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

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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!
 
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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?
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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