Fourier Transform of General Solution for PDE u_{t}= u_{xx} - u

In summary, the Fourier transform of the general solution u(x,t) of the PDE u_{t}= u_{xx} - u can be found by first writing u in terms of \hat u, plugging it into the PDE, and solving. The inverse of the Fourier transform is u(x,t) = \frac{1}{\sqrt{2 \pi}} \int^{\infty}_{- \infty}\hat{u}(w,t)e^{(+ixw)}dw. To get the equation U_t = -w^2 U - U, substitute the formula for u in terms of \hat u into the original PDE.
  • #1
leopard
125
0
Find the Fourier transform [tex]\hat{u}(w,t) = \frac{1}{\sqrt{2 \pi}} \int^{\infty}_{- \infty}u(x,t)e^{(-ixw)}dx[/tex] of the general solution u(x,t) of the PDE [tex]u_{t}= u_{xx} - u[/tex]

Should I start by solving the PDE, or is there another way to do it?
 
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  • #2
Anyone?
 
  • #3
The PDE is much easier to solve when Fourier transformed (it becomes an ODE), so first write [itex]u[/itex] in terms of [itex]\hat u[/itex], plug into the PDE, and solve.
 
  • #4
How du I transform u?
 
  • #5
[tex]u(x,t) = \frac{1}{\sqrt{2 \pi}} \int^{\infty}_{- \infty}\hat{u}(w,t)e^{(+ixw)}dw[/tex]

This is the inverse of the Fourier transform.
 
  • #6
How can i find the inverse when I don't know u?

My book says that the Fourier transform of the PDE is

[tex]U_t = -w^2 U - U[/tex]

How is that achieved?
 
  • #7
First of all, [itex]U=\hat u[/itex].

To get your book's equation, first substitute the formula for [itex]u[/itex] in terms of [itex]\hat u[/itex] (that I gave in my last response) into the original PDE. What do you get?
 

1. What is a Fourier Transform?

A Fourier Transform is a mathematical operation that decomposes a function into its constituent frequencies. It allows us to represent a function as a combination of sinusoidal waves of different frequencies and amplitudes.

2. How is the Fourier Transform used in solving PDEs?

The Fourier Transform is used to transform a partial differential equation (PDE) into an ordinary differential equation (ODE), which can then be solved using traditional methods. This is particularly useful for solving PDEs with periodic boundary conditions, as the Fourier Transform is well-suited for dealing with periodic functions.

3. What is the general solution for the PDE ut = uxx - u?

The general solution for this PDE is u(x,t) = A0 + B0t + ∑n=1(Ancos(nx) + Bnsin(nx))e-n2t, where A0 and B0 are arbitrary constants and An and Bn are coefficients determined by the initial conditions.

4. What is the significance of the Fourier Transform of the general solution for this PDE?

The Fourier Transform of the general solution allows us to see the contributions of each frequency component to the overall solution. It also allows us to easily evaluate the solution at any point in space and time, making it a powerful tool for analyzing the behavior of the system.

5. Are there any limitations to using the Fourier Transform in solving PDEs?

While the Fourier Transform is a powerful tool for solving PDEs, it is not always applicable. It is most useful for problems with periodic boundary conditions and can be challenging to use for problems with non-periodic boundary conditions or in higher dimensions. It also assumes that the solution is smooth and well-behaved, which may not always be the case in real-world applications.

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