjiCarlos 2 роки тому

    I don't actually agree that Fourier analysis is only all about solving PDE's or ODE's. What about image and audio compression, signal processing and stuff?

  • @ugoamaldi8056
    @ugoamaldi8056 2 роки тому

    Great video! Thanks a lot!

  • @lakshya6235
    @lakshya6235 2 роки тому

    I think the use of summation was not fully motivated. So I thought of the following:
    1. the sum of all individual u is also a solution of the laplacian as it is a Linear operator, so that makes it a valid thing to do.
    2. If we don't use summation, then f(y) can only be of form f(y) = sin(ky), this comes from the fact that An should be a constant. Using summation of individual solution of Laplacian allows us to be able to satisfy much richer class of boundary conditions f(y), and in this case any function that doesn't have cosine terms in its Fourier series.
    I learnt Fourier series and Fourier Transform using your videos only :)

  • @ShinjiCarlos
    @ShinjiCarlos 2 роки тому

    I have a really important question: do you write things the other way around as a right hand guy or are you writing lefthandedly and using frame inversion??? Please!!! My wife and I are struggling about that!

  • @tanveer6514
    @tanveer6514 2 роки тому

    0:10 Time travellers....goosebumps...

  • @curtpiazza1688
    @curtpiazza1688 9 місяців тому

    Great lecture! 😂...I'm a little lost.....but I'll try to stay with it! 😮

  • @maudentable
    @maudentable 6 місяців тому

    Awesome!

  • @Cyborg17
    @Cyborg17 2 роки тому

    Congrats !!!

  • @thomasjefferson6225
    @thomasjefferson6225 Рік тому

    Im wondering why theres no strum louville theory here. That really compliments the seperation of variables in my lay opinon. It makes this a much easier problem to solve. I know this is sin instantly buy looking at the boundry condtions. If you had von neuman BC then it would be Cos, and mixed g(0) = g'(L) = 0 is sine! So idk i think knowing that theory makes this part much more intutative.

  • @xichenjiang7799
    @xichenjiang7799 2 роки тому

    We made an assumption that u can be separated into F(x)G(y), and proceeded to find the solution. How do we know that we found the complete solution for u. That is, how do we know we didn't miss another solution where functions of x and y cannot be separated?

    • @batu9049
      @batu9049 Рік тому

      if you supposed that u =f(x)g(y) but solution maybe cannot be like this. we cant know this u just need to try.almost everytime that you cant find exact solutions for pdes

  • @StaticMusic
    @StaticMusic 2 роки тому

    Hey Steve! This was a fantastic video, thanks!
    Couple of questions: (1) Why does n == m? I didn't follow this part. (2) Couldn't lambda be < 0 (negative) if G(y) = sin(i*sqrt(lambda)*y)? Thanks so much!

    • @basics5427
      @basics5427 2 роки тому +2

      (1) because for every F_n(x) you evaluate G_n(y) s.t. u_n(x,y) = F_n(x)G_n(y) is a solution of the problem (before you prescribe the non homogeneous boundary conditions)
      (2) if you already have some experience, the choice of the sign of lambda come from the boundary conditions: here u(x,y)=0 on the horizontal boundaries of the domain, so you need some functions that is identically null there, and thus the sin(ny/pi) functions (this results as the basis of a Fourier series of the function u(x,y) in y direction).
      If you have little experience, you're right that you need to add the contributions you get with lambda

    • @StaticMusic
      @StaticMusic 2 роки тому

      @@basics5427 awesome thanks so much for taking the time to write that out. Much appreciated

    • @basics5427
      @basics5427 2 роки тому

      @@StaticMusic you're welcome. enjoy it

  • @nouralhuda3530
    @nouralhuda3530 8 місяців тому

    Thank you

  • @arvindp551
    @arvindp551 2 роки тому

    Hardly is there anything in mathematics that's not one of your favourites. Please make a video on Python plotting as well 🥺🥺🥺

  • @ahmed-abdalhaleem
    @ahmed-abdalhaleem Рік тому

    U(x,y)=x^2+y^2
    exp(U)=exp(x^2).exp(y^2)=f(x).g(y)
    Can be separated😊

  •  2 роки тому +1

    Muito bom!

  • @sitrakaforler8696
    @sitrakaforler8696 2 роки тому +1

    I wanna do a PhD in maths and CS !

  • @carultch
    @carultch Рік тому

    Separable DiffEQ's .... or "how I learned to stop worrying, and treat the Leibnitz notation as a fraction."

  • @basics5427
    @basics5427 2 роки тому

    Dr. Strangelove ;)

  • @onenotesm9076
    @onenotesm9076 2 роки тому

    all we need is translate another (all) Langwitch 😀

  • @carlosgalois6268
    @carlosgalois6268 Рік тому

    f4

  • @hakimal-hakim8890
    @hakimal-hakim8890 Рік тому

    و ابتلينا في هذا القرن بخدعة جديدة هي ما أسموه "الربيع العربي" ... الذي ليس عربيا و لا ربيعا....

  • @AdoptedPoo
    @AdoptedPoo Рік тому

    very well done.

  • @kemiakinnola
    @kemiakinnola 2 роки тому

    44:00 lol my university teaches PDEs and Fourier's series together, as a single module (class). That's why I'm here now. Wish me luck🫠