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

[Markov2]

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!

 

[ThePerfectHacker]

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 $x$.

 

[Markov2]

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?

 

[Markov2]

Can anyone confirm this please?

 

[blue_raver22]

To apply the Fourier transform, we first define the Fourier transform of a function $f(x)$ as:

$\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)e^{-2\pi i \xi x}dx$

Applying this definition to our equation, we get:

$\hat{t^2u_t}(\xi) = \int_{-\infty}^{\infty} t^2u_t(x,t)e^{-2\pi i \xi x}dx$

Using the product rule for differentiation, we can rewrite this as:

$\hat{t^2u_t}(\xi) = \int_{-\infty}^{\infty} t^2e^{-2\pi i \xi x}\frac{\partial u}{\partial t}(x,t)dx$

Now, we can use integration by parts to further simplify this expression. Letting $v = t^2e^{-2\pi i \xi x}$ and $u = \frac{\partial u}{\partial t}(x,t)$, we get:

$\hat{t^2u_t}(\xi) = \left[ t^2e^{-2\pi i \xi x}\frac{\partial u}{\partial t}(x,t) \right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{\partial}{\partial t}\left( t^2e^{-2\pi i \xi x} \right)\frac{\partial u}{\partial x}(x,t)dx$

Since we are given that $u(x,1) = 0$, the first term in the above expression is zero. We can also rewrite the second term using the Fourier transform of $\frac{\partial u}{\partial x}(x,t)$:

$\hat{t^2u_t}(\xi) = -\int_{-\infty}^{\infty} \frac{\partial}{\partial t}\left( t^2e^{-2\pi i \xi x} \right)\frac{\partial}{\partial x}\hat{u}(\xi)e^{2\pi i \xi x}dx$

Substituting this into our original equation, we get:

$-\int_{-\infty}^{\infty} \frac{\partial}{\partial