This is by far the best professor I have come across on UA-cam. Going through all the steps and explaining every detail thoroughly. Wow! Just amazing. All your videos are amazing sir. You are helping a million of students out here👏🏾
Dr. Thanks for your videos, it is helping me seriously in this semester. This is the first time our outfit is taking this course and we are the first group. Your video is giving us more explanations. Good work done.
But on a serious note can someone tell me if we always have a cosine Fourier series when we have Neumann initial conditions and always have a sine Fourier series when we have Dirichlet initial conditions? or did it just kind of work out that way this time?
Hello Dr. I have a question regarding the heat equation when it is somewhat different than the normal kUxx = Ut, my question is how would you do the separation of variables if the equation is kUxx -hu = Ut or Uxx-3u=Ut. Thanks and your videos are very helpful.
whenever an aussie speaks, i cant help but trust in what he's saying lol. Good video. Ive found that the only tricky part with the H.E. is to not make a mistake with the integral when solving An
I have this problem ut-uxx+2u=0 boundary conditions; u(0,t)=250ºC, u(L,t)=135ºC, u(x,0)=20ºC, any help about this problem or maybe where hay can find information thanks
ut-uxx+2u=0 is separable, so let u=f(x)g(t). by differentiating we have f*g' - ( f'' - 2f) g = 0 g' / g = (f''-2f) /f Since the left hand side depends only on t and the right hand side only on x, we can set both sides equal to a constant C. g' /g = C gives g = g(0) exp (Ct)
and to find B, use the other boundary condition, so that we have f(L) = constant, then we have f(0) cos(pL) + B sin (pL) = f(L) Thus B = f(L) csc (pL) -f(0) cot (pL) Now substitute and you are done
the solution of the ODE of form af''+bf'+cf=0 takes the form f=e^(rx) where r is the 0's of the polynomial ar^2+br+c=0. in this case, it is f''-λf so r^2-λ=0, or r= +/-√λ where lambda cannot be negative or it becomes complex.
This is by far the best professor I have come across on UA-cam. Going through all the steps and explaining every detail thoroughly. Wow! Just amazing. All your videos are amazing sir. You are helping a million of students out here👏🏾
My right ear is gunna nail this exam
That just cracked me up. Ha ha.
Left ear for me.
My pleasure and hope you are finding the ebook of some use also.
Our f is originally defined on the interval [0,\pi]. Thus we can extend f as an even or odd function on [-\pi,\pi] as we wish.
I appreciate that I put subtitles since I do not speak English. Thank you very much for the explanation and for the encouragement. Greetings from Peru
awesome! like the pace you went with, slow and elaborate!
Dr. Thanks for your videos, it is helping me seriously in this semester. This is the first time our outfit is taking this course and we are the first group. Your video is giving us more explanations. Good work done.
Yup, we do! I'm taking heat transfer this semester, and it was derived in class. We also had a few homework problems related to it.
that laugh at 25:37 hilarious!
Great to hear! Hope you enjoy my new ebook as well - the link is in the description.
Thank you so much, this video has saved my life!
Why, at 22:30, is it not F(x)=Acos(px)+Bisin(px)? Why doesn't he have the i?
in ODE if a*F "a constant" is a solution then F is a solution too in your case "a=i" and we need a real solution not a complex one
Hi - thank you very much. If you enjoyed this lesson then please consider my new ebook, which is freely available - the ilnk is in the description.
you sir, are a mathemagician!
Thank you so much!
This was so helpful.. thank you so much
Hello sir. Will you solve a PDE if I send you?
For 38:22 isn't the integral doubled only if f(X) is even as well? How do we know f(X) is doubled
But on a serious note can someone tell me if we always have a cosine Fourier series when we have Neumann initial conditions and always have a sine Fourier series when we have Dirichlet initial conditions? or did it just kind of work out that way this time?
When Landa is 0 (at video time=15:00) so A is arbitrary (not B) and B should be 0 (not A). Am i right?
Why are n = -1, -2, -3, etc not solutions as well? They would still satisfy -n^20. Another great video as well.
then it would be linearly dependent
How does the radius of the cross section affect the diffusion?
That is a good question. We assume the bar is thin, so the cross-sectional area is small and has no effect.
very very nice lecture you cleared everything boss
awesome video helped me a lot in PDE thank you so much :) ...
Can we get a soft copy of that answer sheet of yours?? it would be easier for us to understand instead of rewinding and forwarding.
Is there way to solve nonlinear heat equation by separation of varibles ? I am trying to do that. I know this case is called fast diffusion equation.
I've been viewing your work sir, and you have helped me a lot! much thanks for the book!!
My pleasure. Glad you are enjoying the book.
Dr Tisdell, I love you !
your accent is a little easier to understand than my prof. Thank you
Thank you, Dr. Tisdell!
thanks, its awesome to have this available
Hello Dr. I have a question regarding the heat equation when it is somewhat different than the normal kUxx = Ut, my question is how would you do the separation of variables if the equation is kUxx -hu = Ut or Uxx-3u=Ut. Thanks and your videos are very helpful.
Thanks for the feedback.
Thanks for the wonderfull class!
thank u sir. u made it too easy to understand
I don't understand because my English isn't good. But this video is intersting.
whenever an aussie speaks, i cant help but trust in what he's saying lol. Good video. Ive found that the only tricky part with the H.E. is to not make a mistake with the integral when solving An
Do mechanical engineers learn about heat equation?
Thank you so much sir,it now makes sense to me.
thank you so much sir...
thank you so much this video is perfect
I have this problem ut-uxx+2u=0 boundary conditions; u(0,t)=250ºC, u(L,t)=135ºC, u(x,0)=20ºC, any help about this problem or maybe where hay can find information thanks
ut-uxx+2u=0
is separable, so let u=f(x)g(t).
by differentiating we have
f*g' - ( f'' - 2f) g = 0
g' / g = (f''-2f) /f
Since the left hand side depends only on t and the right hand side only on x, we can set both sides equal to a constant C.
g' /g = C gives g = g(0) exp (Ct)
now consider f" = (2+C) f
Like what Dr Tisdell did, write lambda = 2 +C =-p^2 for some real p.
This is justified because if 2 + C
Here A = f(0)
and to find B, use the other boundary condition, so that we have f(L) = constant, then we have
f(0) cos(pL) + B sin (pL) = f(L)
Thus B = f(L) csc (pL) -f(0) cot (pL)
Now substitute and you are done
I thank u for this video.
Can someone please explain where (r^2 - Lamda) for Lamda>0 coming from ?
at 15:35
the solution of the ODE of form af''+bf'+cf=0 takes the form f=e^(rx)
where r is the 0's of the polynomial ar^2+br+c=0. in this case, it is f''-λf so r^2-λ=0, or r= +/-√λ where lambda cannot be negative or it becomes complex.
Awesome, many thanks.
yesterday?
Thanks a lot!!!
Thanks
How to solve the heat equation in 3d.
Worthful!!
thanks man !
Thanks a lot
25:40 amazing
Yes they do. Almost all engineers do I think.
ThanKsssss :) :) :)
Thank god I have my right ear...