22
$\begingroup$

Given an integral equation is there always a differential equation which has the same (say smooth) solutions? It seems like not but can one prove this in some example?

Edit: Naively I'm hoping for some algorithm which takes an integral equation and applies some operations like taking derivatives, substituting variables for some new ones, adding additional differential equations etc... such that after this procedure you have made all integral signs vanish and obtained a differential equation which has the same solutions as the integral eqution. (maybe similarly to how one can transform any system of PDEs into a system of first order equations)

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ It might be more reasonable to ask this in a more definite context, such as Fredholm theory. The operators considered in the abstract theory of integral equations, for a given class of kernels, are very different in nature from differential operators. But the two theories are related, in some cases, by a type of inversion. You may be asking the question "how extensive is that relationship"? $\endgroup$
    – Charles Matthews
    Commented Jun 1, 2010 at 10:37
  • 2
    $\begingroup$ Any smooth function is the solution of a differential equation: given f(x), we have that f(x) is the solution of the equation y'(x)=f'(x) for y(x) $\endgroup$
    – Guy Katriel
    Commented Jun 1, 2010 at 10:39
  • $\begingroup$ Indeed, taking Guy's example further towards absurdum, any function f(x) is the unique solution of the equation y(x) = f(x)! The question becomes interesting once you choose some restrictions on the coefficients involved (and perhaps other aspects of the form of the equations). @Michael: can you give some examples that you had in mind? $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Jun 1, 2010 at 10:47
  • 1
    $\begingroup$ @ Guy and Peter: sorry I fail to see how this is related to the question. $\endgroup$
    – Michael Bächtold
    Commented Jun 1, 2010 at 11:19
  • 3
    $\begingroup$ The term "integral equation" is perhaps too vague. Any chance you want to indicate the most general form you need? Also, any chance you want to say a little about why you would want to do this. In general, we prefer to convert differential equations into integral equations and not vice versa. $\endgroup$
    – Deane Yang
    Commented Jun 1, 2010 at 11:44

5 Answers 5

Reset to default
10
$\begingroup$

While I second Deane's comment that the author should be a bit more specific about the kind of equations he is interested in, in general the answer is no for integral and, more broadly,integro-differential equations. However, the latter can be reduced to functional-differential equations rather than to purely differential ones. For more details, see Section 6.6 of the book Symmetries and Conservation Laws for Differential Equations of Mathematical Physics (unfortunately the relevant pages appear to be excluded from Google preview).

For instance, I greatly doubt that one could reduce the Smoluchowski coagulation equation from Example 6.5 of the above book to a differential (as opposed to functional-differential) equation or system thereof.

Improve this answer
$\endgroup$
17
$\begingroup$

In general, no. An integral equation can be non-local, whereas a differential equation is local (in the sense that it can be described by a function over the jet-bundle). As an illustration

Let $K(x) = \delta_0(x) + \delta_1(x)$ be an integral kernel, where $\delta_i$ are the Dirac delta's supported at $i$. Consider the integral equation, for some fixed smooth $f$ $$ f(x) = \int K(x-y) \phi(y) dy $$ for the unknown $\phi$. The equation reduces to $\phi(x) + \phi(x+1) = f(x)$. Any continuous function $g(x)$ on $[0,1]$ satisfying $g(0) + g(1) = f(0)$ generates a continuous solution of the equation. I challenge you to find a differential equation whose solution set can be thus generated.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ I wonder if we could produce this equation in differential form by going up a dimension. I've seen non-local conditions like this (maybe not exactly the same) arise when taking a 1D slice through a solution to a PDE in a 2D domain. In this case, maybe the equation could be contrived to arise from considering solutions to a wave equation reflected from a domain boundary. $\endgroup$
    – Dan Piponi
    Commented Jun 1, 2010 at 18:43
  • $\begingroup$ Sometimes, yes. It is well-known, for example, that the square-root of the Laplacian operator, which is non-local, can be obtained from the Dirichlet-Neumann map of a boundary value problem. Caffarelli and Silvestre examined other fractions of the Laplacian in their seminal paper arxiv.org/abs/math/0608640 I don't doubt that in cases with geometric intuition such a procedure is possible, and I second Deane's comment above asking for more information on why the OP is interested. I'd be interested if you can come up with a good characterization of the example I gave. $\endgroup$
    – Willie Wong
    Commented Jun 2, 2010 at 14:09
  • $\begingroup$ (I did think, briefly, about reduction from higher dimensions. So I may just not be clever enough. But keep in mind that I constructed the example to have its solution set a really large chunk of $C^0[0,1]$.) $\endgroup$
    – Willie Wong
    Commented Jun 2, 2010 at 14:11
  • 4
    $\begingroup$ By going up a dimension as Dan suggests, one can at least get an equivalent PDE system with boundary conditions for your integral equation: $\partial_y F(x,y) = K(x-y) \Phi(x,y)$, $\partial_y \Phi(x,y) = 0$, $F(x,-\infty)=0$, $F(x,+\infty)= f(x)$. Looks like a similar construction can be made for other integral equations (after adding sufficiently many new variables and dimensions). $\endgroup$
    – Terry Tao
    Commented May 23, 2015 at 14:50
  • $\begingroup$ Replying to Willie Wong, it seems to me that a differential equation can be produced from that integral equation with a similar result, at least formally: $$\frac{d f}{dx}=\int \frac{d}{dx}\delta(x-y)\phi(y)dy+\int\frac{d}{dx}\delta(x-y-1)\phi(y)dy=-\int \frac{d}{dy}\delta(x-y)\phi(y)dy-\int\frac{d}{dy}\delta(x-y-1)\phi(y)dy=\int \delta(x-y)\frac{d}{dy}\phi(y)dy+\int\delta(x-y-1)\frac{d}{dy}\phi(y)dy=\frac{d}{dx}\phi(x)+\frac{d}{dx}\phi(x-1) $$ which differs from your answer by a constant... $\endgroup$
    – João Esteves
    Commented Jan 10, 2017 at 19:41
4
$\begingroup$

With Charles Matthews comments in perspective, these are some notes I made sometime ago on this topic. I dont have the books in front of me so I can't look up the details right now.

1) In Zabreyko's book Integral equations (902860393X), there is the method based on Green's functions in Appendix A.

2) Bellman in Perturbation techniques Sec 10 points out that the other way (ODE to integral equation) is actually better

Conversion of differential eqn to integral equation is one of the powerful devices in approximation theory. Its potency is due to the fact that integration is a smoothing op, while differentiation accentuates small variations. If u(t) and v(t) are close together, then ∫u(s)ds and ∫v(s)ds will be comparable in value, but du/dt and dv/dt may be arbitrarily far apart. Consequently, when carrying out successive approximations, we prefer integral operators to differential operators. On the other hand, in numerical solutions, we prefer differential operators to integral operators.

3) You can also look up Handbook of Integral Equations by Polyanin. Sec 8.4.5, Sec 9.7 and sec 9.3.3 are three situations where the method reduces a specific integral equation to an ODE

Improve this answer
$\endgroup$
4
$\begingroup$

Well, the answer, if there is any, must be subtle. Take the following example. Let $H$ be the Hilbert transform. Then $$\frac{d}{dx}Hu=f$$ is an integral equation, where the kernel is the Fourier transform of $\xi\mapsto|\xi|$. It cannot be rewritten as a set of differential equations (ODEs) within $\mathbb R$. Nevertheless, it can be recast in terms of PDEs, using an harmonic extension in the upper half-plane: $$-\Delta_{x,y}\phi=0\hbox{ in }\{y>0\},\qquad \frac{\partial\phi}{\partial y}(x,0)=f(x),\quad u(x)=\phi(x,0).$$

Improve this answer
$\endgroup$
3
$\begingroup$

No, when it comes to stochastic differential equations these are only a shortcut - there is only a meaning to the integral representation, the paths are non-differentiable.

See e.g. here: http://en.wikipedia.org/wiki/Ito_calculus

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .