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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 $\Gamma > 0$

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  • $\begingroup$ 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
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    $\begingroup$ 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
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Let

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

then

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

and

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

thus,

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

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:

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

hence,

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

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

(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.

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  • $\begingroup$ 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

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