104 lines
3.4 KiB
Typst
104 lines
3.4 KiB
Typst
#import "@preview/physica:0.9.6": *
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recall that a function $f: D -> V$ is lipschitz continuous iff
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$
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forall y_1, y_2 in D, quad underbracket(
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abs(f(y_2) - f(y_1)) <= L abs(y_2 - y_1),
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"lipschitz condition"
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)
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$
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for some $L in RR$.
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== a)
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let $f: RR^2 -> RR,quad (t, y) |-> y/t^2$.
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converting the function to it's polar coordinate form using
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$
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t = r cos(theta) quad "and" quad y = r sin(theta)
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$
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such that $f[r, theta] = sin(theta)/(r cos^2(theta)) = 1/r sec(theta) tan(theta)$
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the magnitude of the gradient of the function tells us how much the function
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changes in a point
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$
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grad f[r, theta] & = vecrow(pdv(f, r), pdv(f, theta))^TT \
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& = vecrow(
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-1/r^2 sec(theta) tan(theta),
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1/r sec(θ) (sec^2(θ) + tan^2(θ))
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)^TT \
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==> abs(grad f[r, theta]) & = sqrt(
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1/r^4 sec^2(theta) tan^2(theta) + 1/r^2
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sec^2(theta) (sec^2(theta) + tan^2(theta))^2
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) \
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& = 1/r^2 sec(theta) sqrt(
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tan^2(theta) + r^2
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(sec^2(theta) + tan^2(theta))^2
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)
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$
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if we look at $lim_(r -> oo) abs(grad f[r,theta])$ we observe that the function
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is well-behaved -- i.e. lipschitz continuous -- for values of $theta$ that do
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not yield $tan(theta) = plus.minus oo$, since it converges to $0$ due to
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$1 slash r^2$.
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we need to inspect the expression more closely for those other values of
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$theta$. let $phi = pi slash 2 - theta$ so we can observe
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$
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sec(theta) & = sec(pi/2 - phi) = csc(phi) approx 1/phi \
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tan(theta) & = tan(pi/2 - phi) = cot(phi) approx 1/phi
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$
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such that
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$
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#let L = $lim_(phi -> 0^+)$
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#L abs(grad f[r, phi]) & = #L 1/(r^2 phi) sqrt(
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1/phi^2 + r^2(1/phi^2
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+ 1/phi^2)^2
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) \
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& = #L 1/(r^2 phi) sqrt(1/phi^2 + r^2 4/phi^4) \
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& = #L 1/(r^2 phi) sqrt(
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4r^2 / phi^4 (1 + phi^2 / (4
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r^2))
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) \
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& = #L 2 / (r phi^3) sqrt(1 + phi^2 / (4 r^2)) \
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& = underbracket(#L 2 / (r phi^3), oo) dot underbracket(
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#L sqrt(
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1 + phi^2 /
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(4 r^2)
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),
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sqrt(1) = 1
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) \
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& -> oo
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$
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we can see that the derivative diverges for such problematic angles. this shows
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that it is not lipschitz continuous for all its domain, since its rate of change
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becomes unbounded for certain input values. #h(1fr) $qed$
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it was not really worth it going via polar coordinates.
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== b)
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let $f(t, y) = sin(t) / t y$ on $RR$.
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the only point that may be problematic here is $f(0, 0)$.
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recall that $lim_(x -> 0) sin(x) / x = 1$, such that
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$
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#let L = $lim_((t, y) -> (0, 0))$
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#L f(t, y) = #L sin(t) / t y = #L y = 0
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$
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we need to check the rate of change in this point, so let
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$
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f_t (0 + epsilon) = sin(t)/t epsilon
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$
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but
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$
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sin(t)/t epsilon <= L epsilon
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$
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so the rate of change is bounded in $y$-direction. similarly for $t$-direction,
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we observe that the function is well-behaved. it is thus lipschitz continuous on
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$RR$.
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