# Confused by pseudo-derivative

I'm studying the following differential equation:

$-(pu')'+qu={\lambda}wu$

where $p,q,w$ are $[a,b]{\longrightarrow}{\mathbb R}, \lambda\in{\mathbb C}$ is a spectral parameter (not important), $p,w>0$ and $\frac{1}{p},q,w{\in}L_{\it loc}^1[a,b]$ (that is to say, locally absolutely integrable).

Now, $u$ and $u'$ aren't necessarily supposed to be continuously differentiable: it is only the case that $u$ and $pu'$ must be absolutely continuous*. On the other hand, $pu'$, which two authors call the 'pseudo-derivative', does have to be differentiable. I have searched the internet to learn more about pseudoderivatives but the word doesn't bring up anything on any of the usual mathematical sites.

I would really appreciate it if somebody were able to answer these two questions that are on my mind.

• If $u$ is not supposed to be differentiable, how can we write $pu'$, which clearly has a differentiated $u$ in it?
• If you know more about this type of differential equation, does ensuring $pu'$ is differentiable tend to be a heavy constraint?

Sorry if these seem naive questions. I am writing an essay which does not really deal with pseudo-derivatives in a big way and mentions these constraints in passing. I am trying not to plunge headlong into the theory of weak solutions which would seriously derail my work.

Thank you if anybody is able to advise.

• Initially I did not make clear that $pu'$ was required to be absolutely continuous. This is relevant to one of the points raised in the answer below.
-
 Another 'strange' thing is that one author also represents the above equation in matrix form as $\begin{pmatrix}u\\pu'\end{pmatrix}'$=$\begin{pmatrix}0&\frac{1}{p}\\q-{\lambda}w&‌​0\end{pmatrix}$$\begin{pmatrix}u\\pu'\end{pmatrix}$. Here again he appears to be differentiating u, judging by the top line. – Josef K. May 8 '11 at 2:59

There is no problem differentiating $u$ if $u$ is assumed to be absolutely continuous. A particular application of the Radon-Nikodym Theorem says that absolutely continuous functions can be identified as integrals of Lebesgue integrable functions. So by assumption $u'$ is defined as a measurable function.
Again, if you assume that $u'$ is absolutely continuous, its derivative can also be defined. The problem is that assuming $1/p$ to be just $L^1_{\mathrm{loc}}$ is generally too weak to guarantee that $(pu')'$ exists in the weak sense. So some assumption is necessary for the equation to make sense. I'm not familiar enough with this particular equation to tell you the strength of the assumptions.
The assumption is that $pu'$ is absolutely continuous (I was trying to keep it brief, but obviously too brief!). I can iterate your argument to answer that bit now! Thanks a lot - this all makes sense now. – Josef K. May 8 '11 at 3:12
Also, I'm fairly sure that the terminology "pseudo-derivative" is non-standard. Most likely the authors use the name because it is "at the level" of a derivative, and because of the first-order Hamiltonian/matrix form of the equation you wrote in your comments. Generally a linear ODE can be converted to first order by writing it as a first order matrix equation on the vector formed by $u$ and its derivatives. In this case the expression is cleaner if you write it in terms of $u$ and $pu'$. – Willie Wong May 8 '11 at 3:13