- #1
Coelum
- 97
- 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')\rVert^2<br /> \leq C_\alpha^2\lVert w-w'\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')|^2&=|u_x(y-tu_y(z))-u_x(y'-tu_y(z'))|^2 \\<br /> &\leq C_\gamma^2|(y-tu_y(z))-(y'-tu_y(z'))|^{2\gamma} \\<br /> &\leq C_\gamma^2|(y-y')-t(u_y(z)-u_y(z'))|^{2\gamma} \\<br /> &\leq C_\gamma^2|(y-y')-tC_\beta|z-z'|^\beta|^{2\gamma} \\<br /> |u_y(t,w)-u_y(t,w')|^2&=|u_y(z)-u_y(z')|^2 \\<br /> &\leq C_\beta^2|z-z'|^{2\beta} \\<br /> |u_z(t,w)-u_z(t,w')|^2&=|c-c|^2 \\<br /> &=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')\rVert^2 <br /> \leq C_\gamma^2[(y-y')-tC_\beta|z-z'|^\beta]^{2\gamma} + C_\beta^2|z-z'|^{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', \; Z=|z-z'|.
Unfortunately, the last inequality is wrong - as it is easy to spot by letting Z=0.
\begin{align*}<br /> \lVert u(t,w)-u(t,w')\rVert^2<br /> \leq C_\alpha^2\lVert w-w'\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')|^2&=|u_x(y-tu_y(z))-u_x(y'-tu_y(z'))|^2 \\<br /> &\leq C_\gamma^2|(y-tu_y(z))-(y'-tu_y(z'))|^{2\gamma} \\<br /> &\leq C_\gamma^2|(y-y')-t(u_y(z)-u_y(z'))|^{2\gamma} \\<br /> &\leq C_\gamma^2|(y-y')-tC_\beta|z-z'|^\beta|^{2\gamma} \\<br /> |u_y(t,w)-u_y(t,w')|^2&=|u_y(z)-u_y(z')|^2 \\<br /> &\leq C_\beta^2|z-z'|^{2\beta} \\<br /> |u_z(t,w)-u_z(t,w')|^2&=|c-c|^2 \\<br /> &=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')\rVert^2 <br /> \leq C_\gamma^2[(y-y')-tC_\beta|z-z'|^\beta]^{2\gamma} + C_\beta^2|z-z'|^{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', \; Z=|z-z'|.
Unfortunately, the last inequality is wrong - as it is easy to spot by letting Z=0.