Loss of Hölder continuity by solutions of the Euler equations

Coelum
Messages
97
Reaction score
32
Homework Statement
Show that if the initial conditions [itex]u(0,x), x\in\mathbb R^3[/itex] for a three-dimensional shear flow are such that [itex]u_x(0,y,z)[/itex] is [itex]\gamma[/itex]-Holder continuous and [itex]u_y(z)[/itex] is [itex]\beta[/itex]-Holder continuous, then the solution [itex]u(t,x)[/itex] for [itex]t\neq 0[/itex] need not be any smoother than [itex]\alpha=\beta\gamma<\min\{\beta,\gamma\}[/itex].
[This is part of project 8.5 from Craig's "A Course on Partial Differential Equations", AMS 2018]
Relevant Equations
A function [itex]v(y):\mathbb R^d\rightarrow\mathbb R^d [/itex] is a Holder continuous function of [itex]y[/itex] with Holder exponent [itex]0<\beta<1[/itex] if [itex]\exists C_\beta\in\mathbb R: \forall y,y'\in\mathbb R^d[/itex]
[tex]\begin{equation*}
\lVert v(y)-v(y')\rVert\leq C_\beta\lVert y-y'\rVert^\beta.
\end{equation*}[/tex]
A shear flow in 3D takes the form
[tex]\begin{align*}
u_x(t,x,y,z)&=u_x(y - t u_y(z)) \\
u_y(t,x,y,z)&=u_y(z) \\
u_z(t,x,y,z)&=c
\end{align*}[/tex]
where c is a constant.
Our thesis can be restated as follows: \exists C_\alpha\in\mathbb R_+ s.t. \forall w\in\mathbb R^2_+
\begin{align*}<br /> \lVert u(t,w)-u(t,w&#039;)\rVert^2<br /> \leq C_\alpha^2\lVert w-w&#039;\rVert^{2\alpha}<br /> \end{align*}
where w=(y,z) and \alpha=\beta\gamma.
We get an upper bound for each (squared) component of u(w) by applying the Holder continuity conditions:
\begin{align*}<br /> |u_x(t,w)-u_x(t,w&#039;)|^2&amp;=|u_x(y-tu_y(z))-u_x(y&#039;-tu_y(z&#039;))|^2 \\<br /> &amp;\leq C_\gamma^2|(y-tu_y(z))-(y&#039;-tu_y(z&#039;))|^{2\gamma} \\<br /> &amp;\leq C_\gamma^2|(y-y&#039;)-t(u_y(z)-u_y(z&#039;))|^{2\gamma} \\<br /> &amp;\leq C_\gamma^2|(y-y&#039;)-tC_\beta|z-z&#039;|^\beta|^{2\gamma} \\<br /> |u_y(t,w)-u_y(t,w&#039;)|^2&amp;=|u_y(z)-u_y(z&#039;)|^2 \\<br /> &amp;\leq C_\beta^2|z-z&#039;|^{2\beta} \\<br /> |u_z(t,w)-u_z(t,w&#039;)|^2&amp;=|c-c|^2 \\<br /> &amp;=0.<br /> \end{align*}
The RHS of the inequality we want to prove is bound by the sum of the bounds on each component:
\begin{equation*}<br /> \lVert u(t,w)-u(t,w&#039;)\rVert^2 <br /> \leq C_\gamma^2[(y-y&#039;)-tC_\beta|z-z&#039;|^\beta]^{2\gamma} + C_\beta^2|z-z&#039;|^{2\beta} .<br /> \end{equation*}
Now, in order to prove our thesis, we need to show that \exists C_\alpha\in\mathbb R_+ such that
\begin{equation*}<br /> C_\gamma^2[Y-tC_\beta|Z|^\beta]^{2\gamma} + C_\beta^2|Z|^{2\beta} <br /> \leq C_\alpha^2[Y^2+Z^2]^{\beta\gamma}<br /> \end{equation*}
where we let Y=y-y&#039;, \; Z=|z-z&#039;|.
Unfortunately, the last inequality is wrong - as it is easy to spot by letting Z=0.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top