TITLE: Find coefficients for harmonic oscillator ODE
QUESTION [0 upvotes]: Related to Small mismatch in theoretical vs MMA solutions of a damped harmonic oscillator differential equation, given the ODE
$$
\ddot y + 2\xi \omega \dot y + \omega^2 y = fo \cos( \omega_{dr}x), \quad y(0)=1,\ \dot y(0)=1 \quad(3)
$$
can anyone give me the expressions for the coefficients $A, \phi$ in the solution
$$
y(x)= A e^{-\xi\omega x}\sin(\omega_d x + \theta) + A_0 \cos(\omega_{dr} x - \phi) \quad(4)
$$
I believe they are
$$
A = \sqrt{[1+\xi \omega -A_0(\sin(\phi)+\xi \omega \cos(\phi))]^2+(1-A_0 \cos(\phi))^2} \quad (5)
$$
$$
\theta = atan2(1-A_0 \cos(\phi),1+\xi \omega -A_0(\sin(\phi)+\xi \omega \cos(\phi))) \quad (6)
$$
but I got a mismatch between my results and the ones from a software. $atan2$ is the four-quadrant inverse tangent, as in Matlab software.
EDIT The theory of ODE says you can evaluate coefficients $A_0, \phi$ of the steady state solution from the coefficients of equation (3),
$$
A_0 = \frac{fo}{\sqrt{(\omega^2 - \omega_{dr}^2)^2 + (2\xi \ \omega \ \omega_{dr})^2}}
$$
$$
\phi = atan2(\omega^2 - \omega_{dr}^2,2\xi \ \omega \ \omega_{dr})
$$
while the coefficients $A, \theta$ should be found from the initial conditions.

REPLY [0 votes]: You almost have it. The total solution is $y(t) = y_h(t)+y_p(t)$, where $y_h(t)$ is the solution to the homogenous equation and $y_p(t)$ is a particular solution to the equation.
The full derivation of the answer can be find in  Rao, Mechanical Vibrations. It follows like that for the underdamped system:
$\ddot{y}+2\xi\omega_n\dot{y}+\omega_n^2y=f_0\cos(\omega t)$
The particular solution is
$y_p(t)=Y\cos(\omega t-\phi)$
Substituting it in the original equation, we get
$$Y=\frac{f_0}{\sqrt{(\omega_n^2-\omega^2)^2+(2\xi\omega_n\omega)^2}}\\
\phi= atan2(\omega_n^2-\omega^2,2\xi\omega_n\omega)$$
which is exactly what you wrote.
The homogeneous solution is the solution to
$\ddot{y}_h+2\xi\omega_n\dot{y}_h+\omega_n^2y_h=0$ . For the undamped case is:
$y_h(t) = Y_0e^{-\xi\omega_n t}\cos\left( \sqrt{1-\xi^2}\omega_n t-\phi_0 \right)$
To ease the equations, we call $\sqrt{1-\xi^2}\omega_n=\omega_d$.
The complete solution can be written as $y(t)=y_h(t)+y_p(t)$:
$y(t)=Y_0e^{-\xi\omega_nt}\cos(\omega_dt-\phi_0)+Y\cos(\omega t-\phi)$
Substituting the initial conditions $y(0)=y_0$ and $\dot{y}(0)=\dot{y}_0$, we get:
$y_0 = Y_0\cos\phi_0+Y\cos\phi\\
\dot{y}_0=-\xi\omega_n Y_0\cos\phi_0+\omega_d Y_0\sin\phi_0+\omega Y\sin\phi$
Solving this set of equations, we get:
$Y_0 = \left[ (y_0-Y\cos\phi)^2 + \frac{1}{\omega_d^2}(\xi\omega_n y_0+\dot{y}_0 - \xi\omega_n Y \cos\phi - \omega Y\sin\phi)^2\right]^2\\
\tan\phi_0=\frac{ \xi\omega_n y_0 + \dot{y}_0 - \xi\omega_n Y\cos\phi-\omega Y\sin\phi}{\omega_d(y_0-Y\cos\phi)}$


TITLE: Reduction to standard form.
QUESTION [7 upvotes]: I was wondering whether this ODE has been studied yet or whether there is anything we can say about its solutions?
$$(1-t^2)u_{tt}-tu_t+4\left[n\beta (2t^2-1)+  \beta^2 (2t^2-1)^2+C\right]u=0$$
$C$ is a free parameter. So if you know a function that would fulfill this equation only for particular $C$, this would be perfectly fine. I am interested in its solutions on $[-1,1]$.
I should include the motivation/reference here: The equation is motivated by Physics (Quantum Mechanics) and you might want to see this great answer that gives us (meanwhile complete hints) about the structure of the solution (due to O.L. (thank you!))
The problem is that the approach taken by O.L. does not offer an analytical representation of the solution. He was able to show that some types of polynomials will give you the solution, but still you have to throw this ' guess' into the equation. I suspect that the solutions form a nice orthogonal basis of $L^2[-1,1]$, but was incapable of constructing them by recursion or explicit representation.
Now, I also got the hint to consider symmetries in my ODE and separately consider even and odd solutions so that I could reduce my ODE to a confluent Heun's equation.
If I follow this hint and substitute $s=t^2$ and $u(t)= v(s)$ I get
$$v''(s)+\frac{1}{2}\left(\frac{1}{s}+ \frac{1}{s-1}\right) v'(s) + \left(n \beta \frac{(2s-1)}{s(1-s)} + \frac{\beta^2(2s-1)^2}{s(1-s)} + C\right) v(s)=0.$$
This is very close to the confluent Heun's equation but not exactly the form that we are looking for (since $s^2$ is appearing in the nominator of the right term in the parenthesis. So probably we need to substitute even more(anything like $v(s) = exp(\alpha s)w(s)$ might help), but actually I don't see how to go further.
So is there anybody who knows how to finish this and who is able to construct the solutions from here?(and how $C$ has to be chosen in order to find a solution)
If anything is unclear, please let me know and just to point this out: If you are able to reduce this ODE to the confluent Heun equation, then this answers my question totally.

REPLY [6 votes]: The equation is indeed equivalent to confluent Heun. This follows already from the analysis of its singularity structure: 2 regular singular points and an irregular point of Poincaré rank $1$.
Specifically, setting $v(s)=e^{2\gamma s}u(s)$ with $$\gamma=\sqrt{\beta^2-\frac{C}{4}},$$
transforms your second equation into
$$u''(s)+\frac12\left(\frac1s+\frac{1}{s-1}+8\gamma\right)u'(s)+\frac{2(\gamma-n\beta)s+(n\beta-\beta^2-\gamma)}{s(s-1)}u(s)=0.$$
(Compare with 31.12.1 here).
However in my opinion this does not add much new knowledge about the solutions except that they have a name.


TITLE: Find the area of the region enclosed by $x=2-y^2$ and $x+y=0$
QUESTION [0 upvotes]: When I change the equations to equal $x$. I get $\frac{11}{2}$
However, when I change the equations to equal $y$ I don't get the same answer.
Here is my work so far:
Change the equations to equal $y$:
$y=\pm\sqrt{-x+2}$
$y=-x$
Find the points of intersection:
$-x=\sqrt{-x+2}$
$x^2=-x+2$
$x^2+x-2=0$
$(x+2)(x-1)$
$x=-2$ and $x=1 $
$\int_{-2}^{0} (x-\sqrt{-x+2})dx$
Get the antiderivative:
$[\frac{1}{2}x^2+\frac{2}{3}(-x+2)^\frac{3}{2}]|_{-2}^{0}$
$=\frac{2}{3}2^\frac{3}{2} + 2$
Do the same for the next integral:
$\int_{0}^{1} (-x--\sqrt{-x+2})dx$
$\int_{0}^{1} (-x+\sqrt{-x+2})dx$
Get the antiderivative:
$[-\frac{1}{2}x^2-\frac{2}{3}(-x+2)^\frac{3}{2}]|_{0}^{1}$
$=\frac{1}{6}+\frac{2}{3}2^\frac{3}{2}$
I'm not sure where I'm going wrong here because when I add these two answers up from both integrals, I don't get same same answer as $\frac{11}{2}$

REPLY [1 votes]: Finding the area with respect to y is easier.
If you want to calculate it your way, then set up the integrals with respect to x differently.
Referring to the the general graph below:

$$\int_{-2}^1 \sqrt{2-x}-(-x)dx+\int_1^2 \sqrt{2-x}-(-\sqrt{2-x})dx=
[-\frac{2}{3}(2-x)^\frac{3}{2}+\frac{1}{2}x^2]|_{-2}^1+2[-\frac{2}{3}(2-x)^\frac{3}{2}]|_1^2=[(-\frac{2}{3}+\frac{1}{2})-(-\frac{2}{3}\cdot8+\frac{1}{2}\cdot4)]+2[0-(-\frac{2}{3})]=-\frac{1}{6}+\frac{10}{3}+\frac{4}{3}=\frac{27}{6}=\frac{9}{2}.$$
Note: The general rule for the integrand is to subtract from the above function the below function.