Problem
Heat pipe's inner radius is $1$ and outer radius is $2$. It is also known that heat pipe's inner surface temperature is $100^{\circ}C$ and outer surface has temperature of $0^{\circ}C$. Define heat pipe's temperature $u=u(R)$ in function of radius $R$ when $1 \le R \le 2$. Temperature sa...
@AkivaWeinberger theuy say as for all $x\in E$ there exists a unique $a_k\in K$ such that $||x-a_k||=\inf_{k\in K}||x-k||$ then $\pi_{k}(x)=a_k$ define a map which is the identity on $K$ , can you tel me why it is the identity ?
@TedShifrin: Fix $x_0 \in X$. Consider $d(x_0,p)$, for $p \in X$. Since $X$ is connected and $d$ is continuous, we have that $im(d) \subset \mathbb{R}$ is also connected. Thus, we know $im(d)$ is some interval of $\mathbb{R}$. Since intervals of $\mathbb{R}$ are uncountable, and we have a bijection between $X$ and the interval, $X$ is also uncountable.
@TedShifrin Could I ask you a question (still regarding my problem of yesterday), given that you seem to be only of the few that really understands this stuff?
@nbro: I don't want to look at all that. I would recommend you first change the standard result from $[-\pi,\pi]$ to $[0,2\pi]$ (by extending the function periodically).
@Semiclassical $c_1$ seems to be more difficult. I got $$ c_1=\ln(-\frac{100}{\ln(2)})=\text{non real solution} $$ it does have solution in complex plane though.
@Ted so I know you didn't do quite as much research since you preferred teaching, and you seem to have leaned to asking questions that were of a more classical or differential nature, but I am curious what methods you usually found yourself using from related areas
We did polar coordinates. Then complex numbers, roots of unity, etc., and then lots of linear algebra (actually talking about geometric things with linear maps!).
the point is that, once you've found a solution which satisfies the ODE and the boundary conditions, you're basically done. (in a formal course one would probably appeal to some theorem in order to conclude that the above solution is unique, but that's all.)
