Boundary-Value Problems

[@cadmio9413]

Always loved your emotion when explaining something, it makes me feel even more exited for what you show us every video, thanks for all ^v^, specially when you are in front of the witheboard, thanks for all :>

[@jyotsanabenpanchal7271]

Happy Diwali 🪔 🪔 🎇 🎇 🎇 to you Dr. Peyam

[@drpeyam]

Awwwww happy Diwali to you too :3

[@husseinmir9869]

Dr Peyam could you do a video over initial boundary value problems for diffusion equations, and maybe also for wave equations? Covering one dimensional, 2-D, and 3-D cases would also be amazing. Just a suggestion from a PDE student. Love your videos sir!

[@drpeyam]

In fact I have already done this!! Check out my PDE playlist :)

[@JarogniewBorkowski]

Could You explain in any video case 2? Where solution y=te^wt comes from? Why there is no t^2e^wt then and other powers of t?

[@drpeyam]

No video required for this, please check out my video on repeated roots in the playlist!!

[@carultch]

When the characteristic equation has a repeated solution, that's when you get y=t*e^(w*t) as one of the solutions. You only get t^2*e^(w*t) if you have a thrice-repeated solution, or when you have resonance with a forcing function of t that matches one of your homogeneous solutions. You start with e^(w*t), and you multiply by t for every repetition of the solution to the characteristic equation you have. You can show that t^2*e^(w*t) can't be a solution, by doing the following. Consider the following diffEQ, which has a repeated root in its characteristic equation: y" + 6*y' + 9*y = 0 Part 1: show that e^(-3*t) is a possible solution y = e^(-3*t) y' = -3*e^(-3*t) y" = 9*e^(-3*t) Substitute: 9*e^(-3*t) - 6*3*e^(-3*t) + 9*e^(-3*t) =?= 0 Add up the coefficients, and you see it is zero, confirming y = e^(-3*t) is a possible solution. Part 2: show that t*e^(-3*t) is a possible solution: y = t*e^(-3*t) y' = -3*t*e^(-3*t) + e^(-3*t) y" = 9*t*e^(-3*t) - 6*e^(-3*t) Substitute: 9*t*e^(-3*t) - 6*e^(-3*t) + 6*[-3*t*e^(-3*t) + e^(-3*t)] + 9*t*e^(-3*t) =?= 0 Add up coefficients on t*e^(-3*t): 9 - 18 + 9 = 0 Add up coefficients on e^(-3*t): -6 + 6 = 0 As you can see, these both add to zero, confirming y = t*e^(-3*t) is a possible solution. The general solution will be any linear combination of t*e^(-3*t) and e^(-3*t), that is consistent with the initial conditions and/or boundary conditions. Now try y = t^2*e^(-3*t): y = t^2*e^(-3*t) y' = -3*t^2*e^(-3*t) + 2*t*e^(-3*t) y" = 9*t^2*e^(-3*t) - 12*t*e^(-3*t) + 2*e^(-3*t) Substitute: 9*t^2*e^(-3*t) - 12*t*e^(-3*t) + 2*e^(-3*t) + 6*[-3*t^2*e^(-3*t) + 2*t*e^(-3*t)] + 9*t^2*e^(-3*t) =?= 0 Add up coefs on t^2*e^(-3*t): 9 - 18 + 9 = 0 Add up coefs on t*e^(-3*t): -12 + 12 = 0 Add up coefs on e^(-3*t): 2 + nothing else to add = 2 You can see that while it passes the first two tests, it fails to add up to zero for the remaining test, which shows that t^2*e^(-3*t) cannot be a solution to the original diffEQ.

[@JarogniewBorkowski]

@@drpeyam thanks

[@renesperb]

The way you wrote the problem with zero boundary condition it has of course no real solution for real lamda . This can be seen by just mutiplying both sides by y' and integrating by parts. It has of course the important meaning of an eigenvalue problem (e.g. vibrating string ),as you pointed out.

[@YouTube_username_not_found]

What are you talking about? Of course there are real solution for real lamda which are sine functions. Perhaps you meant to say something else?

[@renesperb]

@@KZbin_username_not_found If the equation is y''=c(x) *y ,y = 0 on the boundary and c(x) > 0 ,then you have no real solution.

[@YouTube_username_not_found]

@@renesperb Ummm, why are you using c(x) ? , that's not the equation we have. Lamda is constant.

[@YouTube_username_not_found]

I got it you just forgot to say when lamda is positive.

[@ohanneskamerkoseyan3157]

λ is negative. √λ is the exponent, which is not real.

[@phill3986]

👍👏😊👏👍

[@renesperb]

Take the more general case of the equation y''=c (x) *y in (0 , a) with y(0)=0 ,y(a) =0 . Assume c(x) >0 in (0 ,a). Multiply both sides by y and integrate over (0 , a) .If you integrate by parts on the left side and use the boundary conditions you find that - ∫ y' 2 = ∫c*y^2 ,which means that y = 0 . This is the trivial solution which is always possible, but of no interest. This reasoning would also hold for the boundary value problem ∆u = c* u in G with u= 0 on ∂G .Here ∆ is the Laplacian , G a domain with boundary ∂G , and c a nonnegative function.

[@jaybae8056]

How is my s.i.g.n convention is wrong. I got (-) sine