Apply Fourier Transform to Solve t2u_t-u_x=g(x)

Click For Summary

Discussion Overview

The discussion revolves around applying the Fourier transform to solve the partial differential equation $t^2u_t - u_x = g(x)$, with specified conditions. The focus is on the mathematical techniques involved in using the Fourier transform in this context.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant inquires about applying the Fourier transform specifically to the term $t^2u_t$ in the equation.
  • Another participant suggests taking the Fourier transform with respect to $x$ and treating $t$ as a constant, leading to the expression $t^2 \frac{\partial U}{\partial t}$, where $U(\omega, t)$ denotes the Fourier transform of $u(x,t)$.
  • A subsequent post indicates that this leads to the ordinary differential equation $t^2 \frac{\partial U}{\partial t} - iwU = F(g)$, and seeks confirmation on the correctness of this formulation.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the correctness of the derived ordinary differential equation and seek confirmation, indicating that the discussion remains unresolved.

Contextual Notes

The discussion does not clarify assumptions regarding the existence or uniqueness of solutions, nor does it address potential limitations of the Fourier transform method in this context.

Markov2
Messages
149
Reaction score
0
I need to apply Fourier transform to solve the following: $t^2u_t-u_x=g(x),$ $x\in\mathbb R,$ $t>0$ and $u(x,1)=0,$ $x\in\mathbb R.$
How do I apply the Fourier transform for $t^2u_t$ ?

Thanks!
 
Physics news on Phys.org
インテグラルキラー;437 said:
I need to apply Fourier transform to solve the following: $t^2u_t-u_x=g(x),$ $x\in\mathbb R,$ $t>0$ and $u(x,1)=0,$ $x\in\mathbb R.$
How do I apply the Fourier transform for $t^2u_t$ ?

Thanks!

Take the Fourier transform with respect to $x$ then you simply treat $t$ as a constant.

$$ \int \limits_{-\infty}^{\infty} t^2 \frac{\partial u}{\partial t} e^{-i\omega x} dx = t^2 \frac{\partial }{\partial t}\int \limits_{-\infty}^{\infty} u(x,t) e^{-i\omega x} dx = t^2 \frac{\partial U}{\partial t}$$

Here we differenciated under the integral sign, and $U(\omega , t)$ is the notation for the Fourier transform of $u(x,t)$ with respect to $x$.
 
Ah, then I have ${{t}^{2}}\dfrac{\partial U}{\partial t}-iwU=F(g)$ and I need to solve that ODE, first I need a particular solution.

Does this look right?
 
Last edited:
Can anyone confirm this please?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K