Partial Differential Equations of the second order

[mooeypoo]

Okay, I took an ODE class and we barely touched PDEs. I am trying to figure it out now, but I am having troubles understanding how to solve this seemingly very simple equation. I could use a bit of a guidance, even to sites or texts that explain the concept.

 

[math]a\frac{d^{2}y}{dx^2}-b\frac{d^{2}y}{dt^2}=0[/math]

Where a and b are constants.

 

If this was on the first order, I'd be doing separation of variables, and integrating both sides (I think) but this is second order.. I'm a bit lost.

 

Thanks!

 

~mooey

[timo]

I'm afraid I can't tell you how to solve this; I just happen to know what the solutions look like. But maybe it helps you to know that the equation you're showing frequently appears in many physical scenarios as a "wave equation" (and one possible set of solutions are plane waves, then).

[mississippichem]

I often use this site: Paul's Online Math Notes: Differential Equations whenever I get over my head with some nasty DE's. I've also been known to whip out the TI-89 as well though. If you do a significant amount of work with non-homogeneous, higher order, or partial DE's the TI-89 "partials" package is worth it's weight in gold.

[Bignose]

Okay, I took an ODE class and we barely touched PDEs. I am trying to figure it out now, but I am having troubles understanding how to solve this seemingly very simple equation. I could use a bit of a guidance, even to sites or texts that explain the concept.

 

[math]a\frac{d^{2}y}{dx^2}-b\frac{d^{2}y}{dt^2}=0[/math]

Where a and b are constants.

 

If this was on the first order, I'd be doing separation of variables, and integrating both sides (I think) but this is second order.. I'm a bit lost.

 

Thanks!

 

~mooey

 

the first step is to rearrange the equation at hand:

 

[math]c\frac{d^{2}y}{dx^2}=\frac{d^{2}y}{dt^2}[/math] (I'm going to glom together a and b as c = a/b)

 

and note that the left hand side is only differentiated with respect to x. and the right hand side is differentiated with respect to t. This suggests a solution of the form y(x,t) = X(x)T(t). That is, the final solution is the product of two functions, each of only 1 variable.

 

Now, if you let ' represent differentiation with respect to the only variable of the function (i.e. X' is differentiation with respect to x, and T' is with respect to t), the equation can be written:

 

cX''T = XT''

 

or

 

cX''/X = T''/T

 

now, since the left hand side is ONLY a function of x, and since the right hand side is ONLY a function of t, we know that the only way this is so is if both sides are equal to a constant.

 

cX''/X = k

 

and

 

T''/T = k

 

Now you have two ODEs you should be able to solve, and when you get each of X(x) and T(t) the final solution is y(x,t) = X(x)T(t).

 

You should be able to check out any book on PDEs from the library as a reference for this procedure of breaking PDEs down to ODEs. It only works for certain cases of PDEs, but it is the standard procedure for these separable-like ones.

Edited April 2, 2011 by Bignose

[mooeypoo]

I'm afraid I can't tell you how to solve this; I just happen to know what the solutions look like. But maybe it helps you to know that the equation you're showing frequently appears in many physical scenarios as a "wave equation" (and one possible set of solutions are plane waves, then).

 

Oh, I know that, timo, it looked like the stuff we had done in Quantum. But most of the processes to solve these type of question that I've seen so far go something like "Let us assume a wave-like solution" and go on from there.

 

I wanted to see if I can understand how the solution was done so I can see *why* the assumption is made.

 

the first step is to rearrange the equation at hand:

 

[math]c\frac{d^{2}y}{dx^2}=\frac{d^{2}y}{dt^2}[/math] (I'm going to glom together a and b as c = a/b)

 

and note that the left hand side is only differentiated with respect to x. and the right hand side is differentiated with respect to t. This suggests a solution of the form y(x,t) = X(x)T(t). That is, the final solution is the product of two functions, each of only 1 variable.

 

Now, if you let ' represent differentiation with respect to the only variable of the function (i.e. X' is differentiation with respect to x, and T' is with respect to t), the equation can be written:

 

cX''T = XT''

 

or

 

cX''/X = T''/T

 

now, since the left hand side is ONLY a function of x, and since the right hand side is ONLY a function of t, we know that the only way this is so is if both sides are equal to a constant.

 

cX''/X = k

 

and

 

T''/T = k

 

Now you have two ODEs you should be able to solve, and when you get each of X(x) and T(t) the final solution is y(x,t) = X(x)T(t).

 

You should be able to check out any book on PDEs from the library as a reference for this procedure of breaking PDEs down to ODEs. It only works for certain cases of PDEs, but it is the standard procedure for these separable-like ones.

 

Awesome, thanks!

This looks awesomely familiar... I think I've seen it in E/M when we tried to consider boundary problems, like a charged box. I might be mistaken, but I think so. We had that with three variables (X'', Y'' and Z'') and did this to separate them.

 

I don't quite remember what this method is called, which is slightly annoying 'cause I'd like to read more about it. Any ideas?

 

I'm working it out now and will post what I did so I can see if I got it right.

 

Thanks!!

 

~mooey

 

K, here's my process. Is this right?

 

[math]\frac{d^{2}y}{dx^2}=\frac{xk}{c}[/math]

[math]dy=\frac{k}{c}\int x dx^2 =\frac{k}{c} \int \frac{x^2}{2}+C dx[/math]

[math]y(x)=\frac{k}{2c} \left( \frac{x^3}{3}+Cx+D \right)[/math]

 

AND

 

 

[math]\frac{d^{2}y}{dt^2}=tk[/math]

[math]y(t)=k\frac{t^2}{2}+E dt=\frac{k}{2} \left( \frac{t^3}{3}+Et+F \right)[/math]

 

and the final equation is:

 

[math]y(x,t)=\frac{k^2}{4c} \left( \frac{x^3}{3}+Cx+D \right)\left( \frac{t^3}{3}+Et+F \right)[/math]

 

 

So now, I should open this up and use the initial conditions to solve for C, D, E and F, right? I should have 4 initial conditions in this case, y0(t), y0(x) and those two as primes.

 

Am I doing this right?

 

Meh, this makes no sense, I'm supposed to be getting something "sin/cos" or exponential at least.

 

What am I doing wrong?

[Bignose]

Ah, my memory isn't perfect. Usually, we set the constant to be negative, and it is often convenient to make it a square, too. So, let us let [math]k=-\beta^2[/math].

 

Then you get [math]X'' + \beta^2 X = 0[/math] and [math]T'' + \beta^2 T =0[/math]

 

The solution of these is: [math]X(x) = C_1 \cos \beta x + C_2 \sin \beta x[/math] and similar for T.

 

When you apply the BCs, you should get an infinite number of solutions because cos and sin are periodic.

 

I pulled out my copy of Strauss' Partial Differential Equations An Introduction to refresh my memory on these, but again most any text on PDEs will cover this technique. I believe it also goes by the name of separation of variables.