First Order PDE
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- เผยแพร่เมื่อ 16 ต.ค. 2024
- First-order constant coefficient PDE
In this video, I show how to solve the PDE 2 u_x + 3 u_y = 0 by just recognizing it as a directional derivative and using some simple multivariable calculus. This is part of a technique called the method of characteristics. Enjoy!
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Thanks for making this geometric method so easy to understand. I did not understand when I read the textbook. Your video certainly helped.
it shows at (10:42) that you have had already a lot of hours in front of a camera under your belt. to come back with such an eloquent style. bravo! :)
i can only add, that once you start doing physics, in case of any doubt, you can always quickly check in your mind whether the units match up.
This will be one of the best series of your channel, I’m looking forward the next video 🤩
You know.... A fan when you hear them. "9001 because it's over 9000" - without going saiyan. Control demonstrated.
Finally! Between this video and the other method of characteristics video, I get where we are going throw in Fourier series. This is easily the most helpful method of characteristics video. I am eternally grateful
Thank you!!!!
I want to say that over 9000 joke really caught me off-guard haha. Great video though, helps a lot!
Hahaha, my pleasure! 😊
All the f(x-ct) is super cool and very important in Physics, as it's one of the solutions to the wave equation, which is probable one of the most important equations is Physics. It can be shown that a solution U(x,t) for the second order linear pde Utt = c^2 Uxx can be written as U(x,t) = f(x-ct) + g(x+ct). Here U would be the wave profile, the perturbation of the medium we're studying and c the perturbation's speed in such medium. Now the cool thung is that f and g can be thought of as "travelling waves" because of the "moving profile" perspective Peyam talked about. This is extremely important because we know most of the waves "in nature" are moving (f,g) and are not standing waves (U) like the ones you might find inside a wind instrument, for instance. The way to get those f and g is via D'Alambert's solution, of which I think Peyam talked about. And if he didn't, he might in the future, as I think it's a kind of important point in PDE courses
Yeah I think we can expect a video on the wave equation in the coming weeks, as you pointed out it is the natural progression of the f(x-ct) solution
There’s already a video like that on my playlist
Thanks for your video, Sir. It is really helpful for me. My major is aerospace engineering. I have studied ACFD and PDE. I passed them but they are still confused for me. Now I think I have known more things such as relationship about them. Thanks. Looking forward further video. Thanks
Can you create a video where you explain that how you, given a problem like the transport problem, can express it as a PDE so how you get to the approach U_t+cU_x=0?
Brilliant video, perfect explanation! I love your channel!!
Best intuitive explanation, love these videos, thanks!
Congrats to your efforts to show how beautiful math is. You are well successful in this journey. But what the equation says is that the grad of u is orthogonal to v and not necessarily zero over the line. Also if you multiply v by alpha, any Real number, you will have a line passing through the origin ... but you will get the same PDE ... then what the PDE is saying is that the grad of u is orthogonal to alpha v.
luv u bro! saved my life! almost died in the PDE class :(
Thank you for your video,
I have a difficulty understanding why the line u and the vector v are not perpendicular instead of parallel .
Since their dot product is 0 shouldn’t they be perpendicular instead of parallel ?
You just made this so easy THANK YOU!
My lecturer has used it for the linear advection equation. My exams are over but I still has a question. He has done it the same way as you.
I know that as variables, we have x and t and grad is a function x,y and z. Then, how can we dot the grad of u and the other vector to get zero when grad isn't even a function of t?
Given a function of n variables
f(x1,x2,…,xn) the gradient (unless you specify that you're doing something different, like working in polar coordinates) is just the vector represented by each partial derivative.
It doesn't matter if you label your variables with x, y, z or t; and it also doesn't matter if it's two or three variables.
On the other hand, and maybe this is the actual confusion:
U(x,t) is exactly the same as U(x,y): just consider the change of variables y=t.
All letters are just labels, it is the same concept whatever you choose.
That's the reason why you study this U(x,t) but then we apply the same result on v(z,t) in another problem.
You can even have two identical variables but with different names, like x and y=x in e^(x*y), they are called mute variables.
Thank for the clarification. I was doing these kind of things in numerical solution of pdes which is really interesting.
Still waiting for the final exam unboxing .....
Oooooh, I forgot to send her an email 😅
freaking love you, thanks you so much
Tell me if I am wrong but we could imagine wave propagating in the direction perpendicular to v and the parallel lines .
Thank you dr for this awesome co-relation you shared. One doubt I have...
Sir, like df/dx (d = partial derivative)... like how you made df/dx = df/dy as you canceled the terms in 8:16
Somewhere intuitively that looks, it holds for linear combinations. Please let me know sir, I'd be grateful to you.
So the 3D plot of such a U is some function plot f(z) where z is on the line perpendicular to v, extruded to infinity in the directions parallel to v
Love your content! I would appreciate if you could do a video on PDE for vibration string and d'Alembert's solution.
Already done
Sir how we will solve a system of partial differential equation (linear of first order and 2nd order)
Is there anything to assume about the regularity of the function or about its domain? Or can it be actually any function of the form f(ax-by)? Sorry if the question is stupid
A perfectly valid question. Well, you do need to be able to differentiate u at least once in order to talk about it being a solution (or "classical solution", as people call it) of the equation. So f needs to be differentiable, too. But if you consider this last "moving graph" intuition at about 12:05, it kinda seems that f might be just about any function - even non-continuous one. And there's a sensible way we can call such functions solutions to diff equations (they are called "weak" or "generalized" solutions). But you'll generally want your candidate solutions to PDEs to heave at least some regularity to be able to take derivatives (in some sense). You can look up Sobolev spaces if you're interested.
@@Qwerty13146 thank you very much for the kind answer!
The Jacobean of u wet (x,y) gives the row vector
The gradient is the common vector
dr peyam, could you shed a light on what to do if we have 2 u_x + 3 u_y = const, and what does it mean? particularily, what would u_t + c u_x = const in your physical interpretation mean?
I barely know any physics 😅
excellent video !! incredible
I cannot believe the seriousness with which you wrote "AYYY LMAO" on your blackboard.
i always get confused to when i do this type of comparison😂
I think I'm going to need to watch this two or three times....
Recently, I found that the half derivative of e^x is not e^x. I thought this fact is suprising! Are you interested in making a video to talk about it and other fractional derivative of e^x?
Already done that
@@drpeyam Do you mean this video ( th-cam.com/video/k2T0YilPrWw/w-d-xo.html )? The result for fractional derivative of e^kx in this video is not consistent with the definition in th-cam.com/video/eB3OUl5TVSY/w-d-xo.html .
cursive b is gone?
cursive f as well?
The cursive b always looked like a v to me
@@MagicGonads I think it looked more like h
@@gordonchan4801 yeah I might be thinking of a different confusing letter
Hahaha
Could you use e^(3x-2y) as well?
Any function of 3x-2y, so yes
physically, x 'goes with' ct (length with speed times time) so x-ct is physically consistent.
Wow, interesting, I haven’t thought about it this way!
ayy lmao
Im late,i had exams at M and P ,well done
The First Order PDE is pretty good, but I'm a bit more partial to The Empire PDE.
Did you just say "you or" solution? Boy, I miss living in south India.
Applaud the raw unedited version.
9:20~10:43 I think you were gonna edit this part?
btw link of next video: th-cam.com/video/35Wj1NtmywA/w-d-xo.html
Thank you! I was looking for that part, haha
Personally i solved this one by letting u(x,y)=h(x-y,3x-2y).
Getting 2u_x+3u_y=-h_x(x-y,3x-2y)=0
So h_x(x-y,3x-2y)=0 for all x and y
By replacing x with -2x+y,y with -3x+y,
We get that for all x and y
H_x(x,y)=0
Which means h(x,y)=f(y) for f a diff function of y.
u(x,y)=f(3x-2y). Is this a common thing in P∂Es, making substitutions like this.
Ok
PDE gang rise up.
You know why maths is hard... Its because 90% of the teacher we have are bad