My whole time at UNSW I've been hoping to have you for the third year PDEs class, Ah well missed my chance at least I have these videos. I must say I am truly thankful of you and Wildberger for continuing to support free education. You have both inspired me to one day do the same.
Definitely, I think I should take this opportunity to thank you for your videos over the years. I have found them very useful for my distance education degree in maths!!! KUDOS to you!
Respected Dr Chris Please make a video on method of characteristics for quasi linear problem. I was going thru elements of pde by Ian N Sneddon pg no 62 method of characteristics, but didn't understood the method. I will be thankful to you if you can explain that Regards
Good question. In that case we interpret the du / f part of the characteristics equation as leading to u = constant. Hope this is helpful and you can see a few homogeneous PDE solved in this manner in my PDE playlist.
I can apply this method to all the usual PDE-s like wave eq, Laplace eq and so on right? And also, would this method also work for a stochastic PDE, like the Langevin equation?
I have now thought some more about this. Apologies for the length of this comment, but I was struggling with the ideas for while and this at last made sense of it, for me at least. Let a solution to the PDE (the “solution surface”) be a surface in (xyz)-space described by z = u(x, y). This can be written u(x, y) - z = 0, and the normal to this is the gradient vector grad(u(x,y) - z) = (∂u/∂x, ∂u/∂y, -1) ...(*). The PDE itself tells us that the vector (a, b, f) is normal to (*). So it lies in the tangent plane to the surface. This is true at every point so if you join the vector field elements (a, b, f) to make a family of curves in (xyz)-space, the solution surface is comprised entirely of these curves. These are the “characteristic curves”. How do you find the equation of a particular characteristic curve? By definition, along the curve, dx/a = dy/b = dz/f. Solving the first two on their own, gives a whole family of characteristic curves which together comprise a surface in (xyz)-space. It will contain an arbitrary constant of integration c1. The equation will be of form g(x, y, z) = c1. The taking the second and third, or the first and the third, will give another surface formed of characteristic curves, something of the form h(h, y, z) = c2. If you take a fixed value for the c’s, and find the intersection of the two surfaces formed by g and h, you will get a single characteristic curve (in general). If you now vary the c’s, you can get all the characteristic curves in turn. Each one is the intersection of the g and h surfaces for a particular pair of values of the c’s. In fact the c’s act like a pair of coordinates for the whole set of characteristic curves. We know that a given solution surface is a set of characteristic curves. In general, this will be a one-parameter set out of the whole two-parameter set of characteristic curves, as indexed by the c’s. How do you describe a one-parameter set of points in a two-dimensional space? In general, a function k of the form c2 = k(c1) will do it. This is where the arbitrary function comes in. Adding the Cauchy condition determines k uniquely, in general. Prof Tisdell's video is brilliant, but it left me with the odd gap in understanding which maybe this comment will help fill for others, too.
Hi, I have just one doubt, what happen when 'c1' and 'c2' has the value 'u' in them?? how do you use the I.F.T (implicit function theorem?). Thanks you so much for the video, It's really useful!!
Excellent video. Can I suggest just one other thought which may help to understand this, at least it helped me. But I'm not sure if I have got this right. Perhaps someone can comment. The characteristic curves with constants c1 and c2 can be seen as providing a way of coordinatizing the surface along which the solution u travels ('travels', as a function of t, that is). The set of points comprising the positions of u as time proceeds, forms a curve on this surface. If c1 and c2 are looked on as coordinates of the points u, then the points have 'coordinates' c1 and c2 which together form a curve in this two-dimensional space. One way of describing a curve in two-dimensional space is as the graph of a function, which can be written in the form c2 = g(c1), which is the result Dr Tisdell arrives at as the solution. (a) is that right, and (b) does it help to understand it?
Thanks a lot for your effort. I understand from your illustration more than I usually understand from my lecturer's in college. I have a Final exam tomorrow morning in PDEs. Wish me luck.. ^_^
Thank you. I downloaded the book and, even that is a very useful book, it doesn't talk about the method of characteristics. Some examples to solve PDEs with the method of separation of variables but nothing about the method of characteristics. However, thank you for your video.
Makes 0 sense to me why the normal becomes (ux,uy,-1) since i try to verify this with the chain rule for the given equation before and it isnt the same.
how would you solve du/dx with an y. for example: dx/1 = dy/y = du/(-u+x+y). you mentioned something about how to figure it out, but I cant solve this one. Would be awesome, if you have a solution or a tip how to do it. I have more of those in my homework. And thank you for your video!
Are quasi-linear PDEs not defined as PDEs which are linear in the highest order partial derivatives appearing in the PDE (but arbitrary in the other appearing partial derivatives)?
Not all but from what I’ve seen the heat, Laplace and wave equation( linear) can be solved through separation of variables and then assuming a power series solution
@@chymoney1 so there is the Laplace equation , heat equation , and wave equation. Are there any other pde's? Do u think pde's are easier to learn if you got an A in multivariable calculus
Mohammad Ali Nabeel there are many others. And yes calc 3 does help but you also need a course in differential equations before you do partial differential equations
My whole time at UNSW I've been hoping to have you for the third year PDEs class, Ah well missed my chance at least I have these videos.
I must say I am truly thankful of you and Wildberger for continuing to support free education. You have both inspired me to one day do the same.
It was an amazing video Dr. Tisdell. It made things crystal clear. Thank you SO much!
Definitely, I think I should take this opportunity to thank you for your videos over the years. I have found them very useful for my distance education degree in maths!!! KUDOS to you!
Respected Dr Chris
Please make a video on method of characteristics for quasi linear problem. I was going thru elements of pde by Ian N Sneddon pg no 62 method of characteristics, but didn't understood the method.
I will be thankful to you if you can explain that
Regards
I think this is what you're looking for: ua-cam.com/video/5ZrwxQr6aV4/v-deo.html
It is my pleasure and I wish the very very best with your studies.
HI, i am facing a lot of difficulties to plot characteristic net from MOC. Can you please help me
Thank you so much, I was wondering if I could find the power point you used in the video?
Hi - the notes will be made available (for free) in a new ebook later this year.
Thank you sir! My teacher didn't explain this fully like how we got the actual formula and all!
Thanks for your videos! Do you have a playlist with videos about PDE's which are solved only by separation of variables ??
Thanks a lot!
Great quotes made me feel better
Thank you!, you saved my life
you are the best, i truly liked the way u explained and did everything, i think im gonna try to suscribte to you 10.000 times ;D
Plz help for characteristics of dalembert solution
what happens when f(x,y,u)=0?
Good question. In that case we interpret the du / f part of the characteristics equation as leading to u = constant. Hope this is helpful and you can see a few homogeneous PDE solved in this manner in my PDE playlist.
In minute 2:47 - I think it must be u(0,y)=u_1(y) not u_1(x)
Thanks!
And at 15:16, it should be du/dx=exp(+u) instead of exp(-u) .
But the solution works though...
Thanks
It's interesting to get insights.
Really enjoyed
I can apply this method to all the usual PDE-s like wave eq, Laplace eq and so on right? And also, would this method also work for a stochastic PDE, like the Langevin equation?
No
Thanks. Vert helpful
Sorry Chris, I still don't understand but got used to it anyway.
Watching this at 1.25x the original speed is a blast
This looked like a torture when I finished the video
I have now thought some more about this. Apologies for the length of this comment, but I was struggling with the ideas for while and this at last made sense of it, for me at least.
Let a solution to the PDE (the “solution surface”) be a surface in (xyz)-space described by z = u(x, y). This can be written u(x, y) - z = 0, and the normal to this is the gradient vector grad(u(x,y) - z) = (∂u/∂x, ∂u/∂y, -1) ...(*).
The PDE itself tells us that the vector (a, b, f) is normal to (*). So it lies in the tangent plane to the surface. This is true at every point so if you join the vector field elements (a, b, f) to make a family of curves in (xyz)-space, the solution surface is comprised entirely of these curves. These are the “characteristic curves”.
How do you find the equation of a particular characteristic curve? By definition, along the curve, dx/a = dy/b = dz/f. Solving the first two on their own, gives a whole family of characteristic curves which together comprise a surface in (xyz)-space. It will contain an arbitrary constant of integration c1. The equation will be of form
g(x, y, z) = c1. The taking the second and third, or the first and the third, will give another surface formed of characteristic curves, something of the form
h(h, y, z) = c2.
If you take a fixed value for the c’s, and find the intersection of the two surfaces formed by g and h, you will get a single characteristic curve (in general). If you now vary the c’s, you can get all the characteristic curves in turn. Each one is the intersection of the g and h surfaces for a particular pair of values of the c’s. In fact the c’s act like a pair of coordinates for the whole set of characteristic curves.
We know that a given solution surface is a set of characteristic curves. In general, this will be a one-parameter set out of the whole two-parameter set of characteristic curves, as indexed by the c’s. How do you describe a one-parameter set of points in a two-dimensional space? In general, a function k of the form c2 = k(c1) will do it. This is where the arbitrary function comes in. Adding the Cauchy condition determines k uniquely, in general.
Prof Tisdell's video is brilliant, but it left me with the odd gap in understanding which maybe this comment will help fill for others, too.
Great explanation. Thanks.
It's amazing that constants of integration play a role as coordinates on the solution surface.
Just wanted to say thank you for this video. I am teaching myself PDEs and this was very helpful!
Hi, I have just one doubt, what happen when 'c1' and 'c2' has the value 'u' in them?? how do you use the I.F.T (implicit function theorem?).
Thanks you so much for the video, It's really useful!!
Thank you Dr Tisdell. Your videos helped me earn an A in PDE after 30 years after my undergrad.
Actually I had almost given up on method of characteristics but your tutorials have tremendously blessed me thanks Dr.
Excellent video. Can I suggest just one other thought which may help to understand this, at least it helped me. But I'm not sure if I have got this right. Perhaps someone can comment.
The characteristic curves with constants c1 and c2 can be seen as providing a way of coordinatizing the surface along which the solution u travels ('travels', as a function of t, that is). The set of points comprising the positions of u as time proceeds, forms a curve on this surface.
If c1 and c2 are looked on as coordinates of the points u, then the points have 'coordinates' c1 and c2 which together form a curve in this two-dimensional space. One way of describing a curve in two-dimensional space is as the graph of a function, which can be written in the form c2 = g(c1), which is the result Dr Tisdell arrives at as the solution. (a) is that right, and (b) does it help to understand it?
and also you are good when it comes to explain please keep it like that
Hi. Take the function and calculate its derivatives (perhaps by using implicit differentiation) and the see if the PDE holds.
Plz help
Need ur help
Thanks a lot for your effort. I understand from your illustration more than I usually understand from my lecturer's in college. I have a Final exam tomorrow morning in PDEs. Wish me luck.. ^_^
Good luck!!
Thank you. I downloaded the book and, even that is a very useful book, it doesn't talk about the method of characteristics. Some examples to solve PDEs with the method of separation of variables but nothing about the method of characteristics. However, thank you for your video.
kawaka2323 took a whole class on pdes and all we did was separation of variables, Fourier transform solutions and Cauchy-weistress solution
Why doesn't the final solution of u not satisfy the formula dx/x=dy/y=du/xe^u again ??
Thanks a lot for helping me understand this method😁
Kindly tell me the name of book which are you following plz
I'm afraid i still dont understand
Makes 0 sense to me why the normal becomes (ux,uy,-1) since i try to verify this with the chain rule for the given equation before and it isnt the same.
Nvm. U is a variable independent of x haha.
Better Understanding !! I can now get the actual idea behind the steps for solving pde.
Amazing video sir ❤ and yes we understand the method of charisteristic 😅
It's like watching Fun with Flags with Dr. Sheldon Cooper.
how would you solve du/dx with an y. for example: dx/1 = dy/y = du/(-u+x+y). you mentioned something about how to figure it out, but I cant solve this one. Would be awesome, if you have a solution or a tip how to do it. I have more of those in my homework. And thank you for your video!
Nice acedamy
Are quasi-linear PDEs not defined as PDEs which are linear in the highest order partial derivatives appearing in the PDE (but arbitrary in the other appearing partial derivatives)?
I was asked to solve a practical problem which is water hammer ysing this method and I'm stuck
Prof will you be expanding on this amy time soon?
When I teach another PDE course then I'll try to expand on this.
Hi I have a question. Which are the characteristics curves in your instance? Thank you
John Von Neumann got it right for many methods in maths!
Please can you solve some problem , how i can attach for you?
Thank you very much Dr. Chris. I have really enjoyed the video.
Best video out there on the topic, thanks a lot!!
So now we're used to it, but we don't understand it.
Awesome! very clear, very much appreciated!
does this also work with constant coefficients?
Thank you so much for this lecture. I was trapped by this MOC when solving practical problem, but I am totally clear now.
You look like a young Sean Pertwee. :o
It helped me a lot , thank you ❤️
A lot of thanks sir, you made my life simpler
Thank you a lot, great video!
like
best
i love the sentence
Great work..
very helpful
2 years after graduating from unsw and I'm still seeking for your help. You are the true hero Chris
is it true any pde can be solved by fourier series or analysis? i only took caculus 2 till now
not really, but it is powerful
Not all but from what I’ve seen the heat, Laplace and wave equation( linear) can be solved through separation of variables and then assuming a power series solution
@@chymoney1 so there is the Laplace equation , heat equation , and wave equation. Are there any other pde's? Do u think pde's are easier to learn if you got an A in multivariable calculus
Mohammad Ali Nabeel there are many others. And yes calc 3 does help but you also need a course in differential equations before you do partial differential equations
Mohammad Ali Nabeel tutorial.math.lamar.edu/Classes/DE/DE.aspx
I recommend you look at this
He was clever, wasn't he!
Professor are you using the two camera for lectures.
@@grelearners8048 yes i m
So clear and helpful, think I might have a chance at passing my finals now!
Great!