## frank0520 one year ago Let w be a positive real number. Find the kernel of D^2 + w^2 I

1. anonymous

is $$D^2$$ the second derivative operator?

2. anonymous

in that case, consider: $$D^2+w^2I=(D-iwI)(D+iwI)$$ and the kernels of these are spanned simply by $$e^{\pm iwt}$$. the kernel of $$D^2+w^2I$$ is then simply $$\operatorname{span}\{e^{iwt},e^{-iwt}\}=\{c_1e^{iwt}+c_2e^{-iwt}:c_1,c_2\in\mathbb{R}\}$$

3. anonymous

i guess that should be $$c_1,c_2\in\mathbb{C}$$ actually

4. anonymous

using Euler's formula $$e^{iwt}=\cos(wt)+i\sin(wt)$$ we find that $$\cos(wt),\sin(wt)$$ are another perfectly suitable basis for the kernel and so we could also write our kernel as consisting of complex linear combinations $$c_1\cos(wt)+c_2\sin(wt)$$ as well

5. frank0520

@oldrin.bataku here is a picture if it makes any difference.

6. anonymous

okay, so D is a differential operator, and the work I wrote above is correct

7. anonymous

the kernel of $$D^2+\omega^2I$$ is the solution space to the following idfferential equation: $$(D^2+\omega^2I)y=0\\\frac{d^2y}{dt^2}+\omega^2y=0\\\implies y(t)=c_1\cos(\omega t)+c_2\sin(\omega t)$$ which is what I wrote above

8. anonymous

to find the image of $$D-\lambda I$$, consider we get every possible inhomogeneous part $$f(t)$$ to the equation $$\frac{dy}{dt}-\lambda y=f(t)$$ which seems to be more or less the entire function space we're looking at

9. frank0520

@oldrin.bataku thanks for the help!