I'm looking for eigenvalues of operator L such as:

$$ L=\frac{d^2}{dt^2} + \Gamma \frac{d}{dt} $$

I know that I need to solve equation:

$$ f''(t) + \Gamma f'(t) = \lambda f(t) $$

But I have no idea even where to start.

Edit 1: $\Gamma$ is a constant, not a function of $t$ and $\Gamma > 0$.

Edit 2: I tried to solve a differential equation which I wrote. I got something like this:

$$ f(t) = C_1 e^{-\frac{\Gamma t}{2}} + C_2 t e^{-\frac{\Gamma t}{2}} $$

for $$ \lambda = - \frac{1}{4} \Gamma^2 $$ Is this lambda my eigenvalue?

But this is solution for $\Delta=0$ of a characteristic equation. What about other values of delta?

As far as I know $\Gamma > 0$

  • $\begingroup$ What is $\Gamma$? $\endgroup$
    – Bernard W
    Jul 20 '17 at 23:49
  • $\begingroup$ $\Gamma$ is just a constant. $\endgroup$
    – user464980
    Jul 20 '17 at 23:51
  • 7
    $\begingroup$ So...this is a just a second order differential equation with constant coefficients. Your text should cover those in detail. $\endgroup$
    – lulu
    Jul 20 '17 at 23:53


$f = e^{\mu t}; \tag{1}$


$f' = \mu e^{\mu t}, \tag{2}$


$f'' = \mu^2 e^{\mu t}; \tag{3}$


$f'' + \Gamma f' = \mu^2 e^{\mu t} + \Gamma \mu e^{\mu t} = \lambda e^{\mu t}; \tag{4}$

since there is no $t$ such that $e^{\mu t} = 0$, we may cancel it out, yielding

$\mu^2 + \Gamma \mu = \lambda; \tag{5}$

we complete the square:

$(\mu + \dfrac{\Gamma}{2})^2 = \mu^2 + \Gamma \mu + \dfrac{\Gamma^2}{4} = \lambda + \dfrac{\Gamma^2}{4}; \tag{6}$


$\mu + \dfrac{\Gamma}{2} = \pm \sqrt{\lambda + \dfrac{\Gamma^2}{4}}; \tag{7}$

$\mu = -\dfrac{\Gamma}{2} \pm \sqrt{\lambda + \dfrac{\Gamma^2}{4}}; \tag{8}$

(1)-(8) show that for every $\lambda \in \Bbb C$ there is at least one and at most two $\mu \in \Bbb C$ such that $\lambda$ is an eigenvalue of

$\dfrac{d}{dt^2} + \Gamma \dfrac{d}{dt} \tag{9}$

with eigenfunction $e^{\mu t}$.

The sign of $\Gamma$ doesn't affect this result.

Note Added Thursday 20 July 2017 5:55 PM PST: In answer to OP user464980's question, posted in the comment stream to this answer, I must confess I chose $e^{\mu t}$ as a sample eigenfunction because I knew, after many years of experience, that it would work. But there are good theoretical reasons as well; one is that, letting $D_t = d/dt$. any constant-coefficient linear operator $\sum a_i D_t^i$ will simply have the effect of multiplying an exponential function by a polynomial: we have $D_t e^{\mu t} = \mu e^{\mu }$, $D_t^2 e^{\mu t} = \mu^2 e^{\mu t}$, and so forth, $D_t^n e^{\mu t} = \mu^n e^{\mu t}$, leading to

$(\sum a_i D_t^i) e^{\mu t}= (\sum a_i \mu^i)e^{\mu t}; \tag{10}$

so I knew that the operator (9) would convert a calculus/analysis problem into an algebraic one which I knew how to solve. Another theoretical reason is that ODEs (once boundary/initial conditions are supplied), have unique solutions. So I knew that $e^{\mu t}$ was really the only choice. End of Note.

  • $\begingroup$ Why did you choose $f(t)=e^{\mu t}$? $\endgroup$
    – user464980
    Jul 21 '17 at 0:28
  • $\begingroup$ Okay, I get it but still - you didn't found exact eigenvalues of this operator. I wish they were somehow connected with Gamma - just like in my answer. $\endgroup$
    – user464980
    Jul 21 '17 at 1:39

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.