       discretization and plotting pde system

• To: mathgroup at smc.vnet.net
• Subject: [mg40034] discretization and plotting pde system
• From: john boy <johnboy98105 at yahoo.com>
• Date: Sun, 16 Mar 2003 02:48:17 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```hello.

i have been trying to discretize second order
derivatives in following system and then solve and
plot the solution of the discretized version, using
the mathematica software. Being a novice, i  have been
having some difficulties...

first, the actual system looks like this. and this is
the one of the systems i would like to learn how to
solve...

{dS[t,x] = 1 d^2S[t,x]/dx^2 - 10 S[t,x] C[t,x] + 1
SC[t,x] - 10 SC[t,x],
dC[t,x] = 0.1 d^2C[t,x]/dx^2 - 10 S[t,x] C[t,x] + 1
SC[t,x] + 10 SC[t,x],
dSC[t,x] = 1 d^2SC[t,x]/dx^2 + 10 S[t,x] C[t,x] - 1
SC[t,x] - 10 SC[t,x]}

when i make steady state assuptioms( as many
mathematicians do)  then i can change this system into
somethign that doesn't involve time derivatives. ( i
this?)

ssd= {0 = 1 d^2S[x]/dx^2 - 10 S[x] C[x] + 1 SC[x] - 10
SC[x],
0 = 0.1 d^2C[x]/dx^2 - 10 S[x] C[x] + 1 SC[x] + 10
SC[x],
0 = 1 d^2SC[x]/dx^2 + 10 S[x] C[x] - 1 SC[x] - 10
SC[x]}

at this point the system isn't pde anymore, its just
ode's so when i try to solve using,

NDSolve[{ssd, S== 0, C == 0, SC== 0, S''[x]
== 0, C''[x] ==0, SC''[x]==0}, {S[x], C[x], SC[x]}.
{x,-1, 1}]

then it gives errors about initial conditions...

then if I discretize the second order derivatives, I
think it should look somethign like this? (This way i
will just have algebraic expression? )

ssrd = {0 = 1 (S[x+1] - 2S[x] -S[x-1])/(Abs[xmas -
xmin)/n)^2 - 10 S[x] C[x] + 1 SC[x] - 10 SC[x],
0 = 0.1 (C[x+1] - 2C[x] -C[x-1])/(Abs[xmas -
xmin)/n)^2 - 10 S[x] C[x] + 1 SC[x] + 10 SC[x],
0 = 1 (SC[x+1] - 2SC[x] -SC[x-1])/(Abs[xmas -
xmin)/n)^2 + 10 S[x] C[x] - 1 SC[x] - 10 SC[x]}

But if I use NSolve[ssrd, {S[x], C[x], SC[x]}, {x, -1,
1}]  i get some errors about valid variables..

i could really use soem help in solving these
equations, I know some of you have natural talent for
things like this, but I have been struggling a lot
with it.

I hope someone knows what i'm talking about and hope
is willing to give me soem working codes to do this.
I'm pretty new at it and i'm spending a lot of time
without getting anywhere.

i'll paste the notebook cell expressions below.

thank you all veru much in advance.

In:=

\!\(d = \[IndentingNewLine]{\[PartialD]\_t s[t, x]\
== \
Ds\ \ \[PartialD]\_{x, 2}\ s\ [t, x]\  - \ f\
s[t, x]\ c[t, x]\  + \
r\ sc[t, x]\  - \
a\ t[t, x]\ s[t, x], \
\[IndentingNewLine]\[PartialD]\_t c\ [t,
x]\  == \
Db\ \ \[PartialD]\_{x, 2}\ c\ [t, x]\  - \ f\
s[t, x]\ c[t, x]\  + \
r\ sc[t, x]\  + \
l\ t[t, x]\ sc[t, x],
\[IndentingNewLine]\[PartialD]\_t\
sc\ [t, x]\  == \
Dc\ \ \[PartialD]\_{x, 2}\ sc\ [t, x]\  + \ f\
s[t, x]\ c[t, x]\  - \
r\ sc[t, x]\  - \ l\ t[t, x]\ sc[t, x]}\)

In:=

\!\(ssd = \[IndentingNewLine]{0\  == \
Ds\ \ \[PartialD]\_{x, 2}\ s\ [x]\  - \ f\
s[x]\ c[x]\  + \
r\ sc[x]\  - \ 10\ s[x], \
\[IndentingNewLine]0\  == \
Db\ \ \[PartialD]\_{x, 2}\ c\ [x]\  - \ f\
s[x]\ c[x]\  + \
r\ sc[x]\  + \ 10\ sc[x],
\[IndentingNewLine]0\  == \
Dc\ \ \[PartialD]\_{x, 2}\ sc\ [x]\  + \ f\
s[x]\ c[x]\  - \
r\ sc[x]\  - \ 10\ \ sc[x]}\)

In3:=

\!\(\*
RowBox[{"NDSolve", "[",
RowBox[{
RowBox[{"{",
RowBox[{
"ssd", ",", \(s == 0\), " ", ",", " ",
\(c == 0\), ",",
" ", \(sc == 0\), ",", "  ",
RowBox[{
RowBox[{
SuperscriptBox["s", "\[Prime]\[Prime]",
MultilineFunction->None], "[", "x",
"]"}], " ", "==", " ",
"0"}], ",", "  ",
RowBox[{
RowBox[{
SuperscriptBox["c", "\[Prime]\[Prime]",
MultilineFunction->None], "[", "x",
"]"}], "==", "0"}], ",",
" ",
RowBox[{
RowBox[{
SuperscriptBox["sc", "\[Prime]\[Prime]",
MultilineFunction->None], "[", "x",
"]"}], "==", " ", "0"}]}],
"}"}], ",", " ", \({s[x], \ c[x], \ sc[x]}\),
",",
" ", \({x, \ \(-1\), \ 1}\)}], "]"}]\)

In4:=

\!\(ssd = \[IndentingNewLine]{0\  == \ \ f\ s[x]\
c[x]\  + \ r\ sc[x]\  - \
a\ t[x]\ s[x], \ \[IndentingNewLine]0\  == \

Db\ \ \[PartialD]\_{x, 2}\ c\ [x]\  - \ f\
s[x]\ c[x]\  + \
r\ sc[x]\  + \ l\ t[x]\ sc[x],
\[IndentingNewLine]0\  == \
Dc\ \ \[PartialD]\_{x, 2}\ sc\ [x]\  + \ f\
s[x]\ c[x]\  - \
r\ sc[x]\  - \ l\ t[x]\ sc[x]}\)

In5:=

\!\(ssd\  = {0\  == \
Ds\ \((\ \(\(\ \)\(s\ [\((n + 1)\)]\  - \ 2\
s[\((n)\)] - \ s[\((n - \
1)\)]\)\)\/\((dx)\)^2)\)\  - \ f\ s[x]\ c[x]\  + \ r\
sc[x]\  - \
a\ t[x]\ s[x], \ \[IndentingNewLine]0\  == \

Db\ \ \((\ \(\(\ \)\(c\ [\((n + 1)\)]\  - \ 2\
c[\((n)\)] - \ c[\((n \
- 1)\)]\)\)\/\((dx)\)^2)\)\  - \ f\ s[x]\ c[x]\  + \
r\ sc[x]\  + \
l\ t[x]\ sc[x], \[IndentingNewLine]0\  == \
Dc\ \((\ \ \(sc\ [\((n + 1)\)]\  - \ 2\
sc[\((n)\)] - \ sc[\((n - \
1)\)]\)\/\((dx)\)^2)\)\  + \ f\ s[x]\ c[x]\  - \ r\
sc[x]\  - \
l\ t[x]\ sc[x]}\)

In6 :=
NSolve[ssrd, {s[x], c[x], sc[x]}, {x, -1, 1}]

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