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

[leopard]

Find the Fourier transform $$\hat{u}(w,t) = \frac{1}{\sqrt{2 \pi}} \int^{\infty}_{- \infty}u(x,t)e^{(-ixw)}dx$$ of the general solution u(x,t) of the PDE $$u_{t}= u_{xx} - u$$

Should I start by solving the PDE, or is there another way to do it?

 

[leopard]

Anyone?

 

[Avodyne]

The PDE is much easier to solve when Fourier transformed (it becomes an ODE), so first write $u$ in terms of $\hat u$, plug into the PDE, and solve.

 

[leopard]

How du I transform u?

 

[Avodyne]

$$u(x,t) = \frac{1}{\sqrt{2 \pi}} \int^{\infty}_{- \infty}\hat{u}(w,t)e^{(+ixw)}dw$$

This is the inverse of the Fourier transform.

 

[leopard]

How can i find the inverse when I don't know u?

My book says that the Fourier transform of the PDE is

$$U_t = -w^2 U - U$$

How is that achieved?

 

[Avodyne]

First of all, $U=\hat u$.

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