0
0.5
1
1.5
2
2.5
3
3.5
4
0
0.5
1
1.5
2
2D Stagnation Point Similarity Solution
eta
f'', f', and f
Z
n2
,
Z
n3
,
Z
n4
,
Z
n1
,
n
1 num_steps
..
:=
Now generate a plot of the similarity variables:
Note that I have iterated with various guesses until the correct Y1guess value was found (close enough).
Z
num_steps 1
+
3
,
1.00000403
=
Check the value of Y2 at "infinity" to see if it is equal to 1:
Z
Rkadapt YBC
η
start
,η
end
,
num_steps
,
D
,
()
:=
num_steps
2000
:=
η
end
10
:=
η
start
0
:=
Here the function Rkadapt is used, which is similar to rkfixed except it internally uses adaptable spacing
instead of fixed spacing (more accuracy where needed). It reports at fixed spacing however.
Now calculate the solution as
η
marches from
η
start to
η
end. Here Z is the solution matrix, where the first
column is
η
, the second column is Y
1
, the third column is Y
2
, and the fourth column is Y
3
.
YBC
1.2325877
0
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
=
YBC
Y1guess
Y2BC
Y3BC
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
:=
Set up a vector of the boundary conditions for vector Y at
η
= 0 and verify:
Y3BC
0
:=
Y2BC
0
:=
Y1guess
1.2325877
:=
Define the boundary conditions at
η
= 0. Note that Y
1
at
η
= 0 must be guessed since it is not known.
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 Fall '07
 CIMBALA
 Derivative, Vector Space, Boundary value problem, J. M. Cimbala, stagnation point similarity

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