frank0520
  • frank0520
Let w be a positive real number. Find the kernel of D^2 + w^2 I
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
is \(D^2\) the second derivative operator?
anonymous
  • 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 $$\operatorname{span}\{e^{iwt},e^{-iwt}\}=\{c_1e^{iwt}+c_2e^{-iwt}:c_1,c_2\in\mathbb{R}\}$$
anonymous
  • anonymous
i guess that should be \(c_1,c_2\in\mathbb{C}\) actually

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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
frank0520
  • frank0520
@oldrin.bataku here is a picture if it makes any difference.
anonymous
  • anonymous
okay, so D is a differential operator, and the work I wrote above is correct
anonymous
  • 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
anonymous
  • 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
frank0520
  • frank0520
@oldrin.bataku thanks for the help!

Looking for something else?

Not the answer you are looking for? Search for more explanations.