Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Pardon me if this question sounds silly. Consider any differential equation without any boundary conditions. Does the solution of it is always smooth ?

share|cite|improve this question
of course not. Just take $y'(x) = g(x)$ for some prescribed function $g$. If $g$ is not $C^{k}$, then $y$ cannot be $C^{k+1}$. I assume you need some assumptions on the coefficients. – Willie Wong Jul 26 '11 at 17:44
@Willie : Is there any notion as differentiability of a differential equation and something like a smooth differential equation. – Rajesh Dachiraju Jul 26 '11 at 17:59
Like I said, smoothness of the coefficients. Give a (scalar) ODE $y'(x) = F(x,y)$, you can formally take derivatives $y''(x) = \partial_xF(x,y) + \partial_yF(x,y)y'$ etc. If at the level of each derivative you have that the RHS is Lifschitz continuous, then you can apply the Picard-Lindelof theorem to gain derivatives. Of course, the solution may exist only for a small time. For example, $y' = y^2$ is an ODE with real analytic coefficients, and has real analytic solution for short time. But if you take positive initial data, the solution will blow-up in finite time. – Willie Wong Jul 26 '11 at 18:10
@Willie : By coefficients is it assumed that the differential equation looks like a polynomial (eg. $D^2-xD+1 = 0$). What if the DE cannot be represented as a polynomial in $D$ with finite terms ? – Rajesh Dachiraju Jul 26 '11 at 18:25
No. Consider the ODE $y'(x) = F(x,y)$. By coefficients we refer to the function $F:\mathbb{R}^2\to\mathbb{R}$. The terminology is inherited from the case where $F(x,y)$ is a polynomial in $y$, in which case $F$ is real analytic in the $y$ variable, and the regularity of $F$ depends essentially only on the regularity of the coefficients of the polynomial. – Willie Wong Jul 26 '11 at 19:49
up vote 4 down vote accepted

One particular sub-class of PDE deserves mention, namely, "elliptic" PDE with smooth coefficients, such as $(\Delta+\lambda)u=f$ where $\Delta$ is the usual Laplacian on $\mathbb R^n$, for example. A simple smoothness assertion here is that, if $f$ is smooth, then $u$ is smooth, and, in particular, any solution of $(\Delta+\lambda)u=0$ is smooth. This is a simple case of "elliptic regularity" (a reasonable keyword/phrase to google).

For contrast, another standard: the wave operator $\Delta-\frac{\partial^2}{\partial t^2}$ is definitely not elliptic, but, instead, is "hyperbolic". Without giving that definition either, note that in 1-D, any function of the form $F(x,t)=f(x+t)+g(x-t)$ seems to satisfy the wave equation, even when $f,g$ are not differentiable, etc. This example was known before 1800, to Euler and others, and was some of the ammunition in arguments over the notion of "function" and the notion of "solution" of a differential equation. In particular, even tho' the coefficients are smooth, there can be very non-smooth solutions (if we allow distributional derivatives...)

share|cite|improve this answer

No, "smooth" means infinitely differentiable. Solutions to differential equations of finite order only need to have finitely many derivatives.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.