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frank0520

  • one year ago

Let w be a positive real number. Find the kernel of D^2 + w^2 I

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  1. anonymous
    • one year ago
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    is \(D^2\) the second derivative operator?

  2. anonymous
    • one year ago
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    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
    • one year ago
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    i guess that should be \(c_1,c_2\in\mathbb{C}\) actually

  4. anonymous
    • one year ago
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    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
    • one year ago
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    @oldrin.bataku here is a picture if it makes any difference.

  6. anonymous
    • one year ago
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    okay, so D is a differential operator, and the work I wrote above is correct

  7. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    @oldrin.bataku thanks for the help!

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