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A: Solution for Cauchy Problem $u_t-u_{xx} = 0$ belongs to the Gevrey class of order $1/2$

Calvin KhorIn this answer, all norms are spatial norms (i.e. there is no integration/supremum wrt the parameter $t_0$ that was already fixed in the question, and I have dropped the subscript $0$). The unique solution to the heat equation with nice initial data is well known as the convolution with the Gaus...

Lucas Zanella
Lucas Zanella
Can you be more specific in the norm of the derivative of a convolution and Cauchy Schwarz being applied to it? I don't understand how these $L^2$ norms appear, I think I don't have background on these things
Calvin Khor
Calvin Khor
@LucasZanella I've added some lines, does it help?
Lucas Zanella
Lucas Zanella
what do you mean when you write $L_x^2$ for example? What is the sub $x$ for?
Calvin Khor
Calvin Khor
@LucasZanella It is to emphasise that the integral defining the norm is taken in the $x$ variable. e.g. if $y,z$ is fixed and $g(x) = f(x,y,z)$ then $$ \|f(x,y,z)\|_{L^p_x} := \|g\|_{L^p}$$ ps - If someone has better notation that does not involve replacing variables with dots or $\mapsto$ or defining another function (the first two are horrendously ugly) I'm all ears
Lucas Zanella
Lucas Zanella
Thanks, I understoodd. But now what does $L^2$ alone mean? In which variable?
Calvin Khor
Calvin Khor
14:35
@LucasZanella In this answer, if I didn't subscript, then its always in the spatial variable (and in each one there is a unique spatial variable so there is no ambiguity). Which line are you talking about?
Lucas Zanella
Lucas Zanella
From line $4$ to $5$ you go from $L_y^2$ to $L^2$
I was in doubt because $t$ can also vary so I could take the norm in $t$, but apparently you made in $x$
Calvin Khor
Calvin Khor
@LucasZanella this is the translation symmetry of the Lebesgue integral $\int_{\mathbb R^n} f(y+c) dy =\int_{\mathbb R^n} f(y) dy $. For the second point fair enough, since you were using $\sup$ over $K\subset \mathbb R$ i thought it was obvious that $L^\infty = L^\infty(\mathbb R)$ but I can see that this is not clear. Will edit this in
Lucas Zanella
Lucas Zanella
I finally understood all norms, but how did you do that Chain Rule? I can't reproduce here where the $\sqrt{}$ appears and where $\phi_{1/4}$ appears (I know what $\phi$ is)
Calvin Khor
Calvin Khor
@LucasZanella ignoring the constant in the front $C(t) = $something to do with $t$ and $\pi$, we have derivatives of $\phi_{1/4} ( \lambda x)$. Each derivative pulls out one copy of $\lambda$, e.g. $$ \phi_{1/4}(\lambda x) ' = \lambda \phi_{1/4}'(\lambda x) $$ I may have gotten the precise value 1/4 wrong...but its not really important what this number is
Lucas Zanella
Lucas Zanella
How does $\pi$ appears? I think I understood it terribly wrong, because the first derivative with respect to $x$ is $-\frac{2x}{4t}e^{-\frac{x^2}{4t}}$
Calvin Khor
Calvin Khor
14:35
@LucasZanella I do not recommend actually expanding out what the derivative is, write the function $\phi_t$ in terms of $\phi_{1/4}$ and use chain rule. The constant $C(t)$ is the $1/\sqrt{4\pi t}$ thing times also $\sqrt{\pi}$ to get it to match with $\phi_{1/4}$...so you're right actually there's no $\pi$ but it really doesn't matter
Lucas Zanella
Lucas Zanella
I arrived at $\|\partial^n_x\phi_t\|_{L^2} = \|\partial_x^n \phi_{1/4}\|_{L^2}$ does that mean I can simply say this is less than $C^{n+1}\|\partial_x^n \phi_{1/4}\|_{L^2}$ for some $C^{n+1}$ greater than $1$?
Calvin Khor
Calvin Khor
@LucasZanella really? Well the worst it can do is pull out an nth power. I don't think it should exactly cancel for every n but as long as you don't get something that grows faster it's ok yes
Lucas Zanella
Lucas Zanella
Take a look at my update, I don't know what you said about canceling for every $n$ but if I do the rule $\int_{\mathbb R} f(\lambda x)^2 dx = \frac1\lambda \int_{\mathbb R} f^2(y) dy$ I get $\phi_{1/4}$
$\phi_{1/4} = e^{-x^2}/\sqrt{\pi}$ right?
Calvin Khor
Calvin Khor
Yes, and you compare with $$\phi_t(x) = \frac1{\sqrt{\pi}}\frac1{\sqrt{4 t}} \exp\left(-(x/\sqrt{4t})^2\right)$$
@LucasZanella your computation in the update is as if you were integrating $$ (\partial_x^n \phi_{1/4} )(\lambda x) $$ which is not the same as integrating $$ (\partial_x^n \phi_{t} )(x) $$
@LucasZanella I should have said, "not the same as integrating $\partial_x^n [\phi_t(x)]$". I think this is clearer. Note $$ \phi_t(x) = \phi_{1/4}(\lambda x)$$ where $\lambda = 1/\sqrt{4t}$. Then chain rule gives $$ \partial_x\phi_t(x)=\partial_x[\phi_{1/4}(\lambda x)] = \lambda [\partial_x\phi_{1/4}](\lambda x)$$ do it $n$ times and then do the $L^2$ norm scaling
Lucas Zanella
Lucas Zanella
I've just arrived now at $(\frac{1}{\sqrt{4\pi t}}\int|\partial_x^n e^{-x^2}|^2)^{1/2}$ which is almost $||\partial_x^n(x)\phi_{1/4}||$, but I don't get how $C^{n+1}$ appears
Calvin Khor
Calvin Khor
14:35
@LucasZanella, no, that's not right, please see my edit. In words: the rescaling of the derivative is not the derivative of the rescaling. This is the job of chain rule. $\phi_t = \phi_{1/4} \circ L$. $\phi_t' = \phi_{1/4}'\circ L $, times $L'$. Here $L$ is a linear scaling.
@LucasZanella $$ d/dx (e^{-(cx)^2}) = -2c (cx) e^{-(cx)^2} $$ $$ d/d\tilde x ( e^{-\tilde x^2} )|_{\tilde x= cx} = 2 cx e^{-(cx)^2} $$ The two functions differ by a multiplicative constant, as predicted by chain rule. you perform the same $L^p$ norm rescaling on both sides after this, so there is no hope to fix this for any $p$. If you don't understand this, I'm sorry but I feel like I have run out of explanations, please try to compute some examples.
@LucasZanella I lied, I thought of a simple example to compute that should convince you. set $f(x) = \exp(-2x)$. $f'(x) = -2\exp(2x)$ and set $g(x) = \exp(-x)$ so that $f = g(2x)$. on $(0,\infty)$, $$\int_0^\infty \exp(-kx) dx = \frac1k$$ just Lebesgue scaling for $n=0$ is ok, $$ \| f\|^2_{L^2} = \int_0^\infty g(2x)^2 dx = \frac14 = \frac12\int_0^\infty g(x)^2 dx$$ chain rule needed for $n=1$. this is like $\|\partial_x \phi_t\|^2_{L^2}:$ $$\|\partial_x f\|_{L^2}^2 = \int_0^\infty |\partial_x f(x)|^2 dx = 2^2 \int_0^\infty f^2 dx = 1 $$
@LucasZanella but if I proceed like your update, (in red is wrong) $$\|\partial_x f\|^2_{L^2} = \int_0^\infty |\partial_x f(x)|^2 dx \color{red}{\overset{?}{=} \int_0^\infty |(\partial_x g)(2x)|^2 dx} = \frac1{2}\int_0^\infty |\partial_x g(x)|^2 dx = \frac12$$
(too late to edit, that last equal sign should be $= 1/4$)

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