I am interested in Airy's equation $$\frac{\partial u}{\partial t}(t,x)=-\frac{\partial^3 u}{\partial x^3}(t,x)$$ on a bounded or semi-bounded domain, e.g. on $(-\infty,0)$. In order to obtain a group of isometries as propagator in $L^2(-\infty,0)$, two boundary conditions have to imposed - e.g., $u'(0)=0$ and either $u(0)=0$ or $u''(0)=0$. But then it seems that the spectrum of the operator is far too large, so I doubt that this operator does not generate a semigroup in $L^2(-\infty,0)$ at all, regardless of the imposed boundary conditions. Is it so? Has anybody ever proved well-posedness of the above equation in some reasonable function or distribution space?
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2$\begingroup$ On a bounded interval, $A=iD^3$ has self-adjoint realizations and then $e^{itA}$ is a unitary group. On a half line, $D^3$ has unequal deficiency indices $(2,1)$, so this approach won't work. $\endgroup$– Christian RemlingCommented Jan 13, 2016 at 18:46
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$\begingroup$ On a bounded interval, the deficiency index is $3$, so you need this many boundary conditions (not two). $\endgroup$– Christian RemlingCommented Jan 13, 2016 at 18:54
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$\begingroup$ @ChristianRemling Indeed! There are several works suggesting that two b.c. have to be imposed on the right and one on the left endpoints; but in these works I can never find a "clean" well-posedness theorem in a "nice" function space (ideally, a Sobolev or Lebesgue space, but I'd be happy even with one space from Triebel's zoo; ideally, by means of a unitary $C_0$-group). Remarkably, the equation on $\mathbb R$ is governed by a propagator that is just the convolution of an integral kernel based on Airy's function and the Fourier transform of the initial data, so a group might exist after all. $\endgroup$– Delio MugnoloCommented Jan 13, 2016 at 20:04
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$\begingroup$ Yes, on $\mathbb R$ the $L^2$ theory is straightforward: $A=iD^3$ is essentially self-adjoint on $C_0^{\infty}$, so again $e^{itA}$ is a unitary group. $\endgroup$– Christian RemlingCommented Jan 13, 2016 at 21:57
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2 Answers
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The operators $A=-\,\partial_x^3$ with domain $\mathscr D(A) = \left(H^3\cap H_0^2\right)(\mathbb R_-)$ and $A^\ast=\partial_x^3$ with domain $\mathscr D(A^\ast) = \left(H^3\cap H_0^1\right)(\mathbb R_-)$ are $m$-dissipative on $L^2(\mathbb R_-)$. This follows from $\Re\langle Au,u\rangle =0$ for $u\in\mathscr D(A)$ and $\Re\langle A^\ast v,v\rangle =-\,\frac12\,|v'(0)|^2\leq 0$ for $v\in\mathscr D(A^\ast)$. Therefore, $A$ and $A^\ast$ generate $C_0$-contraction semigroups on $L^2(\mathbb R_-)$, by the Lumer-Phillips theorem. Moreover, the semigroup $\{e^{tA}\}_{t\geq0}$ consists of isometries (though not of unitary operators for $t>0$, as $iA$ is not selfadjoint, see Christian's comment above).
One can also directly solve the initial-boundary value problem $$ \left\{ \enspace \begin{aligned} & u_t + u_{xxx} = f(t,x), & (t,x)\in \mathbb R_+ \times \mathbb R_-, \\ & u\bigr|_{x=0} = h_0(t), \enspace u_x\bigr|_{x=0} = h_1(t), \\ & u\bigr|_{t=0} = u_0(x) \end{aligned} \right. $$ (and similar for the operator $\partial_t-\partial_x^3$ with one in place of two boundary conditions), e.g., by using the Laplace transform with respect to $x$. See N. Hayashi, E. Kaikina, Nonlinear Theory of Pseudodifferential Equations on a Half-line, North-Holland, 2004, which contains a chapter on linear problems.
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This is a very interesting question and I do not know the answer. I would start at something like
N. A. Larkin, Correct initial boundary value problems for dispersive equations, J. Math. Anal. Appl. 344 (2008), no. 2, 1079--1092.
answered Jan 13, 2016 at 18:42