Operators and Linear ODEs (Part 2)

算子与线性常微分方程(下) 苏剑林 · 2012-11-30 · Mathematics differential equation linear operator

Non-commutativity

It's natural to want to extend this method to solving equations with variable coefficients, and perhaps some readers went off and tinkered with this themselves, only to end up confused because they couldn't find a suitable solution. This is where the non-Abelian nature of the group makes itself felt. Let's start with an example to illustrate this — consider the composition of operators

$$(D-x)(D+x)=D^2-x^2+(Dx-xD)$$

We need to be careful about using the commutative law here. Let us write $[P,Q]=PQ-QP$

where P and Q are two operators; this is the "commutator" from quantum mechanics, which measures how far operators P and Q are from commuting with each other. Of course, it is itself an operator too. Let's first work out what $[D,x]$ gives (if it turns out to be 0, that would mean the two operations commute). What exactly does it equal? We can't tell just by looking at it, so let's apply it to a function:

$$[D,x]y=(Dx-xD)y=D(xy)-xDy=yDx+xDy-xDy=y$$

Since "the pavilion nearest the water enjoys the moonlight first" (i.e., we apply things in the order closest to the function first), $Dxy$ means x acts on y first, and then D acts on (xy); while $xDy$ means D acts on y first, and then x acts on Dy. In the end we get

$$[D,x]y=y=1\times y$$

That is to say, [D,x] acting on y is equivalent to 1 times y, so

$$[D,x]=1$$

Then $(D-x)(D+x)=D^2-x^2+1$

This way, the general solution of a class of second-order variable-coefficient equations $D^2 y-(x^2-1)y=f(x)$ is

$$y=(D+x)^{-1} (D-x)^{-1} f(x)$$

and using formula (3) from the previous section, we can then find its explicit solution.

Linear ODEs with variable coefficients

In principle, general linear ODEs with variable coefficients could also be split and solved using a similar approach. Unfortunately, though, because operators do not commute, it is very difficult to pin down the explicit functional form of the split-up pieces. Since the derivative operator commutes with constants, we can factor constant-coefficient equations directly by finding roots; but as illustrated by the example above, once we mix a general function with the derivative operator, non-commutativity makes root-finding infeasible.

Consider the second-order linear ODE

$$[D^2+g(x)]y=f(x)\tag{12}$$

and consider the operator

$$[D-a(x)][D+a(x)]=D^2-a^2 (x)+a'(x)$$

We would like to find an a(x) satisfying

$$-a^2 (x)+a'(x)=g(x)\tag{13}$$

If we can do this, then we will have successfully factored (12). Moreover, we only need a single particular solution of (13). An equation of the form (13) is called a Riccati equation; although it is first-order, it is essentially still a second-order linear ODE with variable coefficients. Indeed, setting $a(x)=-\frac{u'}{u}$, we have

$$a'(x)=-\frac{u''}{u}+(\frac{u'}{u})^2=-\frac{u''}{u}+a^2 (x)$$

so that (13) can be turned into (14):

$$u''+g(x) u=0\tag{14}$$

which is again a second-order linear ODE with variable coefficients — so it seems we've fallen into a loop. Hence this method doesn't extend very far in general; however, it can be used for "post hoc analysis," meaning we first fix an operator, then determine a particular form of equation matching it, and then use that operator to solve that specific class of equations. This is often a workable approach — the harmonic oscillator problem in quantum mechanics is handled in a similar way.

We can also observe something interesting: knowing one particular solution of (14) is equivalent to knowing the general solution of (12). In other words, a particular solution of $[D^2+g(x)]y=0$ is equivalent to the general solution of $[D^2+g(x)]y=f(x)$. This really makes us appreciate the wonder of mathematics: a local property can imply a global one. In a broad sense, one could even regard this as a kind of "analogy." It reminds me of something with a similar flavor in quantum mechanics: once you know the initial form of a wavefunction, Fourier analysis lets you obtain its general form at any later time. Mathematics keeps revealing its unexpected beauty in all sorts of places.