Mean value theorem and heuristics from differential equation

A Footbridge in Beijing (from Wikimedia)
A Footbridge in Beijing (from Wikimedia)

This entry is a retrospective summary of heursitics I occasionally came up with when I was at senior high school. Part of the idea presented in this article can be found in this post in Baidu Tieba (in Chinese only).

I taught myself calculus and linear algebra in senior high school, during which I did extensive exercises and I was already familiar many elementary calculus and linear algebra techniques then.

Let’s get to the topic. A typical kind of problems posed in exams of elementary calculus is the following: given a function that is continuous on a closed interval and (perhaps multiple-time) differentiable internally, and the condition of some values of the function on the boundary, prove that some designated equation of the function should hold for some point inside the interval. The solution is usually to apply the appropriate mean value theorems to some function, either given in the problem or an auxiliary one carefully constructed.

Example Let f(x)f(x) be a function differentiable on [a,b][{a,b}] and twice differentiable in (a,b)({a,b}). Given f(a)=f(b)=0f(a)=f(b)=0 and f+(a)f(b)>0f_{+}'(a)f_{-}'(b)>0, show that there exists ξ(a,b)\xi\in({a,b}) such that f(ξ)2f(ξ)3f(ξ)=0f''(\xi)-2f'(\xi)-3f(\xi)=0.

I gave a, presumably elegant, solution:

Solution By the condition we already know there exists some c(a,b)c\in({a,b}) such that f(c)=0f(c)=0 (this step is skipped for brevity, as it is not the concern here). Consider g(x)=f(x)exg(x)=f(x)e^x: by Rolle’s theorem g(x)g'(x) has two (distinct) zeros in (a,b)({a,b}). Finally let h(x)=g(x)e4xh(x)=g'(x)e^{-4x}: again by Rolle’s theorem h(x)h'(x) has a zero in (a,b)({a,b}). When expanded and simplified, this is exactly what we want.

You may be curious how I knew how to transform the function. It’s inspired by solving the differential equation y2y3y=0y''-2y'-3y=0. Its solution is y=C1e3x+C2exy=C_1e^{3x}+C_2e^{-x}. On the other hand, one can rewrite the differential equation as ((yex)e4x)=0\left(\left(ye^x\right)'e^{-4x}\right)'=0. It is now very clear where the mysterious constructions come from.

If you really think for a moment, how is Rolle’s theorem similar to the differential equation y=0y'=0? Think for a moment before moving your eyeballs down.

My interpretation The solution to y=0y'=0 is y=Cy=C. Rolle’s theorem says that, if a intervally continuous and internally differentiable function yy is constant on the boundary (similar to the solution to y=0y'=0, just on the boundary), then its derivative vanishes somewhere inside the interval (similar to the equation y=0y'=0, just somewhere inside the interval).

If the function satisfies the solution a bit, it also satisfies the equation a bit.

Applying the heuristics to general differential equations gives the hint. To solve an exam problem relating conditions of a function to some interior property of its (high-order) derivatives, solve the target equation as a differential function, and rewrite the equation by eliding the constants in the general solution with differentiation, then try constructing the functions all the way to see if it works.

I was not sure if the heuristics always worked, at least not when I was in senior high school. I searched the Internet after I discovered the magical trick. There was little written work on this, but still there was some. I cannot find the original document I once found, but an example of others’ discovery of the technique can be found here (in Chinese).

Rethinking it at this moment, it’s easy to give a proof for ‘factorable’ equations. Let L=(D+u1)(D+un)L=({D+u_1})\cdots({D+u_n}) be a linear differential equation operator and ff a continuous function on [a,b][{a,b}] that is internally differentiable up to nn times. Suppose ff has n+1n+1 distinct zeros on the interval. We would like to prove that (Lf)(ξ)=0(Lf)(\xi)=0 for some ξ(a,b)\xi\in({a,b}). Consider the most nested factor: We set g(x)=f(x)exp(axun(t)dt),g(x)=f(x)\exp\left(\int_a^x{u_n(t)\mathrm{d}t}\right), and apply Rolle’s theorem to obtain nn distinct zeros of f1(x)=g(x)exp(axun(t)dt).f_1(x)=g'(x)\exp\left({-\int_a^x{u_n(t)\mathrm{d}t}}\right). Let L1=(D+u1)(D+un1)L_1=({D+u_1})\cdots({D+u_{n-1}}), then f1(x)f_1(x) is a solution to L1y=0L_1y=0 and has nn distinct zeros. We then continue the process inductively.

The requirement of ‘factorisable’ cannot be removed (hence not immediately obvious is how to relax it): Consider y=cosx12y=\cos x-\frac12 on (π3,2ππ3)({-\frac\pi3,2\pi-\frac\pi3}), we have y+y=120y''+y=-\frac12\neq 0, but yy clearly has 3 distinct zeros. To further inspect the case, if we want to factor D2+1=(D+u)(D+v)=D2+(u+v)D+(uv+v),D^2+1=({D+u})({D+v})=D^2+({u+v})D+({uv+v'}), we will need u+v=0,uv+v=1u+v=0,uv+v'=1, i.e., v=u=tan(x+C)v=-u=\tan({x+C}) for some CC. For this factorisation to be successful, il faut et il suffit que the length of the interval (the domain) be less than π\pi. Otherwise said, the function needs 3 distinct zeros that are close enough.

Just mentioning, it is easy to extend this technique to the case of non-homogenous linear equation. For other equations, the trick serves purely as heuristics and a careful formal validation is still needed.

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