Equazione Trasporto

[Gp741]

Salve a tutti, ho il seguente problema da risolvere:

"Risolvere mediande una function creata col Matlab l'equazione di trasporto monodimensionale \( \displaystyle \frac{{\partial\phi}}{{\partial{t}}}+{a}\frac{{\partial\phi}}{{\partial{x}}}-{k}\frac{{{\partial}^{{2}}\phi}}{{\partial{{x}}^{{2}}}}=-{270}\cdot{{x}}^{{4}}+{522}\cdot{{x}}^{{3}}-{316.5}\cdot{{x}}^{{2}}+{63.5}\cdot{x}+{1} \) ( con \( \displaystyle {0}\le{x}\le{L} \) e \( \displaystyle {0}\le{t}\le{L} \) ) assumendo che
-\( \displaystyle \phi{\left({x},{0}\right)}={0} \) (condizione iniziale)
-\( \displaystyle \phi{\left({0},{t}\right)}=\phi{\left({L},{t}\right)}={0} \) (condizioni al contorno)
-\( \displaystyle {L}={1} \)
-\( \displaystyle {a}={1} \)
-\( \displaystyle {k}={10} \)
-\( \displaystyle {T}{f{=}}{0.9} \)
Si utilizzi un metodo alle differenze finite che assume i seguenti schemi relativi alla derivata prima spaziale e derivata seconda temporale :

(Si ingnorino le o)
Convezione
x-----x n+1
|
* (l'asterisco rappresenta il p.to di collocazione; Si valutino le derivate spaziali nel p.to di collocazione come media di quelle relative ai punti \( \displaystyle {i} \) dello stencil )
|
x-----x n
i-1oooi

Diffusione
oooooox-----x------x n+1
ooooooooooo|
ooooooooooo* (l'asterisco rappresenta il p.to di collocazione; Si valutino le derivate spaziali nel p.to di collocazione come media di quelle relative ai punti \( \displaystyle {i} \) dello stencil)
ooooooooooo|
oooooox-----x------x n
ooooo(i-1)oo iooooo i+1

Il metodo alle differenze finite implementato nella funzione di matlab deve innazittutto fornire due matrici A e B relative agli schemi tali che \( \displaystyle {A}{\Phi}^{{{n}+{1}}}={B}{\Phi}^{{n}}+{C} \)
e successivamente utilizzarle per risolve l'equazione."

Allora ho provato a determinare le matrici dividendo il dominio spaziale \( \displaystyle {L} \) in 6 intervalli uguali:
\( \displaystyle {A}={\left(\matrix{\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}+\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{\left.{d}{x}\right.}}^{{2}}}+{1}&-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&{0}&{0}&{0}\\-\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}+\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{\left.{d}{x}\right.}}^{{2}}}+{1}&-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&{0}&{0}\\{0}&-\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}+\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{\left.{d}{x}\right.}}^{{2}}}+{1}&-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&{0}\\{0}&{0}&-\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}+\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{\left.{d}{x}\right.}}^{{2}}}+{1}&-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}\\{0}&{0}&{0}&\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}+\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{\left.{d}{x}\right.}}^{{2}}}+{1}&-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}}\right)} \)

\( \displaystyle {B}={\left(\matrix{{1}-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{\left.{d}{x}\right.}}^{{2}}}-\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}&\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&{0}&{0}&{0}\\\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}+\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&{1}-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{\left.{d}{x}\right.}}^{{2}}}-\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}&\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&{0}&{0}\\{0}&\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}+\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&{1}-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{\left.{d}{x}\right.}}^{{2}}}-\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}&\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&{0}\\{0}&{0}&\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}+\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}&{1}-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{\left.{d}{x}\right.}}^{{2}}}-\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}&\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}\\{0}&{0}&{0}&{1}-\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{\left.{d}{x}\right.}}^{{2}}}-\frac{{{a}\cdot{\left.{d}{t}\right.}}}{{{2}\cdot{\left.{d}{x}\right.}}}&\frac{{{\left.{d}{t}\right.}\cdot{k}}}{{{2}\cdot{{\left.{d}{x}\right.}}^{{2}}}}}\right)} \)

Assumo poi di dividere \( \displaystyle {L} \) in 29 intervalli invece di 6 ( dx=1/30) e e che dt= (0.001).Sostituendo tali valori nelle matrici ottengo nelle matrici \( \displaystyle {A} \) e \( \displaystyle {B} \) si ottiene:
(Sono due matrici sparse tridiagonali. il forum non le visualizza correttamente ma si capisce cmq qual'è la subdiagonale, la diagonale e la superdiagonale)
A=
Columns 1 through 7

1.8425 -0.4205 0 0 0 0 0
-0.4219 1.8425 -0.4205 0 0 0 0
0 -0.4219 1.8425 -0.4205 0 0 0
0 0 -0.4219 1.8425 -0.4205 0 0
0 0 0 -0.4219 1.8425 -0.4205 0
0 0 0 0 -0.4219 1.8425 -0.4205
0 0 0 0 0 -0.4219 1.8425
0 0 0 0 0 0 -0.4219
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0

Columns 8 through 14

0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
-0.4205 0 0 0 0 0 0
1.8425 -0.4205 0 0 0 0 0
-0.4219 1.8425 -0.4205 0 0 0 0
0 -0.4219 1.8425 -0.4205 0 0 0
0 0 -0.4219 1.8425 -0.4205 0 0
0 0 0 -0.4219 1.8425 -0.4205 0
0 0 0 0 -0.4219 1.8425 -0.4205
0 0 0 0 0 -0.4219 1.8425
0 0 0 0 0 0 -0.4219
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0

Columns 15 through 21

0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
-0.4205 0 0 0 0 0 0
1.8425 -0.4205 0 0 0 0 0
-0.4219 1.8425 -0.4205 0 0 0 0
0 -0.4219 1.8425 -0.4205 0 0 0
0 0 -0.4219 1.8425 -0.4205 0 0
0 0 0 -0.4219 1.8425 -0.4205 0
0 0 0 0 -0.4219 1.8425 -0.4205
0 0 0 0 0 -0.4219 1.8425
0 0 0 0 0 0 -0.4219
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0

Columns 22 through 28

0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
-0.4205 0 0 0 0 0 0
1.8425 -0.4205 0 0 0 0 0
-0.4219 1.8425 -0.4205 0 0 0 0
0 -0.4219 1.8425 -0.4205 0 0 0
0 0 -0.4219 1.8425 -0.4205 0 0
0 0 0 -0.4219 1.8425 -0.4205 0
0 0 0 0 -0.4219 1.8425 -0.4205
0 0 0 0 0 -0.4219 1.8425

B=

Columns 1 through 7

0.1575 0.4205 0 0 0 0 0
0.4219 0.1575 0.4205 0 0 0 0
0 0.4219 0.1575 0.4205 0 0 0
0 0 0.4219 0.1575 0.4205 0 0
0 0 0 0.4219 0.1575 0.4205 0
0 0 0 0 0.4219 0.1575 0.4205
0 0 0 0 0 0.4219 0.1575
0 0 0 0 0 0 0.4219
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0

Columns 8 through 14

0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0.4205 0 0 0 0 0 0
0.1575 0.4205 0 0 0 0 0
0.4219 0.1575 0.4205 0 0 0 0
0 0.4219 0.1575 0.4205 0 0 0
0 0 0.4219 0.1575 0.4205 0 0
0 0 0 0.4219 0.1575 0.4205 0
0 0 0 0 0.4219 0.1575 0.4205
0 0 0 0 0 0.4219 0.1575
0 0 0 0 0 0 0.4219
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0

Columns 15 through 21

0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0.4205 0 0 0 0 0 0
0.1575 0.4205 0 0 0 0 0
0.4219 0.1575 0.4205 0 0 0 0
0 0.4219 0.1575 0.4205 0 0 0
0 0 0.4219 0.1575 0.4205 0 0
0 0 0 0.4219 0.1575 0.4205 0
0 0 0 0 0.4219 0.1575 0.4205
0 0 0 0 0 0.4219 0.1575
0 0 0 0 0 0 0.4219
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0

Columns 22 through 28

0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0.4205 0 0 0 0 0 0
0.1575 0.4205 0 0 0 0 0
0.4219 0.1575 0.4205 0 0 0 0
0 0.4219 0.1575 0.4205 0 0 0
0 0 0.4219 0.1575 0.4205 0 0
0 0 0 0.4219 0.1575 0.4205 0
0 0 0 0 0.4219 0.1575 0.4205
0 0 0 0 0 0.4219 0.1575

Tuttavia quando le utilizzo per risolvere l'equazione di trasporto risolvendo il sistema \( \displaystyle {\Phi}^{{{n}+{1}}}={{A}}^{{-{1}}}{B}{\Phi}^{{n}}+{{A}}^{{-{1}}}{C} \) ottengo un risultato diverso da quello della soluzione stazionaria(nel senzo che il grafico della soluzione stazionaria ottenuta con dsolve è 1/1000 piu piccolo(lungo l'asse delle ascisse) da quello ottenuto con la soluzione numerica anche se la forma è pressochè identica) facilmente ottenibile col comando dsolve in matlab risolvendo \( \displaystyle {a}\frac{{\partial\phi}}{{\partial{x}}}-{k}\frac{{{\partial}^{{2}}\phi}}{{\partial{{x}}^{{2}}}}=-{270}\cdot{{x}}^{{4}}+{522}\cdot{{x}}^{{3}}-{316.5}\cdot{{x}}^{{2}}+{63.5}\cdot{x}+{1} \).
Qualcuno potrebbe verificare la correttezza delle matrici A e B che ho ottenuto? Grazie mille per le risposte.