Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G\in \mathcal{F}(\mathbb{R}^{n+1})'$ be a distribution on the space of spatial Fourier transform'able function, ie an $L^1_{\mathrm{loc}}(\mathbb{R^{n+1}})$ function, $G = G(t,\xi)$. Assume $a(\xi)\in C^{\infty}(\mathbb{R}^n)$ and for a.e. $\xi\in \mathbb{R^n}$ fixed, $G(\cdot, \xi)$ satisfies: $$ \frac{dG}{dt}+a(\xi)G(t, \xi) = \delta(t) \quad \text{ in } C^{\infty}_c(\mathbb{R})' $$

Then prove $G(\cdot,\cdot)$ satisfies:

$$ \frac{dG}{dt}+a(\xi)G = 1_{\xi} \otimes \delta(t) \quad \text{ in } \mathcal{F}(\mathbb{R}^{n+1})' $$

After using the definition of the distribution for the first differential equation, ie multiplying a function $g(s)$ and integrate along the $\mathbb{R}$, then multiply another $C^{\infty}_c(\mathbb{R})$ function $f(x)$, since I would like to use the density argument, saying the sums of products like $f(x)g(t)$ are dense in $C^{\infty}_c$, hence dense in the Fourier transformable function space. Then integrate and do change of variable somehow I ended up with $a(\xi-x)$ instead of $a(\xi)$.

Thanks a lot!

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.