I'm self studying DE's and found some notes on the Internet. If I understood well, S-L problem is to find the solution to [itex]Ly=(py')'-qy=f[/itex] where y is a function of [itex]x[/itex]. Getting to the point and skipping tons of details, if [itex]u_1 (x)[/itex] and [itex]u_2 (x)[/itex] are linearly independent solutions to the homogeneous S-L problem, i.e. [itex]Ly=0[/itex], then a solution to [itex]Ly=f[/itex] is given by ... [itex]y(x)=\int _a^b g(x,t)f(t)dt[/itex] where [itex]g(x,t)[/itex] is the green function associated to the operator L, namely [itex]g(x,t)=\frac{1}{p(a)W(a)}[/itex] multiplied by [itex]u_1 (x) u_2 (t)[/itex] if a [itex]\leq x \leq t[/itex] or [itex]u_1 (t) u_2 (x)[/itex] if [itex]t <x<b[/itex]. Where [itex]W(a)[/itex] is the Wronksian of [itex]u_1 (a)[/itex] and [itex]u_2 (a)[/itex]. So it seems that in order to solve the original S-L DE, I must first solve the corresponding homogeneous DE. Then, calculate the Wronskian of these 2 L.I. solutions. Then form [itex]g(x,t)[/itex] and finally perform the integral that has no reason to be easy to solve. My question is... is this method (via Green function+Wronskian) really simpler than just using variation of coefficient method, once I solved the homogeneous S-L DE? In other words, once I get [itex]u_1 (x)[/itex] and [itex]u_2(x)[/itex], is seeking a solution to the non homogeneous DE of the form [itex]y(x)=u_1 (x) v_1(x)+u_2(x)v_2 (x)[/itex] really harder than using Green function method?! P.S.:I attach the document I studied on.
Your PDF doens't seem to mention the more interesting case where there is a parameter in the ODE, and satisfying the boundary conditions leads to an eignevalue problem for the parameter. In that situation S-L theory says interesting things about the general properties of solutions, the distribution of the eigenvalues, etc, quite apart from solving particular equations. Given the usual caveats about Wikipedia, see http://en.wikipedia.org/wiki/Sturm–Liouville_theory
Rather than opening a new thread, I think it's ok to ask 2 new questions here. 1)I've read that any second order differential equation can be put into Sturm-Liouville form, namely [itex]P(x)u''+Q(x)u'+[R(x)+\lambda ]u=0[/itex] can be put under the form [itex]\frac{d}{dx}\left [ p(x)\frac{du}{dx} \right ]+[q(x)+\lambda \rho (x)]u=0[/itex]. The way to do this is to divide the general form of a second order DE by [itex]P(x)[/itex] and then to mutiply it by [itex]p(x)=e^{\int ^x \frac{Q}{P}dt}[/itex]. So I'm trying this for the Legendre differential equation [itex](1-x^2)y''-2xy'+\mu (\mu +1)y=0[/itex]. So I divide by [itex]P(x)=(1-x^2)[/itex] and mutiply by [itex]e^{\int ^x \frac{-2x}{1-x^2}dt}[/itex] to get [itex]e^{\int ^x \frac{-2x}{1-x^2}dt}y''- \left ( \frac{2x}{1-x^2} \right ) e^{\int ^x \frac{-2x}{1-x^2}dt}y'+e^{\int ^x \frac{-2x}{1-x^2}dt} \left [ \frac{\mu (\mu +1) }{1-x^2} \right ] =0[/itex]. I'm a bit lost on understanding the integral. It looks only half definite so I don't know how to compute it. I think that [itex]\frac{d}{dx} e^{\int ^x \frac{-2x}{1-x^2}dt}=- \left ( \frac{2x}{1-x^2} \right ) e^{\int ^x \frac{-2x}{1-x^2}dt}[/itex] so that I can write the equation as [itex]\frac{d}{dx}\left [ p(x)\frac{dy}{dx} \right ]+e^{\int ^x \frac{-2x}{1-x^2}dt} \left [ \frac{\mu (\mu +1) }{1-x^2} \right ] =0[/itex] and this is where I'm stuck at finding [itex]q(x)[/itex], [itex]\rho (x)[/itex] and [itex]\lambda[/itex] in order to write the DE into the desired form. 2)It seems like any second order DE have infinitely many eigenvalues [itex]\lambda _n[/itex] that satisfies the Sturm-Liouville's problem. What does this mean? The solutions (eigenfunctions) are infinitely many to satisfy the S-T problem?
I think you might be over thinking this. First, S-L are in the form -d/dx[p(x)dy/dx] +q(x)y = whatever The Legendre equation is (1-x^2)d^2/dx^2'-2xdy/dx+n(n+1)y=0 D[1-x^2] = -2x So we already have the S-L you are looking for: d/dx[(1-x^2)dy/dx]+n(n+1)y=0 You'll have to do the whole integrating factor jazz with more complex equations, but for the Legendre it's really as simple as stepping back and looking at it for a second. As for your integral, I don't know how you ended up with that half definite, never seen it, but then again, I'm self-taught so what do I know. As for question number 2 As far as I know there exist an infinite number (countable) of eigenvalues and each one corresponds to an eigenfunction (I think) but I really haven't gotten a chance to see what this really means.
First of all, thank you for your answer. Ok that's the same method used as in wikipedia. In that case it was easy but I wanted to see how to use the general way for finding it for any second order DE. I'm also self teaching this to me. I found that on http://www.math.osu.edu/~gerlach.1/math/BVtypset/node63.html. I think you are right. I have no idea what they represent.
Well, as far as I know the integrating factor should end up being ln(1-x^2). Simply integrate -2x/(1-x^2). Not sure why they used the x there in the upper limit, but i've never seen it like that. Try it my way and i'm sure you'll get the same result.
If I consider the ST problem in wikipedia and for example I take the 2nd order DE [itex]y''+\omega ^2 y=0[/itex] (appears a lot in physics). By inspection I get that [itex]p(x)=1[/itex] and [itex]\lambda w(x)-q(x)=0[/itex]. Maybe with a little work I could find the [itex]\lambda_n[/itex]. What do these eigenvalues mean exactly? And for example the DE [itex]y''(x)=0[/itex]? What does the eigenvalues represent?
Eigenvalues have physical significance. Take a vibrating plate. Depending on the frequency of the stimulus, it will deform differently. As long as all the boundary conditions remain the same and all that changes is the frequency, all these different modes of vibration are described by the various eigenvalues. More clearly, in this case, each eigenvalue corresponds to a different frequency.
Ok but is it because the DE already represent something physical like deformation of the plate? What about a "random" 2nd order ODE? It seems like any 2nd order ODE can be reduced to a Sturm-Liouville problem.
True, however differential equations inherently describe physical phenomena right? That is why initial or boundary conditions make sense. Also, keep in mind that in the general case (for a random 2nd order ODE), if you just write it down in Sturm-Liouville formulation, the eigenvalue will be 0. It is non-zero only if: 1. The problem the equation describes has this property. 2. Even if it does, the boundary conditions must be such that the eigenvalue is not zero. But from a more abstract standpoint, although I may not be the most qualified person to discuss it, eigenvalues are simply parameters. Take the Helmholtz equation: [tex]y(x)''+λy(x)=0[/tex] What this tells you, is that for a given set of boundary conditions, you can have various (often infinite) solutions. Practically, as many as the possible values of λ. That is why these are called eigenvalues (=values of self), they are an intrinsic property of the equation (or, by extension, physical problem) itself, and not dependent on the value of either x or y. Edit: To help visualise it a little better, eigenvalue problems arise often in wave equations, where the eigenvalue relates to the period (or frequency) of the wave.
Hmm ok, I'll need some time to digest all of this. I do know about the eigenvalues of the wave equation. For the radial part of the Schrödinger's equation when considering the hydrogen atom, we also get a 2nd order DE where the eigenvalues represent the total energy of the system. There are also infinitely many possible values for [itex]\lambda _ n[/itex]'s, I think there should always be infinitely many of them according to wikipedia. In the case of the wave equation, the eigenvalues are "inside" a sin or cos function or something like that. So that a particular solution to that equation implies a particular lambda. Thus the general solution is the sum of all the particular solutions, so the sum over all [itex]\lambda _n[/itex]'s. All the sine/cosine terms involving the [itex]\lambda _n[/itex] are therefore linearly independent and the dimension of the space generated by the solutions is therefore... infinite? But this cannot be right since we're talking about a 2nd order DE. The dimension of the general solution function must be at most 2. Where is my error of thinking?
The dimension of the solution function in cartesian space, would indeed always be u(x,y). What you must realise is that eigenvalues are not connected in any way to the variables. With this in mind, let's say that your final solution is: [tex]u(x,y)=\sum{u_i(x,y)}[/tex] [tex]u(x,y)=u_1(x,y)+u_2(x,y)+u_3(x,y)+u_4(x,y)+u_5(x,y)[/tex] where every [itex]u_i[/itex] corresponds to a different eigenvalue. Since you are merely superposing solutions in 2D space, isn't this solution still 2-dimensional? It's basically adding different contours to get a desired contour. Qualitatively, Fourier theory suggests that you can fit certain functions of sines and cosines to your equation's solution, if you superpose infinite different (but appropriate) combinations of them. In other words, you need infinitely many terms of the series to have it converge to the desired function (the solution). This actually happens in nature. For instance, here is my understanding of the vibrating string: The vibration has some harmonics. While we may only perceive a limited amount of harmonics, theoretically the sound produced can be the superposition of infinite modes of vibration, each with a different eigenvalue (btw, that is why the instrument material is important in music). If the string was vibrating at a single eigenfrequency, we wouldn't be able to measure more than one harmonic. However, it turns out that if we analyse the sound spectrum of a vibrating string, we can measure a lot of harmonics. I'm no expert, but I can imagine that this may be where people got the idea to superpose the solutions :tongue2: See the picture here: http://en.wikipedia.org/wiki/Vibrating_string The real-life situation can be the superposition of some of all these modes of vibration. Interestingly, the math would suggest that you have to add them in a specific sequence (never thought of it that way before..!). Practically, after a certain amount of summations, the rest are damped and cannot be discerned. Also, sometimes, the boundary conditions may dictate a single value for the eigenvalue, for instance, λ=1.