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10.1: Úvod do systémov diferenciálnych rovníc - matematika


Mnoho fyzikálnych situácií je modelovaných systémami (n ) diferenciálnych rovníc v (n ) neznámych funkciách, kde (n ge2 ). Nasledujúce tri príklady ilustrujú fyzikálne problémy, ktoré vedú k systémom diferenciálnych rovníc. V týchto príkladoch a v tejto kapitole označíme nezávislú premennú pomocou (t ).

Príklad ( PageIndex {1} ): Tok

Nádrže (T_1 ) a (T_2 ) obsahujú 100 galónov, respektíve 300 galónov soľných roztokov. Soľné roztoky sa do obidvoch nádrží pridávajú z externých zdrojov, čerpajú sa z každej nádrže do druhej a z obidvoch nádrží sa odvádzajú (obrázok ( PageIndex {1} )). Roztok s (1 ) soľou na galón sa čerpá do (T_1 ) z externého zdroja rýchlosťou (5 gal) / min a roztok s (2 ) soľami na galón je čerpané do (T_2 ) z externého zdroja rýchlosťou (4 ) gal / min. Roztok z (T_1 ) sa čerpá do (T_2 ) rýchlosťou 2 gal / min a roztok z (T_2 ) sa čerpá do (T_1 ) rýchlosťou (3 ) gal / min. (T_1 ) sa vypúšťa pri (6 ) gal / min a (T_2 ) sa vypúšťa pri 3 gal / min. Nech (Q_1 (t) ) a (Q_2 (t) ) je počet libier soli v (T_1 ), respektíve (T_2 ) v danom čase (t> 0 ). Odvodiť systém diferenciálnych rovníc pre (Q_1 ) a (Q_2 ). Predpokladajme, že obe zmesi sú dobre premiešané.

Rovnako ako v časti 4.2 písm sadzba v a ohodnotiť označte rýchlosti (lb / min), pri ktorých soľ vstupuje do nádrže a opúšťa ju; teda

[ begin {aligned} Q_1 '& = ( mbox {rate in}) _ 1 - ( mbox {rate out}) _ 1, [4pt] Q_2' & = ( mbox {rate in}) _ 2- ( mbox {rate out}) _ 2. end {aligned} ]

Upozorňujeme, že objemy roztokov v (T_1 ) a (T_2 ) zostávajú konštantné na hodnote 100 galónov, respektíve 300 galónov.

(T_1 ) prijíma soľ z externého zdroja v množstve

[ mbox {(1 lb / gal)}} krát mbox {(5 ~ gal / min)} = mbox {5 lb / min}, nonumber ]

a od (T_2 ) vo výške

[ mbox {(lb / gal in} T_2) times mbox {(3 ~ gal / min)} = {1 over300} Q_2 times3 [4pt] = {1 over100} Q_2 mbox { lb / min}. nonumber ]

Preto

[ label {eq: 10.1.1} mbox {(sadzba v)} _ 1 = 5+ {1 over100} Q_2. ]

Listy roztoku (T_1 ) rýchlosťou 8 gal / min, pretože sú vypúšťané 6 gal / min a 2 gal / min sú čerpané do (T_2 ); teda

[ label {eq: 10.1.2} ( mbox {rate out}) _ 1 = ( mbox {lb / gal v T} _1) times mbox {(8 ~ gal / min)} = {1 nad100} Q_1 times8 = {2 nad25} Q_1. ]

Rovnice ref {eq: 10.1.1} a Rovnica ref {eq: 10.1.2} to naznačujú

[ label {eq: 10.1.3} Q_1 '= 5 + {1 over100} Q_2- {2 over25} Q_1. ]

(T_2 ) prijíma soľ z externého zdroja v množstve

[ mbox {(2 lb / gal)}} krát mbox {(4 ~ gal / min)} = mbox {8 lb / min}, nonumber ]

a od (T_1 ) vo výške

[ mbox {(lb / gal in} T_1) times mbox {(2 ~ gal / min)} = {1 over100} Q_1 times2 [4pt] = {1 over50} Q_1 mbox { lb / min}. nonumber ]

Preto

[ label {eq: 10.1.4} mbox {(sadzba v)} _ 2 = 8+ {1 over50} Q_1. ]

Ponecháva roztok (T_2 ) rýchlosťou (6 ) gal / min, pretože (3 ) gal / min sú vyčerpané a (3 gal) za minútu sú čerpané do (T_1 ); teda

[ label {eq: 10.1.5} ( mbox {rate out}) _ 2 = ( mbox {lb / gal v T} _2) times mbox {(6 ~ gal / min)} = {1 nad 300} Q_2 times6 = {1 over50} Q_2. ]

Rovnice ref {eq: 10.1.4} a ref {eq: 10.1.5} to naznačujú

[ label {eq: 10.1.6} Q_2 '= 8 + {1 over50} Q_1 - {1 over50} Q_2. ]

Hovoríme, že rovnice ref {eq: 10.1.3} a ref {eq: 10.1.6} tvoria a sústava dvoch rovníc prvého poriadku v dvoch neznámycha napíšte ich spolu ako

[ begin {align *} Q_1 '& = 5- dfrac {2} {25} Q_1 + dfrac {1} {100} Q_2 Q_2' & = 8+ dfrac {1} {50} Q_1 - dfrac {1} {50} Q_2. end {zarovnať *} ]

Príklad ( PageIndex {2} ): spojené pružiny

Hmota (m_1 ) je zavesená z tuhej podpery na pružine (S_1 ) a druhá hmotnosť (m_2 ) je zavesená z prvej na pružine (S_2 ) (Obrázok ( PageIndex { 2} )). Pružiny sa riadia Hookovým zákonom s pružinovými konštantami (k_1 ) a (k_2 ). Vnútorné trenie spôsobuje, že pružiny vyvíjajú tlmiace sily úmerné rýchlosti zmien ich dĺžok s tlmiacimi konštantami (c_1 ) a (c_2 ). Nech (y_1 = y_1 (t) ) a (y_2 = y_2 (t) ) sú posuny dvoch hmôt z ich rovnovážnych polôh v čase (t ), merané pozitívne smerom nahor. Odvodiť systém diferenciálnych rovníc pre (y_1 ) a (y_2 ) za predpokladu, že hmotnosti pružín sú zanedbateľné a že na objekty pôsobia aj vertikálne vonkajšie sily (F_1 ) a (F_2 ).

Riešenie

V rovnováhe podporuje (S_1 ) obidva (m_1 ) a (m_2 ) a (S_2 ) podporuje iba (m_2 ). Preto ak ( Delta ell_1 ) a ( Delta ell_2 ) sú predĺženia pružín v rovnováhe, potom

[ label {eq: 10.1.7} (m_1 + m_2) g = k_1 Delta ell_1 quad mbox {a} quad m_2g = k_2 Delta ell_2. ]

Nech (H_1 ) je Hookova zákonná sila pôsobiaca na (m_1 ) a nech (D_1 ) je tlmiaca sila pôsobiaca na (m_1 ). Podobne nech sú (H_2 ) a (D_2 ) Hookov zákon a tlmiace sily pôsobiace na (m_2 ). Podľa druhého Newtonovho zákona pohybu

[ label {eq: 10.1.8} begin {array} {ccl} m_1y_1 '' = - m_1g + H_1 + D_1 + F_1, [4pt] m_2y_2 '' = - m_2g + H_2 + D_2 + F_2. end {pole} ]

Keď sú posuny (y_1 ) a (y_2 ), zmena dĺžky (S_1 ) je (- y_1 + Delta ell_1 ) a zmena dĺžky (S_2 ) je (-y_2 + y_1 + Delta ell_2 ). Obidve pružiny pôsobia Hookeove zákonné sily na (m_1 ), zatiaľ čo iba (S_2 ) pôsobia Hookeove zákonné sily na (m_2 ). Tieto sily sú v smeroch, ktoré majú tendenciu obnovovať pružiny v ich prirodzenej dĺžke. Preto

[ label {eq: 10.1.9} H_1 = k_1 (-y_1 + Delta ell_1) -k_2 (-y_2 + y_1 + Delta ell_2) quad mbox {a} quad H_2 = k_2 (-y_2 + y_1 + Delta ell_2). ]

Keď sú rýchlosti (y_1 ') a (y_2' ), (S_1 ) a (S_2 ) menia dĺžku pri rýchlostiach (- y_1 ') a (- y_2' + y_1 '), v uvedenom poradí. Obe pružiny vyvíjajú tlmiace sily na (m_1 ), zatiaľ čo iba (S_2 ) vyvíja tlmiacu silu na (m_2 ). Pretože sila spôsobená tlmením vyvinutá pružinou je úmerná rýchlosti zmeny dĺžky pružiny a v smere, ktorý je proti zmene, vyplýva z toho, že

[ label {eq: 10.1.10} D_1 = -c_1y_1 '+ c_2 (y_2'-y_1') quad mbox {a} quad D_2 = -c_2 (y_2'-y_1 '). ]

Z rovníc ref {eq: 10.1.8}, ref {eq: 10.1.9} a ref {eq: 10.1.10},

[ label {eq: 10.1.11} begin {array} {m_1y_1 ''} & {= -m_1g + k_1 (-y_1 + Delta ell_1) -k_2 (-y_2 + y_1 + Delta ell_2) -c_1y_1 '+ c_2 (y_2'-y_1') + F_1} {} & {= - (m_1g-k_1 Delta ell_1 + k_2 Delta ell_2) -k_1y_1 + k_2 (y_2-y_1) -c_1y_1 '+ c_2 (y_2'-y_1 ') + F_1} end {array} ]

a

[ label {eq: 10.1.12} begin {array} {ccl} m_2y_2 '' & = -m_2g + k_2 (-y_2 + y_1 + Delta ell_2) -c_2 (y_2'-y_1 ') + F_2 [4pt] & = - (m_2g-k_2 Delta ell_2) -k_2 (y_2-y_1) -c_2 (y_2'-y_1 ') + F_2. end {pole} ]

Z rovnice ref {eq: 10.1.7},

[m_1g-k_1 Delta ell_1 + k_2 Delta ell_2 = -m_2g + k_2 Delta ell_2 = 0. nonumber ]

Preto môžeme prepísať Rovnica ref {eq: 10.1.11} a Rovnica ref {eq: 10.1.12} ako

[ begin {align *} m_1y_1 '' & = - (c_1 + c_2) y_1 '+ c_2y_2' - (k_1 + k_2) y_1 + k_2y_2 + F_1 m_2y_2 '' & = c_2y_1'-c_2y_2 '+ k_2y_1- k_2y_2 + F_2. quad end {align *} ]

Príklad ( PageIndex {3} ): Gravitácia

Nech ({ bf X} = { bf X} (t) = x (t) , { bf i} + y (t) , { bf j} + z (t) , {) bf k} ) je vektor polohy v čase (t ) objektu s hmotou (m ), vo vzťahu k obdĺžnikovému súradnicovému systému s počiatkom v strede Zeme (obrázok ( PageIndex {3} )) .

Podľa Newtonovho gravitačného zákona je gravitačná sila Zeme ({ bf F} = { bf F} (x, y, z) ) na objekt nepriamo úmerná druhej mocnine vzdialenosti objektu od stredu Zeme a smerujúce do stredu; teda

[ label {eq: 10.1.13} bf {F} = dfrac {K} {|| bf {X} || ^ 2} zľava (- dfrac { bf {X}} {|| bf {X} ||} doprava) = -K {x , { bf i} + y , { bf j} + z , { bf k} nad doľava (x ^ 2 + y ^ 2 + z ^ 2 doprava) ^ {3/2}}, ]

kde (K ) je konštanta. Aby sme určili (K ), pozorujeme, že veľkosť ({ bf F} ) je

[ | { bf F} | = K { | { bf X} | nad | { bf X} | ^ 3} = {K nad | { bf X} | ^ 2} = {K over (x ^ 2 + y ^ 2 + z ^ 2)}. nonumber ]

Nech (R ) je polomer Zeme. Pretože ( | { bf F} | = mg ), keď je objekt na povrchu Zeme,

[mg = {K nad R ^ 2}, quad mbox {so} quad K = mgR ^ 2. nonumber ]

Preto môžeme rovnicu ref {eq: 10.1.13} prepísať na

[{ bf F} = - mgR ^ 2 {x , { bf i} + y , { bf j} + z , { bf k} nad mbox {} vľavo (x ^ 2 + y ^ 2 + z ^ 2 vpravo) ^ {3/2}}. nonumber ]

Teraz predpokladajme, že ({ bf F} ) je jediná sila pôsobiaca na objekt. Podľa druhého Newtonovho zákona pohybu ({ bf F} = m { bf X} '' ); to je,

[m (x '' , { bf i} + y '' , { bf j} + z '' , { bf k}) = -mgR ^ 2 {x , { bf i } + y , { bf j} + z , { bf k} nad doľava (x ^ 2 + y ^ 2 + z ^ 2 doprava) ^ {3/2}}. nonumber ]

Zrušením spoločného faktora (m ) a rovníckych zložiek na dvoch stranách tejto rovnice vznikne systém [ label {eq: 10.1.14} begin {array} {ll} {x ''} & {= - dfrac {gR ^ 2x} {(x ^ 2 + y ^ 2 + z ^ 2) ^ {3/2}}} {y ''} & {= - dfrac {gR ^ 2y} {(x ^ 2 + y ^ 2 + z ^ 2) ^ {3/2}}} {z ''} & {= - dfrac {gR ^ 2z} {(x ^ 2 + y ^ 2 + z ^ 2 ) ^ {3/2}}.} End {pole} ]

Prepisovanie systémov vyšších rád na systémy prvého rádu

Systém formulára

[ label {eq: 10.1.15} begin {array} {ccl} y_1 '& = g_1 (t, y_1, y_2, dots, y_n) y_2' & = g_2 (t, y_1, y_2, dots, y_n) & vdots & y_n '& = g_n (t, y_1, y_2, dots, y_n) end {pole} ]

sa nazýva a systém prvého rádu, pretože jediné deriváty, ktoré sa v ňom vyskytujú, sú prvé deriváty. Derivát každej z neznámych môže závisieť od nezávislej premennej a všetkých neznámych, nie však od derivátov iných neznámych. Ak chceme zdôrazniť počet neznámych funkcií v rovnici ref {eq: 10.1.15}, povieme, že rovnica ref {eq: 10.1.15} je (n krát n ) systém.

Systémy zahrnujúce deriváty vyššieho rádu možno často preformulovať ako systémy prvého rádu zavedením ďalších neznámych. Nasledujúce dva príklady to ilustrujú.

Príklad ( PageIndex {4} )

Prepíšte systém

[ label {eq: 10.1.16} begin {array} {ll} {m_1y_1 ''} & {= - (c_1 + c_2) y_1 '+ c_2y_2' - (k_1 + k_2) y_1 + k_2y_2 + F_1} {m_2y_2 ''} & {= c_2y_1'-c_2y_2 '+ k_2y_1-k_2y_2 + F_2.} end {pole} ]

odvodený v príklade ( PageIndex {2} ) ako systém rovníc prvého rádu.

Riešenie

Ak definujeme (v_1 = y_1 ') a (v_2 = y_2' ), potom (v_1 '= y_1' ') a (v_2' = y_2 '' ), tak rovnica ref {eq : 10.1.16} sa stáva

[ begin {aligned} m_1v_1 '& = - (c_1 + c_2) v_1 + c_2v_2- (k_1 + k_2) y_1 + k_2y_2 + F_1 [4pt] m_2v_2' & = c_2v_1-c_2v_2 + k_2y_1-k_2y_2 + F_2. end {zarovnané} nonumber ]

Preto ( {y_1, y_2, v_1, v_2 } ) vyhovuje systému (4 times4 ) prvého rádu

[ label {eq: 10.1.17} begin {array} {ll} {y_1 '} & {= v_1} {y_2'} & {= v_2} {v_1 '} & {= dfrac {1} {m_1} left [- (c_1 + c_2) v_1 + c_2v_2- (k_1 + k_2) y_1 + k_2y_2 + F_1 right]} {v_2 '} & {= dfrac {1} {m_2} left [c_2v_1-c_2v_2 + k_2y_1-k_2y_2 + F_2 right].} end {array} ]

Poznámka

Rozdiel vo forme medzi rovnicou ref {eq: 10.1.15} a rovnicou ref {eq: 10.1.17} nie je dôležitý kvôli spôsobu, akým sú neznáme označované v týchto dvoch systémoch; Rovnica ref {eq: 10.1.17} je systém prvého poriadku v tom zmysle, že každá rovnica v rovnici ref {eq: 10.1.17} vyjadruje prvú deriváciu jednej z neznámych funkcií spôsobom, ktorý nezahŕňa derivácie niektorú z ďalších neznámych.

Príklad ( PageIndex {5} )

Prepíšte systém

[ begin {array} {ccc} x '' & = f (t, x, x ', y, y', y '') [4pt] y '' '& = g (t, x, x ', y, y'y' ') end {pole} nonumber ]

ako systém prvého rádu.

Riešenie

Považujeme (x ), (x '), (y ), (y' ) a (y ' ) za neznáme funkcie a premenujeme ich

[x = x_1, ; x '= x_2, quad y = y_1, quad y' = y_2, quad y '= y_3. nonumber ]

Tieto neznáme systémy uspokojujú

[ begin {aligned} x_1 '& = x_2 [4pt] x_2' & = f (t, x_1, x_2, y_1, y_2, y_3) [4pt] y_1 '& = y_2 [4pt] y_2 '& = y_3 [4pt] y_3' & = g (t, x_1, x_2, y_1, y_2, y_3). end {zarovnané} nonumber ]

Prepisovanie skalárnych diferenciálnych rovníc ako systémov

V tejto kapitole sa budeme venovať diferenciálnym rovniciam zahŕňajúcim iba jednu neznámu funkciu skalárny diferenciálne rovnice. Skalárne diferenciálne rovnice je možné prepísať ako systémy rovníc prvého rádu metódou ilustrovanou v nasledujúcich dvoch príkladoch.

Príklad ( PageIndex {6} )

Prepíš rovnicu

[ label {eq: 10.1.18} y ^ {(4)} + 4y '' '+ 6y' '+ 4y' + y = 0 ]

ako (4 times4 ) systém prvého rádu.

Riešenie

Považujeme (y ), (y '), (y' ) a (y '' ') za neznáme a premenujeme ich

[y = y_1, quad y '= y_2, quad y' '= y_3, quad text {a} quad y' '' = y_4. nonumber ]

Potom (y ^ {(4)} = y_4 '), takže rovnica ref {eq: 10.1.18} môže byť napísaná ako

[y_4 '+ 4y_4 + 6y_3 + 4y_2 + y_1 = 0. nonumber ]

Preto ( {y_1, y_2, y_3, y_4 } ) vyhovuje systému

[ begin {align *} y_1 '& = y_2 y_2' & = y_3 y_3 '& = y_4 y_4' & = -4y_4-6y_3-4y_2-y_1. end {zarovnať *} ]

Príklad ( PageIndex {7} )

Prepísať

[x '' '= f (t, x, x', x '') nečíslo ]

ako sústava rovníc prvého rádu.

Riešenie

Považujeme (x ), (x ') a (x' ') za neznáme a premenujeme ich

[x = y_1, quad x '= y_2, quad text {a} quad x' '= y_3. nonumber ]

Potom

[y_1 '= x' = y_2, quad y_2 '= x' '= y_3, quad text {a} quad y_3' = x '' '. nonumber ]

Preto ( {y_1, y_2, y_3 } ) vyhovuje systému prvého rádu

[ begin {align *} y_1 '& = y_2 [4pt] y_2' & = y_3 [4pt] y_3 '& = f (t, y_1, y_2, y_3). end {zarovnať *} ]

Pretože systémy diferenciálnych rovníc zahrnujúcich vyššie deriváty možno prepísať ako systémy prvého rádu metódou použitou v príkladoch ( PageIndex {5} ) - ( PageIndex {7} ), budeme brať do úvahy iba systémy prvého rádu.

Numerické riešenie systémov

Numerické metódy, ktoré sme študovali v kapitole 3, je možné rozšíriť na systémy a väčšina softvérových balíkov diferenciálnych rovníc obsahuje programy na riešenie systémov rovníc. Nebudeme sa podrobne zaoberať numerickými metódami pre systémy; pre ilustráciu však popíšeme Runge-Kuttovu metódu pre numerické riešenie problému počiatočných hodnôt

[ begin {aligned} y_1 '& = g_1 (t, y_1, y_2), quad y_1 (t_0) = y_ {10}, [4pt] y_2' & = g_2 (t, y_1, y_2), quad y_2 (t_0) = y_ {20} end {zarovnané} ]

v rovnako rozmiestnených bodoch (t_0 ), (t_1 ), ..., (t_n = b ) v intervale ([t_0, b] ). Teda

[t_i = t_0 + ih, quad i = 0,1, dots, n, nonumber ]

kde

[h = {b-t_0 n). nonumber ]

Približné hodnoty (y_1 ) a (y_2 ) v týchto bodoch budeme označovať (y_ {10}, y_ {11}, dots, y_ {1n} ) a (y_ {20 }, y_ {21}, dots, y_ {2n} ). Metóda Runge-Kutta počíta tieto približné hodnoty takto: dané (y_ {1i} ) a (y_ {2i} ), vypočítať

[ begin {aligned} I_ {1i} & = g_1 (t_i, y_ {1i}, y_ {2i}), [4pt] J_ {1i} & = g_2 (t_i, y_ {1i}, y_ { 2i}), [4pt] I_ {2i} & = g_1 doľava (t_i + {h over2}, y_ {1i} + {h over2} I_ {1i}, y_ {2i} + {h over2 } J_ {1i} vpravo), [4pt] J_ {2i} & = g_2 doľava (t_i + {h over2}, y_ {1i} + {h over2} I_ {1i}, y_ {2i} + {h over2} J_ {1i} vpravo), [4pt] I_ {3i} & = g_1 doľava (t_i + {h over2}, y_ {1i} + {h over2} I_ {2i} , y_ {2i} + {h over2} J_ {2i} vpravo), [4pt] J_ {3i} & = g_2 doľava (t_i + {h over2}, y_ {1i} + {h over2 } I_ {2i}, y_ {2i} + {h over2} J_ {2i} vpravo), [4pt] I_ {4i} & = g_1 (t_i + h, y_ {1i} + hI_ {3i} , y_ {2i} + hJ_ {3i}), [4pt] J_ {4i} & = g_2 (t_i + h, y_ {1i} + hI_ {3i}, y_ {2i} + hJ_ {3i}), end {zarovnané} ]

a

[ begin {aligned} y_ {1, i + 1} & = y_ {1i} + {h over6} (I_ {1i} + 2I_ {2i} + 2I_ {3i} + I_ {4i}), [4pt] y_ {2, i + 1} & = y_ {2i} + {h over6} (J_ {1i} + 2J_ {2i} + 2J_ {3i} + J_ {4i}) end {zarovnaný} ]

pre (i = 0 ), ..., (n-1 ). Za vhodných podmienok na (g_1 ) a (g_2 ) je možné preukázať, že globálna chyba skrátenia pre metódu Runge-Kutta je (O (h ^ 4) ), ako v skalárnom prípade uvažovanom v Oddiel 3.3.


Matematické modelovanie a výpočet

Diferenciálne rovnice sú základom pre modely akýchkoľvek fyzikálnych systémov, ktoré vykazujú plynulú zmenu. Táto kniha kombinuje väčšinu materiálu nájdeného v tradičnom kurze bežných diferenciálnych rovníc s úvodom do modernejšej teórie dynamických systémov. Aplikácia tejto teórie na fyziku, biológiu, chémiu a inžinierstvo je demonštrovaná na príkladoch v takých oblastiach, ako je populačné modelovanie, dynamika tekutín, elektronika a mechanika.

Diferenciálne dynamické systémy začína pokrytím lineárnych systémov vrátane maticovej algebry, potom sa zameranie posúva k základnému materiálu na nelineárnych diferenciálnych rovniciach, čím sa výrazne využíva veta o kontrakčnom mapovaní. Nasledujúce kapitoly sa osobitne zaoberajú pojmami dynamických systémov - tokom, stabilitou, nemennými rozdeľovačmi, fázovou rovinou, bifurkáciou, chaosom a hamiltonovskou dynamikou.

V celej knihe autor obsahuje cvičenia, ktoré majú študentom pomôcť rozvíjať analytické a geometrické chápanie dynamiky. Mnohé z týchto cvičení a príkladov sú založené na aplikáciách a niektoré zahŕňajú výpočty. Príloha ponúka jednoduché kódy napísané v softvéroch Maple®, Mathematica® a MATLAB®, vďaka ktorým si študenti môžu precvičiť výpočty aplikované na problémy dynamických systémov.

Na jednej úrovni je možné tento text považovať za vhodný pre tradičný kurz bežných diferenciálnych rovníc (ODR). Pretože diferenciálne rovnice sú základom pre modely akýchkoľvek fyzikálnych systémov, ktoré vykazujú plynulú zmenu, študenti všetkých oblastí matematických vied a inžinierstva potrebujú nástroje na pochopenie metód riešenia týchto rovníc. Tradične sa táto expozícia začína v priebehu druhého roku výcviku počtu, kde sa študujú základné metódy riešenia jednorazových a dvojrozmerných (primárne lineárnych) ODR. Typický čitateľ tohto textu bude mať taký kurz, ako aj úvod do analýzy, kde budú objasnené teoretické základy (ε a δ) kalkulu. Materiál pre tento text bol vyvinutý desaťročie v kurze pre vysokoškolákov a začínajúcich študentov postgraduálneho štúdia aplikovanej matematiky, inžinierstva a fyziky na Coloradskej univerzite. V jednomsemestrálnom kurze obvykle pokryjem väčšinu materiálu z kapitol 1–6 a pridám výber častí z neskorších kapitol.

Existuje napríklad niekoľko klasických textov pre kurz tradičných diferenciálnych rovníc (Coddington a Levinson 1955 Hirsch a Smale 1974 Hartman 2002). Takéto kurzy zvyčajne začínajú štúdiom lineárnych systémov, ktoré začneme aj v kapitole 2. Maticová algebra je pre toto liečenie základom, preto poskytujeme krátku diskusiu o vlastných vektorových metódach a rozsiahlom spracovaní maticového exponenciálu. Ďalšou etapou tradičného kurzu je poskytnúť základ pre štúdium nelineárnych diferenciálnych rovníc tým, že sa ukáže, že za určitých podmienok majú tieto rovnice riešenia (existenciu) a že existuje iba jedno riešenie, ktoré spĺňa danú počiatočnú podmienku (jedinečnosť) . Teoretickým základom tohto výsledku, ako aj mnohých ďalších výsledkov v aplikovanej matematike, je majestátna veta o mapovaní kontrakcie. Kapitola 3 poskytuje samostatný úvod do analytických základov potrebných na pochopenie tejto vety. Len čo je tento nástroj konkrétne pochopený, študenti vidia, že veľa dôkazov rýchlo podľahne jeho sile. Je možné vynechať body §3.3–3.5, pretože väčšina materiálu sa v ďalších kapitolách príliš nepoužíva, aj keď je potrebné podporiť aspoň absolvovanie oboznámenia sa s vetami 3.10 a Lemma 3.13 (Grönwall).


10.1: Úvod do systémov diferenciálnych rovníc - matematika

Pretože budeme pracovať takmer výlučne so systémami rovníc, v ktorých sa počet neznámych rovná počtu rovníc, obmedzíme našu kontrolu na tieto druhy systémov.

Všetko, čo tu budeme robiť, sa dá ľahko rozšíriť na systémy s viac neznámymi ako rovnice alebo s viacerými rovnicami ako neznáme, ak je to potrebné.

Začnime s nasledujúcim systémom (n ) rovníc s (n ) neznámymi, (x_ <1> ), (x_ <2> ), ..., (x_).

Upozorňujeme, že v dolných indexoch koeficientov v tomto systéme (a_), (i ) zodpovedá rovnici, v ktorej je koeficient, a (j ) zodpovedá neznámej, ktorá sa vynásobí koeficientom.

Aby sme mohli na riešenie tohto systému použiť lineárnu algebru, najskôr si zapíšeme rozšírená matica pre tento systém. Rozšírená matica sú skutočne iba všetky koeficienty systému a čísla pre pravú stranu systému napísané v maticovej podobe. Tu je rozšírená matica pre tento systém.

Na vyriešenie tohto systému použijeme elementárne operácie s riadkami (ktoré trochu definujeme) na prepísanie rozšírenej matice do trojuholníkového tvaru. Matica bude v trojuholníkovom tvare, ak budú všetky záznamy pod hlavnou uhlopriečkou (uhlopriečka obsahujúca (a_ <11> ), (a_ <22> ), ..., (a_)) sú nuly.

Akonáhle je to hotové, môžeme si spomenúť, že každý riadok v rozšírenej matici zodpovedá rovnici. Potom prevedieme našu novú rozšírenú maticu späť na rovnice a v tomto okamihu bude riešenie systému veľmi jednoduché.

Pred prácou na príklade si najskôr zadefinujeme operácie so základnými riadkami. Sú tri z nich.

    Zamieňajte dva riadky. To je presne to, čo hovorí. Zameníme riadok (i ) za riadok (j ). Zápis, ktorý použijeme na označenie tejto operácie, je: ( ľavá šípka )

Je vždy o niečo jednoduchšie pochopiť tieto operácie, ak ich uvidíme v akcii. Poďme teda vyriešiť niekoľko systémov.

Prvým krokom je zapísanie rozšírenej matice pre tento systém. Nezabudnite, že koeficienty výrazov, ktoré nie sú prítomné, sú nulové.

Teraz chceme, aby položky pod hlavnou uhlopriečkou boli nulové. Hlavná uhlopriečka bola zafarbená na červeno, aby sme ju mohli sledovať počas tohto prvého príkladu. Z dôvodov, ktoré budú zrejmé nakoniec, by sme radšej dostali všetky diagonálne vstupy aj všetky.

Jeden môžeme získať na najvyššom hornom mieste tak, že si všimneme, že ak vymeníme prvý a druhý rad, dostaneme jeden na najvyššom mieste zadarmo. Poďme na to.

Teraz musíme dostať posledné dva záznamy (-2 a 3) do prvého stĺpca tak, aby boli nulové. Môžeme to urobiť pomocou operácie tretieho radu. Všimnite si, že ak vezmeme 2-krát prvý riadok a pridáme ho do druhého riadku, dostaneme nulu v druhom zázname v prvom stĺpci a ak vezmeme -3-krát prvý riadok do tretieho riadku, dostaneme 3 k byť nula. Môžeme robiť obe tieto operácie súčasne, tak poďme na to.

Pred pokračovaním v ďalšom kroku sa uistite, že ste postupovali podľa toho, čo sme práve urobili. Pozrime sa na prvú operáciu, ktorú sme vykonali. Táto operácia hovorí, že sa vynásobí položka v riadku 1 číslom 2 a pridá sa k zodpovedajúcemu záznamu v riadku 2, potom sa nahradí starý záznam v riadku 2 týmto novým záznamom. Nasledujú štyri jednotlivé operácie, ktoré sme vykonali, aby sme to dosiahli.

[začať2 ľavé (1 pravé) + ľavé (<- 2> pravé) & = 0 2 ľavé (2 pravé) + 1 & = 5 2 ľavé (3 pravé) + ľavé (<- 1> pravý) & = 5 2 ľavý (<13> pravý) + 4 & = 30 koniec]

Dobre, ďalší krok je voliteľný, ale opäť je vhodný. Technicky je 5 v druhom stĺpci v poriadku na opustenie. Avšak uľahčí nám život na ceste, ak je to 1. Môžeme sa o to postarať pomocou operácie v druhom rade. Celý riadok môžeme rozdeliť na 5. Toto dáva,

Ďalším krokom je potom použitie operácie v treťom rade na premenu hodnoty -6 v druhom stĺpci na nulu.

Teraz sme oficiálne hotoví, ale opäť je celkom pohodlné dostať všetky na hlavnú uhlopriečku, takže urobíme posledný krok.

Teraz môžeme previesť späť na rovnice.

V tomto okamihu je riešenie celkom jednoduché. Dostaneme (x_ <3> ) zadarmo a akonáhle dostaneme, môžeme to zapojiť do druhej rovnice a dostať (x_ <2> ). Potom môžeme použiť prvú rovnicu na získanie (x_ <1> ). Tiež si uvedomte, že mať tento proces pozdĺž hlavnej uhlopriečky 1-tku.

Riešením tohto systému rovníc je

Proces použitý v tomto príklade sa nazýva Gaussova eliminácia. Pozrime sa na ďalší príklad.

Najskôr si zapíšte rozšírenú maticu.

Pri práci na tomto príklade nebudeme uvádzať toľko slov. Tu je práca pre túto rozšírenú maticu.

V tomto príklade nepôjdeme ďalej. Vráťme sa k rovniciam, aby sme zistili, prečo.

Posledná rovnica by mala spôsobiť určité obavy. Je tu jedna z troch možností. Najprv sa nám nejako podarilo dokázať, že 0 sa rovná 8 a vieme, že to nie je možné. Po druhé, urobili sme chybu, ale po návrate k našej práci sa nezdá, že by sme urobili chybu.

Zostáva teda tretia možnosť. Keď dostaneme niečo ako tretia rovnica, ktorá jednoducho nedáva zmysel, okamžite vieme, že neexistuje riešenie. Inými slovami, neexistuje žiadna sada troch čísel, ktoré by umožnili splniť všetky tri rovnice súčasne.

Poďme si uviesť ďalší príklad. Získame systém pre tento nový príklad vykonaním veľmi malej zmeny v systéme z predchádzajúceho príkladu.

Jediný rozdiel medzi týmto systémom a systémom z druhého príkladu je, že sme zmenili 1 na pravej strane znaku rovnosti v tretej rovnici na -7.

Teraz si zapíšte rozšírenú maticu pre tento systém.

Kroky pre tento problém sú identické s krokmi pre druhý problém, takže si ich nebudeme zapisovať všetky. Po vykonaní rovnakých krokov prídeme k nasledujúcej matici.

Tentokrát sa posledná rovnica redukuje na

a na rozdiel od druhého príkladu to nie je problém. Nula sa v skutočnosti rovná nule!

Mohli by sme sa tu zastaviť a vrátiť sa k rovniciam, aby sme dostali riešenie, a v tomto prípade existuje riešenie. Ak však pôjdeme ešte jeden krok a dostaneme nulu nad úrovňou v druhom stĺpci aj pod ňou, náš život bude o niečo jednoduchší. Toto dáva,

Ak sa teraz vrátime k rovnici, dostaneme nasledujúce dve rovnice.

Máme dve rovnice a tri neznáme. To znamená, že môžeme vyriešiť dve z premenných z hľadiska zostávajúcej premennej. Pretože (x_ <3> ) je v oboch rovniciach, budeme ich z tohto dôvodu riešiť.

Toto riešenie znamená, že môžeme zvoliť hodnotu (x_ <3> ) ako čokoľvek, čo by sme chceli, a potom nájsť hodnoty (x_ <1> ) a (x_ <2> ) . V týchto prípadoch obvykle napíšeme riešenie nasledovne,

Týmto spôsobom získame nekonečné množstvo riešení, jedno pre každú hodnotu (t ).

Tieto tri príklady nás vedú k peknému faktu o sústavách rovníc.

Vzhľadom na sústavu rovníc ( eqref), budeme mať jednu z troch možností počtu riešení.

Pred prechodom na ďalšiu časť sa musíme pozrieť na ešte jednu situáciu. Sústava rovníc v ( eqref) sa nazýva nehomogénny systém, ak aspoň jeden z bis nie je nula. Ak však všetky (b_) sú nulové, systém nazývame homogénny a systém bude,

Teraz si všimnite, že v homogénnom prípade máme zaručené nasledujúce riešenie.

Toto riešenie sa často nazýva triviálne riešenie.

Pre homogénne systémy je možné vyššie uvedenú skutočnosť upraviť takto.

Vzhľadom na homogénny systém rovníc, ( eqref), budeme mať jednu z dvoch možností počtu riešení.

    Presne jedno riešenie, triviálne riešenie

V druhej možnosti môžeme povedať nenulové riešenie, pretože ak bude mať nekonečne veľa riešení a vieme, že jedno z nich je triviálne riešenie, potom všetky ostatné musia mať aspoň jedno z (x_) budú nenulové, a preto dostaneme nenulové riešenie.


Úvod do systémov

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Po zvyšok tohto obdobia budeme študovať nielen jednu diferenciálnu rovnicu súčasne, ale skôr to, čo sa nazýva systémy diferenciálnych rovníc.

To sú ako systémy lineárnych rovníc.

Musia byť riešené súčasne, inými slovami, nielen po jednom.

Ako teda vyzerá systém, keď si ho zapíšete?

Pretože budeme hovoriť o systémoch bežných diferenciálnych rovníc, bude tu stále iba jedna nezávislá premenná, ale bude tam niekoľko závislých premenných. Chystám sa zavolať, povedzme dva. Závislé premenné budú, budem ich nazývať x a y, a potom bude systém prvého rádu, niečo zahŕňajúci iba prvé derivácie, vyzerať takto. Inými slovami, na ľavej strane bude x prime. Na pravej strane budú závislé premenné a potom aj nezávislé premenné.

Naznačím to, oddelím to všetko od ostatných vložením bodkočiarky.

A rovnakým spôsobom y prime, derivácia y vzhľadom na t, bude nejaká iná funkcia (x, y) a t. Napíšme si výslovne, že xay sú závislé premenné.

A na čom závisia, je nezávislá premenná t, čas. Systém ako tento sa bude volať prvá objednávka. A budeme uvažovať v zásade iba o systémoch prvého rádu z tajného dôvodu, ktorý vysvetlím na konci obdobia.

Toto je systém prvého rádu, čo znamená, že jediný druh derivátov, ktoré sú tu hore, sú deriváty prvého rádu.

Takže x prime je dx nad dt a tak ďalej.

Teraz existuje ešte viac terminológie.

Samozrejme, prakticky všetky rovnice po začiatku pojmu, prakticky všetky rovnice, o ktorých sme uvažovali, sú lineárne rovnice, takže musí platiť, že lineárne systémy sú najlepším druhom.

A chlapče, určite sú. Kedy budeme nazývať systém lineárnym? Myslím si, že na začiatku by ste sa mali naučiť trochu terminológie, skôr ako začneme, a skutočne sa pokúsiť začať tieto veci riešiť.

X a y, závislé premenné sa musia vyskytovať lineárne. Inými slovami, musí to vyzerať takto, sekera plus by.

Teraz môže byť neporiadok. A tak tam vhodím extra funkciu t. A y prime bude nejaká iná lineárna kombinácia x a y, plus nejaká iná chaotická funkcia t. Ale aj a, b, c a d môžu byť funkciami t.

Môžu to byť ľudia nad kockami alebo sínusmi alebo niečo podobné. Musím teda rozlišovať tie prípady. Prípad, keď a, b, c a d sú konštanty, ktoré zavolám - No, dá sa to nazvať rôznymi spôsobmi.

Zjednodušene to nazvime sústava s konštantnými koeficientmi.

Systém s koeficientmi by bol pravdepodobne lepšou angličtinou. Na druhej strane, a, b, c a d bude tento systém stále nazývaný lineárny, ak ide o funkcie t.

Môžu to byť aj funkcie t.

Bol by to teda úplne dobrý lineárny systém, keď x prvočísel rovná tx plus sínus t krát y plus e na mínus t na druhú.

Nikdy by ste niečo také nevideli, ale je to v poriadku.

Čo ešte potrebujete vedieť? Čo by to bol za homogénny systém? Homogénny systém je systém bez týchto ďalších ľudí. To neznamená, že v ňom nie je t. Mohlo by byť t v a, b, c a d, ale tieto výrazy, ktoré v nich nemajú x a y, sa nesmú vyskytnúť. Takže lineárne homogénne.

A to je druh, ktorý začneme študovať najskôr rovnakým spôsobom, keď sme študovali rovnice vyššieho rádu.

Študovali sme najskôr homogénne. Najprv ste museli vedieť vyriešiť tieto a potom ste sa mohli naučiť riešiť všeobecnejší druh. Lineárna homogénna teda znamená, že r1 je nula a r2 je nula po celú dobu.

Sú identicky nulové. Nie sú tam.

Ty ich nevidíš. Vynechal som niečo?

Áno, počiatočné podmienky. Pretože je to dosť všeobecné, povedzme si, ako by vyzerali počiatočné podmienky?

Všeobecne platí, že dôvod, prečo musíte mať počiatočné podmienky, je získať hodnoty pre ľubovoľné konštanty, ktoré sa objavia v riešení.

Otázkou je, koľko ľubovoľných konštánt sa objaví v riešeniach týchto rovníc?

No, len ti dám odpoveď.

Dva. Počet ľubovoľných konštánt, ktoré sa objavia, je celkové poradie systému.

Napríklad keby to bola druhá derivácia a toto by bola prvá derivácia, čakal by som v systéme tri ľubovoľné konštanty - - pretože súčet, súčet dvoch a jednej robí tri. Musíte teda mať toľko počiatočných podmienok, koľko máte v riešení ľubovoľných konštánt. A to samozrejme vysvetľuje, keď sme študovali rovnice druhého rádu, museli sme mať dve počiatočné podmienky.

Musel som určiť počiatočný východiskový bod a počiatočnú rýchlosť. A dôvod, prečo sme museli mať dve podmienky, bol ten, že všeobecné riešenie malo v sebe dve ľubovoľné konštanty. The same thing happens here but the answer is it is more natural, the conditions here are more natural. I don't have to specify the velocity. Why not?

Well, because an initial condition, of course, would want me to say what the starting value of x is, some number, and it will also want to know what the starting value of y is at that same point.

Well, there are my two conditions.

And since this is going to have two arbitrary constants in it, it is these initial conditions that will satisfy, the arbitrary constants will have to be picked so as to satisfy those initial conditions.

In some sense, the giving of initial conditions for a system is a more natural activity than giving the initial conditions of a second order system.

You don't have to be the least bit cleaver about it.

Anybody would give these two numbers.

Whereas, somebody faced with a second order system might scratch his head. And, in fact, there are other kinds of conditions.

There are boundary conditions you learned a little bit about instead of initial conditions for a second order equation.

I cannot think of any more general terminology, so it sounds like we are going to actually have to get to work.

Okay, let's get to work. I want to set up a system and solve it. And since one of the things in this course is supposed to be simple modeling, it should be a system that models something.

In general, the kinds of models we are going to use when we study systems are the same ones we used in studying just first-order equations. Mixing, radioactive decay, temperature, the motion of temperature.

Heat, heat conduction, in other words.

Diffusion. I have given you a diffusion problem for your first homework on this subject.

What else did we do? That's all I can think of for the moment, but I am sure they will occur to me.

When, out of those physical ideas, are we going to get a system? The answer is, whenever there are two of something that there was only one of before. For example, if I have mixing with two tanks where the fluid goes like that.

Say you want to have a big tank and a little tank here and you want to put some stuff into the little tank so that it will get mixed in the big tank without having to climb a big ladder and stop and drop the stuff in. That will require two tanks, the concentration of the substance in each tank, therefore, that will require a system of equations rather than just one. Or, to give something closer to home, closer to this backboard, anyway, suppose you have dah, dah, dah, don't groan, at least not audibly, something that looks like that. And next to it put an EMF there. That is just a first order.

That just leads to a single first order equation.

But suppose it is a two loop circuit.

Now I need a pair of equations. Each of these loops gives a first order differential equation, but they have to be solved simultaneously to find the current or the charges on the condensers. And if I want a system of three equations, throw in another loop.

Now, suppose I put in a coil instead.

What is this going to lead to? This is going to give me a system of three equations of which this will be first order, first order. And this will be second order because it has a coil. You are up to that, right? You've had coils, inductance? Good.

So the whole thing is going to count as first-order, first-order, second-order.

To find out how complicated it is, you have to add up the orders. That is one and one, and two. This is really fourth-order stuff that we are talking about here.

We can expect it to be a little complicated.

Well, now let's take a modest little problem.

I am going to return to a problem we considered earlier in the problem of heat conduction. I had forgotten whether it was on the problem set or I did it in class, but I am choosing it because it leads to something we will be able to solve.

And because it illustrates how to add a little sophistication to something that was unsophisticated before.

A pot of water. External temperature Te of t.

I am talking about the temperature of something. And what I am talking about the temperature of will be an egg that is cooking inside, but with a difference. This egg is not homogenous inside. Instead it has a white and it has a yolk in the middle. In other words, it is a real egg and not a phony egg.

That is a small pot, or it is an ostrich egg.

[LAUGHTER] That is the yoke. The yolk is contained in a little membrane inside. And there are little yucky things that hold it in position. And we are going to let the temperature of the yolk, if you can see in the back of the room, be T1. That is the temperature of the yolk. The temperature of the white, which we will assume is uniform, is going to be T2.

Oh, that's the water bath. The temperature of the white is T2, and then the temperature of the external water bath.

In other words, the reason for introducing two variables instead of just the one variable for the overall temperature of the egg we had is because egg white is liquid pure protein, more or less, and the T1, the yolk has a lot of fat and cholesterol and other stuff like that which is supposed to be bad for you. It certainly has different conducting. It is liquid, at the beginning at any rate, but it certainly has different constants of conductivity than the egg white would.

And the condition of heat through the shell of the egg would be different from the conduction of heat through the membrane that keeps the yoke together.

So it is quite reasonable to consider that the white and the yolk will be at different temperatures and will have different conductivity properties.

I am going to use Newton's laws but with this further refinement. In other words, introducing two temperatures. Whereas, before we only had one temperature. But let's use Newton's law.

Let's see. The question is how does T1, the temperature of the yolk, vary with time?

Well, the yolk is getting all its heat from the white.

Therefore, Newton's law of conduction will be some constant of conductivity for the yolk times T2 minus T1.

The yolk does not know anything about the external temperature of the water bath. It is completely surrounded, snug and secure within itself. But how about the temperature of the egg white? That gets heat and gives heat to two sources, from the external water and also from the internal yolk inside.

So you have to take into account both of those.

Its conduction of the heat through that membrane, we will use the same a, which is going to be a times T1 minus T2. Remember the order in which you have to write these is governed by the yolk outside to the white. Therefore, that has to come first when I write it in order that a be a positive constant.

But it is also getting heat from the water bath.

And, presumably, the conductivity through the shell is different from what it is through this membrane around the yolk. So I am going to call that by a different constant. This is the conductivity through the shell into the white.

And that is going to be T, the external temperature minus the temperature of the egg white.

Here I have a system of equations because I want to make two dependent variables by refining the original problem.

Now, you always have to write a system in standard form to solve it. You will see that the left-hand side will give the dependent variables in a certain order.

In this case, the temperature of the yolk and then the temperature of the white.

The law is that in order not to make mistakes -- And it's a very frequent source of error so learn from the beginning not to do this. You must write the variables on the right-hand side in the same order left to right in which they occur top to bottom here. In other words, this is not a good way to leave that.

This is the first attempt in writing this system, but the final version should like this.

T1 prime, I won't bother writing dT / dt, is equal to -- T1 must come first, so minus a times T1 plus a times T2.

And the same law for the second one.

It must come in the same order. Now, the coefficient of T1, that is easy. That's a times T1.

The coefficient of T2 is minus a minus b, so minus (a plus b) times T2.

But I am not done yet. There is still this external temperature I must put into the equation.

Now, that is not a variable. This is some given function of t. And what the function of t is, of course, depends upon what the problem is.

So that, for example, what might be some possibilities, well, suppose the problem was I wanted to coddle the egg. I think there is a generation gap here. How many of you know what a coddled egg is? How many of you don't know?

Well, I'm just saying my daughter didn't know.

I mentioned it to her. I said I think I'm going to do a coddled egg tomorrow in class. And she said what is that?

And so I said a cuddled egg? She said why would someone cuddle an egg? I said coddle.

And she said, oh, you mean like a person, like what you do to somebody you like or don't like or I don't know. Whatever.

I thought a while and said, yeah, more like that.

[LAUGHTER] Anyway, for the enrichment of your cooking skills, to coddle an egg, it is considered to produce a better quality product than boiling an egg. That is why people do it.

You heat up the water to boiling, the egg should be at room temperature, and then you carefully lower the egg into the water. And you turn off the heat so the water bath cools exponentially while the egg inside is rising in temperature. And then you wait four minutes or six minutes or whatever and take it out.

You have a perfect egg. So for coddling, spelled so, what will the external temperature be?

Well, it starts out at time zero at 100 degrees centigrade because the water is supposed to be boiling.

The reason you have it boiling is for calibration so that you can know what temperature it is without having to use a thermometer, unless you're on Pike's Peak or some place.

It starts out at 100 degrees. And after that, since the light is off, it cools exponential because that is another law. You only have to know what K is for your particular pot and you will be able to solve the coddled egg problem. In other words, you will then be able to solve these equations and know how the temperature rises. I am going to solve a different problem because I don't want to have to deal with this inhomogeneous term. Let's use, as a different problem, a person cooks an egg. Coddles the egg by the first process, decides the egg is done, let's say hardboiled, and then you are supposed to drop a hardboiled egg into cold water. Not just to cool it but also because I think it prevents that dark thing from forming that looks sort of unattractive. Let's ice bath.

The only reason for dropping the egg into an ice bath is so that you could have a homogenous equation to solve.

And since this a first system we are going to solve, let's make life easy for ourselves.

Now, all my work in preparing this example, and it took considerably longer time than actually solving the problem, was in picking values for a and b which would make everything come out nice. It's harder than it looks.

The values that we are going to use, which make no physical sense whatsoever, but a equals 2 and b equals 3. These are called nice numbers.

What is the equation? What is the system?

Can somebody read it off for me?

It is T1 prime equals, what is it, minus 2T1 plus 2T2.

I think this is 2T1. And the other one is minus a plus b, so minus 5.

This is a system. Now, on Wednesday I will teach you a fancy way of solving this. But, to be honest, the fancy way will take roughly about as long as the way I am going to do it now. The main reason for doing it is that it introduces new vocabulary which everyone wants you to have. And also, more important reasons, it gives more insight into the solution than this method. This method just produces the answer, but you want insight, also.

And that is just as important. But for now, let's use a method which always works and which in 40 years, after you have forgotten all other fancy methods, will still be available to you because it is method you can figure out yourself. You don't have to remember anything. The method is to eliminate one of the dependent variables. It is just the way you solve systems of linear equations in general if you aren't doing something fancy with determinants and matrices.

If you just eliminate variables.

We are going to eliminate one of these variables.

Let's eliminate T2. You could also eliminate T1.

The main thing is eliminate one of them so you will have just one left to work with. How do I eliminate T2?

Beg your pardon? Is something wrong?

If somebody thinks something is wrong raise his hand.

Why do I want to get rid of T1? Well, I can add them.

But, on the left-hand side, I will have T1 prime plus T2 prime. What good is that?

[LAUGHTER] I think you will want to do it my way.

[APPLAUSE] Solve for T2 in terms of T1. That is going to be T1 prime plus 2T1 divided by 2.

Now, take that and substitute it into the second equation.

Wherever you see a T2, put that in, and what you will be left with is something just in T1.

To be honest, I don't know any other good way of doing this. There is a fancy method that I think is talked about in your book, which leads to extraneous solutions and so on, but you don't want to know about that. This will work for a simple linear equation with constant coefficients, always. Substitute in.

What do I do? Now, here I do not advise doing this mentally. It is just too easy to make a mistake. Here, I will do it carefully, writing everything out just as you would.

T1 prime plus 2T1 over 2, prime, equals 2T1 minus 5 time T1 prime plus 2T1 over two.

I took that and substituted into this equation. Now, I don't like those two's.

Let's get rid of them by multiplying.

And now write this out. What is this when you look at it? This is an equation just in T1.

It has constant coefficients. And what is its order?

Its order is two because T1 prime primed.

In other words, I can eliminate T2 okay, but the equation I am going to get is no longer a first-order.

It becomes a second-order differential equation.

And that's a basic law. Even if you have a system of more equations, three or four or whatever, the law is that after you do the elimination successfully and end up with a single equation, normally the order of that equation will be the sum of the orders of the things you started with. So two first-order equations will always produce a second-order equation in just one dependent variable, three will produce a third order equation and so on. So you trade one complexity for another. You trade the complexity of having to deal with two equations simultaneously instead of just one for the complexity of having to deal with a single higher order equation which is more trouble to solve.

It is like all mathematical problems.

Unless you are very lucky, if you push them down one way, they are really simple now, they just pop up some place else. You say, oh, I didn't save anything after all.

That is the law of conservation of mathematical difficulty.

[LAUGHTER] You saw that even with the Laplace transform.

In the beginning it looks great, you've got these tables, take the equation, horrible to solve.

Take some transform, trivial to solve for capital Y.

Now I have to find the inverse Laplace transform.

And suddenly all the work is there, partial fractions, funny formulas and so on. It is very hard in mathematics to get away with something. It happens now and then and everybody cheers. Let's write this out now in the form in which it looks like an equation we can actually solve.

Just be careful. Now it is all right to use the method by which you collect terms.

There is only one term involving T1 double prime.

It's the one that comes from here.

How about the terms in T1 prime?

There is a 2. Here, there is minus 5 T1 prime. If I put it on the other side it makes plus 5 T1 prime plus this two makes 7 T1 prime.

And how many T1's are there? Well, none on the left-hand side. On the right-hand side I have 4 here minus 10. 4 minus 10 is negative 6.

Negative 6 T1 put on this left-hand side the way we want to do makes plus 6 T1.

There are no inhomogeneous terms, so that is equal to zero.

If I had gotten a negative number for one of these coefficients, I would instantly know if I had made a mistake. Prečo?

Why must those numbers come out to be positive?

It is because the system must be, the system must be, fill in with one word, stable.

And why must this system be stable?

In other words, the long-term solutions must be zero, must all go to zero, whatever they are.

Why is that? Well, because you are putting the egg into an ice bath. Or, because we know it was living but after being hardboiled it is dead and, therefore, dead systems are stable.

That's not a good reason but it is, so to speak, the real one. It's clear anyway that all solutions must tend to zero physically.

That's obvious. And, therefore, the differential equation must have the same property, and that means that its coefficients must be positive.

All its coefficients must be positive.

If this weren't there, I would get oscillating solutions, which wouldn't go to zero.

That is physical impossible for this egg.

Now the rest is just solving. The characteristic equation, if you can remember way, way back in prehistoric times when we were solving these equations, is this.

And what you want to do is factor it.

This is where all the work was, getting those numbers so that this would factor. So it's r plus 1 times r plus 6 And so the solutions are, the roots are r equals negative 1. I am just making marks on the board, but you have done this often enough, you know what I am talking about.

So the characteristic roots are those two numbers.

And, therefore, the solution is, I could write down immediately with its arbitrary constant as c1 times e to the negative t plus c2 times e to the negative 6t. Now, I have got to get T2.

Here the first worry is T2 is going to give me two more arbitrary constants. It better not.

The system is only allowed to have two arbitrary constants in its solution because that is the initial conditions we are giving it. By the way, I forgot to give initial conditions. Let's give initial conditions.

Let's say the initial temperature of the yolk, when it is put in the ice bath, is 40 degrees centigrade, Celsius. And T2, let's say the white ought to be a little hotter than the yolk is always cooler than the white for a soft boiled egg, I don't know, or a hardboiled egg if it hasn't been chilled too long.

Let's make this 45. Realistic numbers.

Now, the thing not to do is to say, hey, I found T1.

Okay, I will find T2 by the same procedure.

I will go through the whole thing.

I will eliminate T1 instead. Then I will end up with an equation T2 and I will solve that and get T2 equals blah, blah, blah. That is no good, A, because you are working too hard and, B, because you are going to get two more arbitrary constants unrelated to these two. And that is no good.

Because the correct solution only has two constants in it.

Not four. So that procedure is wrong.

You must calculate T2 from the T1 that you found, and that is the equation which does it.

That's the one we have to have. Where is the chalk?

Áno. Maybe I can have a little thing so I can just carry this around with me.

That is the relation between T2 and T1.

Or, if you don't like it, either one of these equations will express T2 in terms of T1 for you.

It doesn't matter. Whichever one you use, however you do it, that's the way you must calculate T2. So what is it?

T2 is calculated from that pink box.

It is one-half of T1 prime plus T1.

Now, if I take the derivative of this, I get minus c1 times the exponential. The coefficient is minus c1, take half of that, that is minus a half c1 and add it to T1. Minus one-half c1 plus c1 gives me one-half c1.

And here I take the derivative, it is minus 6 c2.

Take half of that, minus 3 c2 and add this c2 to it, minus 3 plus 1 makes minus 2.

That is T2. And notice it uses the same arbitrary constants that T1 uses.

So we end up with just two because we calculated T2 from that formula or from the equation which is equivalent to it, not from scratch. We haven't put in the initial conditions yet, but that is easy to do.

Everybody, when working with exponentials, of course, you always want the initial conditions to be when T is equal to zero because that makes all the exponentials one and you don't have to worry about them.

But this you know. If I put in the initial conditions, at time zero, T1 has the value 40.

So 40 should be equal to c1 + c2.

And the other equation will say that 45 is equal to one-half c1 minus 2 c2. Now we are supposed to solve these. Well, this is called solving simultaneous linear equations. We could use Kramer's rule, inverse matrices, but why don't we just eliminate. Let me see.

If I multiply by, 45, so multiply by two, you get 90 equals c1 minus 4 c2.

Then subtract this guy from that guy.

So, 40 taken from 90 makes 50. And c1 taken from c1, because I multiplied by two, makes zero.

And c2 taken from minus 4 c2, that makes minus 5 c2, I guess.

I seem to get c2 is equal to negative 10.

And if c2 is negative 10, then c1 must be 50.

There are two ways of checking the answer.

One is to plug it into the equations, and the other is to peak. Yes, that's right.

[LAUGHTER] The final answer is, in other words, you put a 50 here, 25 there, negative 10 here, and positive 20 there. That gives the answer to the problem. It tells you, in other words, how the temperature of the yolk varies with time and how the temperature of the white varies with time. As I said, we are going to learn a slick way of doing this problem, or at least a very different way of doing the same problem next time, but let's put that on ice for the moment.

And instead I would like to spend the rest of the period doing for first order systems the same thing that I did for you the very first day of the term.

Remember, I walked in assuming that you knew how to separate variables the first day of the term, and I did not talk to you about how to solve fancier equations by fancier methods.

I instead talked to you about the geometric significance, what the geometric meaning of a single first order equation was and how that geometric meaning enabled you to solve it numerically. And we spent a little while working on such problems because nowadays with computers it is really important that you get a feeling for what these things mean as opposed to just algorithms for solving them.

As I say, most differential equations, especially systems, are likely to be solved by a computer anyway.

You have to be the guiding genius that interprets the answers and can see when mistakes are being made, stuff like that. The problem is, therefore, what is the meaning of this system?

Well, you are not going to get anywhere interpreting it geometrically, unless you get rid of that t on the right-hand side. And the only way of getting rid of the t is to declare it is not there.

So I hereby declare that I will only consider, for the rest of the period, that is only ten minutes, systems in which no t appears explicitly on the right-hand side. Because I don't know what to do if it does up here. We have a word for these.

Remember what the first order word was?

A first order equation where there was no t explicitly on the right-hand side, we called it, anybody remember? Just curious.

This is an autonomous system. It is not a linear system because these are messy functions.

This could be x times y or x squared minus 3y squared divided by sine of x plus y.

It could be a mess. Definitely not linear.

But autonomous means no t. t means the independent variable appears on the right-hand side.

Of course, it is there. It is buried in the dx/dt and dy/dt. But it is not on the right-hand side. No t appears on the right-hand side.

Because no t appears on the right-hand side, I can now draw a picture of this.

But, let's see, what does a solution look like?

I never even talked about what a solution was, did I? Well, pretend that immediately after I talked about that, I talked about this.

What is the solution? Well, the solution, maybe you took it for granted, is a pair of functions, x of t, y of t if when you plug it in it satisfies the equation. And so what else is new?

The solution is x equals x of t, y equals y of t.

If I draw a picture of that what would it look like?

This is where your previous knowledge of physics above all 18.02, maybe 18.01 if you learned this in high school, what is x equals x of t and y equals y of t?

How do you draw a picture of that? What does it represent?

A curve. And what will be the title of the chapter of the calculus book in which that is discussed?

Parametric equations. This is a parameterized curve.

So we know what the solution looks like.

Our solution is a parameterized curve.

And what does a parameterized curve look like?

Well, it travels, and in a certain direction.

Why do I have several of those curves?

Well, because I have several solutions.

In fact, given any initial starting point, there is a solution that goes through it.

I will put in possible starting points.

And you can do this on the computer screen with a little program you will have, one of the visuals you'll have.

It's being made right now. You put down starter point, put down a click, and then it just draws the curve passing through that point.

Didn't we do this early in the term?

Áno. But there is a difference now which I will explain. These are various possible starting points at time zero for this solution, and then you see what happens to it afterwards.

In fact, through every point in the plane will pass a solution curve, parameterized curve. Now, what is then the representation of this? Well, what is the meaning of x prime of t and y prime of t?

I am not going to worry for the moment about the right-hand side. What does this mean by itself?

If this is the curve, the parameterized motion, then this represents its velocity vector.

It is the velocity of the solution at time t.

If I think of the solution as being a parameterized motion.

All I have drawn here is the trace, the path of the motion.

This hasn't indicated how fast it was going.

One solution might go whoosh and another one might go rah.

That is a velocity, and that velocity changes from point to point. It changes direction.

Well, we know its direction at each point.

That's tangent. What I cannot tell is the speed. From this picture, I cannot tell what the speed was.

Too bad. Now, what is then the meaning of the system? What the system does, it prescribes at each point the velocity vector.

If you tell me what the point (x, y) is in the plane then these equations give you the velocity vector at that point.

And, therefore, what I end up with, the system is what you call in physics and what you call in 18.02 a velocity field. So at each point there is a certain vector. The vector is always tangent to the solution curve through there, but I cannot predict from just this picture what its length will be because at some points, it might be going slow. The solution might be going slowly. In other words, the plane is filled up with these guys.

So on and so on. We can say a system of first order equations, ODEs of first order equations, autonomous because there must be no t on the right-hand side, is equal to a velocity field. A field of velocity.

The plane covered with velocity vectors.

And a solution is a parameterized curve with the right velocity everywhere.

Now, there obviously must be a connection between that and the direction fields we studied at the beginning of the term.

And there is. It is a very important connection. It is too important to talk about in minus one minute. When we need it, I will have to spend some time talking about it then.


An Introduction to Nonlinear Differential Equations , Second Edition

"This book is well conceived and well written. The author has succeeded in producing a text on nonlinear PDEs that is not only quite readable but also accessible to students from diverse backgrounds."
—SIAM Review

A practical introduction to nonlinear PDEs and their real-world applications

Now in a Second Edition, this popular book on nonlinear partial differential equations (PDEs) contains expanded coverage on the central topics of applied mathematics in an elementary, highly readable format and is accessible to students and researchers in the field of pure and applied mathematics. This book provides a new focus on the increasing use of mathematical applications in the life sciences, while also addressing key topics such as linear PDEs, first-order nonlinear PDEs, classical and weak solutions, shocks, hyperbolic systems, nonlinear diffusion, and elliptic equations. Unlike comparable books that typically only use formal proofs and theory to demonstrate results, An Introduction to Nonlinear Partial Differential Equations, Second Edition takes a more practical approach to nonlinear PDEs by emphasizing how the results are used, why they are important, and how they are applied to real problems.

The intertwining relationship between mathematics and physical phenomena is discovered using detailed examples of applications across various areas such as biology, combustion, traffic flow, heat transfer, fluid mechanics, quantum mechanics, and the chemical reactor theory. New features of the Second Edition also include:

Additional intermediate-level exercises that facilitate the development of advanced problem-solving skills

New applications in the biological sciences, including age-structure, pattern formation, and the propagation of diseases

An expanded bibliography that facilitates further investigation into specialized topics

With individual, self-contained chapters and a broad scope of coverage that offers instructors the flexibility to design courses to meet specific objectives, An Introduction to Nonlinear Partial Differential Equations, Second Edition is an ideal text for applied mathematics courses at the upper-undergraduate and graduate levels. It also serves as a valuable resource for researchers and professionals in the fields of mathematics, biology, engineering, and physics who would like to further their knowledge of PDEs.

Reviews

Author Bios

J. David Logan, PhD, is Willa Cather Professor of Mathematics at the University of Nebraska–Lincoln. He has authored several texts on elementary differential equations and beginning partial differential equations, including Applied Mathematics, Third Edition, also published by Wiley. Dr. Logan's research interests include mathematical physics, combustion and detonation, hydrogeology, and mathematical biology.


Differential Equations

This introductory differential equations textbook presents a convenient way for professors to integrate symbolic computing into the study of differential equations and linear algebra. Mathematica provides the necessary computational power and is employed from the very beginning of the text. Each new idea is interactively developed using it.

After first learning about the fundamentals of differential equations and linear algebra, the student is immediately given an opportunity to examine each new concept using Mathematica. All ideas are explored utilizing Mathematica, and though the computer eases the computational burden, the student is encouraged to think about what the computations reveal, how they are consistent with the mathematics, what any conclusions mean, and how they may be applied.

This new edition updates the text to Mathematica 5.0 and offers a more extensive treatment of linear algebra. It has been thoroughly revised and corrected throughout.

Dr. Clay C. Ross taught mathematics at the university level from 1967 through his retirement in May of 2003. He continues to pursue his interests in mathematics, travel and nature photography, still plays in the University orchestra, and serves as organist at his church. Those activities and much reading keep him productively occupied.

From the reviews of the second edition:

"The introductory differential equations textbook presents a convenient way for professors to integrate … . Each new idea is interactively developed … . After first learning about the fundamentals of differential equations and linear algebra, the student is immediately given an opportunity to examine … . the student is encouraged to think about what the computations reveal, how they are consistent with the mathematics … . The new revised edition updates … and offers a more extensive treatment of linear algebra." (Zentralblatt für Didaktik der Mathematik, January, 2005)

"This text instructs students in solving and using differential equations with both paper-and-pencil techniques and the Mathematica symbolic manipulation program. … This is a good, thorough standard textbook with good explanations." (Steven Dunbar, MathDL, January, 2005)


An Introduction to Delay Differential Equations with Applications to the Life Sciences

This book is intended to be an introduction to Delay Differential Equations for upper level undergraduates or beginning graduate mathematics students who have a good background in ordinary differential equations and would like to learn about the applications. It may also be of interest to applied mathematicians, computational scientists, and engineers. It focuses on key tools necessary to understand the applications literature involving delay equations and to construct and analyze mathematical models. Aside from standard well-posedness results for the initial value problem, it focuses on stability of equilibria via linearization and Lyapunov functions and on Hopf bifurcation. It contains a brief introduction to abstract dynamical systems focused on those generated by delay equations, introducing limit sets and their properties. Differential inequalities play a significant role in applications and are treated here, along with an introduction to monotone systems generated by delay equations. The book contains some quite recent results such as the Poincare-Bendixson theory for monotone cyclic feedback systems, obtained by Mallet-Paret and Sell. The linear chain trick for a special family of infinite delay equations is treated. The book is distinguished by the wealth of examples that are introduced and treated in detail. These include the delayed logistic equation, delayed chemostat model of microbial growth, inverted pendulum with delayed feedback control, a gene regulatory system, and an HIV transmission model. An entire chapter is devoted to the interesting dynamics exhibited by a chemostat model of bacteriophage parasitism of bacteria. The book has a large number of exercises and illustrations. Hal Smith is a Professor at the School of Mathematical and Statistical Sciences at Arizona State University.

Hal Smith is a Professor at the School of Mathematical and Statistical Sciences at Arizona State University.

“This text is designed to be an introduction to the theory of differential equations with delay for advanced undergraduates and beginning graduate students.” (William J. Satzer, The Mathematical Association of America, November, 2010)

“This textbook would serve as an excellent reference text to help a mathematics faculty develop an introductory mathematics course on DDEs … . This textbook clearly places the study of DDEs firmly within the standard curriculum of a graduate program in mathematics and well within the scope of a strong final-year undergraduate mathematics major. … this text will have a positive impact on biomathematics education and be a welcome addition. We certainly find it so.” (John G. Milton and Michael C. Mackey, Mathematical Reviews, Issue 2011 k)

“This book gives a first introduction to delay differential equations that is intended for mathematics students. Thanks to the emphasis on applications to life sciences, it can be recommended also to scientists from this discipline that wish to get a deeper understanding of the theoretical aspects for this widely used class of models.” (Matthias Wolfrum, Zentralblatt MATH, Vol. 1227, 2012)


Methods of Mathematical Modelling

This book presents mathematical modelling and the integrated process of formulating sets of equations to describe real-world problems. It describes methods for obtaining solutions of challenging differential equations stemming from problems in areas such as chemical reactions, population dynamics, mechanical systems, and fluid mechanics.

Chapters 1 to 4 cover essential topics in ordinary differential equations, transport equations and the calculus of variations that are important for formulating models. Chapters 5 to 11 then develop more advanced techniques including similarity solutions, matched asymptotic expansions, multiple scale analysis, long-wave models, and fast/slow dynamical systems.

Methods of Mathematical Modelling will be useful for advanced undergraduate or beginning graduate students in applied mathematics, engineering and other applied sciences.

Thomas Witelski is a Professor of Mathematics at Duke University specializing in nonlinear partial differential equations and fluid dynamics. He is a long-time participant in many study groups on mathematical modelling and industrial problems. He is the co-Editor-in-Chief of the Journal of Engineering Mathematics and also serves on the editorial board for the European Journal of Applied Mathematics. Witelski received his Ph.D. in Applied Mathematics from the California Institute of Technology in 1995 and was a postdoctoral fellow at the Massachusetts Institute of Technology.

Mark Bowen is an Associate Professor in the International Center for Science and Engineering Programs at Waseda University, where he teaches courses in differential equations and nonlinear dynamics. His expertise is in asymptotic analysis, nonlinear differential equations and fluid dynamics. He received his Ph.D. in Applied Mathematics in 1998 from the University of Nottingham.

“The text is well written. The authors have provided a clear and concise presentation of many important topics in a way that should be accessible to students following a first course in differential equations. … More advanced students could easily learn a significant amount of useful mathematics reading the text independently. … Methods of Mathematical Modelling is a welcome addition to the SUMS series and should prove to be useful for many instructors and students.” (Jason M. Graham, MAA Reviews, maa.org, February, 2016)

“The purpose of this text is to introduce the reader to the art of mathematical modeling … . The book provides an account of a number of useful for mathematical modelling techniques which are illustrated with examples and complemented with problems for self study.” (Yuriy V. Rogovchenko, zbMATH 1333.00025, 2016)


Introduction to Differential Equations with Dynamical Systems

Many textbooks on differential equations are written to be interesting to the teacher rather than the student. Introduction to Differential Equations with Dynamical Systems is directed toward students. Táto stručná a aktuálna učebnica sa zameriava na výzvy, ktorým čelia študenti vysokoškolského štúdia matematiky, inžinierstva a prírodných vied počas prvého kurzu diferenciálnych rovníc. A zatiaľ čo pokrýva všetky štandardné časti predmetu, kniha zdôrazňuje rovnice a aplikácie lineárnych konštantných koeficientov vrátane tém dôležitých pre študentov inžinierstva. Stephen Campbell a Richard Haberman & thinsp & # 8212 & thinspusing starostlivo formulovaných derivácií, elementárnych vysvetlení a príkladov, cvičení a obrázkov namiesto viet a dôkazov & thinsp & # 8212 & thinsphave napísali knihu, vďaka ktorej je učenie a výučba diferenciálnych rovníc jednoduchšie a relevantnejšie. Kniha tiež predstavuje elementárne dynamické systémy jedinečným a flexibilným spôsobom, ktorý je vhodný pre všetky kurzy bez ohľadu na dĺžku.

„Títo dvaja skúsení aplikovaní matematici sa snažili poskytnúť typickým študentom ľahko čitateľný úvod do diferenciálnych rovníc ... Písanie je jasné a dobre ilustrované.“—Robert E. O'Mally, Jr., Recenzia SIAM

"Úvod do diferenciálnych rovníc s dynamickými systémami je zameraná na študentov. Táto stručná a aktuálna učebnica sa venuje výzvam, ktorým čelia študenti vysokoškolského štúdia matematiky, inžinierstva a prírodných vied počas prvého kurzu diferenciálnych rovníc. “L'Enseignement Mathematique


Matematické znázornenie lineárnych fyzikálnych systémov

Táto stránka popisuje rôzne spôsoby, ako môžu byť fyzikálne systémy matematicky znázornené. Všeobecne sú systémové rovnice odvodené ako súbor diferenciálnych rovníc. Zvážte napríklad systém zobrazený nižšie (spolu s diagrammi voľného tela a systémovými diferenciálnymi rovnicami).

Schémy voľného tela:

Rovnice: Voľné telo 1 (pri x1) Voľné telo 2 (pri x2)

Toto znázornenie úplne predstavuje systém, jeho použitie je však ťažkopádne. Vyžaduje dve rovnice s viacerými symbolmi a dolnými indexmi. Neovšeobecňuje to dobre na širokú škálu systémov. V tomto dokumente najskôr preskúmame znázornenie systémov s diferenciálnymi rovnicami, potom vyvinieme niekoľko systémov, s ktorými sa vo všeobecnosti dá ľahšie pracovať.

Nezabudnite, že tento dokument popisuje rôzne formy (napr. Prenosovú funkciu, stavový priestor.), Ale nehovorí o ich riešení. Táto diskusia je inde a oddeľuje ju disciplína:


Pozri si video: Sisteme de ecuații diferențiale-1, Alexandru Negrescu, Universitatea Politehnica din București (December 2021).