Lec 2 | MIT 18.085 Computational Science and Engineering I, Fall 2008

[@panda192012]

Professor Strang a gift from heaven to humankind.

[@cronos280]

I missed several chunks on a similar course, but these videos got me back up to speed in no time. Many thanks and keep up the good work.

[@orion2020]

this lecturer is a LEGEND!!i watched many of his lectures!!! thankyou MIT!!

[@davidlovell729]

From 48:45 on, he does something that is misleading on the board. The book gets it right. You don't change the first row of the matrix K, which is the row that corresponds to u1. Instead, you add a new row to the top of the matrix, which corresponds to u0. This should be obvious: since we don't have the boundary condition u0=0 any more, we don't actually know what u0 is, so we have to add a row to the matrix to figure it out. That row has 1 and -1 followed by all zeros, because it is the 2nd order accurate boundary condition he has just discussed. To correspond with that new row, and the addition of u0 to the solution vector, we have to add one element to the right hand side, with the value 1/2. Thus, we should be solving a 6x6 system when before we were solving a 5x5 system.

[@sundarrajn1003]

thanks a lot..i really got confused at that point..

[@saifaddeenal-manaseer6325]

What bothers me is that he ignores the boundary condition u(0) and u(1) (or u_0 and u_6) in the solution and it works out because the boundary conditions are 0. But if they were not then the difference equations for U_1 and U_5 wouldn't be complete (no U_0 and no U_6, respectively)

[@reimorster]

Such amazing lectures. Awesome professor. I wish I had access to these courses when I was on college. But its never too late...

[@Crasshopperrr]

Never knew this about the faster convergence of centered differences at 10:30. Nice add-on for anyone with a basic calculus knowledge who might have to implement numerical derivatives on a computer. A bit reminiscent of the Runge-Kutta idea.

[@davidlovell729]

It's more than a bit reminiscent. It's the exact same idea.

[@antoniolewis1016]

40:50 Good old professor Lewin from Physics.. Well, I miss that guy, and it's a shame he did those things. He was a great lecturer.

[@kstahmer]

“Maybe a good place to sit is over there…” The two women initially waited to sit between to sit between Prof. Strang and the camera, blocking our view. Prof. Strang made the right move. He knows the needs of the many (cyberspace audience) outweigh the needs of the few (MIT audience).

[@Singularitarian]

When he makes an upgrade to 2nd order at 48:55, u0 now becomes one of the variables he is solving for. He should have erased the vector of unknowns on the chalkboard and re-written it to include u0. This is a slight error in the presentation. Wonderful lectures, extremely valuable, please do more.

[@boxu4948]

Daniel O'Connor yes I am confused at this moment. But can you explain it more clearly how to do it. Thank u very much

[@muyuanliu3175]

extend the matrix with an additional row on top for (2*u0-2*u1)/h^2=1

[@jeeveshjuneja445]

@@boxu4948 Instead of erasing the -1 outside the matrix at 48:52, we let it stay as it is. Basically, that row of the matrix ensures that second derivative at u1 is 1. And we add another row(-1 2 -1) on top of it, corresponding to ensuring second derivative at u0 is 1. And then modify this row to ensure that first derivative at u0 is zero, that is using the fact that u_1=u_{-1}, in (-u1+2u0 -u_{-1})/(h*h).

[@jeeveshjuneja445]

@@boxu4948 Instead of erasing the -1 outside the matrix at 48:52, we let it stay as it is. Basically, that row of the matrix ensures that second derivative at u1 is 1. And we add another row(-1 2 -1) on top of it, corresponding to ensuring second derivative at u0 is 1. And then modify this row to ensure that first derivative at u0 is zero, that is using the fact that u_1=u_{-1}, in (-u1+2u0 -u_{-1})/(h*h). Notice that as we are plugging in u_1=u_{-1} in the equation ensuring that second derivative at u0 is zero, we are ensuring both, that the derivative at u0 is 0 and the second derivative is 1.

[@wdlang06]

i am a student in china. i love the lectures so much.

[@luisalbertoluft5306]

At 19:00 it is actually -1+8-9 instead of -4+8-9 that results in -2.

[@antoniolewis1016]

Yes, you are correct.

[@yp1369]

Really appreciate professor Strang! Hope one day I can sit in on his class.

[@jewel6621]

The lecturer is really a well-speaker. everything is trivial after watching the video.

[@Squatchmichael]

I am pleasantly surprised by how good the MIT lectures are; bad ass physicists and mathematicians make good teachers. The Lewin physics lectures are good too.

[@rinaldisabirin7958]

he knows how to explain and bring the people to understand the essentials of mathematics, which comprises of 3 matters, what ? how ? what for ?

[@saifaddeenal-manaseer6325]

The way the equation is at 33:16 to solve for U1 to U5 doesn't give the right answer for U1 and U5 since the difference equation for the end isn't complete. (It's just 2 -1 and -1 2). Isn't that right?

[@vivek37437]

no its correct because due to boundary conditions, u0=u6=0. So 1st column and 7th column are chopped off. so -1 term from 1st column and -1 ter from 7th column are chopped off and the remaining matrix is 5x5

[@InnerBoyka]

In his second example, Strang says he is "freeing up the other end (at x=0)" and then writes the mathematical condition du/dx = 0 (at x=0). Is he just saying that he wants to impose the condition that no force is acting at the beam at x=0 (i.e. like a cable lying on the floor at x=0)?

[@mfamily3772]

7:52 Thank god I'm a normal person

[@vivek37437]

at 45:53 he is referring to a book. Does anybody know which book he is referring to?

[@mitocw]

The textbook for the course is: Strang, Gilbert. Computational Science and Engineering. Wellesley, MA: Wellesley-Cambridge Press, 2007. ISBN: 9780961408817. See the course on MIT OpenCourseWare for more info at: ocw.mit.edu/18-085F08. Best wishes on your studies!

[@OttoFazzl]

Why for the free end at 34:39 the first derivative should be equal to zero?

[@yspaiyogesh]

That's the boundary condition he assumed. Listen to him say, "Change the boundary conditions, change the answer."

[@Crasshopperrr]

For replacing the u'(0)=0 boundary condition around minute 38:00 couldn't you also set the constraint on u'(1)?

[@stearin1978]

what constraint?

[@stearin1978]

Prottoy Nahian 0

[@stearin1978]

Prottoy Nahian parabola with 2 slopes = 0 is no fan as you like:)

[@stearin1978]

Prottoy Nahian just try to draw a parabola with 2 slopes = 0. After it we will talk bro!:)

[@stearin1978]

Prottoy Nahian yes, bro. parabola with 2 slopes at 2 points = 0. Draw! Comon bro!:)

[@studentmele]

h=1/6; x=[h:h:1]; fx=-1/2*x.^2+1/2; plot(x,fx) A=toeplitz([2 -1 zeros(1,1/h-3)],[2 -1 zeros(1,1/h-3)]); A(1,2)=-2; hold on U=A/(h^2)\ones(1/h-1,1); U=[U;0]; plot(x,U,'g') A=toeplitz([2 -1 zeros(1,1/h-3)],[2 -1 zeros(1,1/h-3)]); A(1,1)=1; hold on U=A/(h^2)\ones(1/h-1,1); U=[U;0]; plot(x,U,'r') Why the second one is better!?

[@dhomouz]

You have to change h to 1/5 in the first method. In this approach the first row represents u_0. You can keep h=1/6 by making f(0) = 3 which amounts to adding the zeroth row to 1st row.

[@azizfall5579]

I think the first row should equal 3/2 not 1/2. if you set eqn 1. U-1 = U1 and solve for the system of equations eqn 2. -U2 + 2U1 - U0 = h^2 and eqn 3. -U1 + 2U0 - U-1 = h^2 then if you solve for U0 in terms of U2 and U1 and plug it back into eqn 2. you get (-2U2 + 2U1)/3 = h^2. The book actually adds an extra row and column for U0. where the boundary conditions leads to the constraint 2U0 - 2U1 = h^2

[@imegatrone]

Your Video Is Very Useful Sharing Lecture 02 Difference equations

[@LennyLeonard85]

@giuseppe92205 Just listen to the voice in the beginning. ;)

[@LennyLeonard85]

@giuseppe92205 It's even in the infobox.

[@giuseppe92205]

how get the transcript?

[@sephirothra]

@sikory dont whine...some people have to travel to actually get to the class.....you havent been in a university yet have you?

[@getutadele7326]

How can I get the lecture notes or the book?

[@mitocw]

You can get the course information and materials on MIT OpenCourseWare at ocw.mit.edu/18-085F08. Best wishes on your studies!

[@salardelavarqashqai]

perfect

[@gohan100100]

I disagree