7

What are the finite-dimensional spaces $W$ of differentiable functions with this property: If $f$ is in $W$, then $\frac{df}{dx}$ is in $W$.

Robert Lewis
  • 71,180
  • 3
    One example is all the polynomials of degree less than $;n;$ . – Timbuc Sep 25 '14 at 00:22
  • 1
    You get a two dimensional space generated by $\sin x$ and $\cos x$. – J126 Sep 25 '14 at 00:23
  • 2
    $e^x$ is one too. – Ishfaaq Sep 25 '14 at 00:32
  • 1
    By the way, this is one of my favorite questions on the site. I was recently a TA in a first linear algebra and differential equations course, and I found it very hard to explain intuitively why basic ordinary linear differential equations always had solutions that were so simple, so nice, and always determined by finitely many initial conditions. Answering this question gave me a lot of insight; it makes more sense when you consider the reverse problem!—namely: which differential equations are guaranteed to have finite-dimensional solution sets? – Andrew Dudzik Sep 25 '14 at 19:13
  • 1
    This is a great question. One more question: is the bounty still meaningful/available/open now that an answer has been accepted? – Robert Lewis Sep 27 '14 at 20:00
  • I've added some details, which might be what's being asked for in my answer? If there's anything else you feel needs filling in, let me know. – Andrew Dudzik Sep 27 '14 at 21:46
  • See also http://math.stackexchange.com/questions/474034/families-of-functions-closed-under-integration. – lhf Sep 27 '14 at 21:59
  • 1
    In light of the answers given, and on the question's own merits, I've taken the liberty of adding the "differential equations" tag. Hope this is OK. Regards. – Robert Lewis Sep 28 '14 at 05:24

1 Answers1

10

Exactly the ones generated by $x^k e^{cx}$ for various $c\in \mathbb{C}$. (Over $\mathbb{R}$, we should use sines and cosines as well as exponentials)

Finite-dimensionality is the key. We have an operator $D: W\to W$, so $W$ decomposes into generalized eigenspaces. The possible eigenvectors of this operator are exactly $e^{cx}$ for various $c\in\mathbb{C}$, and you can check the generalized eigenspaces are spanned by elements from $\{x^k e^{cx} \mid c\in\mathbb{C}, k=0,1,2,\ldots\}$.

By the way, if you write down the characteristic equation for $D$, you get an ordinary linear differential equation whose solution set is precisely $W$ (this follows from counting dimension). Of course, this is the opposite of how it's usually done in a first differential equations course, where we are given the equation and asked to construct $W$.

To give an example of that last paragraph: if $W$ has basis $\{1, x, e^x\}$, then we can represent $D$ (with respect to this basis) by the matrix $A=\begin{pmatrix}0 & 1 & 0 \\0 & 0 & 0 \\0 & 0 & 1 \end{pmatrix}$. The characteristic polynomial is: $$\det(XI - A) = \det{\begin{pmatrix}X & -1 & 0 \\0 & X & 0 \\0 & 0 & X-1 \end{pmatrix}} = X^2(X-1)$$

By Cayley-Hamilton, $D$ satisfies its own characteristic polynomial, so $D^3 - D^2 = 0$ as operators on $W$. That is, every $f\in W$ satisfies $(\frac{d^3}{dx^3} - \frac{d^2}{dx^2})f = 0$. But this is exactly the linear differential equation $f''' = f''$, and, by standard methods, its solution set is spanned by $\{1,x,e^x\}$—that is, every solution to this equation already lies in $W$. So what we've really done here is described a dictionary between finite-dimensional vector spaces of functions, closed under differentiation, and linear differential equations.

Andrew Dudzik
  • 30,074
  • This is a really neat answer! I have two questions: do you really need sines and cosines, because they're already combinations of $e^{cx}$? And what do you mean by the characteristic equation for $D$, and why is it first order? – Jay Sep 25 '14 at 04:55
  • @Jay Over the complex numbers, you don't need sine and cosine, for the reason you say. Over the reals, they are no longer linear combinations of exponentials. And by the characteristic equation, I mean the equation $P(D)=0$, where $P(X)$ is the polynomial $\det(XI - A)$ for some matrix representation $A$ of $D$. The equation $P(D)f = 0$ is first-order in $f$, and its solution set is the space we started with. – Andrew Dudzik Sep 25 '14 at 08:03
  • @Jay To be more specific about sines and cosines: over the complex numbers, you will see 2-dimensional spaces spanned by ${e^{(a+bi)x},e^{(a-bi)x}}$. The real part of those spaces is spanned by ${e^{ax} \sin(bx), e^{ax}\cos(bx)}$. – Andrew Dudzik Sep 25 '14 at 08:11
  • Oh ok I get why you have to use $e^{ax}\sin(bx)$ over $\mathbb{R}$, thanks! But I'm still confused about the characteristic equation part, if I take linear polynomials as my space with basis ${1,x}$ then isn't the characteristic equation $X^2$, so $D^2f = 0$ is second-order in $f$ right? – Jay Sep 25 '14 at 18:32
  • @Jay Ah yes. I transposed a few terms in my brain, I think. I meant "ordinary", not "first order". Every such equation is equivalent to a first-order system of equations, but you're quite right that my terminology was off. – Andrew Dudzik Sep 25 '14 at 18:51