Timeline for Can an integral equation always be rewritten as a differential equation?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 10, 2017 at 20:41 | comment | added | João Esteves | I see your point... By local I mean that under suitable conditions it will have a solution for $f$ in a neighborhood of x. But in fact it is not local in $\phi$. | |
| Jan 10, 2017 at 20:03 | comment | added | João Esteves | Yes its local if you interpret $\phi(x)$ as a test function, which is $C^\infty$. You could enlarge the space of definition of $\phi$ but then probably this should be interpreted as a relation between distributions. | |
| Jan 10, 2017 at 19:55 | comment | added | Ben McKay | Is that local in $\phi(x)$? | |
| Jan 10, 2017 at 19:41 | comment | added | João Esteves | 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... | |
| May 23, 2015 at 14:50 | comment | added | Terry Tao | 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). | |
| Jun 2, 2010 at 14:11 | comment | added | Willie Wong | (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 $C^0[0,1]$.) | |
| Jun 2, 2010 at 14:09 | comment | added | Willie Wong | 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. | |
| Jun 1, 2010 at 18:43 | comment | added | Dan Piponi | 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. | |
| Jun 1, 2010 at 14:52 | history | answered | Willie Wong | CC BY-SA 2.5 |