Solving a differential equation by integration

[tiredryan]

Hello,

I am reading a paper from George Stokes from 1851, and there is a part that I would like help on. He gives the differential equation,

$$\left(\frac{d^2}{dr^2}-\frac{2}{r^2}\right)^2 f(r) = 0$$

He states that the integral of the equation is

$$f(r) = Ar^{-1} + Br + Cr^2 + Dr^4$$

I am trying to figure out how he went about the integration. Do you have any suggestions? Here is my workflow so far.

$$\left(\frac{d^2}{dr^2}-\frac{2}{r^2}\right)^2 f(r) = \left(\frac{d^2}{dr^2}-\frac{2}{r^2}\right)\left(\frac{d^2}{dr^2}-\frac{2}{r^2}\right)f(r) = 0$$

Multiplying the differentials,

$$0 = \left(\frac{d^4}{dR^4} - \frac{d^2}{dR^2}\frac{2}{R^2} - \frac{2}{R}\frac{d^2}{dR^2} + \frac{4}{R^4}\right)f(x)$$

Simplifying the derivative,

$$0 = f^{IV}(R) - \frac{4}{R^2}f^{II}(R) + \frac{8}{R^3}f^{I}(R) - \frac{8}{R^4}F(R)$$

I can plug in f(r), f'(r), f''(r), and f''''(r) into this differential equation and prove that f(r) is a solution to the differential equation. Now I tried to integrate this equation.

$$\frac{8}{R^4}f(R) = f^{IV}(R) - \frac{4}{R^2}f^{II}(R) + \frac{8}{R^3}f^{I}(R)$$

$$\int\frac{8}{R^4}f(R) = \int f^{IV}(R) - \int\frac{4}{R^2}f^{II}(R) + \int\frac{8}{R^3}f^{I}(R)$$

Now I am stuck with the following section of the equation

$$[...] \int\frac{4}{R^2}f^{II}(R) + \int\frac{8}{R^3}f^{I}(R) [...]$$

I do not know how to integrate an unknown function f(R) that is multiplied by R. I couldn't find an inverse chain rule or inverse product rule for integrals. Is there a trick to doing this? Is there a simpler path to solve the differential equation? Thanks

 

[clustro]

I am pretty sure you need to know what the function F actually is.

You can find the derivative of a product easy (uv' + vu'), but the integration of a product needs a little more massaging.

 

[tiredryan]

Thanks clustro. I've been trying integration by parts, but your point is valid about needing to know the function F. Without knowing the function, then the integration by parts continues on an endless loop. If anyone has any thoughts on how Stokes solved the differential equation, I would appreciate it.

Here is some context. The proof was published in 1851 so it is in the public domain available on Google. I am interested in how he goes from equation 118 to 119. A direct link is:
http://books.google.com/books?id=l2...sical papers&lr=&pg=PA55#v=onepage&q=&f=false
I also attached a pdf to this message of the specific page. Thanks.

I am pretty sure you need to know what the function F actually is.

You can find the derivative of a product easy (uv' + vu'), but the integration of a product needs a little more massaging.

 

[HallsofIvy]

You don't, in general, solve differential equations by simply "integrating".

If you "multiply" that out, you get
$$\frac{d^4 f}{dr^4}- \frac{4}{r^2}\frac{d^2 f}{dr^2}+ \frac{4}{r^4}f= 0$$

Multiply both sides of that by $r^4$ to get
[tex]x^4\frac{d^4 f}{dr^4}- 4r^2\frac{d^2 f}{dr^2}+ 4f= 0[/itex]<br /> NOT what you have.<br /> <br /> That is an "Euler-type" or "equipotential equation"- each derivative is multiplied by a power of r equal to the order of the derivative.<br /> <br /> I you "try" a solution of the form $f= r^s$, then $f'= s r^{s-1}$, $f''= s(s-1) r^{s-2}$, $f'''= s(s-1)(s-2)r^{s-3}$ and $f''''= s(s-1)(s-2)(s-3)r^{s-4}$. Putting those into the equation you get $s(s-1)(s-2)(s-3)r^s- 4s(s-1)r^s+ 4r^s$$= (s^4- 6s^2+ 11s- 6)r^s= 0$. In order for that to be 0 for all r, you must have $s^4- 6s^3+ 11s^2- 6s= s(s^3- 6s^2+ 11s- 6)= 0$.<br /> <br /> s= 0 is obviously a solution to that and it is not difficult to see that s= 1 satisfies $s^3- 6s^2+ 11s- 6= 0$. Dividing that polynomial by s-1 gives $s^2- 5s+ 6= (s- 2)(s- 3)$. That is, $s^4- 6s^3+ 11s^2- 6s= 0$ is satisfied by s= 0, s= 1, s= 2, and s= 3.<br /> <br /> That means that each of $f(r)= r^0= 1$, $f(r)= r^1= r$, $f(r)= r^2$ and $f(r)= r^3$ are solutions to the differential equation.<br /> <br /> Since those are four "independent" solutions and the differential equation is a linear, homogenous d.e. of order four, any solution can be written in the form $f(x)= C_1+ C_2r+ C_3r^2+ C_4r^3$.<br /> <br /> Also, with the change of variable x= ln(r), that equation becomes a d.e. with constant coefficients which might be more familiar.[/tex]

 

[g_edgar]

HallsOfIvy is incorrect. That's not how you multiply operators. Tiredryan did expand it correctly (aside from magically changing r to R). Note that $r^3$ is not a solution of the original differential equation, but $1/r$ is.

Here is what you can do to solve this. Observe that it is a 4th degree linear DE and that those four powers of x satisfy it, so the general solution is as given by Stokes.

To actually solve the thing from scratch, you need to solve successively two problems of the form $(d^2y/dr^2 - (2/r^2)y) = g(r)$, first where the right-hand side is o, then where the right-hand side is the solution of the first one.

 

[tiredryan]

Thanks. In case anyone is confused. Here is my logic on how I multiplied out the differential

$$0 = \left(\frac{d^4}{dR^4} - \frac{d^2}{dR^2}\frac{2}{R^2} - \frac{2}{R}\frac{d^2}{dR^2} + \frac{4}{R^4}\right)f(x)$$

Since

$$\frac{d^2}{dR^2}\left(\frac{2}{R^2} f(x)\right) \neq \frac{2}{R}\left(\frac{d^2}{dR^2} f(x)\right)$$I differentiated the $$\frac{2}{R^2}$$ out of the equation using the product rule twice. $$\frac{d^2}{dR^2}\left(\frac{2}{R^2} f(x)\right) = \frac{d}{dR}\left(\frac{d}{dR}\left(\frac{2}{R^2} f(x)\right)\right)$$

$$\frac{d^2}{dR^2}\left(\frac{2}{R^2} f(x)\right) = \frac{d}{dR}\left( \frac{2}{R^2}\frac{\partial}{\partial R}f(x) - \frac{4}{R^3}f(x)\right)\right)$$

Now doing the product rule again,

$$\frac{d}{dR}\left( \frac{2}{R^2}\frac{\partial}{\partial R}f(x) - \frac{4}{R^3}f(x)\right)\right) = -2R^{-2}f''(R) + 4R^{-3}f'(R) - 12R^{-4}f(R) + 4R^{-3}f'(R)$$

The final simplified derivative becomes,

$$0 = f^{IV}(R) - \frac{4}{R^2}f^{II}(R) + \frac{8}{R^3}f^{I}(R) - \frac{8}{R^4}F(R)$$

Note:
I'm sorry for the changing between the r and R. Assume r=R and we are in spherical coordinates. Stokes switches from Cartesian to Cylindrical to finally Spherical to solve his flow around a sphere so I might have mixed it up. All the prior equations occur after he switched to the final spherical coordinates and r should really be R.

 

[tiredryan]

I am having problems solving the equation when "g(r) [...equals...] the solution to the first one"

Let X(R) equal the solution the the differential equation where G(R) = 0.

$$\left(\frac{d^2X}{dR^2} - \frac{2}{R^2}X\right) = G(R)$$

I began with the following equation

$$0 = R^2 X'' - 2X$$

Assuming the solution is X(R)=R^m

$$X(R) = R^m$$

Therefore the first and second differentials are as follows

$$\frac{dX}{dR} = mR^{(m-1)}$$

$$\frac{d^2X}{dR^2} = (m-1)(m)R^{(m-2)}$$

Plugging into the differential equation yields

$$0 = R^2 (m-1)(m)R^{(m-2)} - 2R^m$$

The "R"s are canceled out yielding

$$0 = m^2-m - 2$$

This can be solved with m = -1 and m = 2

This yields the solution of the differential equation in the general form

$$X(R) = \frac{A}{R} + CR^2$$

Now setting X(R) = G(R)

$$\left(\frac{d^2F}{dR^2} - \frac{2}{R^2}F\right) = X(R)$$

Now beginning with

$$R^2 F'' - 2F = R^2X(R)$$

$$R^2 F'' - 2F = AR + CR^4$$

Assuming the solution is F(R)=R^m

$$F(R) = R^m$$

Therefore the first and second differentials are as follows

$$\frac{dF}{dR} = mR^{(m-1)}$$

$$\frac{d^2F}{dR^2} = (m-1)(m)R^{(m-2)}$$

Plugging into the differential equation yields

$$AR + CR^4 = R^2 (m-1)(m)R^{(m-2)} - 2R^m$$

$$AR + CR^4 = (m^2 - m - 2) R^m$$

I am stuck here. Thanks.

PS: On second thought I think the method that HallsofIvy suggested works. I'll try the math now.

 

[tiredryan]

I got it working.

Using the method mentioned by HallsofIvy, I get an equation in the form

$$s^4 - 6s^3 + 7s^2 + 6s - 8 = 0$$

From this the roots, -1, 1, 2, and 4 are obtained.

This was much easier than what I was trying before. Thanks.

 

[clustro]

The horse has been beaten to a pulp, but just for shiggles I plugged it into matlab's dsolve function and got:

$$c_4t + c_2t^{-1} + c_3t^4 + c_5t^2$$

I should have probably caught it last night, but it was late -.-

Good call on the multiplication of the differentials.