Lecture 19

[Aida Razi]

In heat equation: We know IFT of e^ (-(ξ^2)/2) is (√2pi)e (-(x^2)/2). Then when we try to find IFT of e^ (-(ktξ^2)), we scale ξ to √(2kt)ξ and therefore x scale to √(2kt)x. But in the lecture note, it is mentioned that x scale to (2kt)^(-1/2)x. I believe it should be just square root of 2kt and not (-square root) of 2kt.

[Victor Ivrii]

No, if you scale $x\mapsto \sigma x$ you scale $\xi \mapsto \sigma^{-1}\xi$ to keep $x\xi$ the same..

[Miranda Jarvis]

Just to clarify, I'm assuming that in the lead up to equation 25 it was meant to be $-|\xi|^{-1}e^{-|\xi|y}$ not $x$ and $i$  . . .

[Miranda Jarvis]

I'm sorry to nit pick but as someone who is having a hard enough time following much of this material, I was wondering in the lines shown in the screen capture attached (the lead up to equation 39 is there an extra $i$ in the denominator of the first equation? I think this is supposed to be equal to the second term in the right side of equation 38 . . .

[Victor Ivrii]

I'm sorry to nit pick but as someone who is having a hard enough time following much of this material, I was wondering in the lines shown in the screen capture attached (the lead up to equation 39 is there an extra $i$ in the denominator of the first equation? I think this is supposed to be equal to the second term in the right side of equation 38 . . .

You are correct. Fixed

[Miranda Jarvis]

again, not to be a pain, but is it in fact supposed to be $\xi$ not $xi$ in the lead up to equation 25?

[Victor Ivrii]

again, not to be a pain, but is it in fact supposed to be $\xi$ not $xi$ in the lead up to equation 25?

You mean after (5)? Yes there were missing math delimiters--fixed