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3 votes
0 answers
95 views

Deeper reason for why classical orthogonal polynomials have simple generating functions?

Is there a known reason why all classical families of orthogonal polynomials have simple generating functions? I was wondering whether one could get an explanation using the connection with Sturm-...
Plemath's user avatar
  • 312
3 votes
1 answer
216 views

Linear transport equation with Lipschitz conditions

Given the equation here, I would like to ask the following relaxed question: Consider the PDE $$\partial_t f(x,t) = \langle q(x), \nabla \rangle f(t,x) + p(x),$$ with Schwartz initial data $f(0,x) = ...
Pritam Bemis's user avatar
5 votes
2 answers
358 views

Linear transport equation with unbounded coefficients

Consider the PDE $$\partial_t f(x,t) = \langle q(x), \nabla \rangle f(t,x) + p(x),$$ with Schwartz initial data $f(0,x) = f_0(x) \in \mathscr S(\mathbb R^n).$ I am wondering then if $q$ and all its ...
Pritam Bemis's user avatar
4 votes
1 answer
168 views

Method of characteristics beyond the Lipschitz setting

I have come across the following easy-looking problem that is driving me mad. I have a family of measures (on the real line $\mathbb R$) $\{\mu_t\}_{t>0}$ which is uniformly bounded (the measures ...
user111164's user avatar
7 votes
1 answer
391 views

Does $(\nabla \times F) \cdot F= (\nabla \times F) \cdot \nabla f$ have a solution?

A grad student asked me this question during office hours, and I couldn't for the life of me come up with a proof or counterexample: For a given $F:\mathbb{R}^3 \to \mathbb{R}^3$, does $(\nabla \...
Kanye's user avatar
  • 73