I am currently studying Poisson and Laplace equations. This is just a small question that has been causing me some confusion, and I would like some clarification before I resume my study. For example, when we write $\triangle u =0$ in the sense of distributions and u is also a distribution, are we assuming that $u$ is locally integrable. By this, I mean, are we assuming that $u$ is a function that is in $L^1_{loc}$? Thanks very much.
weak solutions need to have local integrability condition?
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analysis
pde
1 Answers
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I don't think that the local integrability is necessarily an a priori assumption in an arbitrary equation when interpreting it in the sense of distributions. It just turns out that any locally integrable function also defines a distribution. However, when we interpret the Poisson equation $\Delta u = f$ in the sense of distributions, the solution is usually locally integrable, at least when $f \in L^p$. Notice that the class of locally integrable functions is HUGE. For instance it contains all the $L^p$ for $1 \le p \le \infty$.
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0Yes, I think so. For example, it is often important to identify the impulse response of a dynamical system as a function. Consider the heat kernel; it arises by looking at $u_t = \Delta u, \quad u(0) = \delta$. – 2012-01-03