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It is well known that the set of solutions $u:\mathbb{R}\rightarrow \mathbb{R}$ of an $n$-th order, linear, homogeneous ordinary differential equation $$a_n(x)\frac{d^n u}{dx^n}+\dots + a_1(x)\frac{du}{dx}+a_0(x)u=0$$ form an $n$-dimensional vectorspace under the usual addition and scalar multiplication of functions. Does this generalise to PDEs, and if so how? For example, the PDE $$\frac{\partial^2u}{\partial x^2}=0$$ has solutions $u=f(y)x+g(x)$ for any functions $f,g$. Clearly the set of solutions has infinite dimension when simply viewed, as in the ODE case, as a vector space under the usual addition and scalar multiplication of functions. However, it is parametrised by two functions. Is there some meaningful sense in which the solution space is $2$-dimensional? And if so, is it true that any set of solutions $u:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$ to a second-order, linear, homogeneous PDE $$a(x,y)\frac{\partial^2 u}{\partial x^2} + b(x,y)\frac{\partial^{2} u}{\partial x\partial y} + c(x,y)\frac{\partial^2 u}{\partial y^2} +d(x,y)\frac{\partial u}{\partial x} + e(x,y)\frac{\partial u}{\partial y} + a_1(x)\frac{\partial u}{\partial x} + a_0(x,y)u=0$$ is $2$-dimensional in that sense?

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3 Answers 3

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In many ways the natural generalization of ODEs are hyperbolic PDEs (in that they admit wellposed initial value problems). What you have here is one of those ways. Roughly speaking for a linear hyperbolic PDE on $\mathbb{R}^{1+n}$ (where $n$ may be zero and in which case we have an ODE), the solution is entirely determined by $k$ free functions prescribed on $\{0\} \times \mathbb{R}^n$, where $k$ is the degree of the PDE. That you have a finite dimensional vector space in the ODE case is just a coincidence due to the triviality of $\mathbb{R}^0 := \{0\}$.

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  • $\begingroup$ Being hyperbolic as well as elliptic linear ODE are a very exceptional class of PDE. $\endgroup$
    – user80744
    Oct 4, 2015 at 13:21
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    $\begingroup$ so in an extremely weak sense it is a finite dimensional free module over the ring of (whatever nice properties you put here) functions. $\endgroup$
    – Fan Zheng
    Oct 19, 2015 at 3:40
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It's not 2-dimensional because, as you observed, the solution contains two arbitrary functions of 1 variable. This means that the solution space has infinite dimension, but the right way to think about "counting" solutions is that the general solution contains a certain number of functions involving a certain number of independent variables. So for the example above, we say that the space of solutions "is parametrized by 2 arbitrary functions of 1 variable." Intuitively, this means that in order to determine a unique solution, you have to specify initial data along some curve in $\mathbb{R}^2$ consisting of two functions along the curve. For your example above, typical initial data would take the form

$$ u(y,0) = g(y), \qquad u_x(y,0) = f(y). $$

Then the corresponding solution would be $$ u(x,y) = x f(y) + g(y). $$

For a general PDE, the story is a little more complicated - for instance, there may or may not exist global solutions for a given initial value problem, and even for local solutions the question of how an initial value problem should be posed (e.g., along which curves in $\mathbb{R}^2$ might it be appropriate to specify initial data?) depends on the PDE---mainly on the configuration of its characteristic curves. But most of the time (and certainly in the linear case) the space of local solutions to a single nondegenerate second-order PDE in a neighborhood of some point $(x,y) \in \mathbb{R}^2$ will be parametrized by 2 arbitrary functions of 1 variable.

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  • $\begingroup$ Right, it's clear what the intuition is, but can this actually be made rigorous the same way that dimensions of vector spaces can be made rigorous? For example, is it clear that the "counts" you obtain in this way are necessarily unique? $\endgroup$ Oct 2, 2015 at 5:55
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    $\begingroup$ Yes, as long as the coefficients of the PDE are real analytic; this follows from the Cartan-Kahler theorem in exterior differential systems. $\endgroup$ Oct 2, 2015 at 12:46
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If coefficients are real analytic, then real analytic solutions can be parametrized locally by their Cauchy data on a non-characteristic hypersurface. This is the Cauchy-Kowalevsky Theorem. However this parametrization is continuous for interesting topologies (essentially) only when the differential operator is hyperbolic with respect to the "initial" surface; see Willie Wong's answer. This excludes - except in the ODE case - the Laplace operator. The solutions of the homogeneous Laplace equation (and of many other elliptic equations) can however be parametrized by a single function, namely the trace on the boundary of a domain (bounded and with sufficiently regular boundary); this is the solution theory of the Dirichlet problem.

Other kinds of parametrization of the general solution of a PDE are given (when coefficients are constant, or on a symmetric space) by the Fourier transform (e.g. the Ehrenpreis' Fundamental Principle) or, e.g. for wave equations, the Radon transform. Here, of course, finite sums are replaced by integrals; so there is no meaningful parameter count. I mention this because I think that, for PDE theory, a focus on parametrization by finitely many parameters is too narrow.

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