Second (forth) order differential equation

  • Thread starter rsq_a
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  • #1
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1
Does anybody know of a way to attack the differential equation:

[tex]
y'''' + f(x)y'' = 0
[/tex]

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

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

Unfortunately, there doesn't seem a way for me to isolate [tex]\widehat{y}[/tex]. 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).
 

Answers and Replies

  • #2
1,796
53
May I propose the following purely formal and unproven approach to this problem:

[tex]y''''+f(x)y''=0[/tex]

Let u=y'' to obtain:

[tex]u''+f(x)u=0[/tex]
[tex](D^2+f(x))u=0[/tex]

How about we factor that?

[tex](D+i\sqrt{f})(D-i\sqrt{f})u=0[/tex]

Now let:

[tex](D-i\sqrt{f})u=g(x)[/tex]

then we have:

[tex](D+i\sqrt{f})g=0[/tex]

[tex]g(x)=e^{-i\int \sqrt{f}+c}=ce^{-i\int\sqrt{f}}[/tex]

Then:

[tex](D-i\sqrt{f})u=ce^{-i\int\sqrt{f}}[/tex]

Turn the crank one more time and I get:

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

Integrate twice and I get:

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

Is that right guys? I'm not sure.

Why don't we move this to the DE sub-forum too?
 
Last edited:
  • #3
107
1
May I propose the following purely formal and unproven approach to this problem:

[tex]y''''+f(x)y''=0[/tex]

Let u=y'' to obtain:

[tex]u''+f(x)u=0[/tex]
[tex](D^2+f(x))u=0[/tex]

How about we factor that?

[tex](D+i\sqrt{f})(D-i\sqrt{f})u=0[/tex]

Now let:

[tex](D-i\sqrt{f})u=g(x)[/tex]

then we have:

[tex](D+i\sqrt{f})g=0[/tex]

[tex]g(x)=e^{-i\int \sqrt{f}}[/tex]

Then:

[tex](D-i\sqrt{f})u=e^{-i\int\sqrt{f}}[/tex]

Turn the crank one more time and I get:

[tex]u=e^{i\int\sqrt{f}}\left(\int e^{-2i\sqrt{f}}+c\right)[/tex]

Integrate twice and I get:

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

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...

[tex](D^2+f(x)) \neq (D+i\sqrt{f})(D-i\sqrt{f})[/tex]

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

[tex]\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[/tex]

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)...
 
  • #4
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5,905
Hmm...I was going to say that this was brilliant. And then I realised that...

[tex](D^2+f(x)) \neq (D+i\sqrt{f})(D-i\sqrt{f})[/tex]

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,

[tex]\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[/tex]

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)...
 
  • #5
107
1
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?
 
  • #6
1,796
53
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:

[tex](D+i\sqrt{g})(D-i\sqrt{g})u=(D^2+(g-d\sqrt{g})u=0[/tex]

so that IF we have the equation:

[tex]u''+(g-d\sqrt{g})u=0[/tex]

then we can use my analysis above and arrive at:

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

Therefore for the equation:

[tex]u''+fu=0[/tex]

then let:

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

and solve for the unknown g(t).

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

[tex]u''+fu=0[/tex]

is:

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

and by association:

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

I believe it's not hard to test that expression numerically in Mathematica.
 
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  • #7
107
1
Hello rsq. How about I try?

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

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

It seems you've defined d to be,

[tex]d\sqrt{g} = -iD\sqrt{g} + i\sqrt{g}D[/tex]

but then afterwards...

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

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

[tex]u'' - xu = (D^2 - x)u[/tex]

...
 
Last edited:
  • #8
1,796
53
Have I missed something? I don't understand.

It seems you've defined d to be,

[tex]d\sqrt{g} = -iD\sqrt{g} + i\sqrt{g}D[/tex]

but then afterwards...

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

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

[tex]u'' - xu = (D^2 - x)u[/tex]

...
I forgot an i factor:

[tex]
\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}
[/tex]

so that I'm defining:

[tex]d\sqrt{g}=-1/2g^{-1/2}g'[/tex]

Do you agree that this is correct?


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

[tex]f=g-1/2i\frac{1}{\sqrt{g}}\frac{dg}{dx}[/tex]

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

[tex]u''+xu=0[/tex]

I would have to solve:

[tex]2x=2g-\frac{i}{\sqrt{g}}\frac{dg}{dx}[/tex]

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

[tex]2=2g-\frac{i}{\sqrt{g}}\frac{dg}{dx}[/tex]

which could be a check for the equation:

[tex]u''+u=0[/tex]

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

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