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.
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
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 ...
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.
@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?
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!
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.
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?
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
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 ;)
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
Hey.
This is exceptionally clear.
Thank you!
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 ...
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.
@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)?
@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
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!
Great video ! Could you please recommend a book to practice ? Perhaps one that has a solution manual ?
+Ali Hariri best of luck
^LOOOL
Great explanations sir..Your videos made me understand PDE 🔥❤️
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?
I guess matters during transportation have energy loss,does that also fit in this decay equation?
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.
Thank you. Your explanation is so clear!
will u explain where the u(x,t) = f(x0) exp(-at) along the characteristic curve come from?
how would you solve the example in 7:30 without the initial values ie. just u_t+3u_x=-u?
I applied this solution technique to an engineering problem governed by that PDE. worked out great.
shouldn't the curve of e^t be lower convex?
Thanks! Will do
Trying to learn all of this before one week my final starts...
Silly me! Thanks for the wonderful videos!
Ur good so good
you rock!
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 !
!!
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