11
$\begingroup$

When I teach constant coefficient linear differential equations, the usual guess of an exponential can be motivated because it is "approximately" a fixed point for the differentiation operator. The aesthetic here being that mathematicians look at fixed points to exploit symmetry to find solutions.

If one considers a 2x2 first order matrix linear differential equation of the form $$X'=AX$$ when $A$ has a repeated eigenvalue with only a one-dimensional eigenspace, one can still discover what needs doing using the approximate fixed point idea while appealing to an analogy with the one-dimensional procedure of "reduction of order". This is done here, for example, in the way most of us probably teach it.

I always feel like I'm making a stretch when teaching this particular material, and was wondering if anyone had a more global explanation for the appearance of generalized eigenvectors here? I always felt that there should be, but can't think of a more natural way to explain this than the process found in the link above.

|improve this question
$\endgroup$
  • $\begingroup$ The common approach is to motivate via guessing e^rx (and xe^rx, etc). Goes back to at least Granville in calc books and is common in ODE books also. Thomas Finney (I have the 81 edition) instead has a derivation where you can actually extract and generate the first root solution using a separable equation and integrating factor. The second solution can be generated then by substitution and will be xe^(rx) if repeated or just e^(rsub2x) if nonrepeating. So this works without the need for LA or for the guessing. $\endgroup$ – guest Apr 8 '18 at 16:56
8
$\begingroup$

So, there is some point where they just propose multiplying the given solution by $t$ and then work out the generalized e-vector of order two condition. Pretty in its way, but I prefer the perspective of the matrix exponential.

Define $e^{tA}=I+tA+(t^2/2)A^2+ \cdots$. It is easy to show that the matrix exponential is a solution matrix and it is an invertible matrix. Thus, the columns of the matrix exponential form a fundamental solution set; contained in $e^{tA}$ are all the solutions of $x'=Ax$.

The question then, how to actually calculate the columns without explicit computation of the series. One technique is suggested by the following formula: $$ e^{tA} = e^{\lambda t}(I+t(A-\lambda I)+(t^2/2)(A-\lambda I)^2+(t^3/3!)(A-\lambda I)^3+ \cdots ) $$ This identity holds for any complex scalar $\lambda$. However, it is most useful when we consider eigenvectors and then generalized e-vectors. It produces the usual cases given in the document linked by the OP as well as an infinite family of possibilities for higher order problems. Of course, the Jordan cannonical form completes the thought of this discussion. I can say more, but this is standard stuff.

All of this said, I'm not sure this is the motivation you seek. Another idea, $y''=0$ has solution $y = c_1+c_2t$ by calculus I. Then do reduction of order and study the structure of the resulting solutions. You'll find our beloved eigenvector solutions as well as generalized e-vectors of order 2. This might motivate. You might look at: https://math.stackexchange.com/q/206967/36530 as the inquiry and discussion there is very much related to your question.

|improve this answer
$\endgroup$
  • $\begingroup$ I was, indeed, wondering if there was a way to see the need for generalized eigenvectors at the point where we tell them to "throw in a t". Whether the students like it or not, though, matrix exponentials are probably the most natural approach... $\endgroup$ – Jon Bannon Apr 20 '14 at 22:42
  • 1
    $\begingroup$ @JonBannon Thanks! Ideally we want to solve $x'=Ax$ for all possible $A$. So, we'll need some structure which covers all possible $A$. Given the structure of the matrix exponential the real Jordan form of the matrix is what we need for an easy solution. So, if we need generality we need the Jordan form. So, do students need generality? Or, how can we better motivate the matrix exponential? That's probably an easier route. $\endgroup$ – James S. Cook Apr 21 '14 at 13:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.