Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know the method. For a linear homogeneous ordinary differential equation with a root of the characteristic polynomial in $\alpha$ has multiplicity $k$, then $y=x^me^{\alpha x}$ with $m=0,1,\cdots k-1$ is a solution. This however was learnt from need of solving a problem and not in a mathematics course, so I do not know the reason for it.

On reading this book (online edition section 2.1.3), I am bit puzzled at the approach by the author. $$z''(t)+\Gamma z'(t)+\omega_0^2 z(t)=0$$ This is readily solved with solutions of the form $z=e^{\alpha t}$ where $$\alpha = -\frac{\Gamma}{2} \pm \sqrt{\frac{\Gamma^2}{4}-\omega_0^2}$$

Defining $\omega^2\equiv \omega_0^2-\Gamma^2/4$

For the case $\Gamma/2<\omega_0$ we have solutions $$z(t) = Ae^{-\Gamma t/2}\cos(\omega t-\theta)\quad -(1)$$or, $$z(t) =e^{-\Gamma t/2}(c\cos\omega t+d\sin\omega t)\quad -(2)$$

For the degenerate case $\Gamma/2=\omega_0$ the author gives the following argument:

.... gives only one solutiopn. One way to find the other solution is to approach the situation from the $\Gamma/2<\omega_0$ case as a limit. Taking $\omega\rightarrow 0$ for (1) gives us $e^{-\Gamma t/2}$ but for (2) we get $0$ However if we divide the second solution by $\omega$ as it does not depend on $t$ we get a non-zero limit $$\lim_{\omega\rightarrow 0}\frac{1}{\omega}e^{-\Gamma t/2}\sin\omega t = te^{-\Gamma t/2}$$

I have two questiotns here :

  1. Is this wrong? As the author seems to to have ignored the cosine part of (2) which diverges
  2. What is the mathemtaical name for this approach as wikipedia does not have any hints
share|improve this question

2 Answers 2

up vote 6 down vote accepted

Instead of trying to figure out what the book is doing, let me show you my preferred way of getting the result. Let's look at $$y''-2ay'+a^2y=0$$ If you let $D$ be the differentiation operator, then this equation can be thought of as $$(D-a)^2y=0$$ If we introduce a new variable $u$ by $u=(D-a)y=y'-ay$, then we have $(D-a)u=0$, that is, $$u'=au$$ which of course has the general solution $u=Ae^{ax}$, $A$ an arbitrary constant. So now we're down to $$y'-ay=Ae^{ax}$$ Multiply through by $e^{-ax}$ and contemplate the left side, arriving at $$(e^{-ax}y)'=A$$ with general solution $e^{-ax}y=Ax+B$, whence $y=Axe^{-ax}+Be^{-ax}$, ta-dum!

share|improve this answer

I believe your reading of the text that you cited isn't quite correct. Equations (1) and (2) that you give are two different ways of expressing the general solution for the underdamped case. But they aren't the two solutions that the author is referring to in the first paragraph of 2.1.3.

Instead, the author is talking about the two particular solutions $$ e^{-\Gamma t/2} \cos \omega t \qquad \mbox{and} \qquad e^{-\Gamma t/2}\sin \omega t $$ These are the two solutions for which you should take a limit (the reason he's taking a limit is to try and recover the solutions in the case $\omega = 0$).

For the first solution, $$ \lim_{\omega \rightarrow 0} \left ( e^{-\Gamma t/2} \cos \omega t \right )= e^{-\Gamma t/2} $$ which is the solution you already knew about from solving the characteristic equation.

For the second solution, $$ \lim_{\omega \rightarrow 0} \left ( e^{-\Gamma t/2} \sin \omega t \right ) = 0 $$ which isn't so helpful in itself, but because of linearity, $e^{-\Gamma t/2} \frac{\sin \omega t}{ \omega}$ is also a solution. Now taking a limit as $\omega \rightarrow 0$, we get $$ \lim_{\omega \rightarrow 0} \left ( e^{-\Gamma t/2} \frac{\sin \omega t}{\omega} \right ) = \lim_{\omega \rightarrow 0} \left ( e^{-\Gamma t/2} \frac{t \cos \omega t}{1} \right ) = e^{-\Gamma t/2} t $$ You can verify that this is indeed a solution by plugging it into the original ode.

share|improve this answer
    
I suspected that. But I still dont have an to my $2$nd question. +1 for pointing it out –  kuch nahi Jul 19 '11 at 18:44
    
@kuch nahi: Sorry, I don't recall ever seeing any specific name for this procedure. –  cch Jul 20 '11 at 1:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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