This trick is applicable to the wave equation as well. For 1D wave-eq you actually get a very nice physical interpretation: The leftward wave travels to the boundary and gets reflected back, making interference with later leftward wave, while the rightward wave remains intact. Beside oddification, you can also 'evenify', making an even extension with the initial condition, which solves the half line PDE with Neumann boundary conditions such as {u_t = ku_xx for x positive, u'(0, t) = 0, u(x, 0) = f(x)}. Even functions automatically satisfy Neumann conditions the same way odd functions satisfy Dirichlet conditions, a pretty neat duality. Excellent video Dr. Peyam!
Very nice! I think we could develop the exponentials a little more, and get something like 1/sqrt(4kpi*t)*exp(-bla*x^2)*integral of Exp(-blabla*y^2)*sinh(something*y)*phi(y)dy. But I know that you often get exp(a*x)sinh(b*y) when solving Laplace eq. Would there be a relationship between the two problems?
Ah, nevermind. The original function would have been f(-x) for x < 0. So, writing -f(-x) would be a correct interpretation of an odd extension (observe that f(-x) for x < 0 would be an even and not an odd extension). Thanks for posting these videos.
You haven't really showed that u(x,0)=phi(x). Doing so would involve the nasty limit t->0 which is not at all trivial (Unless you take as a given that this limit of S(x,t) is just a representation of the delta distribution).
Do you think my way is the easiest to solve the gaussian integral? You just have to know subsitution (calc1) and one easy identity wich is really easy to proff with a product.
@@mauricepanero I mean yes, but if you want to get anywhere in these kinds of videos you have to take some things as known otherwise you're stuck forever explaining basic things. You can just check that it works out. Let f be an even function and g be an odd function. Then f(-x) = x and g(-x) = -g(x) (for all x) so fg(-x) = f(-x)g(-x) = f(x)(-g(x)) = -f(x)g(x) = - fg(x) so fg is an odd function.
This trick is applicable to the wave equation as well. For 1D wave-eq you actually get a very nice physical interpretation: The leftward wave travels to the boundary and gets reflected back, making interference with later leftward wave, while the rightward wave remains intact.
Beside oddification, you can also 'evenify', making an even extension with the initial condition, which solves the half line PDE with Neumann boundary conditions such as {u_t = ku_xx for x positive, u'(0, t) = 0, u(x, 0) = f(x)}. Even functions automatically satisfy Neumann conditions the same way odd functions satisfy Dirichlet conditions, a pretty neat duality.
Excellent video Dr. Peyam!
Proof by laziness - love it! Would you ever get around doing a proof by procrastination? :)
I always keep postponing it 😂
Can we get introduction to the fractional fourier tranform
There’s already a video on that, check out the video Half Derivative and Fourier Transform
Begone, Φ_odd!
GOOD JOB DR. PEYAM!
Very nice!
I think we could develop the exponentials a little more, and get something like 1/sqrt(4kpi*t)*exp(-bla*x^2)*integral of Exp(-blabla*y^2)*sinh(something*y)*phi(y)dy.
But I know that you often get exp(a*x)sinh(b*y) when solving Laplace eq. Would there be a relationship between the two problems?
What prerequisites should I have down so i can understand the heat equation and apply it?
Just watch the playlist from the beginning :)
@@drpeyam Thank you so much and I can't wait to watch more of your videos!!!
As odd functions are defined by f(-x)= -f(x), shouldn't it be -f(x) if x
Ah, nevermind. The original function would have been f(-x) for x < 0. So, writing -f(-x) would be a correct interpretation of an odd extension (observe that f(-x) for x < 0 would be an even and not an odd extension). Thanks for posting these videos.
Dr.Peyam you're badass 🔥
Awesome like always 👌🏻
Good job...👍👌
thank you
You haven't really showed that u(x,0)=phi(x). Doing so would involve the nasty limit t->0 which is not at all trivial (Unless you take as a given that this limit of S(x,t) is just a representation of the delta distribution).
Do you think my way is the easiest to solve the gaussian integral? You just have to know subsitution (calc1) and one easy identity wich is really easy to proff with a product.
Gaussian Integral th-cam.com/play/PLJb1qAQIrmmCgLyHWMXGZnioRHLqOk2bW.html
Thanks
Can be solved via the joint Laplace and Fourier transformation....
Fodd, like the odd fella, lord farquaad, in shrek.
Khan Tran Hahahaha
thank you !!!
Shirly an even PLUS an odd is odd; not even TIMES an odd?
If you are talking about integers, yes. What he has in the video are even/odd *functions* , though, so in this case odd times even is indeed odd.
@@NAMEhzj Why? How do you prove this? Somewhat confusing to the novice don't you think?
@@mauricepanero I mean yes, but if you want to get anywhere in these kinds of videos you have to take some things as known otherwise you're stuck forever explaining basic things.
You can just check that it works out.
Let f be an even function and g be an odd function.
Then f(-x) = x and g(-x) = -g(x) (for all x)
so fg(-x) = f(-x)g(-x) = f(x)(-g(x)) = -f(x)g(x) = - fg(x)
so fg is an odd function.
i can odd!