The Most Important Video in the Series: The Linear Property of the Fundamental PDEs

แชร์
ฝัง
  • เผยแพร่เมื่อ 15 ก.ย. 2024
  • bit.ly/PavelPa...
    lem.ma/LA - Linear Algebra on Lemma
    bit.ly/ITCYTNew - Dr. Grinfeld's Tensor Calculus textbook
    lem.ma/prep - Complete SAT Math Prep

ความคิดเห็น • 6

  • @MathTheBeautiful
    @MathTheBeautiful  5 ปีที่แล้ว +3

    Hi, Pavel here. Someone was very kind to provide closed captions for this and other videos. However, I can't seem to figure out how to determine who made the contribution. Can someone let me know who to determine the author so I can thank them properly?
    Thank you in advance!
    Pavel

  • @georgesadler7830
    @georgesadler7830 ปีที่แล้ว

    Professor MathTheBeautiful, thank you for a solid explanation of The Linear Property of Fundamental PDEs. This is an error free video/lecture on TH-cam TV.

  • @scitwi9164
    @scitwi9164 7 ปีที่แล้ว +7

    06:30 Alan Turing once wrote on a postcard to his friend Robin Gandy:
    "Science is a differential equation. Religion is a boundary condition." ;)

  • @scitwi9164
    @scitwi9164 7 ปีที่แล้ว +1

    02:35 So what is the linear algebra analog for the other two (time-dependent) types of equations? :q
    My guess is that they're actually ALL of the type `A·x=b` (where `b` might be `0`), but their operators now take time into consideration as well. For example, the wave equation can also be written as:
    □² u = 0
    where □² is the d'Alembert operator:
    □² = (1/c²)·∂²/dt² - ∇²

  • @scitwi9164
    @scitwi9164 7 ปีที่แล้ว +2

    03:39 `x³ - 3·x²·y` is not harmonic:
    (∂/∂x)(x³ - 3·x²·y) = 3·x² - 6·x·y, (∂/∂x)(3·x² - 6·x·y) = 6·x - 6·y
    (∂/∂y)(x³ - 3·x²·y) = -3·x², (∂/∂y)(-3·x²) = 0
    So when added together they don't produce 0 :q
    Perhaps you had in mind this one?: x³ - 3·x·y²
    This one is harmonic:
    (∂/∂x)(x³ - 3·x·y²) = 3·x² - 3·y², (∂/∂x)(3·x² - 3·y²) = 6·x
    (∂/∂y)(x³ - 3·x·y²) = -6·x·y, (∂/∂y)(-6·x·y) = -6·x
    which when added together produce 0.
    Fun fact:
    (x+y·i)² = x² + 2·x·y·i + (y·i)² = x² + 2·x·y·i - y² = (x²-y²) + i·(2·x·y)
    The real and imaginary part are the two harmonic functions you mentioned ;> `x²-y²` and `2·x·y`.
    (x+y·i)³ = x³ + 3·x²·(y·i) + 3·x·(y·i)² + (y·i)³ =
    = x³ + 3·x²·y·i - 3·x·y² - y³·i =
    = (x³ - 3·x·y²) + i·(3·x²·y - y³)
    This gives another two harmonic functions: `x³ - 3·x·y²` and `3·x²·y - y³.
    And you can do the same with higher powers of a complex number `x + y·i` to obtain more of harmonic functions :)
    How cool is that? ;>

  • @sdsa007
    @sdsa007 ปีที่แล้ว

    I'm not use to his notation.... its different from 3blue1brown