ODE - interpretation of solution.

[Benny]

Can someone please help me out with the following question?

Q. A simple harmonic oscillator, of mass m and natural frequency w_0, experiences an oscillating driving force f(t) = macos(wt). Therefore its equation of motion is:

$$\frac{{d^2 x}}{{dt^2 }} + \omega _0 ^2 x = a\cos \left( {\omega t} \right)$$

Given that at t = 0 we have x = dx/dt = 0, find the function x(t). Describe the solution if w is approximately, but not exactly, equal to w_0.

I got:

$$y\left( t \right) = \frac{a}{{\left( {\omega _0 ^2 - \omega ^2 } \right)}}\left( {\cos \left( {\omega t} \right) - \cos \left( {\omega _0 t} \right)} \right)$$

The answer says a couple of things about the behaviour of the solution for w ~ w_0 but I can't figure out how they got it. For instance "for large t it shows beats of maximum amplitude 2((w_0)^2 - w^2)^-1." How is that deduced and how would I determine which are the main characterstics of motion that I need to note. Any help would be appreciated.

 

[saltydog]

Benny. If you saw my earlier post, sorry. I worked the wrong problem (your answer is correct after all). Just cleaner to delete my post which I did.

How about some plots with the solution getting closer and closer to resonance?

Salty

 

[saltydog]

Alright Benny, I'm hesitant to open my mouth now since I tripped you up. Let me attempt to save some face:

Beating occurrs when the natural response and forced response have approximately the same frequency. You can see this by plotting them separately:

$$y1(t)=\frac{a}{\omega_0^2-\omega^2} Cos(\omega t)$$

$$y2(t)=\frac{a}{\omega_0^2-\omega^2} Cos(\omega_0 t)$$

The first plot shows this. They gradually go out of phase and then back again. Thus the difference of the cosines will go from 0 to 2 right. Thus the beating amplitude will be:

$$\frac{2 a}{\omega_0^2-\omega^2}$$

This is seen with the second plot. Ok, I think I better quit while I'm somewhat ahead.

Edit: plotting parameters were:
a=2
$$\omega_0=1$$
$$\omega=0.95$$

 

[Benny]

Thanks for the help Saltydog. Don't worry about your first post, I didn't see it anyway.

I've got another question but I don't really want to start a new thread so I'll post it here. It's an example from my book that I don't understand part of.

$$x^3 \frac{{d^2 y}}{{dx^2 }} - \left( {x^2 + xy} \right)\frac{{dy}}{{dx}} + \left( {y^2 + xy} \right) = 0$$

The equation is homogeneous so substitute y = vx. The equation then becomes:

$$x\frac{{d^2 v}}{{dx^2 }} - \left( {1 - v} \right)\frac{{dv}}{{dx}} = 0$$

I understand how to determine the substitution, the next substitution is what I don't understand the workings of. "Now substitute x = exp(t) into the above equation and we obtain (after some working)"

$$\frac{{d^2 v}}{{dt^2 }} - v\frac{{dv}}{{dt}} = 0$$

My book says that "the change in independent variable x = exp(t) leads to an equation in which the new indepedent variable t is absent except in the form d/dt." Can someone please explain how this is so? I can't see which manipulations to the equation are required to get to what the book says.

 

[saltydog]

Thanks for the help Saltydog. Don't worry about your first post, I didn't see it anyway.

I've got another question but I don't really want to start a new thread so I'll post it here. It's an example from my book that I don't understand part of.

$$x^3 \frac{{d^2 y}}{{dx^2 }} - \left( {x^2 + xy} \right)\frac{{dy}}{{dx}} + \left( {y^2 + xy} \right) = 0$$

The equation is homogeneous so substitute y = vx. The equation then becomes:

$$x\frac{{d^2 v}}{{dx^2 }} - \left( {1 - v} \right)\frac{{dv}}{{dx}} = 0$$

I understand how to determine the substitution, the next substitution is what I don't understand the workings of. "Now substitute x = exp(t) into the above equation and we obtain (after some working)"

$$\frac{{d^2 v}}{{dt^2 }} - v\frac{{dv}}{{dt}} = 0$$

My book says that "the change in independent variable x = exp(t) leads to an equation in which the new indepedent variable t is absent except in the form d/dt." Can someone please explain how this is so? I can't see which manipulations to the equation are required to get to what the book says.

Hello Benny. Let's start with:

$$xv^{''}+(1-v)v^{'}=0$$

and:

$$x(t)=e^t$$

Now:

$$\frac{dv}{dt}=e^t\frac{dv}{dx}$$

and:

$$\frac{d^2v}{dt^2}=e^t\frac{dv}{dx}+e^t\frac{d^2v}{dx^2}e^t$$

Did you remember to differentiate $\frac{dv}{dx}$ with respect to t?

You can finish it now I bet.

Edit: Oh yea, I got a plus up there too. See it?

 

[GCT]

benny, can you explain what you did here (what method you used, what chapter are you on, reduction of order?)

$$x^3 \frac{{d^2 y}}{{dx^2 }} - \left( {x^2 + xy} \right)\frac{{dy}}{{dx}} + \left( {y^2 + xy} \right) = 0$$

The equation is homogeneous so substitute y = vx. The equation then becomes:
$$x\frac{{d^2 v}}{{dx^2 }} - \left( {1 - v} \right)\frac{{dv}}{{dx}} = 0$$

 

[GCT]

actually seems that you've used the $$y=vx,~y'=v'x+v$$, I've seen this for first order differential equations, never applied it to second order. I suppose one can extend it to

$$y''=v''x+2v'$$

(just finished up with differential course, summer, short session...which I aced by the way, nevertheless didn't cover everything adequately I imagine-1 month course)

 

[Benny]

Thanks for the help Saltydog. The first part with the chain rule was what I didn't think about doing, didn't even realize that I needed to use the chain rule in some way.

GCT - I'm using "mathematical methods for physics and engineering" which I got mainly because it gives a rather extensive 'overview' of a lot of topics that I need to know about. The section that I'm on is higher order differential equations - more specifically I'm looking at the part on "isobaric equations" where substitutions of the form y = v(x^m) are made. In the case of m = 1 they call it a homogeneous equation.

Pretty much all of the standard techniques are covered in the chapter so I'm not completely sure about which specific technique is being examined in the example. I used the chain rule in my own attempt to covert the original equation to the next although I probably made algebraic errors as I rushed through it as a quick verification.

I've only started DEs, a bit of preparation for next semester. Good work with your summer course I guess. I don't think I could ever do well in such a short amount of time. Well I never do particularly well but I do worse over short periods of time.:D

 

[saltydog]

Thanks for the help Saltydog. Don't worry about your first post, I didn't see it anyway.

I've got another question but I don't really want to start a new thread so I'll post it here. It's an example from my book that I don't understand part of.

$$x^3 \frac{{d^2 y}}{{dx^2 }} - \left( {x^2 + xy} \right)\frac{{dy}}{{dx}} + \left( {y^2 + xy} \right) = 0$$

The equation is homogeneous so substitute y = vx. The equation then becomes:

$$x\frac{{d^2 v}}{{dx^2 }} - \left( {1 - v} \right)\frac{{dv}}{{dx}} = 0$$

I understand how to determine the substitution, the next substitution is what I don't understand the workings of. "Now substitute x = exp(t) into the above equation and we obtain (after some working)"

$$\frac{{d^2 v}}{{dt^2 }} - v\frac{{dv}}{{dt}} = 0$$

My book says that "the change in independent variable x = exp(t) leads to an equation in which the new indepedent variable t is absent except in the form d/dt." Can someone please explain how this is so? I can't see which manipulations to the equation are required to get to what the book says.

By the way Benny, can you take this one a step further? That is, solve for v(t)? Or even better, solve the original problem through to completion? Say for initial conditions:

$$y(0.1)=0$$

$$y'(0.1)=1$$

Or whatever initial conditions make sense. Note that it's singular at x=0. We can let:

$$v'=p$$

Thus:

$$v''=\frac{dp}{dt}=\frac{dv}{dt}\frac{dp}{dv}=p\frac{dp}{dv}$$

Can you solve for v(t) and then get some expression for y(x)? I'll work on it too.

 

[Benny]

Well I've got the general solution seeing as it's an example from my book. You've suggested to me a few times to try using some initial conditions. I'm just wondering if the ones you've given me were arbitrarily chosen or if you knew that they would satisfy the general solution if I found it correctly. I mean, if I had a general solution for some DE then it's not all that difficult to choose 'nice' ICs which would suit the equation well. Is there a particular reason why ICs should be used? Is it a good check?

 

[saltydog]

Well I've got the general solution seeing as it's an example from my book. You've suggested to me a few times to try using some initial conditions. I'm just wondering if the ones you've given me were arbitrarily chosen or if you knew that they would satisfy the general solution if I found it correctly. I mean, if I had a general solution for some DE then it's not all that difficult to choose 'nice' ICs which would suit the equation well. Is there a particular reason why ICs should be used? Is it a good check?

These are just arbitrarily chosen away from the singularity x=0. This is just to check the results. Another way is to back-substitute the solution into the ODE. You know, just getting the answer is often not good enough. Applying it to a real example (IVP problem) allows one to obtain an intutitive understanding of what's going on with the ODE.

 

[Benny]

Yeah ok, I see what you mean. I'll try it with some other problems that I'll be doing when 2nd semester starts next week. In the mean time I'll just be brushing up on integration and some DEs to prepare.

 

[saltydog]

Just a summary:


Starting with:

$$x^3 \frac{{d^2 y}}{{dx^2 }} - \left( {x^2 + xy} \right)\frac{{dy}}{{dx}} + \left( {y^2 + xy} \right) = 0$$

make the substitution y=xv(x) we obtain:

$$x\frac{{d^2 v}}{{dx^2 }} - \left( {1 - v} \right)\frac{{dv}}{{dx}} = 0$$

Making the substitution:

$$x=e^t$$

give us:

$$\frac{{d^2 v}}{{dt^2 }} - v\frac{{dv}}{{dt}} = 0$$

Letting:

$$v^{'}=p$$

$$v^{''}=p\frac{dp}{dv}$$

gives us:

$$\frac{dp}{dv}-v=0$$

and so:

$$p=\frac{v^2}{2}+c_1$$

Thus:

$$\frac{dv}{dt}=\frac{v^2}{2}+c_1$$

Separating variables and solving:

$$v(t)=\sqrt{c_1}Tan\left[\sqrt{c_1}(t/2+c_2)\right]$$

Since:

$$x=e^t$$

We obtain for y:

$$y(x)=x\left(\sqrt{c_1}Tan\left[\sqrt{c_1}(\frac{ln(x)}{2}+c_2)\right]\right)$$

I back-substituted this into the ODE and obtained the desired result, 0.

I then solved the initial value problem:

$$x^3 \frac{{d^2 y}}{{dx^2 }} - \left( {x^2 + xy} \right)\frac{{dy}}{{dx}} + \left( {y^2 + xy} \right) = 0;\quad y(0.1)=0 \quad y^{'}(0.1)=1$$

solving for $c_1$ and $c_2$, I get:

$$y(x)\approx x\left(\sqrt{0.2}Tan\left[\sqrt{0.2}(\frac{ln(x)}{2}+1.1513)\right]\right)$$

The first plot is a numerical analysis of the ODE. The second plot is y(x) as defined above.

I believe this is correct. Benny, I assume you obtained this answer?

 

[Benny]

I haven't tried doing this question since it is an example but the answer provided agrees with yours apart from a few constants but it's clearly only because they've replaced constants after each integrations(thing like c = 2c_1 etc). Thanks for posting a full solution(my book skips steps in almost all examples), I'll analyse it later on.