Second order differential equations

[Just some guy]

hi,

I have a question showing the 'particle in a box' example of the 1-d Schrödinger equation, and given the initial conditions (walls of infinite potential, zero potential inside the box) the time-independent equation reduces to d^2y/dx^2 = -k^2y, where k is a constant - my text just gives me the answer to this equation and I'm wondering whether it's possible to be solved? My mathematics only goes as far as solving simple first-order differential equations so I'm a bit lost here - I assume one separates the variables to get (1/y)d^2y = (-k^2)dx^2, but what then?

Cheers,
Just some guy

 

[mathmike]

integrate then solve for y

 

[Physics Monkey]

First let me assure you that the equation can be solved, in fact it has the same structure as another equation you have surely seen before: Newton's 2nd Law for a harmonic oscillator.

How can you proceed then?

Well, the most basic approach is educated guessing. Can you think of any functions that are closed under differentiation. That is, do you know some functions that you can differentiate twice and get something like themselves back.

Separation of variables isn't directly applicable, but you can play games under the integral and make progress. A better way to approach the problem is to multiply both sides by $$y'(x)$$ and see what you can do.

The systematic theory of linear differential equations with constant coeffecients gives you a prescription for the solution. You assume a solution of the form $$y(x) = e^{\alpha x}$$, and plug this "solution" into your differential equation (for the motivation behind this procedure, think about what I said above). It isn't hard to convince yourself that this reduces the differential equation in y to an algebraic equation in $$\alpha$$, one which can be solved in this case easily. Don't forget that there is more than one solution (why?).

Other more advanced methods exist including Laplace transform methods and Green's function methods, but they are a bit of overkill for so simple a problem (but fun nevertheless).

I've given you some general sketches of possible methods, see if you can apply one and find the solution. Hope this helps.

 

[HallsofIvy]

No, you cannot "separate" second order derivatives like that.

A standard method is to recognize that the equation is saying that the second derivative on the left side is just a multiple of the function on the right. You should be able to see immediately that it can't be, for example, a logarithm (whose second derivative is -1/x[/sup]2[/sup]) or most other functions. You might think about an exponential: the derivative of e^x is e^x, exactly the same as the function itself. To allow for the k, try y= e^rx where r is some (unknown) number. The derivative of that is re^rx and the second derivivative is r²e^rx. Putting that into the equation gives you r²e^rx= -k²e^rx. Since the exponential is never 0, you can divide through by that to get r²= -k². In order to satisfy that r must be either ki or -ki. Now, you might remember that e^ikx= cos(kx)+ i sin(kx)- and then realize that, of course, if y= sin(kx), y'= k cos(kx), and y''= -k²sin(kx)= -k²y.
Same for y= cos(kx). That's how your textbook got the result.