Commit 923d18c66f8a04cbd39fcf150fa4e521a9d5c32a

some updates to ex. 4.2 and combining ex. 1 and 2
demo4/Harjoitus_4_teht2.lyx
(32 / 9)
  
1#LyX 1.6.4.2 created this file. For more info see http://www.lyx.org/
1#LyX 1.6.7 created this file. For more info see http://www.lyx.org/
22\lyxformat 345
33\begin_document
44\begin_header
8080\begin_inset Graphics
8181 filename h4t1tilanne.eps
8282 display false
83 width 6cm
83 width 10cm
8484
8585\end_inset
8686
125125.
126126 Ennen irtoamista kappale noudattaa seinän liikeyhtälöä
127127\begin_inset Formula \[
128u''(t)=a\,,\quad u(0)=u_{0}\,,\quad u_{t}(0)=v_{0}\,,\quad t\leq t_{cr}\,,\]
128u''(t)=a(t)\,,\quad u(0)=u_{0}\,,\quad u_{t}(0)=v_{0}\,,\quad t\leq t_{cr}\,,\]
129129
130130\end_inset
131131
132132missä
133\begin_inset Formula $a$
133\begin_inset Formula $a(t)$
134134\end_inset
135135
136 on vakiokiihtyvyys ja
136 on tunnettu kiihtyvyys ja
137137\begin_inset Formula $u_{0}$
138138\end_inset
139139
142142\end_inset
143143
144144 on tunnetut (vakio-)alkuarvot kappaleen sijainnille ja nopeudelle.
145 Toisin sanoen
145 Kun oletetaan, että
146\begin_inset Formula $a$
147\end_inset
148
149 on vakio (vastaavasti, jos
150\begin_inset Formula $a$
151\end_inset
152
153 on paloittain vakio), saadaan
146154\begin_inset Formula \begin{equation}
147155u(t)=\frac{1}{2}at^{2}+v_{0}t+u_{0}\,,\quad t\leq t_{cr}\,.\label{eq:t2-seina}\end{equation}
148156
211211
212212\end_inset
213213
214mistä
214mistä (
215\begin_inset Formula $a\neq0$
216\end_inset
217
218)
215219\begin_inset Formula \[
216220t_{cr}=\frac{-kv_{0}+\sqrt{k^{2}v_{0}^{2}-2ka\left[k(u_{0}-u_{s})+F_{cr}\right]}}{ka}\,.\]
217221
244244\end_inset
245245
246246.
247 Jos
248\begin_inset Formula $a=0$
249\end_inset
250
251,
252\begin_inset Formula \[
253t_{cr}=-\frac{k(u_{0}-u_{s})+F_{cr}}{kv_{0}}\,.\]
254
255\end_inset
256
257
247258\end_layout
248259
249260\begin_layout Standard
319319\begin_layout Standard
320320Tarkasteltava malli on siis
321321\begin_inset Formula \begin{align}
322mu''(t) & =ma\,,\quad t\leq t_{cr}\,,\nonumber \\
322mu''(t) & =ma(t)\,,\quad t\leq t_{cr}\,,\nonumber \\
323323mu''(t) & =-ku(t)\,,\quad t>t_{cr}\,.\nonumber \\
324324u(0) & =u_{0}\,,\quad u_{t}(0)=v_{0}\,.\label{eq:t2-malli}\end{align}
325325
353353
354354 regularisoitu muoto on
355355\begin_inset Formula \begin{align*}
356mu''(t) & =\left[1-w(t)\right]ma-w(t)ku(t)\,,\\
356mu''(t) & =\left[1-w(t)\right]ma(t)-w(t)ku(t)\,,\\
357357u(0) & =u_{0}\,,\quad u_{t}(0)=v_{0}\,.\end{align*}
358358
359359\end_inset
demo4/Harjoitus_4_teht2.pdf
(1022 / 706)
  
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demo4/JY1_Harjoitus_04.lyx
(2720 / 0)
  
1#LyX 1.6.7 created this file. For more info see http://www.lyx.org/
2\lyxformat 345
3\begin_document
4\begin_header
5\textclass article
6\use_default_options true
7\language finnish
8\inputencoding auto
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18\graphics default
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24\use_amsmath 1
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29\leftmargin 1in
30\topmargin 1in
31\rightmargin 1in
32\bottommargin 1in
33\secnumdepth 3
34\tocdepth 3
35\paragraph_separation indent
36\defskip medskip
37\quotes_language english
38\papercolumns 1
39\papersides 1
40\paperpagestyle default
41\tracking_changes false
42\output_changes false
43\author ""
44\author ""
45\end_header
46
47\begin_body
48
49\begin_layout Section*
50Jatkuvat mallit - harjoitus 4
51\end_layout
52
53\begin_layout Standard
54\align left
55JY1
56\begin_inset ERT
57status open
58
59\begin_layout Plain Layout
60
61
62\backslash
63
64\backslash
65
66\end_layout
67
68\end_inset
69
70Juha Jeronen, Tytti Saksa ja Tero Tuovinen
71\end_layout
72
73\begin_layout Subsection*
74Tehtävä 1
75\end_layout
76
77\begin_layout Standard
78Tarkastellaan heittoliikettä kimmoisan alustan päällä.
79 Huomioidaan vain pystysuuntainen liike.
80 Säännöllistetään törmäys alustan kappaleeseen kohdistamalla voimatermillä
81\end_layout
82
83\begin_layout List
84\labelwidthstring 00.00.0000
85
86\series bold
87(a)
88\series default
89
90\begin_inset Formula $f(u)=-u^{-}/\varepsilon\;,$
91\end_inset
92
93
94\end_layout
95
96\begin_layout List
97\labelwidthstring 00.00.0000
98
99\series bold
100(b)
101\series default
102
103\begin_inset Formula $f(u)=(u^{-})^{2}/\varepsilon\;,$
104\end_inset
105
106
107\end_layout
108
109\begin_layout Standard
110missä
111\begin_inset Formula $u^{-}:=\min(u,0)$
112\end_inset
113
114.
115 Määritä viive, jonka tehtävän regularisointi aiheuttaa törmäysten väliseen
116 periodiin, ja viiveen suhde vapaaseen lentoaikaan.
117\end_layout
118
119\begin_layout Standard
120\begin_inset Float figure
121placement H
122wide false
123sideways false
124status open
125
126\begin_layout Plain Layout
127\noindent
128\align center
129\begin_inset Graphics
130 filename pomppiva-pallo.png
131 display false
132 width 10cm
133
134\end_inset
135
136
137\end_layout
138
139\begin_layout Plain Layout
140\begin_inset Caption
141
142\begin_layout Plain Layout
143Alkuperäinen (regularisoimaton) systeemi.
144
145\begin_inset CommandInset label
146LatexCommand label
147name "fig:problem-setup"
148
149\end_inset
150
151
152\end_layout
153
154\end_inset
155
156
157\end_layout
158
159\end_inset
160
161
162\end_layout
163
164\begin_layout Standard
165\begin_inset Float figure
166placement H
167wide false
168sideways false
169status open
170
171\begin_layout Plain Layout
172\noindent
173\align center
174\begin_inset Graphics
175 filename pomppiva-pallo-2.png
176 display false
177 width 10cm
178
179\end_inset
180
181
182\end_layout
183
184\begin_layout Plain Layout
185\begin_inset Caption
186
187\begin_layout Plain Layout
188Regularisoitu systeemi.
189 Törmäykset ovat edelleen täysin elastisia, mutta jokainen kimpoaminen aiheuttaa
190 viiveen
191\begin_inset Formula $\tau$
192\end_inset
193
194.
195
196\begin_inset CommandInset label
197LatexCommand label
198name "fig:problem-setup-reg"
199
200\end_inset
201
202
203\end_layout
204
205\end_inset
206
207
208\end_layout
209
210\end_inset
211
212
213\end_layout
214
215\begin_layout Subsubsection*
216Ratkaisu
217\end_layout
218
219\begin_layout Standard
220Ratkaistaan ensin, miten systeemi käyttäytyy vapaan lennon aikana.
221 Kiinnitetään koordinaatisto siten, että
222\begin_inset Formula $u$
223\end_inset
224
225 ja
226\begin_inset Formula $v$
227\end_inset
228
229 kasvavat ylöspäin.
230 Systeemin liikeyhtälöt ovat
231\begin_inset Formula \begin{align}
232u_{t} & =v\label{eq:ut}\\
233v_{t} & =-g\label{eq:vt}\\
234u & \ge0\label{eq:uge0}\\
235v_{-}^{2} & =v_{+}^{2}\quad\text{törmäyshetkellä,}\label{eq:ke}\end{align}
236
237\end_inset
238
239missä
240\begin_inset Formula $u$
241\end_inset
242
243 on kappaleen korkeus alustan tasosta,
244\begin_inset Formula $v$
245\end_inset
246
247 on pystysuuntainen nopeus,
248\begin_inset Formula $g$
249\end_inset
250
251 kuvaa vakiopainovoimakentän vaikutusta, ja alaindeksit
252\begin_inset Formula $-$
253\end_inset
254
255 ja
256\begin_inset Formula $+$
257\end_inset
258
259 tarkoittavat eri suunnista otettuja raja-arvoja ajan
260\begin_inset Formula $t$
261\end_inset
262
263 suhteen.
264\end_layout
265
266\begin_layout Standard
267Alkuehdot ovat
268\begin_inset Formula \begin{align}
269u(0) & =u_{0}\label{eq:ICu}\\
270v(0) & =v_{0}\;.\label{eq:ICv}\end{align}
271
272\end_inset
273
274Koska mallissa energia säilyy (esim.
275 ilmanvastusta tai alustaan törmäyksen vaimentavaa vaikutusta ei ole huomioitu),
276 systeemin käytös muuttuu periodiseksi ensimmäisen törmäyksen jälkeen.
277 Olkoon ensimmäinen törmäyshetki
278\begin_inset Formula $t=t_{0}$
279\end_inset
280
281.
282 Ks.
283 kuva
284\begin_inset CommandInset ref
285LatexCommand ref
286reference "fig:problem-setup"
287
288\end_inset
289
290.
291\end_layout
292
293\begin_layout Standard
294Ratkaisu kannattaa tehdä kahdessa osassa: ratkaistaan ensin systeemin käytös
295 ensimmäiseen törmäykseen asti, ja
296\begin_inset Quotes eld
297\end_inset
298
299käynnistetään uudelleen
300\begin_inset Quotes erd
301\end_inset
302
303 tehtävä tästä saadun ratkaisun mukaisella korkeudella (
304\begin_inset Formula $=0$
305\end_inset
306
307) ja nopeudella.
308 Tällöin alkuehtojen vaikutus voidaan huomioida helposti yhden nopeuden
309 avulla, ja voidaan helposti tarkastella pelkkää periodista osaa.
310\end_layout
311
312\begin_layout Standard
313Integroidaan
314\begin_inset CommandInset ref
315LatexCommand eqref
316reference "eq:vt"
317
318\end_inset
319
320:
321\begin_inset Formula \begin{equation}
322v=v_{0}-gt\;,\label{eq:intv}\end{equation}
323
324\end_inset
325
326missä on käytetty alkuehtoa
327\begin_inset CommandInset ref
328LatexCommand eqref
329reference "eq:ICv"
330
331\end_inset
332
333 integrointivakion määräämiseen.
334 Sijoitetaan
335\begin_inset CommandInset ref
336LatexCommand eqref
337reference "eq:intv"
338
339\end_inset
340
341 yhtälöön
342\begin_inset CommandInset ref
343LatexCommand eqref
344reference "eq:u"
345
346\end_inset
347
348, integroidaan uudelleen, ja käytetään alkuehtoa
349\begin_inset CommandInset ref
350LatexCommand eqref
351reference "eq:ICu"
352
353\end_inset
354
355:
356\begin_inset Formula \begin{equation}
357u=u_{0}+v_{0}t-\frac{1}{2}gt^{2}\;.\label{eq:u}\end{equation}
358
359\end_inset
360
361Milloin on
362\begin_inset Formula $u=0$
363\end_inset
364
365? Yhtälö
366\begin_inset CommandInset ref
367LatexCommand eqref
368reference "eq:u"
369
370\end_inset
371
372 on toisen asteen polynomi ajan
373\begin_inset Formula $t$
374\end_inset
375
376 suhteen, mistä saamme ratkaisut
377\begin_inset Formula \[
378t=\frac{1}{g}\left(v_{0}\mp\sqrt{v_{0}^{2}+2u_{0}g}\right)\;.\]
379
380\end_inset
381
382Koska neliöjuurilauseke sisältää vain epänegatiivisia termejä, se on yhtä
383 suuri tai suurempi kuin
384\begin_inset Formula $v_{0}$
385\end_inset
386
387.
388 Yhtäsuuruus toteutuu vain, jos alkukorkeus on
389\begin_inset Formula $0$
390\end_inset
391
392, jolloin miinusmerkkiä vastaavaksi ratkaisuksi tulee
393\begin_inset Formula $t=0$
394\end_inset
395
396.
397 Muissa tapauksissa tämä ratkaisu olisi negatiivinen, joten se on hylättävä.
398\end_layout
399
400\begin_layout Standard
401Siispä fysikaalisesti järkevän ratkaisun saamme plusmerkillä, eli
402\begin_inset Formula \begin{equation}
403t_{0}=\frac{1}{g}\left(v_{0}+\sqrt{v_{0}^{2}+2u_{0}g}\right)\;.\label{eq:t0}\end{equation}
404
405\end_inset
406
407Mikä on nopeus törmäyshetkellä? Tämä selviää sijoittamalla
408\begin_inset CommandInset ref
409LatexCommand eqref
410reference "eq:t0"
411
412\end_inset
413
414 yhtälöön
415\begin_inset CommandInset ref
416LatexCommand eqref
417reference "eq:intv"
418
419\end_inset
420
421:
422\begin_inset Formula \[
423v(t_{0})=v_{0}-gt_{0}=-\sqrt{v_{0}^{2}+2u_{0}g}\;.\]
424
425\end_inset
426
427Tulemme tarvitsemaan tätä arvoa myöhemmin, joten merkitään
428\begin_inset Formula \begin{equation}
429\hat{v}_{0}:=-\sqrt{v_{0}^{2}+2u_{0}g}\label{eq:v0hat}\end{equation}
430
431\end_inset
432
433(huomaa miinusmerkki).
434\end_layout
435
436\begin_layout Standard
437Nyt voimme laskea vapaan lentoajan.
438 Vapaa lentoaika
439\begin_inset Formula $\Delta t$
440\end_inset
441
442 on ensimmäisen ja toisen törmäyksen välinen aika (ks.
443 kuva
444\begin_inset CommandInset ref
445LatexCommand ref
446reference "fig:problem-setup"
447
448\end_inset
449
450).
451 Käytetään kaavaa
452\begin_inset CommandInset ref
453LatexCommand eqref
454reference "eq:t0"
455
456\end_inset
457
458 alkuarvoilla
459\begin_inset Formula $\tilde{u}_{0}=0$
460\end_inset
461
462 ja
463\begin_inset Formula $\tilde{v}_{0}=-\hat{v}_{0}$
464\end_inset
465
466 (pistemäinen, liike-energian säilymisehdon
467\begin_inset CommandInset ref
468LatexCommand eqref
469reference "eq:ke"
470
471\end_inset
472
473 toteuttava kimpoaminen).
474\end_layout
475
476\begin_layout Standard
477Huomaa, että tämän uuden tehtävän hetkeksi
478\begin_inset Formula $t=0$
479\end_inset
480
481 on valittu ensimmäinen törmäyshetki, joka oli alkuperäisessä tehtävässä
482
483\begin_inset Formula $t=t_{0}$
484\end_inset
485
486.
487 Tästä syystä saamme ratkaisuksi erotuksen
488\begin_inset Formula $\Delta t:=t_{1}-t_{0}$
489\end_inset
490
491.
492 Lasketaan:
493\begin_inset Formula \begin{align}
494\Delta t & =\frac{1}{g}\left(-\hat{v}_{0}+\sqrt{\hat{v}_{0}^{2}+2\hat{u}_{0}g}\right)\nonumber \\
495 & =\frac{1}{g}\left(-\hat{v}_{0}+\sqrt{\hat{v}_{0}^{2}}\right)\nonumber \\
496 & =\frac{1}{g}\left(-\hat{v}_{0}-\hat{v}_{0}\right)\nonumber \\
497 & =-\frac{2\hat{v}_{0}}{g}\;.\label{eq:freetime}\end{align}
498
499\end_inset
500
501Toisessa muodossa on huomioitu
502\begin_inset Formula $\hat{u}_{0}=0$
503\end_inset
504
505 ja kolmannessa
506\begin_inset Formula $\hat{v}_{0}<0$
507\end_inset
508
509.
510 Toinen ratkaisu
511\begin_inset Formula $\Delta t=0$
512\end_inset
513
514 voidaan jälleen jättää huomiotta, koska se kuvaa lähtötilannetta (nythän
515 alkukorkeus oli
516\begin_inset Formula $0$
517\end_inset
518
519).
520\end_layout
521
522\begin_layout Standard
523\begin_inset Newpage pagebreak
524\end_inset
525
526
527\end_layout
528
529\begin_layout Standard
530Tarkastellaan nyt tehtävän regularisoitua muotoa yleisesti.
531 Ks.
532 kuva
533\begin_inset CommandInset ref
534LatexCommand ref
535reference "fig:problem-setup-reg"
536
537\end_inset
538
539.
540\end_layout
541
542\begin_layout Standard
543Ennen ensimmäistä törmäystä regularisoitu ratkaisu käyttäytyy kuten alkuperäinen
544kin.
545 Lähdetään siis tilanteesta
546\begin_inset Formula $u_{0}=0$
547\end_inset
548
549,
550\begin_inset Formula $v_{0}=\hat{v}_{0}$
551\end_inset
552
553.
554 Huomaa, että lähtönopeus suuntautuu alaspäin, koska nyt emme idealisoi
555 törmäystä pistemäiseksi.
556\end_layout
557
558\begin_layout Standard
559Kimpoamisen aikana systeemin liikeyhtälöt ovat
560\begin_inset Formula \begin{align}
561u_{t} & =v\label{eq:ut-bounce}\\
562v_{t} & =-g+f(u)\label{eq:vt-bounce}\\
563u & \le0\label{eq:ule0}\\
564v(t_{0})^{2} & =v(\tau)^{2}\;,\label{eq:ke2}\end{align}
565
566\end_inset
567
568missä
569\begin_inset Formula $\tau$
570\end_inset
571
572 kuvaa irtoamishetkeä.
573 Alkuehdot ovat
574\begin_inset Formula \begin{align*}
575u(0) & =0\\
576v(0) & =\hat{v}_{0}=-\sqrt{v_{0}^{2}+2u_{0}g}\;.\end{align*}
577
578\end_inset
579
580Kuten edelläkin, aikamuuttuja
581\begin_inset Formula $t$
582\end_inset
583
584 kuvaa nyt ensimmäisestä törmäyksestä kulunutta aikaa.
585 (Huomaa, että kuvissa aikamuuttuja kuvaa alkuperäisestä lähtötilanteesta
586 kulunutta aikaa.)
587\end_layout
588
589\begin_layout Standard
590Kumpikin annettu regularisointi on toteutettu konservatiivisen voimakentän
591 avulla.
592 Molemmilla on (triviaalisti) olemassa potentiaali, jonka gradientti voima
593 on.
594 Tätä potentiaalia vastaa potentiaalienergia.
595 Sama pätee vakiopainovoimakentälle.
596 Näin ollen sama kappaleen korkeus (kimpoamisen aikana) vastaa samassa tehtäväss
597ä aina samaa potentiaalienergiaa.
598 Koska systeemin kokonaisenergia säilyy, tällöin myös liike-energia ja siten
599 myös nopeus on samalla korkeudella aina (etumerkkiä vaille) sama.
600 Täten kappaleen nopeus irtoamishetkellä (
601\begin_inset Formula $u=0$
602\end_inset
603
604, koska ei adheesiota) on yhtä suuri kuin se oli törmäyshetkellä (jolloin
605 myös
606\begin_inset Formula $u=0$
607\end_inset
608
609).
610\end_layout
611
612\begin_layout Standard
613Annetut regularisoinnit siis toteuttavat ehdon
614\begin_inset CommandInset ref
615LatexCommand eqref
616reference "eq:ke2"
617
618\end_inset
619
620, ja ne eivät vaikuta systeemin käyttäytymiseen alueessa
621\begin_inset Formula $u>0$
622\end_inset
623
624.
625 Vapaa lentoaika säilyy siis samana, ja systeemin käyttäytyminen säilyy
626 periodisena.
627 Regularisointi luo kuitenkin periodiin viiveen, koska nyt törmäys ei tapahdu
628 pistemäisesti, vaan kappaleen palaaminen tasolle
629\begin_inset Formula $u=0$
630\end_inset
631
632 kestää jonkin aikaa.
633 Tämän viiveen pituus määräytyy regularisoinnin tyypistä sekä tehtäväparametreis
634ta.
635\end_layout
636
637\begin_layout Standard
638Oleellinen kysymys siis on, milloin kimpoamisen alettua jälleen
639\begin_inset Formula $u=0$
640\end_inset
641
642.
643 Nimitetään tätä kimpoamisaikaa (nollatasolle palautumishetkeä) symbolilla
644
645\begin_inset Formula $\tau$
646\end_inset
647
648.
649\end_layout
650
651\begin_layout Standard
652\begin_inset Newpage pagebreak
653\end_inset
654
655
656\end_layout
657
658\begin_layout List
659\labelwidthstring 00.00.0000
660
661\series bold
662(a)
663\series default
664 Regularisointi
665\begin_inset Formula $u^{-}/\varepsilon$
666\end_inset
667
668.
669 Viive?
670\end_layout
671
672\begin_layout Standard
673Olkoon
674\begin_inset Formula $f(u)=-u^{-}/\varepsilon$
675\end_inset
676
677.
678 Yhtälö
679\begin_inset CommandInset ref
680LatexCommand eqref
681reference "eq:vt-bounce"
682
683\end_inset
684
685 muuntuu muotoon
686\begin_inset Formula \[
687v_{t}=-g-\frac{1}{\varepsilon}u\;.\]
688
689\end_inset
690
691Derivoimalla
692\begin_inset CommandInset ref
693LatexCommand eqref
694reference "eq:ut-bounce"
695
696\end_inset
697
698 kerran voidaan yhtälöt
699\begin_inset CommandInset ref
700LatexCommand eqref
701reference "eq:ut-bounce"
702
703\end_inset
704
705 ja
706\begin_inset CommandInset ref
707LatexCommand eqref
708reference "eq:vt-bounce"
709
710\end_inset
711
712 yhdistää.
713 Järjestellään samalla termit:
714\begin_inset Formula \begin{equation}
715u_{tt}+\frac{1}{\varepsilon}u=-g\;.\label{eq:2ndorder}\end{equation}
716
717\end_inset
718
719Havaitaan, että yhtälö
720\begin_inset CommandInset ref
721LatexCommand eqref
722reference "eq:2ndorder"
723
724\end_inset
725
726 kuvaa vakiokuormitettua, lineaarista yhden vapausasteen jousi-massasysteemiä.
727 Se on lineaarinen toisen kertaluvun vakiokertoiminen tavallinen differentiaaliy
728htälö.
729 Sen ratkaisu on muotoa
730\begin_inset Formula \begin{equation}
731u=u_{h}+u_{p}\;,\label{eq:uuhup}\end{equation}
732
733\end_inset
734
735missä
736\begin_inset Formula $u_{h}$
737\end_inset
738
739 on vastaavan homogeenisen (kuormittamattoman) tehtävän ratkaisu (general
740 solution) ja
741\begin_inset Formula $u_{p}$
742\end_inset
743
744 on kuormitetun tehtävän yksittäisratkaisu (particular solution).
745\end_layout
746
747\begin_layout Standard
748Huomataan, että
749\begin_inset Formula \begin{equation}
750u_{p}=-\varepsilon g\label{eq:up}\end{equation}
751
752\end_inset
753
754sopii yksittäisratkaisuksi.
755 Homogeenisen tehtävän ratkaisu puolestaan on
756\begin_inset Formula \begin{equation}
757u_{h}=A\sin(t/\sqrt{\varepsilon})+B\cos(t/\sqrt{\varepsilon})\;.\label{eq:uh}\end{equation}
758
759\end_inset
760
761Vakiot
762\begin_inset Formula $A$
763\end_inset
764
765 ja
766\begin_inset Formula $B$
767\end_inset
768
769 saadaan alkuehdoista:
770\begin_inset Formula \begin{align*}
771u(0) & =\left(A\sin(t/\sqrt{\varepsilon})+B\cos(t/\sqrt{\varepsilon})-\varepsilon g\right)|_{t=0}=0\\
772v(0)=u_{t}(0) & =\left(\frac{A}{\sqrt{\varepsilon}}\cos(t/\sqrt{\varepsilon})-\frac{B}{\sqrt{\varepsilon}}\sin(t/\sqrt{\varepsilon})\right)|_{t=0}=\hat{v}_{0}\;.\end{align*}
773
774\end_inset
775
776Siis
777\begin_inset Formula \begin{align*}
778B-\varepsilon g & =0\quad\Leftrightarrow\quad B=\varepsilon g\\
779\frac{A}{\sqrt{\varepsilon}} & =\hat{v}_{0}\quad\Leftrightarrow\quad A=\hat{v}_{0}\sqrt{\varepsilon}\;.\end{align*}
780
781\end_inset
782
783Huomaa, että
784\begin_inset Formula $A<0$
785\end_inset
786
787.
788 Ratkaisuksi saatiin siis
789\begin_inset Formula \begin{equation}
790u=\hat{v}_{0}\sqrt{\varepsilon}\sin(t/\sqrt{\varepsilon})+\varepsilon g\left[\cos(t/\sqrt{\varepsilon})-1\right]\;,\label{eq:u-reg1}\end{equation}
791
792\end_inset
793
794joka pätee ensimmäisestä törmäyshetkestä (
795\begin_inset Formula $t=0$
796\end_inset
797
798) hetkeen
799\begin_inset Formula $t=\tau$
800\end_inset
801
802.
803 Merkitään
804\begin_inset Formula $x:=t/\sqrt{\varepsilon}\quad\Leftrightarrow\quad t=\sqrt{\varepsilon}x$
805\end_inset
806
807.
808\end_layout
809
810\begin_layout Standard
811Ratkaistaan, milloin jälleen
812\begin_inset Formula $u=0$
813\end_inset
814
815:
816\begin_inset Formula \begin{equation}
817\hat{v}_{0}\sqrt{\varepsilon}\sin(x)+\varepsilon g\left[\cos(x)-1\right]=0\;.\label{eq:u-reg1-is0}\end{equation}
818
819\end_inset
820
821Yhtälö
822\begin_inset CommandInset ref
823LatexCommand eqref
824reference "eq:u-reg1-is0"
825
826\end_inset
827
828 on transkendentaaliyhtälö, jolla ei ole suljetun muodon ratkaisua.
829 Funktio
830\begin_inset CommandInset ref
831LatexCommand eqref
832reference "eq:u-reg1"
833
834\end_inset
835
836 on
837\begin_inset Formula $2\pi$
838\end_inset
839
840-periodinen.
841 Koska
842\begin_inset Formula $u(0)=0$
843\end_inset
844
845, yhtälön nollakohtia ovat siis ainakin
846\begin_inset Formula $x=2k\pi$
847\end_inset
848
849,
850\begin_inset Formula $k\in\mathbb{Z}$
851\end_inset
852
853.
854\end_layout
855
856\begin_layout Standard
857Koska
858\begin_inset Formula $\varepsilon$
859\end_inset
860
861 on pieni parametri, jälkimmäinen termi muodostaa pienen korjauksen ensimmäiseen
862, jonka nollakohdat ovat
863\begin_inset Formula $x=k\pi$
864\end_inset
865
866,
867\begin_inset Formula $k\in\mathbb{Z}$
868\end_inset
869
870.
871 Voidaan siis odottaa yhden nollakohdan löytymistä myös arvon
872\begin_inset Formula $x\approx\pi$
873\end_inset
874
875 läheltä.
876\end_layout
877
878\begin_layout Standard
879Lausekkeen
880\begin_inset CommandInset ref
881LatexCommand eqref
882reference "eq:u-reg1-is0"
883
884\end_inset
885
886 termien etumerkeistä voidaan päätellä, että kun
887\begin_inset Formula $x\le\pi$
888\end_inset
889
890, niin
891\begin_inset Formula $u<0$
892\end_inset
893
894.
895 Ainoa mahdollisuus ylimääräiselle nollakohdalle on siis periodin jälkimmäinen
896 puolisko,
897\begin_inset Formula $\pi<x<2\pi$
898\end_inset
899
900.
901 Laskemalla derivaatta yhtälöstä
902\begin_inset CommandInset ref
903LatexCommand eqref
904reference "eq:u-reg1-is0"
905
906\end_inset
907
908 saadaan
909\begin_inset Formula $u$
910\end_inset
911
912:n ääriarvoille yhtälö
913\begin_inset Formula \begin{align*}
914\hat{v}_{0}\cos(x)-\sqrt{\varepsilon}g\sin(x) & =0\\
915\Rightarrow\quad\tan(x) & =\frac{\hat{v}_{0}}{\sqrt{\varepsilon}g}\;.\end{align*}
916
917\end_inset
918
919Huomataan, että oikea puoli on negatiivinen ja suuri (koska
920\begin_inset Formula $\varepsilon$
921\end_inset
922
923 on pieni).
924 Siispä funktion
925\begin_inset Formula $u$
926\end_inset
927
928 minimi- ja maksimiarvot löytyvät
929\begin_inset Quotes eld
930\end_inset
931
932hiukan
933\begin_inset Quotes erd
934\end_inset
935
936 tangenttifunktion singulariteettien oikealta puolelta:
937\begin_inset Formula \[
938x=\frac{\pi}{2}+k\pi+h(\varepsilon)\;,\quad k\in\mathbb{N}\;,\]
939
940\end_inset
941
942missä funktion
943\begin_inset Formula $h(\varepsilon)$
944\end_inset
945
946 arvot ovat
947\begin_inset Quotes eld
948\end_inset
949
950pieniä
951\begin_inset Quotes erd
952\end_inset
953
954 ja
955\begin_inset Formula $h>0$
956\end_inset
957
958, mutta funktiota
959\begin_inset Formula $h$
960\end_inset
961
962 ei voida määrittää suljetussa muodossa.
963 Siispä funktion
964\begin_inset CommandInset ref
965LatexCommand eqref
966reference "eq:u-reg1"
967
968\end_inset
969
970 minimi on noin kohdassa
971\begin_inset Formula $x\approx\pi/2$
972\end_inset
973
974 ja maksimi noin kohdassa
975\begin_inset Formula $x\approx3\pi/2$
976\end_inset
977
978; todellisuudessa
979\begin_inset Quotes eld
980\end_inset
981
982hiukan yli
983\begin_inset Quotes erd
984\end_inset
985
986 näistä.
987 (Alkuehdoista ja systeemin periodisesta käytöksestä on selvää, että minimin
988 ja maksimin on oltava näin päin.)
989\end_layout
990
991\begin_layout Standard
992Voitaisiin huomioida termien etumerkit lausekkeessa
993\begin_inset CommandInset ref
994LatexCommand eqref
995reference "eq:u-reg1"
996
997\end_inset
998
999 ja havaita, että minimissä
1000\begin_inset Formula $u<0$
1001\end_inset
1002
1003, mutta tätä ei tarvita, koska tiedetään jo, että
1004\begin_inset Formula $u<0$
1005\end_inset
1006
1007 kaikille
1008\begin_inset Formula $0<x\le\pi$
1009\end_inset
1010
1011.
1012\end_layout
1013
1014\begin_layout Standard
1015Maksimin lähellä, kun
1016\begin_inset Formula $x=3\pi/2$
1017\end_inset
1018
1019, saadaan
1020\begin_inset Formula \[
1021u=-\hat{v}_{0}\sqrt{\varepsilon}-\varepsilon g\;,\]
1022
1023\end_inset
1024
1025missä ensimmäinen termi on positiivinen ja toinen negatiivinen.
1026 Siispä
1027\begin_inset Formula $u\le0$
1028\end_inset
1029
1030, jos
1031\begin_inset Formula \begin{align*}
1032-\hat{v}_{0}\sqrt{\varepsilon}-\varepsilon g & \le0\\
1033-\hat{v}_{0} & \le\sqrt{\varepsilon}g\\
1034\sqrt{\varepsilon} & \ge-\frac{\hat{v}_{0}}{g}\;,\end{align*}
1035
1036\end_inset
1037
1038mikä ei toteudu millekään järkevälle parametrin
1039\begin_inset Formula $\varepsilon$
1040\end_inset
1041
1042 arvolle.
1043 Siispä missä hyvänsä järkevässä tilanteessa
1044\begin_inset Formula $u(3\pi/2)>0$
1045\end_inset
1046
1047, eli maksimi on positiivinen.
1048\end_layout
1049
1050\begin_layout Standard
1051Täten, koska funktio
1052\begin_inset Formula $u$
1053\end_inset
1054
1055 on jatkuva, sillä on ainakin yksi nollakohta pisteiden
1056\begin_inset Formula $x=\pi$
1057\end_inset
1058
1059 ja
1060\begin_inset Formula $x=3\pi/2$
1061\end_inset
1062
1063 välissä.
1064 Koska se lisäksi on monotoninen ääriarvojensa välillä (derivaatta on jatkuva,
1065 joten sen etumerkki voi vaihtua vain nollakohdissa), sillä on tarkalleen
1066 yksi nollakohta.
1067\end_layout
1068
1069\begin_layout Standard
1070Tästä päätelmästä nähdään myös, että olemme löytäneet kaikki nollakohdat:
1071 maksimiarvo on positiivinen, ja funktio on jälleen monotoninen seuraavaan
1072 minimiinsä
1073\begin_inset Formula $x\approx5\pi/2$
1074\end_inset
1075
1076 asti.
1077 Tällä välillä on tunnettu nollakohta arvossa
1078\begin_inset Formula $x=2\pi$
1079\end_inset
1080
1081, joten sen on oltava ainoa.
1082\end_layout
1083
1084\begin_layout Standard
1085Nyt kun tiedämme, että välillä
1086\begin_inset Formula $x\in[\pi,3\pi/2]$
1087\end_inset
1088
1089 on aina tarkalleen yksi funktion
1090\begin_inset Formula $u$
1091\end_inset
1092
1093 nollakohta, voimme määrittää sen.
1094 Perustuen
1095\begin_inset Formula $\varepsilon$
1096\end_inset
1097
1098:n kertalukuihin lausekkeessa
1099\begin_inset CommandInset ref
1100LatexCommand eqref
1101reference "eq:u-reg1"
1102
1103\end_inset
1104
1105, nollakohdan on oltava lähellä arvoa
1106\begin_inset Formula $x=\pi$
1107\end_inset
1108
1109.
1110 Merkitään
1111\begin_inset Formula $y:=x-\pi\quad\Leftrightarrow\quad x=\pi+y$
1112\end_inset
1113
1114.
1115\end_layout
1116
1117\begin_layout Standard
1118Jotta voidaan käyttää häiriökehitelmää, tarvitaan polynomeja.
1119 Kehitetään siis
1120\begin_inset Formula $\sin$
1121\end_inset
1122
1123 ja
1124\begin_inset Formula $\cos$
1125\end_inset
1126
1127 Taylorin sarjaksi pisteen
1128\begin_inset Formula $x=\pi$
1129\end_inset
1130
1131 ympäristössä.
1132 Saadaan
1133\begin_inset Formula \begin{align}
1134\sin(\pi+y) & \approx-y+\frac{y^{3}}{6}-\dots\label{eq:sinpix}\\
1135\cos(\pi+y) & \approx-1+\frac{y^{2}}{2}-\frac{y^{4}}{24}+\dots\label{eq:cospix}\end{align}
1136
1137\end_inset
1138
1139Kirjoitetaan
1140\begin_inset Formula \begin{equation}
1141y=y_{1}\sqrt{\varepsilon}+y_{2}\varepsilon+y_{3}\varepsilon^{3/2}+y_{4}\varepsilon^{2}+\dots\label{eq:y}\end{equation}
1142
1143\end_inset
1144
1145Tässä valinta
1146\begin_inset Formula $\sqrt{\varepsilon}$
1147\end_inset
1148
1149 on tehty yhtälön
1150\begin_inset CommandInset ref
1151LatexCommand eqref
1152reference "eq:u-reg1-is0"
1153
1154\end_inset
1155
1156 termien kertalukujen eron mukaisesti.
1157\end_layout
1158
1159\begin_layout Standard
1160Sijoitusten ja termien keräämisen jälkeen (mikä onnistuu helpoiten tietokoneella
1161 symbolisesti) saadaan häiriöyhtälöt
1162\begin_inset Formula \begin{align*}
11632g+y_{1}\hat{v}_{0} & =0\quad\text{(kertaluku }\sqrt{\varepsilon}\text{)}\\
1164y_{2}\hat{v}_{0} & =0\quad\text{(kertaluku }\varepsilon\text{)}\\
11653gy_{1}^{2}-6\hat{v}_{0}y_{3}+\hat{v}_{0}y_{1}^{3} & =0\quad\text{(kertaluku }\varepsilon^{3/2}\text{)}\\
11662gy_{1}y_{2}-2\hat{v}_{0}y_{4}+\hat{v}_{0}y_{1}^{2}y_{2} & =0\quad\text{(kertaluku }\varepsilon^{2}\text{)}\end{align*}
1167
1168\end_inset
1169
1170mistä
1171\begin_inset Formula \begin{align*}
1172y_{1} & =-\frac{2g}{\hat{v}_{0}}\\
1173y_{2} & =0\\
1174y_{3} & =\frac{2g^{3}}{3\hat{v}_{0}^{3}}\\
1175y_{4} & =0\;.\end{align*}
1176
1177\end_inset
1178
1179Sijoittamalla nämä häiriösarjaan
1180\begin_inset CommandInset ref
1181LatexCommand eqref
1182reference "eq:y"
1183
1184\end_inset
1185
1186 ja muuntamalla takaisin (
1187\begin_inset Formula $x=\pi+y$
1188\end_inset
1189
1190,
1191\begin_inset Formula $t=\sqrt{\varepsilon}x$
1192\end_inset
1193
1194, mistä
1195\begin_inset Formula $t=\sqrt{\varepsilon}\pi+\sqrt{\varepsilon}y$
1196\end_inset
1197
1198) saadaan kimpoamisajaksi
1199\begin_inset Formula \begin{equation}