Uniqueness for PDE solutions on the space of distributions

Question

There is an apparently basic question about PDE uniqueness in the distributions framework which I did absolutely not find treated anywhere. Any helpful comment would be very appreciated.

1. Example: To begin with, denote $D_n=\mathscr D'(\mathbb R^n)$ the space of (standard) distributions on $\mathbb R^n$. Question: Is any $u\in D_1$ which satisfies \begin{equation} u' = u \end{equation} (in the sense of distributions) a real multiple of the exponential function?

2. Example: Given two distributions $u,v\in D_{2}$ satisfying both the Kolmogorov equation \begin{equation} \partial_t w - \mathcal A^\ast w + V \cdot w = 0,\\ \qquad w \in\{u,v\}, \end{equation} for a) some continuous function $V\in C^0(\mathbb R^{2})$ bounded from below, and b) $\mathcal A$ being the infinitesimal generator of the solution to the stochastic differential equation $ dX_t = \mu(t,X_t) dt +\sigma(t,X_t) dW_t $ with $\mu,\sigma \in C^\infty(\mathbb R^2)$ Lipschitz and such that the PDE satisfies the uniform "ellipticity" condition, so that \begin{equation} \mathcal A^\ast w= \partial_{xx}(\sigma^2/2 \cdot w) - \partial_x(\mu \cdot w) \end{equation} and such that the above PDE admits at most one weak solution. In order to include initial condition in some way, consider $D_2$ as a $C^\infty(\mathbb R)$-module via the inclusion $C^\infty(\mathbb R) \subset C^\infty(\mathbb R^2), f \mapsto ((t,x) \mapsto f(x))$ and the standard action by multiplication. Question: Are $u$ and $v$ linearly dependent in the module $D_2$?

3. General question: Does the answer to the second question generalize to other second-order PDE satisfying the uniform "ellipticity" condition? Are there restrictions to generalizing them to $\mathbb R^{1+n}$?

I wish you a nice day.

Answer 1

Edit: After the original question was changed in order to ask about a more general concept of linear independence between two given solutions of a first order (in time) operator equation, I added a section to each answer in order to answer it in the required more general set-up: added sections are highlighted by the word "Addendum".

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1. Yes, since the equation $$ u'=u.\label{a}\tag{A} $$is a linear first order ODE and thus the set of its solutions, i.e. the kernel of the operator $L(\cdot)=\frac{\mathrm{d}}{\mathrm{d}x}(\cdot)-1(\cdot)$, is a $1$-dimensional real vector space whose basis is the function $e^t$.
   Addendum: the assertion is easily proved for classical solutions of \eqref{a} (see for example [1], specifically Barrow's formula, §1.5 p. 19, and the related existence and uniqueness theorem, §2.2 p. 36). However, as remarked by Héhéhé in their answer to this part of the question, all generalized solutions of \eqref{a} are classical solutions. To see this, let $u\in C^\infty(\Bbb R)$ be the (unique) classical solution, and consider a distribution $v\in\mathscr{D}'(\Bbb R)$ such that $uv$ is a distributional solution of \eqref{a}, i.e. $(uv)'=uv$: calculating the generalized derivative we have that, for all $\varphi\in C^\infty_0(\Bbb R)$ $$ \begin{split} (uv)'&\triangleq -\langle u v,\varphi'\rangle=-\langle v,u\varphi'\rangle\\ &=-\langle v,(u\varphi)'\rangle+\langle vu',\varphi\rangle\\ &= uv'+vu'=uv'+uv=uv \end{split} $$ Then $uv'=0\iff v'=0\implies v\equiv\mathrm{const.}$ by a standard property of distributions (see for example [2], §1.5.1 pp. 23-24), thus for equation \eqref{a} the notions of classical and generalized solutions coincide.

2. In general no, since the kernel of the operator $$ L(\cdot)=\partial_t (\cdot) - \mathcal A^\ast (\cdot) + V \cdot (\cdot ),\label{1}\tag{1} $$ while still being a linear vector space (of distributions or other generalized functions), is not $2$-dimensional but it is infinite dimensional (at least if \eqref{1} it is not trivial). To see this it is sufficient to consider the Cauchy problem for the operator \eqref{1} $$ \begin{cases} \partial_t w - \mathcal A^\ast w + V \cdot w = 0\\ w|_{t=0}=w_0 \end{cases}\label{2}\tag{2} $$ If we choose the initial condition $w_{0}\in\{u_0,v_0\}$ where $u_0,v_0$ are two non-null non-linearly dependent distributions (for example $u_0=e^{-x^2}\sin x $ and $v_0=e^{-x^2}\cos x $) then we have at two distributional solutions $w_1$ and $w_2$ which are not linearly dependent at least on $\{t\in\Bbb R\:|\:t=0\}\times\Bbb R\subset\Bbb R^2$.
   Addendum: even for the generalized version of the second question, the answer is in general no. Let's see why by restricting to the Schwartz space of slowly increasing distributions $\mathscr{S}'(\Bbb R)$ and applying, at least on a formal level, the theory of pseudo differential operators, ad described for example in the book [1] (the whole chapter IV up to and including §4.7 is particularly relevant to our time evolution problem). By applying the partial Fourier transform respect to the $x\in\Bbb R$ variable to $w$, we get $$ \begin{align} w(t,x)\mapsto\widehat w(t,\xi)&=\mathscr{F}_{x\to\xi}[w(t,x)](x)\\ &\Updownarrow\\ w(t,x)=\frac{1}{2\pi}&\int\limits_{\Bbb R}e^{ix\xi}\widehat w(t,\xi)\mathrm{d}\xi\label{3}\tag{3} \end{align} $$ Now, by substituting \eqref{3} in \eqref{2} we "algebrize" the operator $\mathcal{A}^\ast=\mathcal{A}^\ast(x,\partial_x)$ and we have the following equation, similar to \eqref{a} $$ \frac{\mathrm{d}}{\mathrm{d}t}\widehat w(t,\xi)=\Big(\widehat{\mathcal{A}^\ast}(t,\xi)+V(x)\Big)\widehat{w}(t,\xi) \label{4}\tag{4} $$ where $\widehat{\mathcal{A}^\ast}(x,\xi)=\mathscr{F}_{x\to\xi}[\mathcal{A}^\ast(x,\partial_x)](\xi)$, or more precisely $\widehat{\mathcal{A}^\ast}(x,\xi)$ is the variable coefficient polynomial in $\xi$ obtained by the "formal" substitution $\partial_x\mapsto\xi$, after expanding the terms $\partial_x(f(x)\cdot w(t,x))$ in its expression. Now we are able to obtain the following "formal" solution to \eqref{a} $$ \widehat{w}(t,\xi)=e^{t\left[\widehat{\mathcal{A}^\ast}(t,\xi)+V(x)\right]}\widehat{w}_0(\xi)\label{5}\tag{5} $$ where $\widehat{w}_0$ is the Fourier transform of the initial condition $\widehat{w}_0$, and thus also the "formal" solution of \eqref{2} by substituting it in the right side of \eqref{3}. We have now reached the core of the question: \eqref{4} is not the same as \eqref{a} even in the case when $\mathcal A^\ast$ does not depend on $x$. Apart from the fact that the ordinary product of Fourier transforms becomes a convolution when passing from \eqref{5} to \eqref{3}, thus $D_2$ is never a module respect to ordinary multiplication, but at least could be a module respect to convolution, there is another deeper problem. If $\widehat{u}=\widehat{u}(t,\xi)$ and $\widehat{v}=\widehat{v}(t,\xi)$, in general $$ \widehat{u}(t,\xi)\neq \alpha({\xi} )\widehat{v}(t,\xi) $$ since, in general, for any $\widehat{u}_0,\widehat{v}_0\in \mathscr{S}^\prime(\Bbb R)$ $$ \:\nexists \alpha\in\mathscr{S}^\prime(\Bbb R)\:\text{ such that }\:\widehat{u}_0=\alpha\widehat{v}_0 $$ because $\mathscr{S}^\prime(\Bbb R)$ is not a ring but only an algebra, thus there is no hope to find always a multiplicative inverse. Thus, by restricting the initial data to the space of Schwartz distributions, you can expect at best to work in left module respect to a convolution algebra whit no reasonable linear dependency concept inherited from linear algebra: and passing to the whole space $\mathcal{D}$, the situation surely worsens.

3. The answer is yes: as for the second question you cannot expect the solution of a general uniformly elliptic equation linear second order PDE to be lineraly dependent: this happens again since the kernel of such kind of operator is an infinite dimensional vector space. For example think at the most standard uniformly elliptic operator, i.e. the Laplacian $\Delta$: the equation $$ \Delta u(x)=0\quad x\in \Bbb R^n \text{ (for any $n$)} $$ admits as solutions the whole vector space of solid harmonics, which is infinite dimensional.
   Addendum: the same reasoning used to answer the second question is valid also in this case.

Notes

• The proof of the coincidence of classical and generalized solution of \eqref{a} is equivalent to the one given by Héhéhé: however I preferred to give a version adapted from [1], §1.5.2 pp. 25-26 since that approach can be used with almost no changes to prove that classical and generalized solution coincide also for first order $n\times n$ homogeneous linear system of ODE $$ \pmb{y}'(t)= \mathbf{A}(t)\pmb{y}(t), $$ where $\mathbf{A}(t)$ is a non-singular matrix with $C^\infty$ coefficients.
• Despite the fact that in general we have not a module structure allowing for a concept of "linear dependence between the solutions of the Cauchy problem for a linear evolution equation like \eqref{2}", for some classes of differential operators, this may be possible. A classical source on the theory of algebraic structures related to linear differential operators is Jan Erik Bjork's monograph "Rings of differential operators" published by Elsevier in 1979: if you want to dive deep in this aspect of the theory, it is a nice place to start from since it is still very readable from people with a consistent mathematical analysis background and rudimentary algebraic knowledges.

Bibliography

[1] V. I. Arnol'd, "Ordinary differential equations", various editions from MIT Press and from Springer-Verlag, MR1162307 Zbl 0744.34001.

[2] G. E. Shilov (1968), Generalized functions and partial differential equations, Mathematics and Its Applications, Vol. 7, (English) New York-London-Paris: Gordon and Breach Science Publishers, XII+345, MR0230129, Zbl 0177.36302.

[3] M. E. Taylor (1981), Pseudodifferential operators, Princeton Mathematical Series, 34. Princeton, New Jersey: Princeton University Press. pp XI+452, ZBL0453.47026.

Answer 2

For the question 1, we know that the fonction $t \mapsto e^t$ is a solution of $u^\prime = u$. Now, consider another distribution $v \in \mathscr D^\prime(\mathbb R)$ such that $v^\prime = v$. Then $w=v/u$ belongs to $\mathscr D^\prime(\mathbb R)$ and verifies $w^\prime = 0$ (note that $1/u \in C^\infty$ so $v/u$ makes sense). It implies that $w = \alpha$ where $\alpha \in \mathbb R$ and it follows that $v = \alpha u$, i.e. $v$ is a function in the classical sense and $v(t) = \alpha e^t$ for all $t \in \mathbb R$.

Note that this proof is almost exactly the same that shows the unicity of the exponential.