2
$\begingroup$

solve the following starting value problem for $0\leq \delta < 1$

$\ddot{x}(t)+2\delta \dot{x}(t) + x(t)=\sin(t)$ with $x(0)=0, \dot{x}(t)=1$

and call the solution $x^\delta$ Show that $\lim_{\delta \to 0} x^\delta (t)=x^0(t)$ is.

Solution: The idea to solve that is, splitting it up into a homogenous problem and in a particular one. If think I'm fine if we just discuss solving the homogenous problem.

Let $0<\delta < 1$:

For the homogenous problem, we make the ansatz: $x(t)=e^{\lambda t}$ getting us the equation $\lambda^2+2\delta\lambda+1=0$

Now, I get the solution: $\lambda_{1,2} = -\delta \pm \sqrt{\delta ^2-1}$

Define $\omega :=\sqrt{\delta ^2-1}$

So we get:

$x(t)=Ae^{t(-\delta + i\omega)}+Be^{t(-\delta-i\omega)}$

$=Ae^{-t\delta}e^{i\omega}+Be^{-t\delta}e^{-i\omega}$

$=Ae^{-t\delta}[\cos(\omega)+i\sin(\omega)]+Be^{-t\delta}[\cos(-\omega)+i\sin(-\omega)]$

$=Ae^{-t\delta}[\cos(\omega)+i\sin(\omega)]+Be^{-t\delta}[\cos(\omega)-i\sin(\omega)]$

$=e^{t\delta}[\cos(\omega)(\underbrace{A+B}_{c_1}+sin(\omega)(\underbrace{iA-iB}_{c_2})]$

$=c_1e^{-t\delta}\cos(\omega)+c_2e^{-t\delta}\sin(\omega)$

Question: I think, if I have such a situation, where I have two solutions in the form of $a\pm ib$ I can always write $c_1e^{at}\cos(b) + c_2e^{at}\sin(b)$ right? I didn't really have had complex calculus. I just now barely about it and I know that I used the eulers form here.

Does this have some special name? This kind of differential equation? Can I somehoe kind of generalize this solution? Like what if this diff. eq. had an additional term $\dddot{x}$? So we would get $\lambda^3+\lambda^2+2\delta\lambda + 1 = 0$ How would I solve that?

Can I somehow easier see the resulted format of my solution or is this teh way to go?

$\endgroup$
  • $\begingroup$ It should be $c_1e^{-δt}\cos(ωt)+c_2e^{-δt}\sin(ωt)$ with $ω=\sqrt{1-δ^2}$. $\endgroup$ – LutzL Jan 16 '17 at 18:41
  • $\begingroup$ ah, yeha sure. Sorry about that. Question remains :) $\endgroup$ – xotix Jan 16 '17 at 19:22
  • $\begingroup$ I don't know if this is a better way, but my knee jerk reaction to seeing an $a\pm b i$ in an ODE has always been to write the general form as $y(x)=C_1e^{ax}cos(bx)+C_2e^{ax}sin(bx)$. Probably if you do ODEs everyday as a career this will be a bad strategy, but for the purposes of the physics classes I took, I would always default to that. As for your second questions, the $\lambda^3$ makes me worried. If you could factor by grouping you could still pull it off. In your case, I don't think factoring by grouping would lead to something nice. $\endgroup$ – emka Jan 16 '17 at 19:33
  • $\begingroup$ with "factoring by grouping" you mean e.g. x^2 + 2x + 1 = (x+1)^2? $\endgroup$ – xotix Jan 16 '17 at 19:40
  • $\begingroup$ I mean something like $x^3+x^2+2x+2=x^2(x+1)+2(x+1)=(x^2+2)(x+1)$ $\endgroup$ – emka Jan 16 '17 at 19:42
0
$\begingroup$

After finding $$ y_h=c_1e^{-δt}\cos(ωt)+c_2e^{-δt}\sin(ωt) $$ as the solution of the homogeneous equation, try $$ y_p=d_1\cos(t)+d_2\sin(t) $$ as particular solution for $δ\ne 0$. Then eliminate all constants by inserting the initial values. You should be able to group the terms that they look like difference quotients for the limit $δ\to0$ resp. $ω\to 1$.


If $p$ is a polynomial, then the linear differential operator $L=p(\frac{d}{dt})$ has the same factorization as $p$. Especially it is true that $$ L\bigl(e^{λt}u(t)\bigr)=e^{λt}p\Bigl(λ+\frac{d}{dt}\Bigr)u(t) $$ so that for roots $p(λ)=0$ the order of the ODE can be reduced by simple integration.

$\endgroup$
  • $\begingroup$ yeah, that's the appraoch I used for the particular problem. Thanks $\endgroup$ – xotix Jan 17 '17 at 13:15

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.