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- เผยแพร่เมื่อ 17 เม.ย. 2011
- An introduction to partial differential equations.
PDE playlist: www.youtube.com...
Part 5 topics:
-- the method of characteristics
-- applying the method to the transport equation (6:40)
-- non-homogeneous transport (12:04)
I hate the fact that understanding it in my second language is lot easier than in my native one. Believe me or not, math in English seems much more acquirable. At least I don't fall asleep after 30 seconds. Cheers from Poland. You've just gone international ;)
hello from hk, china o/
Usually I don't like these types of videos since I prefer reading from a textbook, but this video was super clear. Every source I tried to read online for characteristic curves was just needlessly complicated and assumed too much prior knowledge. Well done!
Absolutely loved this. The exercise at the end was so helpful for using the method of characteristics. No-one learns just by watching.
just discovered this and i did not expected this video to be already 9 years old, beautiful graphics and way of presentation
Just want to say, you are awesome. This is the best series on PDEs I've seen on YT. I love the way you use examples, like with a moving cart, to visualize what's behind the equations.
Your videos are awesome, I am graduate student and I use your videos for my PDE cours and I learn each and every of your videos, Thank you so much
Great videos; well explained, easy to understand and good progress pace. Thank you! :-)
Thank you for these videos on method of characteristics. Yours are the only ones I've found that explain this topic clearly and concisely.
This is magic - thank-you for making it so much easier to understand than textbooks make it
I love when someone explain something in math easy, like in this video. Thanks!
this is the only explanation which solved all my doubts. Finally understood the meaning of the phrase "solution propagates along characteristics"
I wasn't talking about burgers equation. What I was trying to say was with
du/dt + c(t)du/dx=0 the constant is changing with time meaning equation 2 in the solution:
du/dt=0 (1)
dx/dt=c(t) (2)
has to be solved numerically using eulers method or another technique (Runge Kutta etc).
Thank you! Your illustrations in the video about observer and the wave are helpful. Before, I was not able to understand what phenomena this PDE was describing
Extremely helpful, commutant! You have the best PDE videos in TH-cam! Thanks!
This was helpful and also wonderful, really really well done. Thanks a million.
Excellent video, the it was smooth as butter to understand
Thanks a lot.
Hi commutant, thanks for a very good video. Doing what you did in this video is not easy, so nice going. I'm trying to learn my engineering mathematics again, but this time properly, so videos like yours are very helpful. What happens in the original PDE when the wave propagation speed, c, is not a constant? If c depends only on time, t, then I think it's not so bad, but what about when c depends on x. Do you have a video for that?
helped me a lot, thanks. It is much clear than any text book explanation.
du/dt + u du/dt=0 is the so-called burgers equation and it's well studied in every pde-textbook. For example Lawrence C. Evans - Partial Differential Equations
Thanks a lot for the video. I was trying to understand this method for a long time and your explanation solved all of my doubts. I would be so thankful if you could make a video about Duhamel's Theorem. Thank you! ;)
Wow, what a great explanation. Somehow our professor made it impossible to understand
You are a very good teacher/explainer/illustrator. Are you a mathematician or a physicist by formal training? Do you have a doctoral degree and are you a tenured faculty member at a college or university in the US? Thank you very much for posting these useful videos. More power to you.
I think the initial condition should be written as u(x,0)=sin(x_0), no? You are using it at 14:31 as sin(x_0) - for me, x_0 describes a fixed x at t=0 whereas u(x,0)=sin(x) doesn't fix x.
I currently think of it as u(x(t),0)=sin(x(t)) and so sin(x_0)=sin(x(0)), this also makes clear that the very last step, where we put x-ct into sin is derived form the fact that dx/dt=c. (Anyway, I like to be pedantic so I think I'll need to think some more about it, when is it actually x and when x(t))
This is a very enlightening method! Thanks for posting this question!
Nice lecture. BTW, which software did you use to write on the blackbaord?
Better expllained than in a Master Degree lecture at ETH Zürich which I took
it seems this is true if we have u(x(t)) but here u(x,t) has an explicit dependence on t, so it's not necessarily u along a path. we could have u(x,t) = 2x(t)+t for example
du/dt = 1 along characteristics lines but in boundary condition u(x,0) = sin(x) --> du/dt = 0 how ???
Ah yes, I appreciate the exercise at the end. I'm studying PDEs as much as I can before I take the class, and would love to be ahead. Thorough lesson. easiest subscription.
Big deal! Stop desperately trying to fill your low self esteem by announcing such worthless details.
hey commutant, which book did you use to study PDEs??
Peter O'Neil, Beginning PDE: An Introdution to therory, 3rd Ed.
Isn't 4:22 to be a total derivative on the rhs and partials on the lhs.
Could someone clear a confusion up for me: Whats the difference between du/dt and u_t ? The way i interpret them is like du/dt = lim_{h->0} [u(x(t+h), t+h) - u(x(t), t)]/h, and u_t = lim_{h->0} [u(x(t), t+h) - u(x(t), t)]/h, But thats just my best guess, i dont know. Can someone explain to me if its not that and what it actually is?
Suppose the initial condition is u(x,0)=U_0 then what would be u(x,t)?
This was excellent, thankyou very much.
I am confused by the initial condition, it says that when t=o, u=sin(x) NOT sin(x0). Could someone explain this because they are different initial value.
shorthand for saying u(x,0) = sin(x)
x0=const.=x-ct and this is the solution of dx/dt=c .So as you see, sin(x0)=sin(x-ct). For every value of t, the x value is a dependent value so as x-ct=x0=const. So ,if t=0, x=x0 .
Love your explanation!
good video. your curve red line describing x(t) does not match dx/dt =c, which is a more particular case.
but if u(x,t) = t + sin(x-ct), then du/dt = -c.cos(x-ct) + 1, what is different than the initial condition du/dt = 1. Where am I mistaken??
+mateusciola when we take du/dt=1,we take at the point when t=0,and as for general case u(x,t)=t+sin(x-ct)..For more assistance see question ,it is written u(x,0)=sinx
+PIYUSH TAILOR . Sale Sab padh liya tune. !
+Satyam Agrawal ha ha ha...bhai bhai..
Satyam Agrawal
thanks for your explanation. is it working just in linear PDI ?
i cannot solve the question. can u help me please?
Given the linear equation: Ux-Uy=0 with the initial conditions: x(0,s)=0, y(0,s)=s, u(0,s)=g(s) where g(s) is an arbitrary differentiable function.
a) Write the Characterist equations: a(x,y,u)Ux+b(x,y,u)Uy=c(x,y,u)
dx = a(x,y,u) (1) dy = b(x,y,u) (2) du = c(x,y,u) (3)
dt dt dt
b) Integrate equations (1-2), use the initial conditions and determine x and y interms of the parameters t and s, then inverting these, write t and s interms of x and y.
c) Integrate equations (3), use the initial conditions and determine u interms of t and s and then write u interms of x and y:u(x,y)
Excelente vídeo!
amazing explanation! Thanks a lot!!!
I was able to solve the Exercise at the end using the characteristic method. However, I still feel confused and don't know why?
I was also confused, but then i realized that du/dt = 1 -> du=dt, integrate both sides , you get u(x_0,t) = t - C, then apply initial conditions to see that x_0 = x at t=0 from x_0 = x-ct , and you also get C = sin(x_0) = sin(x), I use C he uses A, potato potato.
good explanation , thanks you for this great work !
Great job! It's a pleasure to watch you :)
Could you somebody explain the chain rule at 4:24 ?
I watched chain rule and it’s derivating from outside to inside, right?
So how come there’s a addition thing in there?
I think he actually makes a mistake here. If what he has there is true, since u_t is just another way to write du/dt, then you would be left with u_x*dx/dt = 0 since u_t would cancel from both sides of this equation.
When du/dt=1,u=t+A.
Could you tell me what happened when du/dt=other number(like 1/2, 3 etc) or a function?
Why do we never say anything about u_x? It seems like it is just treated as 0, but how do we arrive at this?
I am struggling to solve an advection equation given by [ u_t + u_x = 0 ; u(0,t) = Sin 4pi t and u(x,0) = 0] . any tips?
Very well explained. Thank you
11:37 shouldn’t it be f(x ⁰) instead of f(x)? Thanks a lot for making all this!
Great! Very simply and conceptually introduce PDE
Thanks from the bottom of my heart
Great stuff!
omg too tired sorry for the post if the partial derivative of u with respect to t is 1 minus the product of the constant and the partial derivative of U with respect to x then it has to be 1 by eliminating the
the minus minus the product of the constant and the partial derivative of U with respect to x but I kept thinking there is something off with my thinking.
Why is du/dt essentially equal to 1 in the first hint??
thank you very much! Great video!
hello commutant, you don't discuss the way to solve PDE this way: dx/something=dy/something=dz/something?? I can't see it anywhere. My teacher uses this to solve PDE in class and he doesn't explain anything, So I'm not sure whether I should continue with your videos series or not? Please help. I'm new to PDE! :-( I really love your videos as it is like Khan Academy but I'm not sure whether it has everything I need or not?
沈轲 so does he discuss the dx/p dx/q dx/z method in his PDE video series? Don't call it " my ways " friend, I don't have much idea about it I saw my teacher do it at school that's it. :-P
9:00 . You say " u is constant along the line ." It's not. u is only constant with respect to time. Thus, the surface that u represent can change it's height. But the heights once made would not change with time. Please Explain.
+Satyam Agrawal u is a function of 2 variables, which means you can plot it in a plane with the height or 3rd coordinate being the value of u. Along certain lines in the plane the height is constant, those lines with slope 1/c in this case.
Thanks for all the help. Really appreciated. However, what would one do if given following transport equation:
du/dt (x,t) + du/dx (x,t) = 0; x> 0; t>0;
u(x,0)=0; x>0;
u(0,t)=sin t : t>0
I am slightly confused with how these two come together on the x-t plane and on the line x=t. Thanks
You must solve it with Laplace transform. u(0,t)=sint becomes U(0,s)=1/(s^2-1) .You must also transform the p.d.e equation to Laplace. The answer : u(x,t)= δ(t-x) * sint = integral of 0 to t( δ(τ-x) sin(t-τ)dτ)
Very nice video
isn't the solution to the transport equation just some u(x-ct) because it's just a translation of a graph over time? I don't really see why we need PDE for this.
u=F(x-ct)
Thank you! This was very helpful!
could you explain how u got at 04:13? How did the product rule suddenly come to the picture. I understand u are differentiating (x(t),t), but somehow I got lost at that point
what if the constant is changing with time
du/dt + v(t) du/dx = 0
How would that be solved?
This is Soo Good :) Thanks commutant!
Thanks broseph.
Do you write with your mouse or is it some kind of e-board/pad?
JUST Awesome Buddy!! Keep it up ##$$
Great!
at 6:26 youtube stat shows highest watched!!! so accurate
instead of f(x-ct) if we use U_0(x-ct) it may be more convenient.. as my professor did..
does the fact that dx/dt = c mean that x and t aren't indepedent ?
Yes, indirectly dependent. It is really counter intuitive because there is no obvious entity which is changing the time coordinate when displacement of anything takes place.
I think the solution become u= 2t+sin(x-ct) right.
commutant really took us to the hood with this one
good stuff
Really wish you could tutor me.
what if f(x)=0??
sorry if du/dt=2 solution should be u= 2t+sin(x-ct)
I am looking for tutor in pde
Wait but what is C?
I missed why dU/dt = 1
Thanks :)
goat
R=1/2.
I clicked this video because I have a question. Leave with many more questions.
I have generally liked most of your videos so far, but this was sloppy, I would like to note that both of your so-called hints are vital revelations needed to understand the concept. And you never explain how du/dt some how implies u=t+a? Moreover, why is (a) somehow sin(×0). In the previous section, all you stated was that u(x,t) = f(x-ct) given the transport equation and that u(x,0)=0. This is simply not enough information to gain any insight.
not clear at all
Oh, come on. You're telling me that you are an MIT chemical engineering GRAD student and you couldn't solve that trivial exercise in an instant? You're bullshitting.
You know this is educational, right?