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As I was working on my differential equation homework this week I came across this problem:

Let $y^{(4)} + 16y=0$. Compute the Wronskian of four linearly independent solutions.

It's rather straightforward to find four such solutions solutions:

$\phi_1(x)= e^{\sqrt{2}x} \cos{\sqrt{2}x}$, $\phi_2(x)= e^{\sqrt{2}x} \sin{\sqrt{2}x}$, $\phi_3(x)= e^{-\sqrt{2}x} \cos{\sqrt{2}x}$, $\phi_4(x)= e^{-\sqrt{2}x} \sin{\sqrt{2}x}.$

And from there computing the Wronskian (it is $256$) can be accomplished by trudging through the computation.

However, since the solutions have such a nice symmetry on the complex unit circle, is there an easier way of coming to the solution? I tried to do this problem at first by avoiding taking the determinant, but ended spending more time than I would have anyway in doing so. I want to believe there is an easier way to do this problem, and I feel like there is a trick that I'm not seeing.

nmasanta
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doppz
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  • You posted and deleted a question on $$\sum_{n\geqslant 1}\frac{|\mu(n)|}{n^s}=\frac{\zeta(s)}{\zeta(2s)}$$ There, you claimed that $|\mu(n)|=2^{k(n)}$ where $k(n)$ is the number of prime factors of $n$. This is not correct. $\mu(n)$ is either $1,-1$ or $0$, I don't understand how you're getting $2^{k(n)}$. In fact $|\mu(n)|=1$ if $n$ is squarefree, $=0$ else. – Pedro Nov 03 '13 at 18:09
  • Pedro. Thank you for your response, I deleted the question because I noticed that what I had didn't make sense. I was getting $|\mu(n)|$ confused with $\sum_{d|n} |\mu(d)|$ which was an earlier problem in the packet. – doppz Nov 03 '13 at 18:52

2 Answers2

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We have the Liuville formula: $$ W(x)=W(x_0)\exp\left\{-\int_{x_0}^x\frac{a_1(x)}{a_0(x)}dx\right\} , $$ where $W(x)$ is the Wronskian, and $a_0(x)$ and $a_1(x)$ are $$ a_0(x)y^{(n)}+a_1(x)y^{n-1}+\ldots+a_n(x)y=0. $$ Your $a_1(x)$ is zero, hence $$ W(x)=W(x_0)=\rm const. $$ Since we are free to choose any constant, the answer is, e.g., $$ W(x)=1. $$

Artem
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Added: One can determine the Wronskian up to a multiplicative constant using the equation coefficients, as explained in Artem's answer (my apologies: initially I misunderstood his point). Here this gives simply a constant function. If one is interested in determining this unknown constant explicitly for a particular basis of solutions, then the usual way (e.g. for special function ODEs) is to keep a sufficient number of terms in the asymptotics of solutions. For example here one could replace them by Taylor expansions truncated at 3rd order and compute the Wronskian dropping all non-constant terms. What follows is an alternative to that and to the direct approach.

One option is to compute instead the Wronskian of $\phi_{\pm}=\phi_{1} \pm i \phi_2$, $\psi_{\pm}=\phi_3\pm i\phi_4$. The main point is that $$\phi_{\pm}(x)=e^{(1\pm i)\sqrt{2} x},\qquad \psi_{\pm}(x)=e^{(-1\pm i)\sqrt{2} x},$$ and the derivatives of exponentials are easy to compute. One then finds \begin{align} &W(\phi_+,\phi_-,\psi_+,\psi_-)=-4W(\phi_1,\phi_2,\phi_3,\phi_4)=\\ &=\sqrt{2}\cdot\left(\sqrt{2}\right)^2\cdot\left(\sqrt{2}\right)^3 \operatorname{det}\left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1+i & 1-i & -1+i & -1-i \\ (1+i)^2 & (1-i)^2 & (-1+i)^2 & (-1-i)^2\\ (1+i)^3 & (1-i)^3 & (-1+i)^3 & (-1-i)^3 \end{array}\right)=\\ &=8\prod_{1\leq j<k\leq 4}\left(\lambda_k-\lambda_j\right)=\\ &=8\cdot (-2i)\cdot(-2)\cdot(-2-2i)\cdot(-2+2i)\cdot(-2)\cdot(-2i)=-1024. \end{align} In the 3rd line, we used the expression for the Vandermonde determinant with $$\left(\lambda_1,\lambda_2,\lambda_3,\lambda_4\right)=\left(1+i,1-i,-1+i,-1-i\right).$$

Start wearing purple
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  • Very nice! Thanks for the response! – doppz Oct 25 '13 at 02:21
  • @doppz Thanks, you are welcome! – Start wearing purple Oct 25 '13 at 02:25
  • I do not disagree with your answer :) But you understood the question wrong. – Artem Oct 26 '13 at 00:52
  • @Artem I've just re-read the question. Indeed if one answers the question as it's stated in the assignment, we can say that it's just a constant and stop there. But if one wants to compute the Wronskian of those particular $\phi_{1,2,3,4}$ (I understood that the MSE question asked about this), then, well... Anyhow, my comment towards your post needs to be removed, sorry about that. I will edit my post. – Start wearing purple Oct 26 '13 at 01:04