# Does there exist the applicability limit
of PDE to describe physical phenomena?
— A personal survey of
Quantization, QED, Turbulence —

Atsushi Inoue  Dedication to Masuda san for his warm consideration

(Date: , based on a talk on 19 March 2023)

###### Abstract.

What does it mean to study PDE(=Partial Differential Equation)? How and what
to do “to claim proudly that I’m studying a certain PDE”? Newton mechanic uses
mainly ODE(=Ordinary Differential Equation) and describes nicely movements of
Sun, Moon and Earth etc. Now, so-called quantum phenomenum is described by,
say Schrödinger equation, PDE which explains both wave and particle characters
after quantization of ODE. The coupled Maxwell-Dirac equation is also
“quantized” and QED(=Quantum Electro-Dynamics) theory is invented by
physicists. Though it is said this QED gives very good coincidence between
theoretical111By the way, without “solving” equation mathematically, how they
obtain theoretical value? After H. Lewy’s example, we can’t believe so naively
physicist’s claim that there must exist a solution of that PDE if we may
derive that equation suitably from physical phenomena.and experimental
observed quantities, but what is the equation corresponding to QED? Or, is it
possible to describe QED by “equation” in naive sense?

Author: alias atlom=a tiny little old mathematician, or
suman=superman$-$“per”, “per” pronounces “pa-” in Japanese, means “bone head”
or “stupid”, therefore deleted and finally “suman” stands for “sorry” in
Japanese, curious!

###### Contents

  1. 1 Introduction with a brief personal history of atlom
  2. 2 Quantization and PIM(=Path Integral Method)
    1. 2.1 The beginning of PIM
    2. 2.2 Quantization à la Fujiwara
    3. 2.3 Hamilton formulation of PIM
  3. 3 Another interpretation of method of characteristics, explained through the simplest case
  4. 4 Necessity of superanalysis
    1. 4.1 Reasons of necessity
    2. 4.2 Regarding matrices as differential operators!
    3. 4.3 Chi’s example
      1. 4.3.1 Chi’s proof
      2. 4.3.2 Direct construction of a solution of Hamilton-Jacobi equation
      3. 4.3.3 Continuity equation
      4. 4.3.4 Quantization
      5. 4.3.5 Calculation of $\varphi$
    4. 4.4 An application to Random Matrix Theory
  5. 5 Some examples of FDE(-Functional Derivative Equation)
    1. 5.1 A Schwinger-Dyson equation
      1. 5.1.1 A derivation of A Schwinger-Dyson equation of first order
      2. 5.1.2 Reformulation and calculation
    2. 5.2 Hopf equation
      1. 5.2.1 Differential geometrical expression of Navier-Stokes equation
      2. 5.2.2 Hopf’s motivation
      3. 5.2.3 Derivation of Hopf equation
      4. 5.2.4 How to give the meaning to the trace of 2nd order functional derivative
  6. A Some questions on Euler or Navier-Stokes equations

##  1\. Introduction with a brief personal history of atlom

Though I attended with the TV course of collegial physics222A part of NBC’s
educational program “Continental Classroom” in1958–1963, which is broadcasted
from NHK in 1959–1961 as “physics in nuclear age” with explanation in Japanese
in my high school time, as I felt the score of Shingaku-Furiwake333After
18months of general education, each student should decide to proceed in which
department was insufficient to go to physics, I had no alternative but to
proceed to applied mathematics because for me at that time, two branches seem
to treat analogous objects. On the other hand, my career as a mathematician
has been started rather accidentally. Despite I couldn’t imagine I might be an
“ordinary” salaried man444Seemingly, I’m not good at temporarily going along
with “respectable person” as an amenable member of organization., but
fortunately perhaps, just after finished my master course, I was proposed a
research assistant post of department of mathematics. Such posts made
increased suddenly because of “Science and Engineering Faculty Expansion Plan”
by the Japanese government which began after the Sputnik shock suffered by
USA. By the way, as is said “The teacher of the university is not resigned as
once if I do it” or “Beggars and monks can’t quit after three days”, I
couldn’t only resign from research assistant post but also had an opportunity
to study abroad as a Boursier.

But after several years at around age thirties, I was offered a new position.
At that time, continuously getting salary as researcher, I should do
mathematics more seriously, so I felt, and I started to study not only linear
PDE but also non-linear PDE. Surely, Hörmander’s works on linear PDE are
overwhelming at that time.

So to restart with a little mathematician, I questioned naively why the
Navier-Stokes equation is so famous and I wondered if this equation is
genuinely good enough or worth studying? If it is so good as equation, it
should be invariant under change of variables? At that time, the initial and
boundary value problem in the time dependent domain for the Navier-Stokes
equation is studied by Fujita and Sauer [40] by penalty method. Since I
couldn’t appreciate their method fully555Whether their penalty method works
when $\cup_{0\leq t\leq T}\Omega(t)\times\\{t\\}\subset{\mathbb{R}}^{d+1}$
when $\Omega(t)$ and $\Omega(t^{\prime})$ is not necessarily diffeomorphic
each other but with the same volume?, I find the change of variable formula
for vector field or differential 1-form with the help of Wakimoto [72], under
the condition that “gold fishes in the bowl don’t allow even kissing666That
is, assuming that the volume of $\Omega(t)$ is constant in $t$ and for any
$t,t^{\prime}$, $\Omega(t)$ and $\Omega(t^{\prime})$ are diffeomorphic each
other”! Not only this, even turbulent phenomena are believed to be governed by
this same equation, why so? Though this equation is derived from conservations
of momentum and mass by observing “laminar flow”, why so easily believed the
totally different looking phenomena, “turbulent flows”, are also governed by
the same equation? Not only this, the viscosity occurs after or before viewing
the discrete molecular structure of water as a continuum777A physicist says
naively that viscosity comes from particle structure of water, they don’t
bother when they, to treat fluids, started to regard particles as if
continuum.? Here it is also appropriate quoting Hopf saying:

> Statistical mechanics constructs certain “relevant” phase distributions
> which characterize the “typical” phase motions and which must be used for
> all statistical predictions. $\cdots\cdots$ $\cdots$, in statistical
> hydromechanics – the theory of highly turbulent fluid flow – the small scale
> on which the “fluid elements” interact seems to be of decisive importance,
> The relevant distributions based on this scale – the hypothetical
> Kolmogoroff distributions – must be mathematically very different from
> canonical distribution. So far all attempts to determine the relevant
> hydrodynamical phase distributions, at least to a sufficient degree of
> approximation, have met with considerable mathematical difficulties.

From these consideration, I wonder whether “is there applicability limit for
describing physical phenomena by PDE”? and what is the limit of regarding huge
number of water molecules as continuum?

As is well-known, now called “quantum phenomena” are not so well described
directly by ODE(Newton Mechanics), therefore we need to use PDE, Schrödinger
equation(in some sense, a quantized version of Newton Mechanics). Therefore,
at least logically, there will exist some phenomena assumed to be not well
describable using PDE.

For such phenomena, Gelfand not only proposed to use FDE888Someone wrote
somewhere that, “if there doesn’t exist Riemannian geometry or invariant
theory, theory of relativity, if there doesn’t exist boundary value theory,
wave mechanics, and if there doesn’t exist matrix theory, quantum mechanics
may not be invented? New physical theory is stimulated by precedent
mathematical method, or new mathematical method is invented from newly
recognized phyisical phenomena” (=Functional Derivative Equation) for
understanding turbulence and QED, but also questioned whether our existed
mathematics tools sufficient to describe these phenomena? I completely
empathize with his opinion, and I feel we need to develop theory of FDE999But
in general, we can’t give meaning higher order functional derivatives at each
point. Moreover, in infinite dimensional topological spaces, there doesn’t
exist Lebesgue-like measure which permit integration by parts [92]. This means
any trial to extend “A study of PDE by functional analytic method” seems
breakdown from the outset. In spite of this, physicists, using tools not yet
mathematically justified, get certain theoretical values and experimental
values with complicated and expensive experiments, and astonishingly these
values coincide many digits. This fact seems to imply something exist which is
not yet appreciated mathematically..

Apart from above, to treat initial value problem for linear hyperbolic systems
of PDE at that time, there is a trend to diagonalize that system with posing
conditions on properties of characteristic roots, as mathematical technique.
But I feel strange why we need to diagonalize system of PDE. I feel curious to
such treatise because not only until when such efforts continue but also there
exists not diagonalizable system of PDE. From my point of view, the necessity
of diagnalization is to apply existing theory of pseudo-differential equations
or more precisely, standard symbol calculus is only confined to scalar case.
Therefore, we need to treat matrix structure as it is, this is one motivation
to construct superanalysis(=analysis on superspace $\mathfrak{R}^{d|\ell}$ not
on ${\mathbb{R}}^{d}$).

Even almost 30 years passed after I started to concern with FDE or
superanalysis, these subjects never belong to the main stream of researchs in
mathematical society, at least in Japan. Therefore, it seems natural, at that
time, judges of KAKENHI(=Grants-in-Aid for Scientific Research) disregarded my
proposal concerning FDE and superanalysis almost completely, so I think. But
any way, none of them, even personally, asked me what are them? Finally, I
proposed to “make a room to accept appeals to the judgement of KAKENHI
examiners101010whose names are announced after one year later” which is also
neglected completely. Under these situations, Kŷuya Masuda supported me behind
the scenes.

In this note, mathematics which is disregarded by judges of KAKENHI, are
mentioned, because without doing so, these trials are in vain with my physical
lifespan.

Since originally, the descriptive ability of PDE and its reproducibility of
physical phenomena are main concern, it seems natural we wonder the difference
between the derivation of classical field equation and quantum field
mechanics.

In any way, it seems worth mentioning the following description which is given
in Chapter 9, Functional Methods, of Itzykson and Zuber:“Quantum Field Theory”
1979:

> “The path integral formalism of Feynman and Kac provides a unified view of
> quantum mechanics, field theory, and statistical models. Starting from the
> case of finitely many degrees of freedom it is generalized to include
> fermion systems and then extended to infinite systems. The steepest-descent
> method of integration exhibits the close relationship with classical
> mechanics and allows us to recover ordinary perturbation theory.”

##  2\. Quantization and PIM(=Path Integral Method)

###  2.1. The beginning of PIM

Following explanation is due to Feynman and Hibbs [35] but I cited here from
Albeverio and Hoegh-Krohn [2].

Let consider the representation formula for the solution of Schrödinger
equation on ${\mathbb{R}}^{d}$

$i\hbar\frac{\partial}{\partial
t}\psi(q,t)=-\frac{\hbar^{2}}{2M}\Delta\psi(q,t)+V(q)\psi(q,t),$

with the initial data $\psi(q,0)=\underline{\psi}(q)$. Decomposing

$H=H_{0}+V\quad\text{with}\quad H_{0}=-\frac{\hbar^{2}}{2M}\Delta,$

and assuming that $H$ is selfadjoint in $L^{2}({\mathbb{R}}^{d})$, we have the
solution $\psi(q,t)=e^{-i\hbar^{-1}tH}\underline{\psi}(q)$ by Stone’s theorem.
On the other hand, Lie-Trotter-Kato’s product formula asserts that even though
$[H_{0},V]\neq 0$, we have

$e^{-i\hbar^{-1}tH}=e^{-i\hbar^{-1}t(H_{0}+V)}=\lim_{n\to\infty}(e^{-i\hbar^{-1}(t/n)V}e^{-i\hbar^{-1}(t/n)H_{0}})^{n}.$

Since

$e^{-i\hbar^{-1}tH_{0}}u(q)=(2\pi i\hbar
t/M)^{-d/2}\int_{{\mathbb{R}}^{d}}dq^{\prime}\,e^{iM(q-q^{\prime})^{2}/(2\hbar
t)}u(q^{\prime}),$

we have

$e^{-i\hbar^{-1}(t/n)V}e^{-i\hbar^{-1}(t/n)H_{0}}u(q)=e^{-i\hbar^{-1}(t/n)V(q)}\int_{{\mathbb{R}}^{d}}dq^{\prime}\,e^{iM(q-q^{\prime})^{2}/(2\hbar
t/n)}u(q^{\prime}),$

we get,

$(2\pi i\hbar
t/M)^{-d/2}\int_{{\mathbb{R}}^{d}}dq^{\prime}\,e^{iM(q-q^{\prime})^{2}/(2\hbar
t)}u(q^{\prime})=(2\pi i\hbar
t/M)^{-dn/2}\int_{{\mathbb{R}}^{dn}}dq_{0}{\cdots}dq_{n-1}e^{i\hbar
S^{*}_{t}(q_{n},\cdots,q_{0})}u(q_{0})$

with $q^{\prime}=q_{0}$ and $q_{n}=q$, where

$S^{*}_{t}(q_{n},\cdots,q_{0})=\sum_{j=1}^{n}[\frac{M}{2}\frac{(q_{j}-q_{j-1})^{2}}{(t/n)^{2}}-V(q_{j})]\frac{t}{n}.$

Taking $\gamma^{*}(\tau)$ on $[0,t]$ as a zigzag path from $q_{0}$ at time
$0$, passing through $\gamma(\tau_{j})=q_{j}$ $j=0,{\cdots},n$ where
$\tau_{j}=j\frac{t}{n}$ and $q_{0},{\cdots},q_{n}$ are given points in
${\mathbb{R}}^{d}$. Feynman regards this $S^{*}_{t}(q_{n},\cdots,q_{0})$ as a
Riemann approximation for the classical action $S_{t}(\gamma)$ along the path
$\gamma(\tau)$:

$S_{t}(\gamma)=\int_{0}^{t}d\tau\,[\frac{M}{2}(\frac{d\gamma}{d\tau})^{2}-V(\gamma(\tau))].$

Assuming $\gamma^{*}(\tau)$ is an approximation for any classical path
belonging to the path space

$\Gamma_{(t,\underline{q},\bar{q})}=\\{\gamma\in
AC([0,t]:{\mathbb{R}}^{d});\gamma(0)=\underline{q},\gamma(t)=\bar{q}\\}$

and using assumed “Lebesgue-like measure”$d_{F}\gamma$ on
$\Gamma_{(t,\underline{q},\bar{q})}$, we express its solution as

$\psi(\bar{q},t)=\int_{\gamma(t)=\bar{q}}d_{F}\gamma\,e^{i\hbar^{-1}S_{t}(\gamma)}\underline{\psi}(\gamma(0)),$

called Feynman’s path integral expression for the solution of Schrödinger
equation by Feynman’s time sclicing method.

If we permit this integral representation of the solution with operations
under integral sign admitted, when making $\hbar\to 0$, we have the main
contribution stems from the stationary point $\gamma_{c}(\cdot)$ that is,
$\displaystyle{\frac{\delta
S_{t}(\gamma)}{\delta\gamma}\bigg{|}_{\gamma=\gamma_{c}}}=0$. This expression
with above interpretation is persuasive to claim that classical mechanical
equation appeared from quantum one when making $\hbar\to 0$.

###  2.2. Quantization à la Fujiwara

Using Feynman’s idea rather conversely, Fujiwara [41, 42] constructed a
fundamental solution of Schrödinger equation. From a given Lagrangian
$L(q,\dot{q})$, he found a classical orbit
$\gamma_{c}(\cdot)\in\Gamma_{(t,\underline{q},\bar{q})}$ of that Lagrangian
mechanics:

$\ddot{q}(s)+L(q(s),\dot{q}(s))=0\quad\text{with}\quad
q(0)=\underline{q},\;\dot{q}(0)=\underline{p}^{*}$

such that $\bar{q}=q(t,\underline{q},\underline{p}^{*})$, that is,
$\gamma_{c}(0)=\underline{q},\gamma_{c}(t)=\bar{q}.$ Then, the action integral
corresponding to $\gamma_{c}$ is given

$S_{t}(\gamma_{c})=S_{t}(\bar{q},\underline{q})=\int_{0}^{t}ds\,L(\gamma_{c}(s),\dot{\gamma}_{c}(s)),$

moreover defining van Vleck determinant as

$D(t,\bar{q},\underline{q})=\det\bigg{(}\frac{\partial^{2}}{\partial\underline{q}\partial\bar{q}}S_{t}(\bar{q},\underline{q})\bigg{)},$

he finally defined a short time propagator as a FIOp(=Fourier Integral
Operator)

$T_{t}\underline{\psi}(\bar{q})=\int_{{\mathbb{R}}^{d}}d\underline{q}\,A(t,\bar{q},\underline{q})e^{i\hbar^{-1}S_{t}(\bar{q},\underline{q})}\underline{\psi}(\underline{q})$

where

$A(t,\bar{q},\underline{q})=\sqrt{D(t,\bar{q},\underline{q})}.$

Above Fujiwara’s process is justified when
$\sup_{q}|\partial_{q}^{\alpha}V(q)|\leq C,(|\alpha|\geq 2)$. That is, under
this condition, not only there exists a unique classical orbit, and above
quantities are well-defined, but also $T_{t}$ defines a bounded linear
operator on $L^{2}({\mathbb{R}}^{d})$ with the property

$\|T_{t+s}u-T_{s}T_{t}u\|\leq C(t^{2}+s^{2})\|u\|.$

Moreover, following operator $E_{t}$ is defined by

$\lim_{n\to\infty}\|(T_{t/n})^{n}u-E_{t}u\|=0$

which gives a parametrix of the given initial value problem of Schrödinger
equation [41]. The kernel representation is the fundamental solution such that

$(E_{t}u)(\bar{q})=\int_{{\mathbb{R}}^{d}}\,d\underline{q}K(t,\bar{q},\underline{q})u(\underline{q}).$

Finally, not only how to characterize the long-time behavior111111For example,
how Maslov index appeared in PIM? of $K(t,\bar{q},\underline{q})$ [42], but
also how to use the $\\#$ products of FIOps by Kumano-go group, are the
problems reconsidered.

######  Remark 2.1.

(1)[Feynman-Kac formula]: Sitimulated by Feynman’s idea, M. Kac represents the
solution of

$\frac{\partial}{\partial{t}}u(q,t)=\sigma\Delta u(q,t)-V(q)u(q,t),$

using Wiener measure $dW(\gamma)$ by

$u(q,t)=\int
dW(\gamma)\,e^{-\int_{0}^{t}V(\gamma(s)+q)ds}\underline{u}(\gamma(0)+\underline{q}).$

(2)[Problem of $R/12$]: For a given Riemann manifold $(M,g_{jk})$, applying
Fujiwara’s idea to the heat equation, that is, some type of quantization,
Inoue and Maeda [69] got the notorious term $R/12$ where $R$ is the scalar
curvature of the Riemann metric $g_{jk}(q)dq^{j}dq^{k}$. Curiously, this term
was also derived as the most probable path calculation in probability theory
[39, 95].

Rather recently the term $R/12$ appeared in S. Fukushima [43], but this seems
not the direct consequence of quantization of $g_{jk}(q)dq^{j}dq^{k}$ like
physicists arguments, for example, B. DeWitt [29] or F. Bastianelli et al.
[14]. Fukushima tries to construct a fundamental solution of
$\displaystyle{i\frac{\partial}{\partial{t}}-\frac{1}{2}\Delta_{g}+\frac{R}{12}}$.
(3)[Hamiltonian formulation on manifolds?] Above mentioned works containing
physicist papers, are formulated in Lagrangian case. Since it needs works to
formulate Fourier transformations on manifolds, physicist papers such as Field
[36, 37] claim something interesting without any mathematical estimates.

###  2.3. Hamilton formulation of PIM

For a given Lagrangian $L(q,\dot{q})$, above process is denoted formally as

$K=\int d_{F}\gamma\,e^{i\hbar^{-1}S(\gamma)}\quad\mbox{or}\quad
K(t,\bar{q},\underline{q})=\int_{\Gamma_{(t,\bar{q},\underline{q})}}d_{F}\gamma\,e^{i\hbar^{-1}S(\gamma)}$

where

$S(\gamma)=\int_{0}^{t}ds\,L(\gamma(s),\dot{\gamma}(s)),{\;\;}\mbox{for
each}\;\gamma\in\Gamma_{(t,\bar{q},\underline{q})}.$

What does it become when Hamiltonian $H(q,p)$ is given? Here, as is well-
known, $H(q,p)$ is related to $L(q,\dot{q})$ by Legendre transformation. How
to give meaning to the following formal expression?:

$K=\int d_{F}p\,d_{F}q\,e^{i\hbar^{-1}\int ds(\dot{q}p-H(q,p))}.$

Under same assumption as above, there exists a classical orbit
$(q(t,\underline{q},\underline{p}),p(t,\underline{q},\underline{p}))$ to the
Hamilton equation for the given $H(q,p)$ with initial data
$(\underline{q},\underline{p})$. For any fixed $\underline{p}$, for the map
$\underline{q}\to\bar{q}=q(t,\underline{q},\underline{p})$, there exists an
inverse map $\underline{q}=x(t,\bar{q},\underline{p})$ for sufficiently small
$|t|$:

$\bar{q}=q(t,x(t,\bar{q},\underline{p}),\underline{p})\;\;\mbox{and}\;\;\underline{q}=x(t,q(t,\underline{q},\underline{p}),\underline{p})$

Using these, we put

$S(t,\bar{q},\underline{p})=(\underline{q}\underline{p}-S_{0}(t,\underline{q},\underline{p}))\bigg{|}_{\underline{q}=x(t,\bar{q},\underline{p})}$

and define

$D(t,\bar{q},\underline{p})=\det\bigg{(}\partial_{\bar{q}}\partial_{\underline{p}}S(t,\bar{q},\underline{p})\bigg{)}.$

Finally we define

$T_{t}u(\bar{q})=(2\pi\hbar)^{-d/2}\int_{{\mathbb{R}}^{d}}d\underline{p}\,D(t,\bar{q},\underline{p})^{1/2}e^{i\hbar^{-1}S(t,\bar{q},\underline{p})}\hat{u}(\underline{p}).$

This gives a parametrix for Schrödinger equation [58].

######  Remark 2.2.

In order to consider a classical orbit connecting 2 points $\bar{q}$ and
$\underline{q}$ in $t$, we need 2nd order time derivatives in corresponding
classical mechanical equation. Then, how one treats Weyl or Dirac equation?
Hint is given in the next section.

##  3\. Another interpretation of method of characteristics, explained through
the simplest case

Let check the difference between Lagarangian or Hamiltonian methods, by taking
the simplest example. This is important to treat first order system of PDE
such as Dirac or Weyl equation by applying Feynman’s time slicing method.

Recall m.o.c.(=method of characteristics). On region
$\Omega\subset{{\mathbb{R}}}^{d+1}$, we consider the following initial value
problem: Let solve

(3.1) $\left\\{\begin{aligned}
&\frac{\partial}{\partial{t}}u(t,q)+\sum_{j=1}^{d}a_{j}(t,q){\frac{\partial}{\partial{q_{j}}}}u(t,q)=b(t,q)u(t,q)+f(t,q),\\\
&u({\underline{t}},q)={\underline{u}}(q)\end{aligned}\right.$

The characterisitic equation of (3.1) is given by

$\left\\{\begin{aligned} &\frac{d}{dt}q_{j}(t)=a_{j}(t,q(t)),\\\
&q_{j}({\underline{t}})={\underline{q}}_{j}\quad\text{for}\quad
j=1,\cdots,d,\end{aligned}\right.$

with solution denoted by

$q(t)=q(t,{\underline{t}};{\underline{q}})=(q_{1}(t),\cdots,q_{d}(t))\in{\mathbb{R}}^{d},{\underline{q}}=({\underline{q}}_{1},{\cdots},{\underline{q}}_{d}).$

Using this,

######  Theorem 3.1 (method of characteristics).

In (3.1), assume coefficients $a_{j}\in C^{1}(\Omega:{\mathbb{R}})$, $b,f\in
C(\Omega:{\mathbb{R}})$. Taking any point
$({\underline{t}},{\underline{q}})\in\Omega$, assume ${\underline{u}}$ is
$C^{1}$ in a neighbourhood of ${\underline{q}}$. Then, there exists a unique
solutiion $u(t,q)$ near $({\underline{t}},{\underline{q}})$. Moreover, putting
$B(t,{\underline{q}})=b(t,q(t,{\underline{t}};{\underline{q}}))$,
$F(t,{\underline{q}})=f(t,q(t,{\underline{t}};{\underline{q}}))$ and defining

(3.2)
$U(t,{\underline{q}})=e^{\int_{{\underline{t}}}^{t}d\tau\,B(\tau,{\underline{q}})}\left\\{\int_{{\underline{t}}}^{t}ds\,e^{-\int_{{\underline{t}}}^{s}d\tau\,B(\tau,{\underline{q}})}F(s,{\underline{q}})+{\underline{u}}({\underline{q}})\right\\}$

we have a solution (3.1) given by

(3.3) $u(t,\bar{q})=U(t,x(t,\bar{q})).$

Here, ${\underline{q}}=x(t,{\underline{t}};{\overline{q}})$ is the inverse
function of ${\bar{q}}=q(t,{\underline{t}};{\underline{q}})$.

######  Remark 3.1.

(3.2) satisfies

$\frac{d}{dt}U(t,{\underline{q}})=B(t,{\underline{q}})U(t,{\underline{q}})+F(t,{\underline{q}}).$

##### An example:

With this theorem in mind, we consider the following simplest case:

(3.4) $\left\\{\begin{aligned}
&i\hbar\frac{\partial}{\partial{t}}u(t,q)=a\frac{\hbar}{i}{\frac{\partial}{\partial{q}}}u(t,q)+bqu(t,q),\\\
&u(0,q)={\underline{u}}(q).\end{aligned}\right.$

The symbol of the righthand side of above equation is derived by

$H(q,p)=e^{-i\hbar^{-1}qp}\Big{(}a\frac{\hbar}{i}{\frac{\partial}{\partial{q}}}+bq\Big{)}e^{i\hbar^{-1}qp}=ap+bq.$

From this, we have the characteristic

$\dot{q}(t)=\frac{\partial H(q,p)}{\partial p}=a\quad\text{with}\quad
q(0)=\underline{q},$

whose solution is given

$q(s)={\underline{q}}+as\quad\text{with}\quad{\underline{q}}=x(t,{\overline{q}})={\overline{q}}-at.$

Since above representation (3.2) gives

$U(t,{\underline{q}})={\underline{u}}({\underline{q}})e^{-i\hbar^{-1}(b{\underline{q}}t+2^{-1}ab{t^{2}})},$

from (3.3), we get

$u(t,{\overline{q}})={\underline{u}}({\overline{q}}-at)e^{-i\hbar^{-1}(b{\overline{q}}t-2^{-1}ab{t^{2}})}.$

By this procedure, the information of $p(t)$ isn’t used!

Now, we give another interpretation of this representation by Hamilton path
integral method.

For $(t,{\underline{q}},{\underline{p}})$, putting

$S_{0}(t,{\underline{q}},{\underline{p}})=\int_{0}^{t}ds[\dot{q}(s)p(s)-H(q(s),p(s))]=-b{\underline{q}}t-2^{-1}ab{t^{2}},$

we define the action integral for Hamilton function $H(q,p)$

$S(t,{\overline{q}},{\underline{p}})={\underline{q}}\,{\underline{p}}+S_{0}(t,{\underline{q}},{\underline{p}})\big{|}_{{\underline{q}}=x(t,{\overline{q}})}={\overline{q}}{\underline{p}}-a{\underline{p}}t-b{\overline{q}}t+2^{-1}ab{t^{2}}.$

This $S=S(t,{\overline{q}},{\underline{p}})$ satisfies the following Hamilton-
Jacobi equation:

${\frac{\partial}{\partial{t}}}S+H({\overline{q}},\partial_{\overline{q}}S)=0\quad\text{with}\quad
S(0,{\overline{q}},{\underline{p}})={\overline{q}}{\underline{p}}.$

In this case, van Vleck determinant is a scalar

$\frac{\partial^{2}S(t,{\overline{q}},{\underline{p}})}{\partial{\overline{q}}\partial{\underline{p}}}=1$

and it satisfies the following continuity equation:

${\frac{\partial}{\partial{t}}}D+\frac{1}{2}{\partial_{\overline{q}}}(DH_{p})=0\quad\text{with}\quad
D(0,{\overline{q}},{\underline{p}})=1\quad\text{where}\quad
H_{p}=\frac{\partial H}{\partial p}({\overline{q}},\partial_{\overline{q}}S)$

Using these classical quantities $S$ and $D$, modifying Feynman’s method
slightly121212Though Feynman’s procedure based on Lagrangian, here we changed
to Hamiltonian formulation, we define

$u(t,{\overline{q}})=(2\pi\hbar)^{-1/2}\int
d{\underline{p}}\,D^{1/2}(t,{\overline{q}},{\underline{p}}){\cdot}e^{i\hbar^{-1}S(t,{\overline{q}},{\underline{p}})}\hat{\underline{u}}({\underline{p}}),$

Using $\delta(\overline{q}-at-\underline{q})=(2\pi\hbar)^{-1}\int
d{\underline{p}}e^{i\hbar^{-1}(\overline{q}-at-\underline{q}){\underline{p}}}$,
we have

$\displaystyle u(t,{\overline{q}})$ $\displaystyle=(2\pi\hbar)^{-1/2}\int
d{\underline{p}}\,e^{i\hbar^{-1}S(t,{\overline{q}},{\underline{p}})}\hat{\underline{u}}({\underline{p}})$
$\displaystyle=(2\pi\hbar)^{-1}\iint
d{\underline{p}}d{\underline{q}}\,e^{i\hbar^{-1}(S(t,{\overline{q}},{\underline{p}})-\underline{q}{\underline{p}})}{\underline{u}}({\underline{q}})=\int
d{\underline{q}}\delta({\overline{q}}-at-{\underline{q}}){\underline{u}}(\underline{q})e^{i\hbar^{-1}(-b{\overline{q}}t+2^{-1}ab{t^{2}})}$
$\displaystyle={\underline{u}}({\overline{q}}-at)e^{i\hbar^{-1}(-b{\overline{q}}t+2^{-1}ab{t^{2}})}.$

######  Problem 3.1.

How we interprete the solution of (3.1) by this idea? More generally, how to
apply this interpretation to systems of PDE, for example, Weyl or Dirac
equations131313Inoue[57, 59]?

##  4\. Necessity of superanalysis

###  4.1. Reasons of necessity

There are several reasons at least for me: (i) As is written in p.355 of
R.Feynman & A.R. Hibbs [35], ‘spin’ has been the object outside Feynman’s
procedures at that time (below, underlined by atlom):

> $\cdots$ path integrals suffer grievously from a serious defect. They do not
> permit a discussion of spin operators or other such operators in a simple
> and lucid way. They find their greatest use in systems for which coordinates
> and their conjugate momenta are adequate. Nevertheless, spin is a simple and
> vital part of real quantum-mechanical systems. It is a serious limitation
> that the half-integral spin of the electron does not find a simple and ready
> representation. It can be handled if the amplitudes and quantities are
> considered as quarternions instead of ordinary complex numbers, but the lack
> of commutativity of such numbers is a serious complication.

(ii) E. Berezin [18], who invented the second quantization, proposed to “treat
boson and fermion on equal footing”. That is, instead of ${\mathbb{R}}$ or
$\mathbb{C}$, he claims the necessity to construct a scalar-like, non-
commutative ground field, where lives electron and photon equally.
(iii) E. Witten [100] explained the meaning of supersymmetric idea in physics
to mathematician by new derivation of Morse inequalities using exterior
differential operations. But be careful, in many case, physicists usage of
supersymmetry only means to use fermion or odd variables, not exactly treating
some “symmetry” in ${{\mathfrak{R}}}^{m|n}$, for example K.B. Efetov’s works
[32] and Y.V. Fyodorov [44].

###  4.2. Regarding matrices as differential operators!

A claim “any Clifford algebra has a representation on Grassmann
algebra”141414Though I don’t appreciate fully Chevalley’ theorem“Any Clifford
algebra has a representation on Grassmann algebra”, but following
consideration may be sufficient of my intuitive arguments. is straight
forwardly explained for $2\times 2$ matrices case. Though analogous argument
works for a set of $2^{\ell}\times 2^{\ell}$ matrices because it has Clifford
relation, but for a set of general $N\times N$ matrices, we need to find
suitable algebra on which we need foundation of analysis. I imagine not only
the work of J.L. Martin [79, 80] but also the one by R. Campoamor-Stursberg et
al. [22] give some hint on this. Especially $N=3$ will be interesting when we
consider vorticity equation for Euler or Navier-Stokes equation which is
represented by $3\times 3$ matrices.

Recall Pauli matrices $\\{\boldsymbol{\sigma}_{k}\\}_{k=1}^{3}$:

$\boldsymbol{\sigma}_{1}=\begin{pmatrix}0&1\\\
1&0\end{pmatrix},\quad\boldsymbol{\sigma}_{2}=-i\begin{pmatrix}0&1\\\
-1&0\end{pmatrix},\quad\boldsymbol{\sigma}_{3}=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}.$

These matrices satisfy not only Clifford relation

(4.1)
$\boldsymbol{\sigma}_{j}\boldsymbol{\sigma}_{k}+\boldsymbol{\sigma}_{k}\boldsymbol{\sigma}_{j}=2\delta_{jk},\quad\text{where}\quad
j,k=1,2,3,$

but also the following for any $(j,k,\ell)$, an even permutation of $(1,2,3)$:

(4.2)
$\boldsymbol{\sigma}_{j}\boldsymbol{\sigma}_{k}=i\boldsymbol{\sigma}_{\ell}.$

Decompose any $2\times 2$ matrix $\mathbb{A}$ by
$\\{\boldsymbol{\sigma}_{j}\\}$,

(4.3) $\displaystyle{\mathbb{A}}=\begin{pmatrix}a&c\\\ d&b\end{pmatrix}$
$\displaystyle=\frac{a+b}{2}\begin{pmatrix}1&0\\\
0&1\end{pmatrix}+\frac{a-b}{2}\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}$
$\displaystyle\qquad\qquad+\frac{c+d}{2}\begin{pmatrix}0&1\\\
1&0\end{pmatrix}+\frac{c-d}{2}\begin{pmatrix}0&1\\\ -1&0\end{pmatrix}$
$\displaystyle=\frac{a+b}{2}\mathbb{I}_{2}+\frac{a-b}{2}\boldsymbol{\sigma}_{3}+\frac{c+d}{2}\boldsymbol{\sigma}_{1}+i\frac{c-d}{2}\boldsymbol{\sigma}_{2}.$

That is, a set of all $2\times 2$ matrices has the Clifford algebra structure
with Pauli matrices $\\{\boldsymbol{\sigma}_{k}\\}$ as a basis.

We identify a vector as a function of Grassmann variables. Then, a matrix is
regarded as a differential operator acting on a set of functions composed with
Grassmann variables, which forms something-like field similar to real or
complex number field but non-commutative.

Rather abruptly we prepare151515usage “preparing” odd variables $\theta_{1}$,
$\theta_{2}$ is not frequently used in PDE group two odd variables
$\theta_{1},\theta_{2}$ having the following relations:

(4.4)
$\theta_{1}{\cdot}\theta_{2}+\theta_{2}{\cdot}\theta_{1}=0,\;\theta_{j}{\cdot}\theta_{j}=0,(j=1,2)$

That is, they have product ${\cdot}$ having Grassmann relations and
differentiation such as

$\frac{\partial}{\partial\theta_{1}}1=0,\quad\frac{\partial}{\partial\theta_{1}}\theta_{1}=1,\quad
etc.$

Here, when we differentiate, bringing that variable in front, i.e.

$\frac{\partial}{\partial\theta_{2}}\theta_{1}{\cdot}\theta_{2}=\frac{\partial}{\partial\theta_{2}}(-\theta_{2}{\cdot}\theta_{1})=\frac{\partial}{\partial\theta_{2}}(\theta_{2}{\cdot}(-\theta_{1}))=-\theta_{1}.$

More concretely, for example, taking two variables $z_{1}$ and $z_{2}$ in
${\mathbb{R}}^{2}$, construct differential forms $dz_{1}$, $dz_{2}$ with
exterior product $\wedge$ and interior product $\lfloor$ such that

$dz_{1}\wedge dz_{2}=-dz_{2}\wedge dz_{1},\quad dz_{j}\wedge
dz_{j}=0,\quad\frac{\partial}{\partial{z_{j}}}\lfloor dz_{k}=\delta_{jk}.$

Here, identifying $dz_{j}$ as $\theta_{j}$ for $j=1.2$, $\wedge$ as ${\cdot}$
and $\frac{\partial}{\partial{z_{j}}}\lfloor$ as
$\frac{\partial}{\partial\theta_{j}}$, and abbreviating ${\cdot}$, we continue
to explain.

Using odd variables $\theta_{1}$, $\theta_{2}$, we give the identifying maps
$\\#$, $\flat$ with Grassmann algebras as follows:

(4.5)
$\begin{gathered}\Gamma_{0}=\\{u(\theta)=u_{0}+u_{1}\theta_{1}\theta_{2}\;|\;u_{0},u_{1}\in{\mathbb{C}}\\}\,\begin{matrix}{\overset{\flat}{\rightarrow}}\\\\[-8.0pt]
{\underset{\sharp}{\leftarrow}}\end{matrix}\;{\mathbb{C}}^{2}=\bigg{\\{}{\bf
u}=\binom{u_{0}}{u_{1}}\;\big{|}\;u_{0},u_{1}\in{\mathbb{C}}\bigg{\\}}\\\
\quad\text{with}\quad(\\#{\bf u})(\theta)=\bigg{(}\\#\begin{pmatrix}u_{0}\\\
u_{1}\end{pmatrix}\bigg{)}(\theta)=u_{0}+u_{1}\theta_{1}\theta_{2},\\\
\flat(u_{0}+u_{1}\theta_{1}\theta_{2})=\flat(u(\theta))=\begin{pmatrix}u_{0}\\\
u_{1}\end{pmatrix}.\end{gathered}$

Here, we have $u(0)=u(\theta)|_{\theta=0}=u_{0}$,
$\partial_{\theta_{2}}\partial_{\theta_{1}}u(\theta)|_{\theta=0}=\partial_{\theta_{2}}(\partial_{\theta_{1}}u(\theta))|_{\theta=0}=u_{1}$,
which relates a vector $\bf u$ and a function $u(\theta)$.

Now, we define differential operators w.r.t. odd variables:

(4.6)
$\displaystyle\sigma_{1}(\theta,\partial_{\theta})=\theta_{1}\theta_{2}-\frac{\partial^{2}}{\partial\theta_{1}\partial\theta_{2}},$
$\displaystyle\sigma_{2}(\theta,\partial_{\theta})=i\bigg{(}\theta_{1}\theta_{2}+\frac{\partial^{2}}{\partial\theta_{1}\partial\theta_{2}}\bigg{)},$
$\displaystyle\sigma_{3}(\theta,\partial_{\theta})=1-\theta_{1}\frac{\partial}{\partial\theta_{1}}-\theta_{2}\frac{\partial}{\partial\theta_{2}}.$

Thes operators act on $u(\theta)=u_{0}+u_{1}\theta_{1}\theta_{2}\in\Gamma_{0}$
as follows:

$\displaystyle\sigma_{1}(\theta,\partial_{\theta})(u_{0}+u_{1}\theta_{1}\theta_{2})=u_{0}\theta_{1}\theta_{2}+u_{1},$
$\displaystyle\sigma_{2}(\theta,\partial_{\theta})(u_{0}+u_{1}\theta_{1}\theta_{2})=i(u_{0}\theta_{1}\theta_{2}-u_{1}),$
$\displaystyle\sigma_{3}(\theta,\partial_{\theta})(u_{0}+u_{1}\theta_{1}\theta_{2})=u_{0}-u_{1}\theta_{1}\theta_{2}.$

Here,

$\displaystyle\frac{\partial^{2}}{\partial{\theta_{1}}\partial{\theta_{2}}}\theta_{1}\theta_{2}$
$\displaystyle=\partial_{\theta_{1}}\partial_{\theta_{2}}\theta_{1}\theta_{2}$
$\displaystyle=\partial_{\theta_{1}}(\partial_{\theta_{2}}\theta_{1}\theta_{2})=\partial_{\theta_{1}}(\partial_{\theta_{2}}(-\theta_{2}\theta_{1}))=\partial_{\theta_{1}}(-\theta_{1})=-1.$

They act on $\Gamma_{0}$ as,

$\sigma_{1}(\theta,\partial_{\theta})\sigma_{2}(\theta,\partial_{\theta})=i\sigma_{3}(\theta,\partial_{\theta}),\quad\sigma_{2}(\theta,\partial_{\theta})\sigma_{3}(\theta,\partial_{\theta})=i\sigma_{1}(\theta,\partial_{\theta}),\quad\sigma_{3}(\theta,\partial_{\theta})\sigma_{1}(\theta,\partial_{\theta})=i\sigma_{2}(\theta,\partial_{\theta}).$

######  Remark 4.1.

(1) The action $\sigma_{1}(\theta,\partial_{\theta})$ on $\Gamma_{0}$ has a
matrix representation

$\flat\sigma_{1}(\theta,\partial_{\theta})\sharp\begin{pmatrix}u_{0}\\\
u_{1}\end{pmatrix}=\begin{pmatrix}u_{1}\\\
u_{0}\end{pmatrix},\quad\flat\sigma_{1}(\theta,\partial_{\theta})\sharp=\boldsymbol{\sigma}_{1},\quad
etc.$

(2) Moreover, putting

$\Gamma_{1}=\\{v(\theta)=v_{1}\theta_{1}+v_{2}\theta_{2}\;|\;v_{1},v_{2}\in{\mathbb{C}}\\}\,\begin{matrix}{\overset{\flat}{\rightarrow}}\\\\[-8.0pt]
{\underset{\sharp}{\leftarrow}}\end{matrix}\;{\mathbb{C}}^{2}=\bigg{\\{}{\bf
v}=\binom{v_{1}}{v_{2}}\;|\;v_{1},v_{2}\in{\mathbb{C}}\bigg{\\}}$

we have

$\sigma_{j}(\theta,\partial_{\theta})v(\theta)=0\quad\text{for any}\quad
v\in\Gamma_{1}\;\mbox{and}\;j=1,2,3.$

We define Fourier transformation161616The integration here is very algebraic,
called Berezin integral, which has properties as integral. See, more
precisely, V.S. Vladimirov and I.V. Volovich [99] for odd variables. Taking
number ${{\mathchar 22\relax\mkern-9.0muk}}$ in
${{\mathbb{R}}}^{\times}={{\mathbb{R}}}-\\{0\\}$ or $i{{\mathbb{R}}}^{\times}$
called it as spin constant, we define

$\hat{u}(\pi)={{\mathchar
22\relax\mkern-9.0muk}}\int_{{{\mathfrak{R}}}^{0|2}}d\theta\,e^{-i{{\mathchar
22\relax\mkern-9.0muk}}^{-1}\langle\theta|\pi\rangle}u(\theta),\quad{u}(\theta)={{\mathchar
22\relax\mkern-9.0muk}}\int_{{{\mathfrak{R}}}^{0|2}}d\pi\,e^{i{{\mathchar
22\relax\mkern-9.0muk}}^{-1}\langle\theta|\pi\rangle}\hat{u}(\pi).$

In the above, the integral interval ${{\mathfrak{R}}}^{0|2}$ may be considered
only as a symbol, at least for the time being. Here,
$\langle\theta|\pi\rangle=\theta_{1}\pi_{1}+\theta_{2}\pi_{2}$ and

$\langle\theta|\pi\rangle^{2}=2\theta_{1}\pi_{1}\theta_{2}\pi_{2}=-2\theta_{1}\theta_{2}\pi_{1}\pi_{2},\quad\langle\theta|\pi\rangle^{j}=\overbrace{\langle\theta|\pi\rangle\langle\theta|\pi\rangle{\cdots}\langle\theta|\pi\rangle}^{\text{$j$
times}}=0\;\mbox{if}\;j\geq 3$

we have

$e^{a\langle\theta|\pi\rangle}=\sum_{{\ell}=1}^{\infty}\frac{a^{\ell}\langle\theta|\pi\rangle^{\ell}}{{\ell}!}=1+a\langle\theta|\pi\rangle-a^{2}\theta_{1}\theta_{2}\pi_{1}\pi_{2}.$

######  Remark 4.2.

“Integration” should be considered preferable such that (i) all polynomials
are integrable, (ii) linear with integrand and (iii) translation invariant,
Berezin integration satisfies these properties.

By Fourier transformation, differential operator is regarded as multiplication
in dual space171717This means, must we recognize objects algebraically? Or
four arithmetic operations is the only core to recognize mathematically?, and
theory of pseudo-differential operators treats PDE with variable coefficients
controlling error estimates.

Defining Weyl symbols of differential operators
$\\{\sigma_{j}(\theta,\partial_{\theta})\\}_{j=1}^{3}$ as

(4.7) $\left\\{\begin{aligned}
&\sigma_{1}(\theta,\pi)=\theta_{1}\theta_{2}+{{\mathchar
22\relax\mkern-9.0muk}}^{-2}\pi_{1}\pi_{2},\\\
&\sigma_{2}(\theta,\pi)=i(\theta_{1}\theta_{2}-{{\mathchar
22\relax\mkern-9.0muk}}^{-2}\pi_{1}\pi_{2}),\\\
&\sigma_{3}(\theta,\pi)=-i{{\mathchar
22\relax\mkern-9.0muk}}^{-1}\langle\theta|\pi\rangle=-i{{\mathchar
22\relax\mkern-9.0muk}}^{-1}(\theta_{1}\pi_{1}+\theta_{2}\pi_{2}),\end{aligned}\right.$

we have, for example,

$\displaystyle(\hat{\sigma}_{3}^{\mathrm{w}}(\theta,\partial_{\theta})u)(\theta)$
$\displaystyle=-i{{\mathchar
22\relax\mkern-9.0muk}}\iint_{{{\mathfrak{R}}}^{0|2}\times{{\mathfrak{R}}}^{0|2}}d\pi
d\theta^{\prime}\,e^{i{{\mathchar
22\relax\mkern-9.0muk}}^{-1}\langle\theta-\theta^{\prime}|\pi\rangle}\sigma_{3}(\frac{\theta+\theta^{\prime}}{2},\pi)(u_{0}+u_{1}{\theta^{\prime}}_{1}{\theta^{\prime}}_{2})$
$\displaystyle=u_{0}-u_{1}\theta_{1}\theta_{2}=\sigma_{3}(\theta,\partial_{\theta})(u_{0}+u_{1}\theta_{1}\theta_{2}).\;\;etc.$

Here, we used

$\int_{{{\mathfrak{R}}}^{0|2}}d\pi\,e^{i{{\mathchar
22\relax\mkern-9.0muk}}^{-1}\langle\theta-\theta^{\prime}|\pi\rangle}\sigma_{3}(\frac{\theta+\theta^{\prime}}{2},\pi)=-i{\mathchar
22\relax\mkern-9.0muk}^{-1}(\theta_{1}\theta_{2}-\theta_{1}^{\prime}\theta_{2}^{\prime}).$

######  Remark 4.3.

(i) It takes many time to perceive the meaning of $1$ in
${\sigma}_{3}(\theta,\partial_{\theta})$ appeared in (4.6), that is, $1$ stems
from Weyl quantization!
(ii) Artificially introduced spin constant ${\mathchar 22\relax\mkern-9.0muk}$
gives nice result only when ${\mathchar 22\relax\mkern-9.0muk}\hbar^{-1}=1$,
that is, the action integral obtained by Jacobi method satisfies Hamilton-
Jacobi equation in case ${\mathchar 22\relax\mkern-9.0muk}\hbar^{-1}=1$,
therefore we assume ${\mathchar 22\relax\mkern-9.0muk}=\hbar$!

Therefor, from (4.3), we have

(4.8) $({\sharp}{\mathbb{A}}{\bf
u})(\theta)=[\frac{a+b}{2}+\frac{a-b}{2}{\sigma}_{3}(\theta,\partial_{\theta})+\frac{c+d}{2}{\sigma}_{1}(\theta,\partial_{\theta})+i\frac{c-d}{2}{\sigma}_{2}(\theta,\partial_{\theta})]u(\theta).$

whose Weyl symbol is

(4.9) $\sigma^{\rm
w}({\sharp}{\mathbb{A}}{\flat})=[\frac{a+b}{2}+\frac{a-b}{2}{\sigma}^{\rm
w}_{3}(\theta,\pi)+\frac{c+d}{2}{\sigma}_{1}(\theta,\pi)+i\frac{c-d}{2}{\sigma}_{2}(\theta,\pi)].$

######  Remark 4.4.

In the theory of pseudo-differential operators, symbol calculus is the main
ingredient where algebra is essentially ${\mathbb{R}}$. Whether analogous
procedure works on superspace ${{\mathfrak{R}}}^{m|n}$? More explicitly,
though the symbol of $\sigma_{1}(\theta,\partial_{\theta})$ is given as
$\theta_{1}\theta_{2}+{\mathchar 22\relax\mkern-9.0muk}^{-2}\pi_{1}\pi_{2}$,
but whether the inverse of $\sigma_{1}(\theta,\partial_{\theta})$ is
calculated directly from $\theta_{1}\theta_{2}+{\mathchar
22\relax\mkern-9.0muk}^{-2}\pi_{1}\pi_{2}$?

###  4.3. Chi’s example

The following result is a part of Chi [23] in 1958:

######  Theorem 4.1 (An example of weakly hyperbolic equation).

Let

$L(t)=L(t,\partial_{q})=t^{2}\partial_{q}^{2}+b\partial_{q}.$

Solving the initial value problem

(4.10) $u_{tt}-L(t,\partial_{q})u=0\quad\text{with}\quad
u(0,q)={\underline{u}}_{0}(q),\quad u_{t}(0,q)=0,$

when $b=4k+1$, $k\in{\mathbb{Z}}_{+}$, we have the solution given by

$u(t,q)=\sum_{\ell=0}^{k}\frac{2^{2\ell}\,k!}{(2\ell)!(k-\ell)!}t^{2\ell}{\underline{u}}_{0}^{(\ell)}\bigg{(}q+\frac{t^{2}}{2}\bigg{)}.$

In this section, we derive this result by completely different method from
Chi.

####  4.3.1. Chi’s proof

Applying change of variables,

$\xi=x+\frac{t^{2}}{2},\quad\eta=x-\frac{t^{2}}{2}$

to $u(t,x)$, Chi reduces above problem to the Euler-Darboux equation for
$\displaystyle{\tilde{u}(\xi,\eta)=u\bigg{(}\sqrt{\xi-\eta},\frac{\xi+\eta}{2}\bigg{)}}$
yielding,

(4.11)
$\frac{\partial^{2}\tilde{u}}{\partial\xi\partial\eta}-\frac{1-b}{4(\xi-\eta)}\frac{\partial\tilde{u}}{\partial\xi}+\frac{1+b}{4(\xi-\eta)}\frac{\partial\tilde{u}}{\partial\eta}=0$

with initial condition

$\tilde{u}(\xi,\xi)=\underline{u}_{0}(\xi),\quad\lim_{\xi-\eta\to
0}(\xi-\eta)^{1/2}\bigg{(}\frac{\partial\tilde{u}}{\partial\xi}-\frac{\partial\tilde{u}}{\partial\eta}\bigg{)}=\underline{u}_{1}(\xi).$

######  Remark 4.5.

For

${E}(\alpha,\beta)=\partial_{\xi}\partial_{\eta}-\frac{\beta}{\xi-\eta}\partial_{\xi}+\frac{\alpha}{\xi-\eta}\partial_{\eta},$

with $\alpha=\frac{1+b}{4},\beta=\frac{1-b}{4}$, it gives (4.11). He uses the
fact such that if $0<\alpha,\beta<1$ then the solution of $E(\alpha,\beta)u=0$
is expressed as the convergent Euler-Darboux integral

$\displaystyle u(\xi,\eta)$
$\displaystyle=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}dt\,u_{0}(\xi+(\eta-\xi)t)t^{\beta-1}(1-t)^{\alpha-1}$
$\displaystyle\qquad+\frac{\Gamma(1-\alpha-\beta)}{2\Gamma(1-\alpha)\Gamma(1-\beta)}(\eta-\xi)^{t-\alpha-\beta}\int_{0}^{1}dt\,u_{1}(\xi+(\eta-\xi)t)t^{-\alpha}(1-t)^{-\beta}.$

Moreover, for other $\alpha,\beta$, this integral diverges as it stands, he
gives these as Riemann-Liouville integral.

Seemingly, Euler-Darboux equation has some relations to special functions or
Painleve’s functions, therefore proceeding inversely, we might dream another
perspective to these functions by using superanalysis?

####  4.3.2. Direct construction of a solution of Hamilton-Jacobi equation

Without reducing to Euler-Darboux equation, we give a simple minded proof (at
least as a story) of Chi’s equation applying superanalysis. (See also, [66,
67])

Putting $u_{0}=u$, $u_{1}=u_{t}$, we make (4.10) to a system:

$i\frac{\partial}{\partial{t}}\binom{u_{0}}{u_{1}}=i\begin{pmatrix}0&1\\\
L(t,\partial_{q})&0\end{pmatrix}\binom{u_{0}}{u_{1}}.$

Here, we multiply $i$ to both sides, rather artificially. Changing a vector
representation to non-commutative scalar one by putting
$u(t,x,\theta)=u_{0}(t,x)+u_{1}(t,x)\theta_{1}\theta_{2}$, we have

$(L(t,\partial_{x})\theta_{1}\theta_{2}-\partial_{\theta_{1}}\partial_{\theta_{2}})(u_{0}+u_{1}\theta_{1}\theta_{2})=u_{1}+Lu_{0}\theta_{1}\theta_{2}\sim\begin{pmatrix}0&1\\\
L(t,\partial_{x})&0\end{pmatrix}\binom{u_{0}}{u_{1}}.$

Here, $q\in{\mathbb{R}}$ is imbedded in $x\in{{\mathfrak{R}}}^{1|0}$ such that
$X_{\rm B}=q$. Defining

${\mathcal{H}}(t,\partial_{x},\theta,\partial_{\theta})=iL(t,\partial_{x})\theta_{1}\theta_{2}-i\partial_{\theta_{1}}\partial_{\theta_{2}}$

whose symbol is given by

${\mathcal{H}}(t,\xi,\theta,\pi)=i(-t^{2}\xi^{2}+ib\xi)\theta_{1}\theta_{2}+i\pi_{1}\pi_{2}$

and corresponding Hamilton-Jacobi equation is given by

(4.12)
${\mathcal{S}}_{t}+{\mathcal{H}}(t,{\mathcal{S}}_{x},\theta,{\mathcal{S}}_{\theta})=0$

which is solved with the initial condition
${\mathcal{S}}(0,x,\xi,\theta,\pi)=\langle
x|\xi\rangle+\langle\theta|\pi\rangle$.

In the supersmooth category, since we have

(4.13) $\displaystyle{\mathcal{S}}(t,x,\xi,\theta,\pi)=$ $\displaystyle
S(t,x,\xi)+X(t,x,\xi)\theta_{1}\theta_{2}+Y(t,x,\xi)\theta_{1}\pi_{1}+\tilde{Y}(t,x,\xi)\theta_{2}\pi_{2}$
$\displaystyle\qquad\qquad\quad+V(t,x,\xi)\theta_{1}\pi_{2}+\tilde{V}(t,x,\xi)\theta_{2}\pi_{1}$
$\displaystyle\qquad\qquad\qquad+Z(t,x,\xi)\pi_{1}\pi_{2}+W(t,x,\xi)\theta_{1}\theta_{2}\pi_{1}\pi_{2},$

where

$\displaystyle S(t,x,\xi)$ $\displaystyle={\mathcal{S}}(t,x,\xi,0,0),$
$\displaystyle X(t,x,\xi)$
$\displaystyle=\partial_{\theta_{2}}\partial_{\theta_{1}}{\mathcal{S}}(t,x,\xi,\theta,\pi)\big{|}_{\theta=\pi=0},$
$\displaystyle Y(t,x,\xi)$
$\displaystyle=\partial_{\pi_{1}}\partial_{\theta_{1}}{\mathcal{S}}(t,x,\xi,\theta,\pi)\big{|}_{\theta=\pi=0},\quad\tilde{Y}(t,x,\xi)=\partial_{\pi_{2}}\partial_{\theta_{2}}{\mathcal{S}}(t,x,\xi,\theta,\pi)\big{|}_{\theta=\pi=0},$
$\displaystyle V(t,x,\xi)$
$\displaystyle=\partial_{\pi_{2}}\partial_{\theta_{1}}{\mathcal{S}}(t,x,\xi,\theta,\pi)\big{|}_{\theta=\pi=0},\quad\tilde{V}(t,x,\xi)=\partial_{\pi_{1}}\partial_{\theta_{2}}{\mathcal{S}}(t,x,\xi,\theta,\pi)\big{|}_{\theta=\pi=0},$
$\displaystyle Z(t,x,\xi)$
$\displaystyle=\partial_{\pi_{2}}\partial_{\pi_{1}}{\mathcal{S}}(t,x,\xi,\theta,\pi)\big{|}_{\theta=\pi=0},\quad
W(t,x,\xi)=\partial_{\pi_{2}}\partial_{\pi_{1}}\partial_{\theta_{2}}\partial_{\theta_{1}}{\mathcal{S}}(t,x,\xi,\theta,\pi)\big{|}_{\theta=\pi=0},$

we should seek solution ${\mathcal{S}}$ of (4.12) with this form. Since

${\mathcal{H}}(t,{\mathcal{S}}_{x},\theta,{\mathcal{S}}_{\theta})=i(-t^{2}{\mathcal{S}}_{x}^{2}+ib{\mathcal{S}}_{x})\theta_{1}\theta_{2}+i{\mathcal{S}}_{\theta_{1}}{\mathcal{S}}_{\theta_{2}},$

putting $\theta=\pi=0$ in (4.12), we have readily

$S_{t}(t,x,\xi)=0\quad\text{with}\quad S(0,x,\xi))=\langle x|\xi\rangle.$

This gives $S(t,x,\xi)=\langle x|\xi\rangle$.

Differentiating (4.12) w.r.t. $\theta_{1}$ and $\theta_{2}$ then restricting
to $\theta=\pi=0$, we get

$X_{t}+i(-t^{2}\xi^{2}+ib\xi)+iX^{2}=0.$

Since this is the Riccati type ODE, using $\varphi(t)$, we find
$X=-i\frac{\dot{\varphi}}{\varphi}$, that is,

(4.14)
$\ddot{\varphi}+(t^{2}\xi^{2}-ib\xi)\varphi=0\quad\text{with}\quad\dot{\varphi}(0)=0.$

Regarding $(x,\xi)$ as parameter, we may solve this equation by power series
in $t$, but this procedure is postponed until explaining our construction of
${\mathcal{S}}$ and quantization.

Putting
$Y(t,x,\xi)=\partial_{\pi_{1}}\partial_{\theta_{1}}{\mathcal{S}}(t,x,\xi,\theta,\pi)\big{|}_{\theta=\pi=0}$,
we have from (4.12),

$Y_{t}+iXY=0\quad\text{with}\quad Y(0)=1.$

From above structure of $X$, we have $Y\varphi=1$. Analogously, we have
$\tilde{Y}(t,x,\xi)=\partial_{\pi_{2}}\partial_{\theta_{2}}{\mathcal{S}}(t,x,\xi,\theta,\pi)\big{|}_{\theta=\pi=0}$
which equals to $Y$.

Calculating $V$ and $\tilde{V}$ analogously, we get both equal to $0$.

As
$Z(t,x,\xi)=\partial_{\pi_{2}}\partial_{\pi_{1}}{\mathcal{S}}(t,x,\xi,\theta,\pi)\big{|}_{\theta=\pi=0}$
satisfies

$Z_{t}+iY^{2}=0\quad\text{with}\quad Z(0)=0,$

we have

$Z(t,x,\xi)=-i\int_{0}^{t}ds\,Y^{2}(s)=-i\int_{0}^{t}ds\,\varphi(s)^{-2}.$

Analogously, we get $W=0$.

Therefore using $\varphi$, we have

(4.15) ${\mathcal{S}}=\langle
x|\xi\rangle+X\theta_{1}\theta_{2}+Y\langle\theta|\pi\rangle+Z\pi_{1}\pi_{2}\quad\text{with}\quad
X=-i\frac{\dot{\varphi}}{\varphi},\quad Y=\frac{1}{\varphi},\quad
Z=-i\int_{0}^{t}ds\,\varphi(s)^{-2}.$

####  4.3.3. Continuity equation

Define van Vleck determinant as

${\mathcal{D}}(t,x,\theta,\xi,\pi)=\operatorname{sdet}\begin{pmatrix}\frac{\partial^{2}{\mathcal{S}}(t,x,\theta,\xi,\pi)}{\partial
x\partial\xi}&\frac{\partial^{2}{\mathcal{S}}(t,x,\theta,\xi,\pi)}{\partial
x\partial\pi}\\\
\frac{\partial^{2}{\mathcal{S}}(t,x,\theta,\xi,\pi)}{\partial\theta\partial\xi}&\frac{\partial^{2}{\mathcal{S}}(t,x,\theta,\xi,\pi)}{\partial\theta\partial\pi}\end{pmatrix},$

and abbreviated as ${\mathcal{D}}={\mathcal{D}}(t,x,\theta,\xi,\pi)$. Using
(4.15), we have

${\mathcal{D}}=\varphi^{2}.$

####  4.3.4. Quantization

Using $\mathcal{S}$ and $\mathcal{A}={\mathcal{D}}^{1/2}$, we define

(4.16) $\displaystyle{\mathcal{T}}_{t}\underline{u}(\theta)$
$\displaystyle=(2\pi)^{-1/2}\int
d\xi\,d\pi\,{\mathcal{A}}e^{i{\mathcal{S}}}\hat{\underline{u}}(\pi)$
$\displaystyle=(2\pi)^{-1/2}\int d\xi\,e^{i{\mathcal{S}}_{1}}\bigg{[}\int
d\pi\,Y^{-1}e^{i{\mathcal{S}}_{2}}(\hat{\underline{u}}_{1}+\hat{\underline{u}}_{0}\pi_{1}\pi_{2})\bigg{]}$

where for the sake of notational simplicity, we put

$\begin{gathered}{\mathcal{S}}={\mathcal{S}}_{1}+{\mathcal{S}}_{2},\\\
{\mathcal{S}}_{1}=\langle
x|\xi\rangle+X\theta_{1}\theta_{2},\quad{\mathcal{S}}_{2}=Y\langle\theta|\pi\rangle+Z\pi_{1}\pi_{2}.\end{gathered}$

Since

$\int
d\pi\,Y^{-1}e^{i{\mathcal{S}}_{2}}(\hat{\underline{u}}_{1}+\hat{\underline{u}}_{2}\pi_{1}\pi_{2})=Y^{-1}\hat{\underline{u}}_{0}+Y^{-1}(iZ+Y^{2}\theta_{1}\theta_{2})\hat{\underline{u}}_{1},$

therefore, remarking $e^{iX\theta_{1}\theta_{2}}=1+iX\theta_{1}\theta_{2}$, we
have

$\displaystyle(1+iX\theta_{1}\theta_{2})(Y^{-1}\hat{\underline{u}}_{0}$
$\displaystyle+Y^{-1}(iZ+Y^{2}\theta_{1}\theta_{2})\hat{\underline{u}}_{1})$
$\displaystyle=Y^{-1}\hat{\underline{u}}_{0}+iY^{-1}Z\hat{\underline{u}}_{1}+[iXY^{-1}\hat{\underline{u}}_{0}-(Y+XY^{-1}Z)\hat{\underline{u}}_{1}]\theta_{1}\theta_{2}.$

Finally, we have

(4.17)
$\begin{gathered}({\mathcal{T}}_{t}\underline{u})(\theta)=u_{0}(t,x)+u_{1}(t,x)\theta_{1}\theta_{2}\quad\text{with}\quad\\\
{\begin{aligned} u_{0}(t,x)&=(2\pi)^{-1/2}\int d\xi\,e^{i\langle
x|\xi\rangle}(Y^{-1}\hat{\underline{u}}_{0}+iY^{-1}Z\hat{\underline{u}}_{1}),\\\
u_{1}(t,x)&=(2\pi)^{-1/2}\int d\xi\,e^{i\langle
x|\xi\rangle}[iXY^{-1}\hat{\underline{u}}_{0}-(Y+XY^{-1}Z)\hat{\underline{u}}_{1}].\end{aligned}}\end{gathered}$

Therefore, when ${\underline{u}}_{1}=0$, we have

(4.18) $u_{0}(t,x)=(2\pi)^{-1/2}\int d\xi\,e^{i\langle
x|\xi\rangle}\varphi(t,x,\xi)\hat{\underline{u}}_{0}(\xi).$

####  4.3.5. Calculation of $\varphi$

We try to find a power series solution w.r.t. $t$ of (4.14).
Decomposing181818How to find factor $e^{it^{2}\xi/2}$?
$\varphi(t)=\phi(t)e^{it^{2}\xi/2}$, $\phi$ satisfies below from (4.14):

(4.19)
$\ddot{\phi}+2it\xi\dot{\phi}+i(1-b)\xi\phi=0\quad\text{with}\quad\phi(0)=1,\;\dot{\phi}(0)=0.$

Putting $\alpha=2i\xi$, $\beta=i(1-b)\xi$, we rewrite (4.19) as

(4.20)
$\ddot{\phi}+\alpha\dot{\phi}+\beta\phi=0\quad\text{with}\quad\phi(0)=1,\;\dot{\phi}(0)=0.$

Putting $\phi(t)=\sum_{j=0}^{\infty}c_{j}t^{j}$ into above, comparing
coefficients of each $t^{\ell}$, we have, for any $\ell$, $c_{2\ell+1}=0$ and

$c_{2\ell}=\frac{(-1)^{\ell}(2\alpha)^{\ell}}{(2\ell)!}\big{(}\frac{\beta}{2\alpha}\big{)}_{\ell}\quad\text{where}\quad(x)_{\ell}=x(x+1){\cdots}(x+\ell-1)=\frac{\Gamma(x+\ell)}{\Gamma(x)}.$

######  Remark 4.6.

When $\beta=b-1=4k$ with $k=0,1,2,{\cdots}$, since

$\big{(}\frac{\beta}{2\alpha}\big{)}_{\ell}=(-k)_{\ell}=(-k)(-k+1){\cdots}(-k+\ell-1)=(-1)^{\ell}(k-1){\cdots}(k-\ell+1)=\frac{(-1)^{\ell}k!}{(k-\ell)!}$

above power series becomes a finite sum, we have

(4.21) $\phi(t)=\sum_{j=0}^{k}\frac{4^{j}k!}{(2j)!(k-j)!}(i\xi)^{j}t^{2j}.$

At last, putting (4.21) into (4.18), we get

$\displaystyle u_{0}(t,q)$ $\displaystyle=(2\pi)^{-1/2}\int
dp\,e^{iqp}\bigg{(}\sum_{j=0}^{k}\frac{4^{j}k!}{(2j)!(k-j)!}(ip)^{j}t^{2j}\bigg{)}e^{ipt^{2}/2}\hat{\underline{u}}_{0}(p)$
$\displaystyle=(2\pi)^{-1/2}\int
dp\,e^{ip(q+t^{2}/2)}\bigg{(}\sum_{j=0}^{k}\frac{2^{2j}k!}{(2j)!(k-j)!}(ip)^{j}t^{2j}\bigg{)}\hat{\underline{u}}_{0}(p)$
$\displaystyle=\sum_{j=0}^{k}\frac{2^{2j}k!}{(2j)!(k-j)!}t^{2j}{\underline{u}}^{(j)}_{0}\bigg{(}q+\frac{t^{2}}{2}\bigg{)}.$

###  4.4. An application to Random Matrix Theory

In 1983, K.B. Efetov [32] wrote a paper entitled “Supersymmetry and theory of
disordered metals” where he derived Wigner’s semi-circle law by applying “his
superanalysis”.

Let ${\mathfrak{H}}_{N}$ be a set of $N\times N$ Hermite matrices. Identifying
this topologically with ${\mathbb{R}}^{N^{2}}$, we introduce probability
measure $d\mu_{N}({H})$ on ${\mathfrak{H}}_{N}$.

(4.22) $\begin{gathered}d\mu_{N}({H})=\prod_{k=1}^{N}d(\Re
H_{kk})\prod_{j<k}^{N}d(\Re H_{jk})d(\Im H_{jk})P_{N,J}(H),\\\
P_{N,J}(H)=Z^{-1}_{N,J}\exp\big{[}-{\frac{N}{2J^{2}}}\operatorname{tr\,}{H}^{*}{H}\big{]}.\end{gathered}$

Here, ${H}=(H_{jk})$,
$H^{*}=(H^{*}_{jk})=(\overline{H}_{kj})={}^{t}\overline{H}$,
$\prod_{k=1}^{N}d(\Re H_{kk})\prod_{j<k}^{N}d(\Re H_{jk})d(\Im H_{jk})$ is the
Lebesgue measure on ${\mathbb{R}}^{N^{2}}$, $Z^{-1}_{N,J}$ is the normalized
constant given by $Z_{N,J}=2^{N/2}(J^{2}\pi/N)^{3N/2}$.

Let $E_{\alpha}=E_{\alpha}({H})$ ($\alpha=1,\cdots,N$) be real eigenvalues of
a given matrix ${H}\in{{\mathfrak{H}}_{N}}$. For Dirac’s delta $\delta$, we
put

(4.23)
$\rho_{N}(\lambda)=\rho_{N}(\lambda;{H})=N^{-1}\sum_{\alpha=1}^{N}\delta(\lambda-
E_{\alpha}(H)),$

and for any function $f$ on ${\mathfrak{H}}_{N}$, we consider

$\big{\langle}f\big{\rangle}_{N}=\big{\langle}f(\cdot)\big{\rangle}_{N}=\int_{{\mathfrak{H}}_{N}}d\mu_{N}({H})\,f({H}).$

######  Theorem 4.2 (Wigner’s semi-circle law).

(4.24)
$\lim_{N\to\infty}\big{\langle}\rho_{N}(\lambda)\big{\rangle}_{N}=w_{sc}(\lambda)=\begin{cases}(2{\pi}J^{2})^{-1}\sqrt{4J^{2}-\lambda^{2}}&\quad\text{for}\quad|\lambda|<2J,\\\
0&\quad\text{for}\quad|\lambda|>2J.\end{cases}$

This expression is derived by introducing auxiliary odd variables
$\rho_{1},\rho_{2}$:

(4.25)
$\big{\langle}\rho_{N}(\lambda)\big{\rangle}_{N}=\pi^{-1}\Im\int_{\mathfrak{Q}}dQ\,\big{(}\\{(\lambda-i0)I_{2}-Q\\}^{-1}\big{)}_{bb}\exp{[-N{\mathcal{L}(Q)}]}$

Here $I_{n}$ is the $n\times n$ identity matrix, and

(4.26)
$\begin{gathered}{\mathcal{L}(Q)}=\operatorname{str\,}[(2J^{2})^{-1}Q^{2}+\log((\lambda-i0)I_{2}-Q)],\\\
{\mathfrak{Q}}=\big{\\{}Q=\begin{pmatrix}x_{1}&\rho_{1}\\\
\rho_{2}&ix_{2}\end{pmatrix}\,\big{|}\,x_{1},x_{2}\in{{{\mathfrak{R}}}_{\rm{ev}}},\;\rho_{1},\rho_{2}\in{{{\mathfrak{R}}}_{\rm{od}}}\big{\\}}\cong{\mathfrak{R}}^{2|2},\\\
dQ=\frac{dx_{1}dx_{2}}{2\pi}d\rho_{1}d\rho_{2},\\\
\big{(}((\lambda-i0)I_{2}-Q)^{-1}\big{)}_{bb}=\frac{(\lambda-i0-x_{1})(\lambda-i0-ix_{2})+\rho_{1}\rho_{2}}{(\lambda-i0-x_{1})^{2}(\lambda-i0-ix_{2})}.\end{gathered}$

For $2\times 2$-supermatrix $A$, $(A)_{bb}$ is boson-boson(or even-even) part,
in this case $A_{11}$.

Here, the key point of applying superanalysis is that in (4.25) parameter $N$
appeared in one place191919This fact makes us remind the appearance of $\hbar$
at one place in Feynman Path Integral, and this derivation isn’t justified
mathematically there. In physics literature, they apply formally saddle point
method or steepest descent to (4.25) when $N\to\infty$. Since they get

$\langle\frac{\delta{\mathcal{L}}(Q)}{\delta
Q},\tilde{Q}\rangle=\frac{d}{d\epsilon}{\mathcal{L}}(Q+\epsilon\tilde{Q})\bigg{|}_{\epsilon=0},$

critical point is given by

$\frac{\delta{\mathcal{L}}(Q)}{\delta
Q}=\operatorname{str\,}\big{(}\frac{Q}{J^{2}}-\frac{1}{\lambda-Q}\big{)}=0.$

Defining these as effective saddle points, putting

$Q_{c}=(\frac{1}{2}\lambda+\frac{1}{2}\sqrt{\lambda^{2}-4J^{2}})I_{2},$

they have

$\lim_{N\to\infty}\big{\langle}\rho_{N}(\lambda)\big{\rangle}_{N}=\pi^{-1}\Im(\lambda-
Q_{c})^{-1}_{bb}=w_{sc}(\lambda).\qquad\qquad\square$

######  Remark 4.7.

Though the derivation of expression (4.25) is justified in [71], we can’t
justify the usage of saddle point method in this setting202020Therefore, we
give a proof of a very small portion which is explained in Y.V.Fyodorov [44],
at that time. Since I feel this problem has relation to “Laplace’s method in
function space” in B. Simon [91], so I have an interest.

##  5\. Some examples of FDE(-Functional Derivative Equation)

###  5.1. A Schwinger-Dyson equation

####  5.1.1. A derivation of A Schwinger-Dyson equation of first order

A formally given quadratic Lagrangian, we try to give meaning to the first
order Schwinger-Dyson equation [53]: Let

(5.1) $\displaystyle L(q,v)$
$\displaystyle=\frac{1}{2}\int_{{\mathbb{R}}}dt(\dot{q}(t)^{2}-\omega_{0}^{2}q(t)^{2})$
$\displaystyle\qquad+\frac{1}{2}\int_{{\mathbb{R}}^{4}}dxdt(|v_{t}(x,t)|^{2}-|\nabla
v(x,t)|^{2})-\lambda\int_{{\mathbb{R}}^{4}}dxdt\delta(x)q(t)v(x,t)$

be given, then critical point of $L$ is calculated by

(5.2) $\displaystyle\frac{d}{d\epsilon}$ $\displaystyle
L(q+\epsilon\rho,v)\bigg{|}_{\epsilon=0}=\int_{{\mathbb{R}}}dt(\dot{q}(t)\dot{\rho}(t)-\omega_{0}^{2}q(t)\rho(t))-\lambda\int_{{\mathbb{R}}^{4}}dxdt\delta(x)\rho(t)v(x,t)$
$\displaystyle\qquad\qquad=\int_{{\mathbb{R}}}dt\bigg{(}-\frac{d^{2}}{dt^{2}}q(t)-\omega_{0}^{2}q(t)-\lambda\int_{{\mathbb{R}}^{3}}dx\delta(x)v(x,t)\bigg{)}\rho(t)$
$\displaystyle\qquad\qquad=\langle\frac{\delta L(q,v)}{\delta
q(t)},\rho(t)\rangle=0,$

and

(5.3) $\displaystyle\frac{d}{d\epsilon}$ $\displaystyle
L(q,v+\epsilon\varphi)\bigg{|}_{\epsilon=0}=\int_{{\mathbb{R}}^{4}}dxdt(v_{t}(x,t)\varphi_{t}(x,t)-\nabla
v(x,t)\nabla\varphi(x,t))$
$\displaystyle\qquad\qquad\qquad\qquad-\lambda\int_{{\mathbb{R}}^{4}}dxdt\delta(x)q(t)\varphi(x,t)$
$\displaystyle\qquad=-\int_{{\mathbb{R}}^{4}}dxdt(v_{tt}(x,t)-\Delta
v(x,t)+\lambda\delta(x)q(t))\varphi(x,t)$
$\displaystyle\qquad=\langle\frac{\delta L(q,v)}{\delta
v(x,t)},\varphi(x,t)\rangle=0.$

From these, at least formally, we may suppose the corresponding classical
orbit $(q(t),v(x,t))$ satisfy below:

(5.4) $\left\\{\begin{aligned}
(\frac{d^{2}}{dt^{2}}+\omega_{0}^{2})q(t)&=-\lambda\int_{{\mathbb{R}}^{3}}dx\delta(x)q(t)v(x,t),\\\
\square v(x,t)&=\lambda\delta(x)q(t).\end{aligned}\right.$

Putting the generating functional as

(5.5) $\hat{Z}=\hat{Z}(q,v)=e^{i\hbar^{-1}L(q,v)},$

we define its characteristic functional

(5.6) $\begin{gathered}Z(p,u)=Z_{0}^{-1}\int
d_{F}q\,d_{F}v\,\hat{Z}(q,v)e^{-i\hbar^{-1}(\langle q,p\rangle+\langle
v,u\rangle)}\\\ \quad\text{with}\quad Z_{0}=\int
d_{F}q\,d_{F}v\,\hat{Z}(q,v).\end{gathered}$

This $Z(p,u)$ satisfies formally the following equation:

(5.7) $\left\\{\begin{aligned}
(\frac{d^{2}}{dt^{2}}+\omega_{0}^{2})\frac{\delta Z(p,u)}{\delta
p(t)}&=-i\hbar^{-1}p(t)Z(p,u)-\lambda\int_{{\mathbb{R}}^{3}}dx\delta(x)\frac{\delta
Z(p,u)}{\delta u(x,t)},\\\ \square\frac{\delta Z(p,u)}{\delta
u(x,t)}&=-i\hbar^{-1}u(x,t)Z(p,u)-\lambda\delta(x)\frac{\delta Z(p,u)}{\delta
p(t)}.\end{aligned}\right.$

In fact, since

$\displaystyle\frac{\delta\hat{Z}(q,v)}{\delta q(t)}$
$\displaystyle=-i\hbar^{-1}(\frac{d^{2}}{dt^{2}}+\omega_{0}^{2})q(t)\hat{Z}-i\hbar^{-1}\lambda\int_{{\mathbb{R}}^{3}}dx\delta(x)v(x,t)\hat{Z},$
$\displaystyle\frac{\delta\hat{Z}(q,v)}{\delta v(x,t)}$
$\displaystyle=-i\hbar^{-1}\square
v(x,t)\hat{Z}-i\hbar^{-1}\lambda\delta(x)q(t)\hat{Z},$

assuming that functional integration permit integration by parts, we have

$\displaystyle Z_{0}^{-1}\int d_{F}q\,d_{F}v\,\frac{\delta\hat{Z}(q,v)}{\delta
q(t)}e^{-i\hbar^{-1}(\langle q,p\rangle+\langle
v,u\rangle)}=i\hbar^{-1}p(t)Z(p,u),$ $\displaystyle Z_{0}^{-1}\int
d_{F}q\,d_{F}v\,\frac{\delta\hat{Z}(q,v)}{\delta
v(x,t)}e^{-i\hbar^{-1}(\langle q,p\rangle+\langle
v,u\rangle)}=i\hbar^{-1}u(x,t)Z(p,u),$

Interchanging integration and differentiation formally, we get

$\displaystyle\frac{\delta Z(p,u)}{\delta p(t)}=-i\hbar^{-1}Z_{0}^{-1}\int
d_{F}q\,d_{F}v\,q(t)\hat{Z}(q,v)e^{-i\hbar^{-1}(\langle q,p\rangle+\langle
v,u\rangle)},$ $\displaystyle\frac{\delta Z(p,u)}{\delta
u(x,t)}=-i\hbar^{-1}Z_{0}^{-1}\int
d_{F}q\,d_{F}v\,v(x,t)\hat{Z}(q,v)e^{-i\hbar^{-1}(\langle q,p\rangle+\langle
v,u\rangle)}.$

From these, we have

$\displaystyle Z_{0}^{-1}\int d_{F}q\,d_{F}v$
$\displaystyle\bigg{(}-i\hbar^{-1}(\frac{d^{2}}{dt^{2}}+\omega_{0}^{2})q(t)\hat{Z}-i\hbar^{-1}\lambda\int_{{\mathbb{R}}^{3}}dx\delta(x)v(x,t)\hat{Z}\bigg{)}e^{-i\hbar^{-1}(\langle
q,p\rangle+\langle v,u\rangle)}$
$\displaystyle=(\frac{d^{2}}{dt^{2}}+\omega_{0}^{2})\frac{\delta
Z(p,u)}{\delta p(t)}+\lambda\int_{{\mathbb{R}}^{3}}dx\delta(x)\frac{\delta
Z(p,u)}{\delta u(x,t)}=i\hbar^{-1}p(t)Z(p,u).\qquad/\\!/$

####  5.1.2. Reformulation and calculation

##### Feynman propagator $E(x,t)=\square_{F}^{-1}(x,t)$.

As a formal solution of division problem $\square E(x,t)=\delta(x,t)$ in
$\mathcal{D}^{\prime}({\mathbb{R}}^{4})$, we have

$E(x,t)=\lim_{\epsilon\to 0}(2\pi)^{-4}\int_{{\mathbb{R}}^{4}}d\tau
d\xi\frac{e^{-it\tau+ix\xi}}{-\tau^{2}+|\xi|^{2}-i\epsilon}$

whose precise meanings are given in, for example, Gelfand and Shilov [46].

######  Remark 5.1.

Approximate $\delta(x)$ by $\rho_{\epsilon}(x)=\epsilon^{-2}\rho(x/\epsilon)$,
i.e. for $\rho(x)=\rho(|x|)\in C_{0}^{\infty}({\mathbb{R}}^{3})$, $\rho(x)\geq
0$ and $\int_{{\mathbb{R}}^{3}}dx\rho(x)=1$, making $\epsilon\to 0$ then
$\rho_{\epsilon}(x)\to\delta(x)$ in
${\mathcal{D}}^{\prime}({\mathbb{R}}^{3})$.

######  Lemma 5.1.

For any $u\in{\mathcal{S}}({\mathbb{R}}^{4})$,
$\square_{F}^{-1}u\in{\mathcal{S}}^{\prime}({\mathbb{R}}^{4})$. Moreover, when
$u\in C_{0}^{\infty}({\mathbb{R}}^{4})$, $\lim_{\epsilon\to
0}\langle\rho_{\epsilon}(\cdot),\square_{F}^{-1}u(\cdot,t)\rangle$ exists in
$H^{-1}({\mathbb{R}})$ and formally represented as
$\langle\delta,\square_{F}^{-1}u\rangle(t)$ or
$\langle\delta(\cdot),\square_{F}^{-1}u(\cdot,t)\rangle$.

##### Renormalization.

Applying $\square_{F}^{-1}$ to the second equation of (5.7), and putting this
into the first equation of (5.7), we get

(5.8) $\displaystyle(\frac{d^{2}}{dt^{2}}+\omega_{0}^{2}-\lambda^{2}$
$\displaystyle\int_{{\mathbb{R}}^{3}}dx\,\delta(x)\square_{F}^{-1}(x,t))\frac{\delta
Z(p,u)}{\delta p(t)}$
$\displaystyle=-i\hbar^{-1}p(t)Z(p,u)+i\hbar^{-1}\lambda\int_{{\mathbb{R}}^{3}}dx\,\delta(x)\square_{F}^{-1}u(x,t)Z.$

Here, we approximate $\delta(x)$ by $\rho_{\epsilon}(x)$. In this case,
putting

$(A_{\lambda}^{\epsilon}f)(t)=(\frac{d^{2}}{dt^{2}}+\omega_{0}^{2})f(t)-\lambda^{2}\langle\rho_{\epsilon}(\cdot),(\square_{F}^{-1}(\rho_{\epsilon}f))(\cdot,t)\rangle,$

using Fourier transformation and Plancherel theorem,

$\displaystyle(\widehat{A_{\lambda}^{\epsilon}f})(\tau)$
$\displaystyle=\frac{1}{\sqrt{2\pi}}\int d\tau
e^{it\tau}(A_{\lambda}^{\epsilon}f)(t)$
$\displaystyle=(-\tau^{2}+\omega_{0}^{2})\hat{f}(\tau)+\frac{\lambda^{2}}{4\pi^{2}}\int_{{\mathbb{R}}^{3}}d\xi\frac{|\hat{\rho}_{\epsilon}(\xi)|^{2}}{\tau^{2}-|\xi|^{2}+i0}\hat{f}(\tau)$
$\displaystyle=(-\tau^{2}+\omega_{0}^{2})\hat{f}(\tau)$
$\displaystyle\qquad\qquad+\frac{\lambda^{2}}{4\pi^{2}}\bigg{[}\int_{{\mathbb{R}}^{3}}d\xi|\hat{\rho}_{\epsilon}(\xi)|^{2}\bigg{(}\frac{1}{\tau^{2}-|\xi|^{2}+i0}+\frac{1}{|\xi|^{2}}\bigg{)}-\int_{{\mathbb{R}}^{3}}d\xi\frac{|\hat{\rho}_{\epsilon}(\xi)|^{2}}{|\xi|^{2}}\bigg{]}\hat{f}(\tau).$

Simply disregarding the last term which is $\infty$, making $\epsilon\to 0$,
then we have

$\displaystyle\lim_{\epsilon\to
0}\int_{{\mathbb{R}}^{3}}d\xi|\hat{\rho}_{\epsilon}(\xi)|^{2}$
$\displaystyle\bigg{(}\frac{1}{\tau^{2}-|\xi|^{2}+i0}+\frac{1}{|\xi|^{2}}\bigg{)}$
$\displaystyle=\int_{{\mathbb{R}}^{3}}d\xi\bigg{(}\frac{\tau^{2}+i0}{(\tau^{2}-|\xi|^{2}+i0)|\xi|^{2}}\bigg{)}=\int_{S^{2}}d\omega\int_{0}^{\infty}{\rho}^{2}d{\rho}\frac{\tau^{2}+i0}{(\tau^{2}-{\rho}^{2}+i0){\rho}^{2}}$
$\displaystyle=4\pi\int_{0}^{\infty}d{\rho}\frac{\tau^{2}+i0}{\tau^{2}-{\rho}^{2}+i0}=-4\pi(2\pi
i)\frac{\tau^{2}}{2|\tau|}=-4i\pi^{2}|\tau|$

Therefore

$\lim_{\epsilon\to
0}(\widehat{A_{\lambda}^{\epsilon}f})(\tau)=\\{(-\tau^{2}+\omega_{0}^{2})-i{\lambda^{2}}|\tau|\\}\hat{f}(\tau),$

so we define

$(A_{\lambda}^{R}f)(t)=(\frac{d^{2}}{dt^{2}}+\omega_{0}^{2}-i\lambda^{2}|\frac{d}{dt}|)f(t).$

When $\lambda>0$, for any $\tau\in{\mathbb{R}}$, since
$-\tau^{2}-{i\lambda^{2}}|\tau|+\omega_{0}^{2}\neq 0$, we have

######  Lemma 5.2.

Operator $A_{\lambda}^{R}(\lambda>0)$ is invertible, and
$(A_{\lambda}^{R})^{-1}$ maps $H^{-1}({\mathbb{R}})$ to $H^{1}({\mathbb{R}})$
as bounded operator. Moreover, for any $p\in H^{1}({\mathbb{R}})$, the
quantity $\langle(A_{\lambda}^{R})^{-1}p,p\rangle$ is well-defined.

######  Problem 5.1.

Though I wrote this paper, it isn’t clear the sentence “Simply disregarding
the last term”. For example, whether we may find analogous regularization,
such as zeta regularization for divergent series or divergent integral in [1]
or other summability methods?

##### Renormalized FDE equation and a result

Now, we have the renormalized FDE equation

(5.9) $\displaystyle A_{\lambda}^{R}\frac{\delta Z(p,u)}{\delta p(t)}$
$\displaystyle=-i\hbar^{-1}p(t)Z(p,u)+i\hbar^{-1}\lambda\int_{{\mathbb{R}}^{3}}dx\delta(x)\square_{F}^{-1}u(x,t)Z(p,u),$
$\displaystyle\square\frac{\delta Z(p,u)}{\delta u(x,t)}$
$\displaystyle=[i\hbar^{-1}u(x,t)+i\hbar\lambda^{2}(\delta(x)(A_{\lambda}^{R})^{-1}\langle\delta(\cdot),\square_{F}^{-1}u(\cdot,t)\rangle)$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad-i\hbar^{-1}\lambda(\delta(A_{\lambda}^{R})^{-1}p)(x,t)]Z(p,u)$

whose solution is obtained explicitly as

(5.10) $\displaystyle Z(p,u)=\exp$
$\displaystyle{[i(2\hbar)^{-1}\langle(A_{\lambda}^{R})^{-1}p,p\rangle+i(2\hbar)^{-1}\langle\square_{F}^{-1}u,u\rangle}$
$\displaystyle\qquad{+i(2\hbar)^{-1}\lambda^{2}\langle(A_{\lambda}^{R})^{-1}\langle\delta,\square_{F}^{-1}u\rangle(t),\langle\delta,\square_{F}^{-1}u\rangle(t)\rangle}$
$\displaystyle\qquad\qquad{-i\hbar^{-1}\lambda\langle(A_{\lambda}^{R})^{-1}p,\langle\delta,\square_{F}^{-1}u\rangle(t)\rangle]}$

and all Green function is obtained as bellow:

$\displaystyle G^{(n,m)}$
$\displaystyle(t_{1},\cdots,t_{n},(y_{1},s_{1}),\cdots,(y_{m},s_{m})))$
$\displaystyle\qquad=(i\hbar)^{n+m}\frac{\delta^{n+m}Z(0,0)}{\delta
p(t_{1})\cdots\delta p(t_{n})\delta u(y_{1},s_{1})\cdots\delta
u(y_{m},s_{m})}.$

Small goals: Enlarging Feynman’s idea slightly, following afore mentioned
Itzykson-Zuber’s claim, we dare to imagine “Quantum mechanics(Quantum field
theory) stands for obtaining properties of generating function212121For the
time being, we don’t concern with how these functions relate to the physical
quantities observed by experiments represented by Lagrange or Hamilton
function using Feynman measure”. To relate these quantities to classical
mechanics(classical field theory) and to get rid of the use of non-existing
Feynman measure, we take formally the characteristic function which satisfies
certain FDE222222Since in general, it contains higher order derivatives, we
need new device contrivance such as Colombeau’s generalized functions and give
meaning to it after renormalization and solve it!

As is known, to study PDE without getting explicit solution of it, we fully
use Lebesgue measure to have the existence proof using integration by parts,
change of variables formula under integral sign and Fourier transformations
with the method of functional analysis232323Project called “A study of PDE by
functional analytic method” initiated by K. Yosida in Japan.. As mentioned
before, not only there doesn’t exist a translational invariant, completely
additive measure in function spaces, but also giving meaning to the trace of
higher order functional derivatives is hard. For a simple quadratic
interaction, we give meaning to some devices eliminating $\infty$. This
suggests that as is used by physicists, expand w.r.t. interaction parameter
and performing Gaussian type integration, after summation method, we get
something-like a solution of FDE.

Moreover, in the next paragraph, I give a trial to understand the trace of 2nd
order functional derivatives.

###  5.2. Hopf equation

Let $(M,g_{jk})$ be a compact Riemannian manifold of dimension $d$ with or
without boundary $\partial M$. We denote by
${\overset{\circ}{\mathbf{X}}}_{\sigma}(M)$ and
${\overset{\circ}{\Lambda}}{}_{\sigma}^{1}(M)$, the space of all solenoidal
vector fields on $M$ which vanish near the boundary and that of all divergence
free $1$-forms on $M$ which vanish near the boundary, respectively.
$\tilde{\mathbf{H}}$ (resp. $\mathbf{H}$) stands for the completion of the
space ${\overset{\circ}{\Lambda}}{}_{\sigma}^{1}(M)$ (resp.
${\overset{\circ}{\mathbf{X}}}_{\sigma}(M)$) w.r.t.
$\tilde{\mathbf{L}}^{2}$-norm (resp. ${\mathbf{L}}^{2}$-norm).

####  5.2.1. Differential geometrical expression of Navier-Stokes equation

For a vector field $u=u^{j}\displaystyle{\frac{\partial}{\partial x^{j}}}$, we
have the Navier-Stokes equation

(5.11) $\frac{\partial u^{i}}{\partial t}-\nu(\Delta
u)^{i}+(\nabla_{u}u)^{i}+p^{i}=f^{i},$

with

$\delta u=\frac{1}{\sqrt{g}}\frac{\partial}{\partial
x^{i}}({\sqrt{g}}u^{i})=0,\quad
u(0,x)={\underline{u}}(x)\;\mbox{and}\;u(t,x)\big{|}_{x\in\partial M}=0.$

Here

$(\Delta u)^{i}=\nabla^{k}\nabla_{k}u^{i}-R^{i}_{k}u^{k},etc.$

####  5.2.2. Hopf’s motivation

Following Hopf’s introduction in [52], we quote

> The differential law which governs the motion of a deterministic mechanical
> system has the symbolic form
>
> $\frac{du}{dt}={\mathfrak{F}}(u)$
>
> where $u$ is an instantaneous phase of the system and where the right jand
> side is completely determined by the phase. $\cdots\cdots$
>
> There are mechanical systems the phases of which are characterized by a very
> large number of independent parameters and the phase motions of which are
> tremendously complicated. Two examples are the classical model of a gas with
> its very large number of degrees of freedom and the flow of viscous
> incompressible fluid at a very large value of overall Reynolds number.
>
> In both cases the important task is not the determination of the exact phase
> motion with an exactly given initial phases but the determination of the
> statistical properties of the “typical” phase motion. In order to achieve
> this goal statistical mechanics studies probability distributions of
> simultaneous phases and their evolution in time resulting from the
> individual phase motions. $\cdots\cdots$

Moreover, he claims to find statistical equilibrium of the system, we need to
seek stationary phase distribution. This idea is comparable to find
equilibrium state of Hamilton flow, we try to find a stable solution of
Liouville equation.

####  5.2.3. Derivation of Hopf equation

Let $T_{t}u_{0}$ be a (some kind of) solution of Navier-Stokes equation (NS)
with the initial data $u_{0}$, for a Borel measure $\mu(\omega)$ on ${\bf H}$,
we put $\mu_{t}(\omega)=\mu(T_{t}^{-1}(\omega))$. Here, $\omega$ is a Borel
set on $\bf{H}$. The characteristic function of this measure $\mu_{t}(\omega)$
should satisfy the equation, called Hopf equation. More precisely,

(I) the characteristic function of this measure $d\mu_{t}(\cdot)$ is given by

$W(t,\eta)=\int_{\bf{H}}d\mu_{t}(u)\,e^{i\langle
u,\eta\rangle}=\int_{\bf{H}}d\mu(u)\,e^{i\langle T_{t}u,\eta\rangle}$

which satisfies

(5.12) $\displaystyle\frac{\partial}{\partial{t}}W(t,\eta)$
$\displaystyle=\int_{M}d_{g}x\left[-i\\{\frac{\partial}{\partial
x^{k}}\eta_{j}(x)-\Gamma^{\ell}_{jk}(x)\eta_{\ell}\\}\frac{\delta^{2}W(t,\eta)}{\delta\eta_{j}(x)\delta\eta_{k}(x)}\right.$
$\displaystyle\qquad\qquad\qquad\qquad\left.+\nu(\Delta\eta)_{j}(x)\frac{\delta
W(t,\eta)}{\delta\eta_{j}(x)}+i\eta_{j}(x)f^{j}(t,x)W(t,\eta)\right].$

One of the additional conditions to this equation is

(5.13) $\frac{1}{\sqrt{g(x)}}\frac{\partial}{\partial
x^{j}}\\{\sqrt{g(x)}\frac{\delta W(t,\eta)}{\delta\eta_{j}(x)}\\}=0,$ (5.14)
$W(0,\eta)=W_{0}(\eta)\;\mbox{and}\;W(t,0)=1.$

Here, $t\in(0,\infty)$ and

$\eta=\eta(x)=\eta_{j}(x)dx^{j}\in\overset{\circ}{\Lambda}{}^{1}_{\sigma}(M),\quad
f=f(t,x)=f^{j}(t,x)\partial/\partial
x^{j}\in\overset{\circ}{\bf{X}}{}_{\sigma}(M).$

Given positive definite functional $W_{0}(\eta)$ on $\tilde{\bf H}$ satisfies

(5.15)
$W_{0}(\eta)=1\;\mbox{and}\;\frac{1}{\sqrt{g(x)}}\frac{\partial}{\partial
x^{j}}\\{\sqrt{g(x)}\frac{\delta W_{0}(\eta)}{\delta\eta_{j}(x)}\\}=0.$

This equation is something like Liouville equation corresponding to Navier-
Stokes equation. In fact, let $u(t,x)=T_{t}{\underline{u}}(x)$ be a “solution”
for Navier-Stokes equation with initial data
$\underline{u}=\underline{u}^{j}\partial/\partial x^{j}\in\mathbf{H}$. For
Borel measure $\mu(\omega)$ on ${\bf H}=L^{2}_{\sigma}(\Omega)$, put
$\mu_{t}(\omega)=\mu(T_{t}^{-1}(\omega))$ for any Borel set $\omega$ in
$\bf{H}$. More explicitly, put

(5.16) $\Phi(t,v)=\int e^{i(v,T_{t}u)}\mu(du)=\int e^{i(v,u)}\mu_{t}(du),$

Then, this functional satisfies Hopf equation formally. In the above, for $u$,
we put $\tilde{u}$ as the solenoidal part of $u$.

####  5.2.4. How to give the meaning to the trace of 2nd order functional
derivative

######  Problem 5.2.

Our problems here are, whether does there exist a funtional $W(t,\eta)$
satisfying (5.12), and how to give meaning to the 2nd order functional
derivatives
$\displaystyle{\frac{\delta^{2}W(t,\eta)}{\delta\eta_{j}(x)\delta\eta_{k}(x)}}$?

Above problem (I) is restated as

(II) Find a Borel measure $\mu(t,\cdot)$ on ${\bf H}$ which satisfies

(5.17) $\displaystyle-\int_{0}^{\infty}\int_{\bf H}$ $\displaystyle
d\mu(t,u)\frac{\partial\Phi(t,u)}{\partial t}-\int_{\bf
H}d\mu_{0}(u)\Phi(0,u)$ $\displaystyle=\int_{0}^{\infty}\int_{\bf
H}\int_{M}d_{g}xd\mu(t,u)dt\left[-u^{k}(x)u^{j}(x)\frac{\partial}{\partial
x^{k}}\frac{\delta\Phi(t,u)}{\delta
u^{j}(x)}+\Gamma^{j}_{k\ell}(x)u^{k}(x)u^{\ell}(x)\frac{\delta\Phi(t,u)}{\delta
u^{j}(x)}\right.$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\left.+\nu\nabla_{k}u^{j}(x){\cdot}\nabla^{k}\frac{\delta\Phi(t,u)}{\delta
u^{j}(x)}-f^{j}(t,x)\frac{\delta\Phi(t,u)}{\delta u^{j}(x)}\right]$

for any suitable test functionals $\Phi(t,u)$.

######  Remark 5.2.

There appeared only 1st order Functional Derivatives in (II)!

######  Theorem 5.3 (A part of Theorem A’ of Inoue [54]).

Let $f(\cdot)\in L^{2}((0,\infty);{\bf{V}}^{-1})$ be given and suppose a Borel
measure $\mu_{0}$ on $\bf{H}$ satisfies

$\int_{\bf{H}}d\mu_{0}(u)(1+|u|^{2})<\infty.$

Then, there exists a basic family $\\{\mu(t,\cdot)\\}_{0<t<\infty}$ of Borel
measures on $\bf{H}$ such that

$\displaystyle\int_{0}^{\infty}dt\,\int_{\bf{H}}d\mu(t,u)$
$\displaystyle\Phi_{t}(t,u)+\int_{\bf{H}}d\mu_{0}(u)\Phi_{t}(0,u)$
$\displaystyle=\int_{0}^{\infty}dt\left[\int_{\bf{H}}d\mu(t,u)\left\\{\nu
a(u,\tilde{\Phi}_{u}(t,u))+b(u,u,\tilde{\Phi}_{u}(t,u))-\langle
f(t),{\Phi}_{u}(t,u)\rangle\right\\}\right]$

for any $\Phi\in TF$ with compact support in $t$, i.e. there exists a constant
$T$ depending on $\Phi$ such that $\phi(t,\cdot)=0$ when $t\geq T$.

In the above, forms $a$ and $b$ are defined by

$\begin{gathered}a(u,v)=\int_{M}d_{g}x\,g_{ij}\nabla_{k}u^{i}\nabla^{k}v^{j},\\\
b(u,v,w)=(\nabla_{u}v,w)=\int_{M}d_{g}x\,g_{ij}\\{u^{k}\frac{\partial}{\partial
x^{k}}v^{i}+\Gamma^{i}_{k\ell}u^{k}v^{\ell}\\}w^{j}\end{gathered}$

for $u,v,w\in{\bf{X}}(M)$ with $u=u^{j}\frac{\partial}{\partial x^{j}}$ etc.

######  Definition 5.1 (Definition 2.7 of [54]).

A real functional $\Phi(\cdot,\cdot)$ on $[0,\infty)\times\bf{V}$ is called
TF(=test functionl) denoted by $\Phi\in TF$ if it satisfies the following.

  1. (1)

$\Phi(\cdot,\cdot)$ is continuous on $[0,\infty)\times\bf{V}$ and verifies

$|\Phi_{u}(t,u)|\leq c\;\mbox{and}\;|\Phi_{t}(t,u)|\leq c+c|u|$

where $\Phi_{u}(t,\cdot)$ is regarded as an element in $\tilde{\bf{H}}$.

  2. (2)

$\Phi(\cdot,\cdot)$ is Fréchet $\bf{H}$-differentiable in the direction
$\bf{V}$.

  3. (3)

$\Phi_{u}(\cdot,\cdot)$ is continuous on $[0,\infty)\times\bf{V}$ to
$\tilde{\bf{V}}^{s}$ and is bounded, i.e. there exists a constant $c$
depending on $\Phi$ such that

$\|\Phi_{u}(t,u)\|_{s}\leq c\;\mbox{and}\;|\Phi_{t}(t,u)|\leq
c+c|u|\quad\quad\text{for any}\quad(t,u)\in[0,\infty)\times\bf{V}.$

##  Appendix A Some questions on Euler or Navier-Stokes equations

Though the following results are rather recent one which make me not only
astonish, but also such impression is criticized because “ These results don’t
match physical reality” by physicists and rather bypassed [84, 85] saying,
Euler equation admits various unphysical weak solutions. Even though, one may
construct a weak solution with compact support in a torus dimension 2, a
gravitational equation which is derived from classical idea

On the other hand, by the theory of general relativity, the following
Einstein’s field equation

$G_{\mu\nu}+\Lambda g_{\mu\nu}=\kappa T_{\mu\nu}$

is derived where $G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}$ is the
Einstein tensor, $g_{\mu\nu}$ is the metric tensor, $\Lambda$ is the
cosmological constant and $\kappa$ is the Einstein gravitational constant.
Since from this equation, it is claimed the universe begins abruptly by Big
Bang, I feel some analogy with above result.

We investigate the initial-boundary value problem

(NS) $u_{t}-\nu\Delta u+u\nabla u-\nabla p=f,\quad{\rm div}\,u=0,\quad
u(0,x)=u_{0}(x),\quad u|_{\partial\Omega}=0,$

or

(E) $u_{t}+u\nabla u-\nabla p=0,\quad{\rm div}\,u=f,\quad
u(0,x)=u_{0}(x),\quad u\cdot{\mathbf{n}}=0$

If we introduce the idea of weak solution for the Euler equation by
integration by parts with suitable test functions, we have a result for Euler
equation [27, 28].

######  Theorem A.1 (Theorem 1.7 of Isett and Oh [73]).

(Onsager’s conjecture on manifolds, sharp version). Let $(M,g_{jk})$ be a
compact Riemannian manifold and $I\subseteq{\mathbb{R}}$ an open interval. Let
$(u^{\ell},p)$ be a weak solution to the Euler equations on $I\times M$ such
that $u^{\ell}\in L^{3}_{t}(I;B^{1/3}_{3,c_{0}}(M))\cap C_{t}(I;L^{2}(M))$.
Then, conservation of energy (E) holds.

Analogously, we have

######  Theorem A.2 (Theorem1.2(Non-uniqueness of weak solutions for
(NS))[20]).

There exists $\beta>0$, such that nonnegative smooth function
$e(t):[0,T]\to{\mathbb{R}}_{\geq}$, there exists $v\in
C_{t}^{0}([0,T];H_{x}^{\beta}({\mathbb{T}}^{3}))$ a weak solution of Navier-
Stokes equation (NS), such that $\int_{{\mathbb{T}}^{3}}dx|v(x,t)|^{2}=e(t)$
for all $t\in[0,T]$. Moreover, the associated vorticity $\nabla\times v$ lies
in $C_{t}^{0}([0,T];L_{x}^{1}({\mathbb{T}}^{3}))$.

######  Theorem A.3 (${\mathbb{R}}^{2}$: 1993 Scheffer [87]).

There exists a weak solution, not identically zero, of (E) in
${\mathbb{R}}^{2}$, having compact support in
$(x,t)\in{\mathbb{R}}^{2}\times(0,T)$.

######  Theorem A.4 (${\mathbb{T}}^{2}$: 1997 Shnirelman [89]).

There exists a weak solution, not identically zero, of (E) in
$\mathbb{T}^{2}$, having compact support in $t\in(0,T)$.

######  Theorem A.5 (Theorem 1.3 of Buckmaster et al [19]).

For any $\epsilon>$, there exists a non-trivial continuous weak solution
$v:\mathbb{T}^{3}\times{\mathbb{R}}\to{\mathbb{R}}^{3}$ of (E), with $v\in
L^{1}(C^{1/3-\epsilon})$ with compact support in time.

This stands for the collapse of the uniqueness, that is, if there exists very
rapidly oscillating exterior force, then there exists a weak solution having
space-time compact support.

######  Theorem A.6 ([87, 89, 90]).

There exists a weak solution $u(x,t)\in
L^{2}({\mathbb{R}}^{2}\times{\mathbb{R}})$ such that $u(x,t)\equiv 0$ for
$|x|^{2}+|t|^{2}>1$.

Concerning this, the following paragraph suggests us we need to use new test
functions to recognize anomalous phenomena.

> [Shnirelman [89]] The weak solution constructed by [87, 89] is not in fact a
> solution; very strong external forces are present, but they are infinitely-
> fast oscillating in space, and therefore are indistinguishable from zero in
> the sense of distributions. The smooth test-functions are not “sensitive”
> enough to “feel” these forces. This is the fault of sensors, not of forces.

######  Theorem A.7 (Theorem1.1 of [101]).

Let $\Omega$ be a bounded domain in ${\mathbb{R}}^{3}$ with $C^{2}$-boundary
$\partial\Omega$. Let $u\in L^{\infty}(0,T;L^{2}(\Omega))\cap
L^{2}(0,T;H^{1}(\Omega))$ be a weak solution of the Navier-Stokes equation for
any smooth test function $\varphi\in C^{\infty}({\mathbb{R}}^{+}\times\Omega)$
with compact support, and ${\rm div}\,\varphi=0$. In addition, if

$u\in L^{p}(0,T;L^{q}(\Omega))\cap L^{s}(0,T;B_{s}^{\alpha,\infty}(\Omega))$

for any $\frac{1}{p}+\frac{1}{q}\leq\frac{1}{2},\;q\geq 4,\;s>2$ and for any
$\frac{1}{2}+\frac{1}{s}<\alpha<1$. Then, for any $t\in[0,T]$,

$\int_{\Omega}dx|u(t,x)|^{2}+2\nu\int_{0}^{t}\int_{\Omega}dtdx|\nabla
u|^{2}=\int_{\Omega}dx|u_{0}|^{2}.$

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