Pardon me if this question sounds silly. Consider any differential equation without any boundary conditions. Does the solution of it is always smooth ?
Does the solution of any differential equation (without boundary conditions) is always smooth
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1No. 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. – 2011-07-26
2 Answers
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...)
No, "smooth" means infinitely differentiable. Solutions to differential equations of finite order only need to have finitely many derivatives.