Fourier transform with mixed derivatives/ 2nd order ODE

In summary, Jianxu tried to solve the homework statement, but got stuck on how to solve for \left.\widehat{u}(\omega ,t). He used Fourier transforms, eliminated variables, and found that \left.u(x,t) = \frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}(\widehat{u}(\omega ,t))e^{i\omega x}d\omega.
  • #1
jianxu
94
0

Homework Statement


Hi, So I'm suppose to solve the following problem:

[tex]\left.\frac{d^{2}u}{dt^{2}}-4\frac{d^{3}u}{dt dx^{2}}+3\frac{d^{4}u}{dx^{4}}=0[/tex]

[tex]\left.u(x,0) = f(x)[/tex]
[tex]\left.\frac{du}{dt}(x,0) = g(x)[/tex]

Homework Equations



The Attempt at a Solution


First I use Fourier transform on the given expression so that I get the following:

Fourier transform of [tex]\left.\frac{d^{2}u}{dt^{2}}(x,t) = \frac{d^{2}\widehat{u}}{dt^{2}}(\omega ,t) [/tex]

Fourier transform of [tex]\left.\frac{du}{dt}(x,t) = \frac{d\widehat{u}}{dt}(\omega ,t) [/tex]

Fourier transform of [tex]\left.\frac{d^{2}u}{dx^{2}}(x,t) = \left(i\omega\right)^{2}\widehat{u}(\omega ,t) = -\left(\omega\right)^{2}\widehat{u}(\omega ,t)[/tex]

Fourier transform of [tex]\left.\frac{d^{4}u}{dx^{2}}(x,t) = \left(i\omega\right)^{4}\widehat{u}(\omega ,t) = \left(\omega\right)^{4}\widehat{u}(\omega ,t)[/tex]

Which means me overall expression after transform is:
[tex]\left.\frac{d^{2}\widehat{u}}{dt^{2}}(\omega ,t)+4\left(\omega\right)^{2}\frac{d\widehat{u}}{dt}(\omega ,t)+3\left(\omega\right)^{4}\widehat{u}(\omega ,t)=0[/tex]

Now assuming I did that correctly, the next step I think I should proceed with is to solve for [tex]\left.\widehat{u}(\omega ,t)[/tex]. I don't remember how to solve this type of ODE, I was reading a couple of sites and it says I should use a characteristic equation which would I assume then be, [tex]\left.\lambda^{2}+4\omega^{2} \lambda +3\omega[/tex] where [tex]\lambda[/tex] is just an arbitrary symbol to denote a quadratic equation. I looked for the roots and used it along with the general expression of the 2nd order ODE to get
[tex]\left.\widehat{u}(\omega ,t)=c_{1}+c_{2}e^{-4\omega^{2}t}[/tex]
But it seems to be incorrect since I took the derivative and plugged it back into my Fourier transform expression and did not get a 0 for my answer so...Any guidance would be much appreciated! Thanks!
 
Physics news on Phys.org
  • #2
HI jianxu! :smile:

(have an omega: ω :wink:)
jianxu said:
Which means me overall expression after transform is:
[tex]\left.\frac{d^{2}\widehat{u}}{dt^{2}}(\omega ,t)+4\left(\omega\right)^{2}\frac{d\widehat{u}}{dt}(\omega ,t)+3\left(\omega\right)^{4}\widehat{u}(\omega ,t)=0[/tex]

… I should use a characteristic equation which would I assume then be, [tex]\left.\lambda^{2}+4\omega^{2} \lambda +3\omega^4[/tex] where [tex]\lambda[/tex] is just an arbitrary symbol to denote a quadratic equation. I looked for the roots and used it along with the general expression of the 2nd order ODE to get
[tex]\left.\widehat{u}(\omega ,t)=c_{1}+c_{2}e^{-4\omega^{2}t}[/tex]

No, the roots are ω2 = -1 and -3, so the general solution is c1e2t + c2e-3ω2t :smile:
 
  • #3
Hi TinyTim, thanks for the reply! I just realized where I made my mistake! Thanks very much for the help!
 
  • #4
Hello!

I've been working on this problem and was wondering if someone could check if I've done the rest of this problem correctly!

So after finding the roots, I apply the initial conditions where:
[tex]\left.\widehat{u}\left(\omega,0\right) = \widehat{f}\left(\omega\right)[/tex]

since t = 0, I have:
[tex]\left.\widehat{u}\left(\omega,0\right) = \widehat{f}\left(\omega\right) = C_{1}+C_{2}[/tex]

For [tex]\left.\frac{d\widehat{u}}{dt}[/tex]:
[tex]\left.\frac{d\widehat{u}}{dt}= -\omega C_{1}e^{-\omega^{2}t}-3\omega C_{2}e^{-3\omega^{2}t}[/tex]
applying initial conditions:
[tex]\left.\frac{d\widehat{u}}{dt}\left(\omega,0\right)= -\omega C_{1}-3\omega C_{2} = \widehat{g}\left(\omega\right)[/tex]

Now I solved for the constants using elimination and got:
[tex]\left.C_{1}= \frac{3\widehat{f}(\omega)}{2} + \frac{\widehat{g}(\omega)}{2\omega^{2}}[/tex]
and:
[tex]\left.C_{2}= -\frac{\widehat{f}(\omega)}{2} - \frac{\widehat{g}(\omega)}{2\omega^{2}}[/tex]

Therefore our [tex]\left.\widehat{u}(\omega ,t) = \left(\frac{3\widehat{f}(\omega)}{2} + \frac{\widehat{g}(\omega)}{2\omega^{2}}\right)e^{-\omega^{2}t}+ \left(-\frac{\widehat{f}(\omega)}{2} - \frac{\widehat{g}(\omega)}{2\omega^{2}}\right)e^{-3\omega^{2}t}[/tex]

Now I apply inverse Fourier transformation so that:
[tex]\left.u(x,t) = \widehat{f^{-1}}(\omega)[/tex]

Which means:
[tex]\left.u(x,t) = \frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}(\widehat{u}(\omega ,t))e^{i\omega x}d\omega[/tex]

that would be the solution seeing that we don't know what [tex]\left.\widehat{f}(\omega )[/tex] or [tex]\left.\widehat{g}(\omega )[/tex] are.

Thanks!
 

1. What is the Fourier transform of a function with mixed derivatives?

The Fourier transform of a function with mixed derivatives is a complex-valued function that represents the frequency components of the original function. It is defined as the integral of the function multiplied by a complex exponential function.

2. How is the Fourier transform used for solving 2nd order ordinary differential equations (ODEs)?

The Fourier transform can be used to transform a 2nd order ODE into an algebraic equation in terms of the transformed function. This allows for easier solving of the ODE, as well as providing insight into the frequency components of the solution.

3. What are the advantages of using the Fourier transform for solving 2nd order ODEs?

The Fourier transform allows for the solution of ODEs with mixed derivatives, which can be difficult to solve using other methods. It also allows for the identification of frequency components in the solution, which can be useful in the study of physical systems.

4. Are there any limitations to using the Fourier transform for solving ODEs?

The Fourier transform method is limited to linear ODEs with constant coefficients. It may also not be suitable for problems with discontinuous or non-differentiable functions.

5. How does the inverse Fourier transform relate to the original function with mixed derivatives?

The inverse Fourier transform is the process of transforming the Fourier transform back into the original function. In the case of a function with mixed derivatives, the inverse Fourier transform will yield the original function with the mixed derivatives. This allows for the solution of the ODE to be obtained in its original form.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
733
  • Calculus and Beyond Homework Help
Replies
16
Views
475
  • Calculus and Beyond Homework Help
Replies
5
Views
222
  • Calculus and Beyond Homework Help
Replies
6
Views
131
  • Calculus and Beyond Homework Help
Replies
1
Views
756
  • Calculus and Beyond Homework Help
Replies
3
Views
687
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
98
  • Calculus and Beyond Homework Help
Replies
8
Views
119
Back
Top