Transform Method
ฝัง
- เผยแพร่เมื่อ 16 ต.ค. 2024
- In this video, I solve the PDE a u_x + b u_y + cu = 0 by using the 'transform' method, namely by defining v = u exp(cx/a) and hoping that there's some kind of simplification! Enjoy!
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the yoshi analogy surprised me
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-cu-
I LOVE THIS, will you eventually post some crazy looking pde’s?
I thought this was really cool! I would think a coordinate transfer at first glance, but I like you approach!
It's a nice day when Peyam post a video about PDEs. By the title i thought it was about integral transform method, maybe you can talk about it next?
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This video is about a PDE! a ux + b uy + cu = 0 is a PDE!
Yeah, but i thought you would solve it using a integral transform(Fourier or Laplace) to turn it into a ODE.
Good stuff 👍 PDE's are wonderful. 🙂
I ABSOLUTELY LOVE AND ENJOY THIS, Can we do much harder PDEs in the future?
Check out the playlists
@@drpeyam I’ve been watching them you are saving me in my PDE class
If you had done V=e^(c/b)y * U instead of V=e^(c/a)x * U, the solution would also have that different e^... coefficient. Would it still be the same solution=?
Yeah
The only guy who downvoted this video must have done so by mistake. The video is such a good follow-up to the previous ones in the playlist, given that once Dr Peyam revealed the substitution v(x,y) = exp(cx/a)*U , a viewer could easily solve the PDE on their own.
Pro-Da-Lu 🙂
Hello Dr. Peyam. Your lecture is quite awesome. If you would mind, can you investigate what would be the values of i raise to i raise to i raise to i IN COMPLEX ANALYSIS..
You already solved i raise to i and i root of i. What if i is raise to i and also again the exponent i is raised to i, and again i is also raise to i, what would be the solution?
IM QUITE EXITED :)
thank you and more power
Frm. Philippines.
Can't you do that by substituting v=u*e^(cy/b) as well? Is the idea that the two answers you get differ only by an exponential of the form e^{[c/(ab)](ay-bx)} which is a function of (ay-bx)?
Sure!
I think so!
Is there any videos Dr about the canonical form?
The method is useful to solve partial differential equations
Never heard about it
I tried u=m*v where m is a function in x and y the by substituting [all derivatives are partial]
am dv/dx + bm dv/dy + (a dm/dx + b dm/dy + cm) v = 0
Then we want the coefficient of v to be zero. If we let m be a function in x we get m=exp(-cx/a) , but if we let m be a function in y we get m=exp(-cy/b).
In both cases we get => a dv/dx + b dv/dy = 0 so v=f(ay-bx) for an arbitrary partially differentiable function f and we get u=m*f(ay-bx) where m is any function that makes coefficient of v zero in the PDE [satisfies : a dm/dx + b dm/dy + cm = 0]
but we got 2 representations for m :
1) m=exp(-cx/a) => u=exp(-cx/a) *f(ay-bx)
2) m=exp(-cy/b) => u=exp(-cy/b)*f(ay-bx)
I just can't see how both of them represent the same solution. I also tried linear combinations and it solves the PDE
u = exp(-cx/a) *f1(ay-bx) + exp(-cy/a) *f2(ay-bx) , where f1 and f2 are two arbitrary partially differentiable functions.
Questions :
- If we let m to be a function in both x and y then m also satisfy the PDE so does that mean that the General Soltion is a kind of Product ?
- If we let h be the solution to the homogeneous PDE a du/dx + b du/dy = 0 then The General Soltion is u=h*v where v is some kind of particular solution of the PDE a du/dx + b du/dy + cu =0 ??
- I just noticed that f(ay-bx) can be considered the solution of the homogeneous PDE and both exp(-cx/a) and exp (-by/a) can be considered as particluar solutions .... if that is really legit xD are there any particular solutions and the general solution is just the homogeneous multiplied by any linear combination of the particular solutions ??
Is that the only solution? And if not, how do I get all of them?
It’s the only solution!
Mmm the f(ay - bx) factor is by itself an infinity of solutions
@@drpeyam what is with weak solutions
I was wondering, since the equation is kinda symetric, e^-(c/b)y h(ay-bx) is also a solution. in which case we could consider a linear combination of both. But then taking one of the exponential as factor, we end up with the same form as the original solution...
@@riadsouissi Yes indeed, there are many questions that Dr Peyam might explore, ex. such solution for every Boundary Value problem(Neumann, Dirichlet, Rudin..) or for every bounded subset can there be solution ( For example Problems with different Cauchy Data, Periodic...). Maybe he could do a Video about the Burgers equation and the lax entropy condition :)
Partial by substitution
Great
*thumbnail* : sem cu, meu truta (╯︵╰,)
I thought this was by the method of fourier transform, but it wasn't! so I just tried it myself by using Fourier transform.
what do you think of my attempt? here: i.imgur.com/7ZW2ntU.png