# Second (forth) order differential equation

Does anybody know of a way to attack the differential equation:

$$y'''' + f(x)y'' = 0$$

In this case, you have to assume that f(x) is too complicated to be written down in closed-form. I don't need a closed form solution---an integral equation will do. I can take Fourier Transforms, then get the equation

$$(ik)^4\widehat{y} + (ik)^2 \int_{-\infty}^\infty \widehat{f}(s) \widehat{y}(k-s) \ ds = 0$$

Unfortunately, there doesn't seem a way for me to isolate $$\widehat{y}$$. Is there a standard technique for dealing with these things?

I know, for example, that the Airy equation, y'' = xy has no closed-form solution, but there is a way to put it into integral form. The trick, however, is that you need to know how to take the Fourier transform of xy (which you can).

May I propose the following purely formal and unproven approach to this problem:

$$y''''+f(x)y''=0$$

Let u=y'' to obtain:

$$u''+f(x)u=0$$
$$(D^2+f(x))u=0$$

$$(D+i\sqrt{f})(D-i\sqrt{f})u=0$$

Now let:

$$(D-i\sqrt{f})u=g(x)$$

then we have:

$$(D+i\sqrt{f})g=0$$

$$g(x)=e^{-i\int \sqrt{f}+c}=ce^{-i\int\sqrt{f}}$$

Then:

$$(D-i\sqrt{f})u=ce^{-i\int\sqrt{f}}$$

Turn the crank one more time and I get:

$$u=c_1e^{i\int\sqrt{f}}\left(\int e^{-2i\sqrt{f}}+c_2\right)$$

Integrate twice and I get:

$$y(x)=c_1\iint e^{i\int\sqrt{f}}\left(\int e^{-2i\sqrt{f}}+c_2\right)$$

Is that right guys? I'm not sure.

Why don't we move this to the DE sub-forum too?

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May I propose the following purely formal and unproven approach to this problem:

$$y''''+f(x)y''=0$$

Let u=y'' to obtain:

$$u''+f(x)u=0$$
$$(D^2+f(x))u=0$$

$$(D+i\sqrt{f})(D-i\sqrt{f})u=0$$

Now let:

$$(D-i\sqrt{f})u=g(x)$$

then we have:

$$(D+i\sqrt{f})g=0$$

$$g(x)=e^{-i\int \sqrt{f}}$$

Then:

$$(D-i\sqrt{f})u=e^{-i\int\sqrt{f}}$$

Turn the crank one more time and I get:

$$u=e^{i\int\sqrt{f}}\left(\int e^{-2i\sqrt{f}}+c\right)$$

Integrate twice and I get:

$$y(x)=\iint e^{i\int\sqrt{f}}\left(\int e^{-2i\sqrt{f}}+c\right)$$

Is that right guys? I'm not sure.

Why don't we move this to the DE sub-forum too?
Hmm...I was going to say that this was brilliant. And then I realised that...

$$(D^2+f(x)) \neq (D+i\sqrt{f})(D-i\sqrt{f})$$

I don't think you're allowed to factor operators like that. For example,

$$\frac{d^2}{dx^2} - x \neq \left(\frac{d}{dx} - \sqrt{x}\right)\left(\frac{d}{dx} + \sqrt{x}\right) = \frac{d^2}{dx^2} - \frac{d}{dx}(\sqrt{x}) + \sqrt{x}\frac{d}{dx} - x$$

I don't think the approach can be rescued...

...for a moment, I thought you had presented a method which essentially allows one to solve any linear differential equation (since any linear operator could have been factored into factors with single derivatives)...

Mark44
Mentor
Hmm...I was going to say that this was brilliant. And then I realised that...

$$(D^2+f(x)) \neq (D+i\sqrt{f})(D-i\sqrt{f})$$

I don't think you're allowed to factor operators like that.
Sure you are. For example, consider y'' - 3y' + 2y = 0.
Written using operators, this is (D2 - 3D + 2)y = 0.
Factoring the quadratic, we get (D - 2)(D - 1)y = 0.
So (D - 2)y = 0 or (D - 1)y = 0, which leads us to y = e2t or y = et as basic solutions for this differential equation.
For example,

$$\frac{d^2}{dx^2} - x \neq \left(\frac{d}{dx} - \sqrt{x}\right)\left(\frac{d}{dx} + \sqrt{x}\right) = \frac{d^2}{dx^2} - \frac{d}{dx}(\sqrt{x}) + \sqrt{x}\frac{d}{dx} - x$$

I don't think the approach can be rescued...

...for a moment, I thought you had presented a method which essentially allows one to solve any linear differential equation (since any linear operator could have been factored into factors with single derivatives)...

Sure you are. For example, consider y'' - 3y' + 2y = 0.
Written using operators, this is (D2 - 3D + 2)y = 0.
Factoring the quadratic, we get (D - 2)(D - 1)y = 0.
So (D - 2)y = 0 or (D - 1)y = 0, which leads us to y = e2t or y = et as basic solutions for this differential equation.
Yes, can you provide an example without constant coefficients?

I don't think the approach can be rescued...
Hello rsq. How about I try?

Yes, I made a terrible blunder above by mis-handling the operators. What I believe is:

$$(D+i\sqrt{g})(D-i\sqrt{g})u=(D^2+(g-d\sqrt{g})u=0$$

so that IF we have the equation:

$$u''+(g-d\sqrt{g})u=0$$

then we can use my analysis above and arrive at:

$$u=c_1e^{i\int\sqrt{g}}\left(\int e^{-2i\sqrt{g}}+c_2\right)$$

Therefore for the equation:

$$u''+fu=0$$

then let:

$$f=g-d\sqrt{g}=g-1/2 g^{-1/2}\frac{dg}{dt}$$

and solve for the unknown g(t).

Once we solve for g(t), then the solution for the original equation:

$$u''+fu=0$$

is:

$$u=c_1e^{i\int\sqrt{g}}\left(\int e^{-2i\sqrt{g}}+c_2\right)$$

and by association:

$$y(x)=c_1\iint e^{i\int\sqrt{g}}\left(\int e^{-2i\sqrt{g}}+c_2\right)$$

I believe it's not hard to test that expression numerically in Mathematica.

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Hello rsq. How about I try?

Yes, I made a terrible blunder above by mis-handling the operators. What I believe is:

$$(D+i\sqrt{g})(D-i\sqrt{g})u=(D^2+(g-d\sqrt{g})u=0$$
Have I missed something? I don't understand.

It seems you've defined d to be,

$$d\sqrt{g} = -iD\sqrt{g} + i\sqrt{g}D$$

but then afterwards...

$$f=g-d\sqrt{g}=g-1/2 g^{-1/2}\frac{dg}{dt}$$

This I don't understand. Now you treat d like a regular derivative? Perhaps this is just a problem with notation. Why don't you apply it to a concrete example? Suppose that

$$u'' - xu = (D^2 - x)u$$

...

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Have I missed something? I don't understand.

It seems you've defined d to be,

$$d\sqrt{g} = -iD\sqrt{g} + i\sqrt{g}D$$

but then afterwards...

$$f=g-d\sqrt{g}=g-1/2 g^{-1/2}\frac{dg}{dt}$$

This I don't understand. Now you treat d like a regular derivative? Perhaps this is just a problem with notation. Why don't you apply it to a concrete example? Suppose that

$$u'' - xu = (D^2 - x)u$$

...
I forgot an i factor:

\begin{aligned} (D+i\sqrt{g})(D-i\sqrt{g})w&=(D+i\sqrt{g})(w'-i\sqrt{g}w) \\ &=w''-i\frac{d}{dx}(\sqrt{g}w)+i\sqrt{g}w'+gw \\ &=w''-i(\sqrt{g}w'+1/2wg^{-1/2}g')+i\sqrt{g}w'+gw \\ &=w''-i\sqrt{g}w'-1/2iwg^{-1/2}g'+i\sqrt{g}w'+gw \\ &=w''+(g-1/2ig^{-1/2}g')w\\ &=w''+(g+id \sqrt{g})w \end{aligned}

so that I'm defining:

$$d\sqrt{g}=-1/2g^{-1/2}g'$$

Do you agree that this is correct?

(2) Also, I glossed-over the fact that the equation:

$$f=g-1/2i\frac{1}{\sqrt{g}}\frac{dg}{dx}$$

is tough to solve symbolically. So for your example with plus sign to make it simpler:

$$u''+xu=0$$

I would have to solve:

$$2x=2g-\frac{i}{\sqrt{g}}\frac{dg}{dx}$$

which I don't know how to solve analytically. However, I can solve:

$$2=2g-\frac{i}{\sqrt{g}}\frac{dg}{dx}$$

which could be a check for the equation:

$$u''+u=0$$

that is, does my complicated formula above reduce to the known solution to this equation? Don't know yet. :)

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