My question is to determine whether the PDE $\partial_t u = P(\partial_x) u$, with $2\pi$-periodic boundary conditions, for a polynomial $P$, is well-posed; this depends on the polynomial, and my three specific cases are the following:
- $P(y) = y^3$
- $P(y) = y^2$ (heat equation)
- $P(y) = iy^2$ (Schrödinger equation)
Taking Fourier transforms of the original equation, we obtain the general solution $$ u(t,x) = \sum_{m \in \Bbb Z} e^{tP(im)+imx} \hat u(0,m). $$
I then tried considering two solutions $u$ and $v$, and look at their $L^2$ norm difference, $$ \| u(t) - v(t) \| _{L^2} = \int_{-\pi}^{\pi} |u(t,x) - v(t,x)|^2 dx,$$ and then try to manipulate that (using Parseval-Plancherel and orthogonality of complex exponentials) to get a $\le$-inequality with the above on the LHS and $$c \|u(0)-v(0)\| _{L^2}$$ on the RHS, for some constant $c$; unfortunately, I was unable to. Any help would be most appreciated!
Please note: I realise that there's lots on information on the web about well-posedness of PDEs; I hope that this isn't a duplicate, but I can't check every single question that just says "Is this PDE well-posed?" =P
We have that $\def\norm#1{\left\|#1\right\|}$ \begin{align*} \norm{u(t)-v(t)}^2_{L^2} &= 2\pi\norm{\widehat{u(t)} - \widehat{v(t)}}^2_{\ell^2(\def\Z{\mathbb Z}\Z)}\\ &= 2\pi\sum_{m\in\Z}\def\abs#1{\left|#1\right|}\abs{e^{tP(im)}\bigl(\hat u(0,m)-\hat v(0,m)\bigr)}^2 \end{align*} Now note that if $P$ maps the imaginary axis into itself, that is $P[i\mathbb R]\subseteq i\mathbb R$, as do $y^3$ and $iy^2$, but not $y^2$, we have $P(im) = i\alpha(P, m)$, for some real number $\alpha(P, m)$, giving \begin{align*} \norm{u(t)-v(t)}^2_{L^2} &= 2\pi\sum_{m\in\Z}\abs{e^{tP(im)}\bigl(\hat u(0,m)-\hat v(0,m)\bigr)}^2\\ &= 2 \pi\sum_{m\in\Z}\abs{e^{ti\alpha(P,m)}}^2\abs{\hat u(0,m)-\hat v(0,m)}^2\\ &= 2\pi\sum_{m\in\Z}\abs{\hat u(0,m)-\hat v(0,m)}^2\\ &= \norm{u(0) - v(0)}^2_{L^2} \end{align*} If we have the weaker property $P[i\mathbb R] \subseteq \{z \in \mathbb C \mid \Re z \le M\}$ for some $M$, we get along the same line of thought $$ \norm{u(t)-v(t)}^2_{L^2} \le e^{tM\cdot 2} \norm{u(0)-v(0)}^2_{L^2} $$ This works for the heat equation, as $(ix)^2 = -x^2 \le 0$.
Addendum: For $u \in L^2([0,2\pi])$, $u(x) = \sum_{m\in \Z} e^{imx}\hat u(m)$, we have due to the orthogonality of the exponential functions \begin{align*} \norm{u}_{L^2}^2 &= \norm{\sum_{m\in Z}e^{im\cdot}\hat u(m)}_{L^2}^2\\ &= \sum_{m,n} \int_{0}^{2\pi} e^{imx}\hat u(m)e^{-inx}\overline{\hat u(n)}\, dx\\ &= \sum_n \int_0^{2\pi} e^{0}\hat u(n)\overline{\hat u(n)}\, dx\\ &= 2\pi \sum_ n \abs{\hat u(n)}^2\\ &= 2\pi \norm{\hat u}_{\ell^2(\Z)}^2 \end{align*}