- #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: [itex]\exists C_\alpha\in\mathbb R_+[/itex] s.t. [itex]\forall w\in\mathbb R^2_+[/itex]
[tex]\begin{align*}
\lVert u(t,w)-u(t,w')\rVert^2
\leq C_\alpha^2\lVert w-w'\rVert^{2\alpha}
\end{align*}[/tex]
where [itex]w=(y,z)[/itex] and [itex]\alpha=\beta\gamma[/itex].
We get an upper bound for each (squared) component of [itex]u(w)[/itex] by applying the Holder continuity conditions:
[tex]\begin{align*}
|u_x(t,w)-u_x(t,w')|^2&=|u_x(y-tu_y(z))-u_x(y'-tu_y(z'))|^2 \\
&\leq C_\gamma^2|(y-tu_y(z))-(y'-tu_y(z'))|^{2\gamma} \\
&\leq C_\gamma^2|(y-y')-t(u_y(z)-u_y(z'))|^{2\gamma} \\
&\leq C_\gamma^2|(y-y')-tC_\beta|z-z'|^\beta|^{2\gamma} \\
|u_y(t,w)-u_y(t,w')|^2&=|u_y(z)-u_y(z')|^2 \\
&\leq C_\beta^2|z-z'|^{2\beta} \\
|u_z(t,w)-u_z(t,w')|^2&=|c-c|^2 \\
&=0.
\end{align*}[/tex]
The RHS of the inequality we want to prove is bound by the sum of the bounds on each component:
[tex]\begin{equation*}
\lVert u(t,w)-u(t,w')\rVert^2
\leq C_\gamma^2[(y-y')-tC_\beta|z-z'|^\beta]^{2\gamma} + C_\beta^2|z-z'|^{2\beta} .
\end{equation*}[/tex]
Now, in order to prove our thesis, we need to show that [itex]\exists C_\alpha\in\mathbb R_+[/itex] such that
[tex]\begin{equation*}
C_\gamma^2[Y-tC_\beta|Z|^\beta]^{2\gamma} + C_\beta^2|Z|^{2\beta}
\leq C_\alpha^2[Y^2+Z^2]^{\beta\gamma}
\end{equation*}[/tex]
where we let [itex]Y=y-y', \; Z=|z-z'|[/itex].
Unfortunately, the last inequality is wrong - as it is easy to spot by letting [itex]Z=0[/itex].
[tex]\begin{align*}
\lVert u(t,w)-u(t,w')\rVert^2
\leq C_\alpha^2\lVert w-w'\rVert^{2\alpha}
\end{align*}[/tex]
where [itex]w=(y,z)[/itex] and [itex]\alpha=\beta\gamma[/itex].
We get an upper bound for each (squared) component of [itex]u(w)[/itex] by applying the Holder continuity conditions:
[tex]\begin{align*}
|u_x(t,w)-u_x(t,w')|^2&=|u_x(y-tu_y(z))-u_x(y'-tu_y(z'))|^2 \\
&\leq C_\gamma^2|(y-tu_y(z))-(y'-tu_y(z'))|^{2\gamma} \\
&\leq C_\gamma^2|(y-y')-t(u_y(z)-u_y(z'))|^{2\gamma} \\
&\leq C_\gamma^2|(y-y')-tC_\beta|z-z'|^\beta|^{2\gamma} \\
|u_y(t,w)-u_y(t,w')|^2&=|u_y(z)-u_y(z')|^2 \\
&\leq C_\beta^2|z-z'|^{2\beta} \\
|u_z(t,w)-u_z(t,w')|^2&=|c-c|^2 \\
&=0.
\end{align*}[/tex]
The RHS of the inequality we want to prove is bound by the sum of the bounds on each component:
[tex]\begin{equation*}
\lVert u(t,w)-u(t,w')\rVert^2
\leq C_\gamma^2[(y-y')-tC_\beta|z-z'|^\beta]^{2\gamma} + C_\beta^2|z-z'|^{2\beta} .
\end{equation*}[/tex]
Now, in order to prove our thesis, we need to show that [itex]\exists C_\alpha\in\mathbb R_+[/itex] such that
[tex]\begin{equation*}
C_\gamma^2[Y-tC_\beta|Z|^\beta]^{2\gamma} + C_\beta^2|Z|^{2\beta}
\leq C_\alpha^2[Y^2+Z^2]^{\beta\gamma}
\end{equation*}[/tex]
where we let [itex]Y=y-y', \; Z=|z-z'|[/itex].
Unfortunately, the last inequality is wrong - as it is easy to spot by letting [itex]Z=0[/itex].