ex8: abort

exercise8/main.typ

@@ -17,6 +17,10 @@
   rect(inset: FONT_SIZE / 2)[#it]
 }
 
+#show ref: it => if it.element.func() != heading { it } else {
+  link(it.target, it.element.body)
+}
+
 these are my solutions to the eighth exercise set of TMA4135.
 
 this document was created using
@@ -30,7 +34,6 @@ this document was created using
 
 #pagebreak()
 
-= problem 1
-
 #include "problem1.typ"
 
+#include "problem2.typ"

exercise8/problem1.typ

@@ -1,6 +1,7 @@
 #import "lib.typ": butcher
-#import "@preview/physica:0.9.6": *
 
+= problem 1
+<p1>
 
 == a)
 

exercise8/problem2.typ

@@ -0,0 +1,82 @@
+#import "lib.typ": butcher
+
+= problem 2
+
+== a)
+
+we can read the coefficients from the calculations of the next $y$-value and
+from the stage derivatives $k_i$. additionally, we can tell that there are four
+stages, thus it is a fourth order runge-kutta method.
+
+#butcher(4, (
+  $1 slash 2$,
+  $1 slash 2$,
+  $3 slash 4$,
+  0,
+  $3 slash 4$,
+  1,
+  $2 slash 9$,
+  $3 slash 9$,
+  $4 slash 9$,
+  $7 slash 24$,
+  $1 slash 4$,
+  $1 slash 3$,
+  $1 slash 8$,
+))
+
+
+== b)
+
+reading the loop-conditions we can tell that the loop terminates when the next
+computed value would overstep the bound $T$ which is set to $2.0$. we increment
+the time $t$ by the time step $h = 0.25$ each iteration, such that our values
+for $t$ are as follows (upon entry of the loop)
+$
+  0.0 -> 0.25 -> 0.5 -> 0.75 -> 1.0 -> 1.25 -> 1.5 -> 1.75.
+$
+the next step would yield $t = 2.0$, which is not strictly less than $T$, thus
+we terminate the loop. counting the iteration steps, we get that we compute
+8 different time steps.
+
+
+== c)
+
+we have already established that the order is 4. this can also be seen since the
+pivot elements of the tableau are non-zero.
+
+== d)
+<2d>
+
+noticing that the upper part of this butcher tableau is the same as for
+ralston's method, and we are working on the same initial value problem, we can
+re-use $k_1, k_2$ and $k_3$ from problem @p1.
+
+thus we only need to compute
+$
+  k_4 & = f(t_0 + h, y_0 + h(2 k_1 + 3 k_2 + 4 k_3)/9) \
+      & = 0.48 dot e^(-0.48 dot (2 dot 0 + 3 dot 0.24 + 4 dot 0.3302)/9) \
+      & = #{ 0.48 * calc.exp(-0.48 * (2 * 0 + 3 * 0.24 + 4 * 0.3302) / 9) }
+$
+
+then assemble into
+$
+  y_1 & = h (7/24 k_1 + 1/4 k_2 + 1/3 k_3 + 1/8 k_4) \
+  & = #{ 0.48 * (7 / 24 * 0 + 1 / 4 * 0.24 + 1 / 3 * 0.3302 + 1 / 8 * 0.4305) }
+$
+which is fortunately very reminiscent of what we obtained in @p1.
+
+
+== e)
+
+we can plug our calculated values for $y_1$ from @2d and @p1 into the
+formula
+$
+  hat(epsilon)_(n+1)
+  = abs(y_(n+1) - y^*_(n+1))
+  = h abs(sum_(i=1)^s (b_i - b^*_i) k_i)
+$
+such that
+$
+  hat(epsilon)_1 = abs(y_1 - y^*_1)
+  =
+$