ex7: problem 2

exercise7/main.typ

@@ -34,4 +34,6 @@ this document was created using
 
 #include "problem1.typ"
 
+= problem 2
 
+#include "problem2.typ"

exercise7/problem2.typ

@@ -0,0 +1,103 @@
+#import "@preview/physica:0.9.6": *
+
+recall that a function $f: D -> V$ is lipschitz continuous iff
+$
+  forall y_1, y_2 in D, quad underbracket(
+    abs(f(y_2) - f(y_1)) <= L abs(y_2 - y_1),
+    "lipschitz condition"
+  )
+$
+for some $L in RR$.
+
+== a)
+
+let $f: RR^2 -> RR,quad (t, y) |-> y/t^2$.
+
+converting the function to it's polar coordinate form using
+$
+  t = r cos(theta) quad "and" quad y = r sin(theta)
+$
+such that $f[r, theta] = sin(theta)/(r cos^2(theta)) = 1/r sec(theta) tan(theta)$
+
+the magnitude of the gradient of the function tells us how much the function
+changes in a point
+$
+           grad f[r, theta] & = vecrow(pdv(f, r), pdv(f, theta))^TT \
+                            & = vecrow(
+                                -1/r^2 sec(theta) tan(theta),
+                                1/r sec(θ) (sec^2(θ) + tan^2(θ))
+                              )^TT \
+  ==> abs(grad f[r, theta]) & = sqrt(
+                                1/r^4 sec^2(theta) tan^2(theta) + 1/r^2
+                                sec^2(theta) (sec^2(theta) + tan^2(theta))^2
+                              ) \
+                            & = 1/r^2 sec(theta) sqrt(
+                                tan^2(theta) + r^2
+                                (sec^2(theta) + tan^2(theta))^2
+                              )
+$
+
+if we look at $lim_(r -> oo) abs(grad f[r,theta])$ we observe that the function
+is well-behaved -- i.e. lipschitz continuous -- for values of $theta$ that do
+not yield $tan(theta) = plus.minus oo$, since it converges to $0$ due to
+$1 slash r^2$.
+
+we need to inspect the expression more closely for those other values of
+$theta$. let $phi = pi slash 2 - theta$ so we can observe
+$
+  sec(theta) & = sec(pi/2 - phi) = csc(phi) approx 1/phi \
+  tan(theta) & = tan(pi/2 - phi) = cot(phi) approx 1/phi
+$
+such that
+$
+  #let L = $lim_(phi -> 0^+)$
+  #L abs(grad f[r, phi]) & = #L 1/(r^2 phi) sqrt(
+                             1/phi^2 + r^2(1/phi^2
+                               + 1/phi^2)^2
+                           ) \
+                         & = #L 1/(r^2 phi) sqrt(1/phi^2 + r^2 4/phi^4) \
+                         & = #L 1/(r^2 phi) sqrt(
+                             4r^2 / phi^4 (1 + phi^2 / (4
+                               r^2))
+                           ) \
+                         & = #L 2 / (r phi^3) sqrt(1 + phi^2 / (4 r^2)) \
+                         & = underbracket(#L 2 / (r phi^3), oo) dot underbracket(
+                             #L sqrt(
+                               1 + phi^2 /
+                               (4 r^2)
+                             ),
+                             sqrt(1) = 1
+                           ) \
+                         & -> oo
+$
+
+we can see that the derivative diverges for such problematic angles. this shows
+that it is not lipschitz continuous for all its domain, since its rate of change
+becomes unbounded for certain input values. #h(1fr) $qed$
+
+it was not really worth it going via polar coordinates.
+
+
+== b)
+
+let $f(t, y) = sin(t) / t y$ on $RR$.
+
+the only point that may be problematic here is $f(0, 0)$.
+
+recall that $lim_(x -> 0) sin(x) / x = 1$, such that
+$
+  #let L = $lim_((t, y) -> (0, 0))$
+  #L f(t, y) = #L sin(t) / t y = #L y = 0
+$
+
+we need to check the rate of change in this point, so let
+$
+  f_t (0 + epsilon) = sin(t)/t epsilon
+$
+but
+$
+  sin(t)/t epsilon <= L epsilon
+$
+so the rate of change is bounded in $y$-direction. similarly for $t$-direction,
+we observe that the function is well-behaved. it is thus lipschitz continuous on
+$RR$.