PDE 6 | Transport with decay and nonlinear transport
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- Опубликовано: 20 апр 2011
- An introduction to partial differential equations.
PDE playlist: ruclips.net/user/view_play_list...
Part 6 topics:
-- transport equation with decay, solution using method of characteristics
-- and example of transport with decay (7:06)
-- "non-linear" transport example (9:12)
-- Mathematica movies!
This is awesome, I dont know how this has not gone viral.
Considering that most of the population consider this to be completely arcane knowledge it kind of has gone viral, there are not many 13 year old math videos with this many views.
@@lbgstzockt8493 Exactly! We are the self-selected biased samples...
This videos are pure gold! Crystal clear explanations!!
This is great! It makes me understand at a deeper level so that i can feel confident in my solutions. I will recommend you to my classmates ;)
Thank you. Your explanation is so clear!
Great video! I am a Chinese university student and now i feel hard studying pde. Sincerely, the textbook we use can not make things easy to understand, thought it is correct and full of strict statement. And my teacher get things done quickly(lol). And thank your clear explanation and vivid picture. Those let me feel better and be confident to get further. excuse for my poor English.thanks
feeling exactly the same as you! 666
Hi! Your ideos are amazing. they present an intuitive picture of the concepts which I'd earlier failed to get even after taking a course in PDEs. You present idea in a simple and clear way. It will be nice if you could do videos on non-lineae equations like Burger's equation discussing shock and rarefaction. Thanks!
Sorry ... and I've just flicked through another post that mentioned separation of variables which explains it. Thanks again.
And just echoing everyone else, your videos are awesome.
I applied this solution technique to an engineering problem governed by that PDE. worked out great.
Great explanations sir..Your videos made me understand PDE 🔥❤️
@commutant Very clear video, i really enjoyed watching it and curious to see examples. For the Burger's Eqn u_t + 1/2(u^2)_x =0 I was expecting you to talk about "shocks" who appears after a finite time, but I guess you didn't have time!
Please continue ! Lot of things to do: Systems of hyperbolic eqns, Navier stokes, KdV , Schrodinger's ...
@pdcsv the only function which is equal (up to some constant factor) to its own derivative is the exponential function e^kt
d(e^kt)/dt = k*e^kt
I appreciate if you could help understand why the shape of the wave changes with time in the non-linear case if du/dt=0?
Hi, thanks for the reply. The original question was how did du/dt = -au integrate / arrive to
u(x(t), t) = Ke^-at (understanding this to be a function of t along the characteristic line)?
Naiively I thought it would have been u = -aut + u0. There's another answer posted along the lines that the derivative of the function e^kt is basically itself (k.e^kt), but I can't see how this relates to du/dt = -au ...
Great video ! Could you please recommend a book to practice ? Perhaps one that has a solution manual ?
+Ali Hariri best of luck
^LOOOL
how would you solve the example in 7:30 without the initial values ie. just u_t+3u_x=-u?
I guess matters during transportation have energy loss,does that also fit in this decay equation?
will u explain where the u(x,t) = f(x0) exp(-at) along the characteristic curve come from?
@pdcsv Since u and t are separable, we divide by u and multiply by dt. This gives us du/u = -a*dt. We integrate both sides: Ln |u| = -a*t + constant. Since e is the base of Ln we can say: u = e^(-a*t+constant) = e^( -a*t) * e^(constant) = K * e^(-a*t) where K = e^(constant). Does this help?
can you please tell how you computed u = e^(-a*t+constant) = e^( -a*t) * e^(constant) = K * e^(-a*t)?
Great video, though I must comment that the last equation is not the nonlinear transport equation, nor is it even nonlinear. The generalized nonlinear transport equation is du/dt+c(u) du/dx=0. Here c(u) is the wave speed, which is allowed to change with the value of u.
Thanks! Will do
shouldn't the curve of e^t be lower convex?
you rock!
Silly me! Thanks for the wonderful videos!
Trying to learn all of this before one week my final starts...
Ur good so good
when we established that (1,c) dot (Ux,Ut) is the transport equation we disregarded that fact that (1,c) is not a unit vector because the expression was equal to zero, now it doesn't, it equals -Au so how is this still correct?
He explained that on his later videos.
first of all, I'd like to thank u about these wonderful videos.
and I'd like to ask u if u please could help me to solve:
u_x(x(t),y(t),t) + u_y(x(t),y(t),t) + (f(x(t),y(t),t)+g(t))*u=0
using the method of characteristics
I've got:
u(x(t),y(t),t)=u_0 * exp[-(int (f(x(t),y(t),t)+g(t)))]
could u please confirm if its right or not??
many thanks and best regards
!!
,
please ignore my comment i just saw the answer !