I was up till 3:30. There was no way that I would sleep before the problem would.
As it turns out, the integral equation that I discussed earlier is well-known. It is an example of a linear Volterra equation of the first kind. \[f(c)=\int_{a}^{c}{K(c,x)\rho(x)\textrm{ d}x}\] The function \(K\) is known as the kernel. In our case, the kernel is: \[K(c,x)=c+b-x\] The idea here is to use the Leibniz integral rule (which follows from Fundamental Theorem). By differentiating both sides with respect to \(c\), we obtain: \[\frac{\textrm{d}}{\textrm{d}c}\left(\int_{c}^{\infty}{(c+b-x)\rho(x)\textrm{ d}x}\right)=\frac{\textrm{d}R}{\textrm{d}t}\lim_{R\rightarrow\infty}{\left[(c+b-R)\rho(R)\right]}-b\rho(x)+\int_{c}^{\infty}{\rho(x)\textrm{ d}x}\] Observe that if \(G(c, R)=\int_{c}^{R}{(c+b-x)\rho(x)\textrm{ d}x}\) is well-behaved, then we may interchange the limit and derivative above to obtain: \[\frac{\textrm{d}R}{\textrm{d}t}\lim_{R\rightarrow\infty}{\left[(c+b-R)\rho(R)\right]}=\lim_{R\rightarrow\infty}{\left[(c+b-R)\rho(R)\frac{\textrm{d}R}{\textrm{d}t}\right]}=0\] As cited, the conditions we want for \(G\) for this manipulation to be valid are:
Anyway, we have: \[b\rho(c)=\int_{c}^{\infty}{\rho(x)\textrm{ d}x}\] Let \(P'=\rho\). Then, by the Fundamental Theorem: \[bP'(c)=\lim_{R\rightarrow\infty}{P(R)}-P(c)\] Let \(L=\lim_{R\rightarrow\infty}{P(R)}\). Then, the equation above rearranges to: \[P'+\frac{1}{b}P=\frac{L}{b}\] This is simply a first-order linear ODE. We solve this by using an integrating factor of \(\exp{\int{\frac{1}{b}\textrm{ d}{x}}}\). This yields: \[P(x)=L+Ce^{-x/b}\] Differentiating this, we obtain: \[\boxed{\rho(x)=Ce^{-x/b}}\] Obviously with \(C>0\). Uniform convergence is forced by the fact that this solution satisfies our Volterra equation. It can probably also be proven with \(\epsilon\)-\(\delta\) calculations. That's for another time. And I'm done with mechanics for this weekend I think.
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Here's what I do know.
For brevity, let \(K=c+b\). Let \(u=\rho(x)\) and \(\textrm{d}v=K-x\textrm{ d}x\). Then, integrating by parts: \[\int{(K-x)\rho(x)\textrm{ d}x}=\rho(x)\left(Kx-\frac{1}{2}x^2\right)-\int{\left(Kx-\frac{1}{2}x^2\right)\rho'(x)\textrm{ d}x}\] We continue integrating by parts, letting the polynomial in each new integrand be \(\frac{\textrm{d}v}{\textrm{d}x}\), and the derivatives of \(\rho(x)\) be \(u\). This then yields: \[\int{(K-x)\rho(x)\textrm{ d}x}=\sum_{n=0}^{\infty}{(-1)^n\rho^{(n)}(x)\left(\frac{Kx^{n+1}}{(n+1)!}-\frac{x^{n+2}}{(n+2)!}\right)}\] Ok Andrew, cool. Now what? Ah, simply observe. Our desired integral is the improper integral evaluated from \(c\) to \(\infty\). This means that the limit of the antiderivative (above) as \(x\) approaches \(\infty\) must converge! In other words: \[\lim_{x\rightarrow\infty}{\sum_{n=0}^{\infty}{(-1)^n\rho^{(n)}(x)\left(\frac{Kx^{n+1}}{(n+1)!}-\frac{x^{n+2}}{(n+2)!}\right)}}\] Must converge! For this to occur, every single term must converge! But observe that the polynomials in each term of the summation tend to \(\infty\), hence for convergence to occur, every derivative of \(\rho(x)\), along with \(\rho(x)\) itself, must tend to \(0\) (this follows plainly from L'Hôpital's rule). In other words, we have derived: \[\lim_{x\rightarrow\infty}{\rho^{(n)}(x)}=0\textrm{ }\forall n\in\mathbb{N}_0\] This in and of itself hints at possible functions. For instance, an exponential decay function, or something of the form \(Ax^n\) for some \(n<0\). Ok sure, we can substitute these functions into our original integral equation and find that one of them actually works, but I want a little more insight. And rigor. And what sort of lame solution is that? Deriving one property and guessing and checking? Ew. The next thing I will use is the Leibniz integral rule. That is: \[\frac{\textrm{d}}{\textrm{d}x}\left(\int_{a(x)}^{b(x)}{f(x,t)\textrm{ d}t}\right)=f(x,b(x))\cdot\frac{\textrm{d}}{\textrm{d}x}b(x)-f(x,a(x))\cdot\frac{\textrm{d}}{\textrm{d}x}a(x)+\int_{a(x)}^{b(x)}{\frac{\partial}{\partial x}f(x,t)\textrm{ d}t}\] While this looks complicated, it just follows from the Fundamental Theorem of Calculus. And while I'd love to continue this post with the application of this rule, I should probably do my history for once. The next post will include this, and discussion of uniform continuity, and uniform and pointwise convergence. Stay tuned. I'm stuck on a Morin problem.
Consider a semi-infinite stick (one that extends infinitely in one direction) with a linear mass density \(\rho(x)\). The stick enjoys the property that it can be cut anywhere, have a fulcrum placed at a distance \(b\) away from the cut, and balance on the fulcrum. Find \(\rho(x)\). Here is what I have so far. Suppose that the end of the stick is at \(x=0\). Let a cut be made at \(x=c\). Then, the fulcrum is placed at \(x=c+b\). Balancing torques about the fulcrum: \[\int_{c}^{c+b}{(c+b-x)\rho(x)\textrm{ d}x}=\int_{c+b}^{\infty}{(x-c-b)\rho(x)\textrm{ d}x}\] This can be rearranged to: \[\int_{c}^{\infty}{(c+b-x)\rho(x)\textrm{ d}x}=0\] This is an integral equation. I have no ideas as to how to solve it. I refuse to look at the solution just yet. |
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