Second (forth) order differential equation

In summary, the conversation discusses a way to attack the differential equation y'''' + f(x)y'' = 0, where f(x) is assumed to be too complicated for a closed-form solution. The proposed approach involves using Fourier transforms and factoring operators, but there is some disagreement about the validity of the approach and a need for a concrete example to test it.
  • #1
rsq_a
107
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).
 
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  • #2
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?
 
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  • #3
jackmell said:
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 realized 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
rsq_a said:
Hmm...I was going to say that this was brilliant. And then I realized 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.
rsq_a said:
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
Mark44 said:
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
rsq_a said:
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
jackmell said:
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
rsq_a said:
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. :)
 
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Related to Second (forth) order differential equation

What is a second (forth) order differential equation?

A second (forth) order differential equation is a mathematical equation that involves a function and its first and second (forth) derivatives. It is commonly used to model physical systems in fields such as physics and engineering.

How is a second (forth) order differential equation different from a first order differential equation?

A second (forth) order differential equation involves the second (forth) derivative of a function, while a first order differential equation only involves the first derivative. This means that a second (forth) order differential equation has more complex solutions and can model more complex systems.

What are some real-world applications of second (forth) order differential equations?

Second (forth) order differential equations are commonly used in physics and engineering to model a variety of systems, such as the motion of objects under the influence of forces, electrical circuits, and vibrations of a mass on a spring. They are also used in economics and biology to model population growth and decay.

How do you solve a second (forth) order differential equation?

Solving a second (forth) order differential equation involves finding a function that satisfies the equation. This can be done analytically, using mathematical techniques such as separation of variables and substitution, or numerically, using computer algorithms. The specific method used depends on the form of the equation and the initial conditions given.

Why are second (forth) order differential equations important in science and engineering?

Second (forth) order differential equations are important because they allow us to mathematically model and understand complex systems in the physical world. By studying the behavior of solutions to these equations, we can make predictions and design systems that function efficiently and effectively. They also have applications in a wide range of fields, making them a fundamental tool in scientific research and engineering design.

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