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國立台灣海洋大學 Partial differential equation
作業 4
教授：陳正宗終生特聘教授學生：徐胤承
學號： M98520034
23 3/2 yx uyu x y
3
3/1
3/2
3/2
3
3
1
xy
yx
ydx
dy
yds
dyds
dx
ds
dy
y
u
ds
dx
x
u
ds
du
We could exercise the method of characteristics
We had a PDE like this
0, 1u s s
12),(
12),(
),(
),(
10,
0,
00,
1,0
2,,
2
3
1
23
3
3
3
3/2
3/2
xxystu
tsstu
tssty
sstx
ttu
tty
tx
ttu
susysx
ds
du
yds
dyds
dx
uyu yx
Then we were able to find the solution
The movie that was seen below was form the answer of last page
-5
0
5
10
-5
0
5
10
-5
0
5
10
s0
t0
t1
t2
t1
t2s0.5s1
s1.5s2-5
0
5
10
-5
0
5
10
When T was at 45, we could plot these three figures
-5
0
5
10-5
0
5
10
-5
0
5
10
-5
0
5
10
-5
0
5
10
-5
0
5
10
-5
0
5
10
s0
t0
t1
t2
t1
t2s0.5s1
s1.5s2-5
0
5
10
-5
0
5
10
-5
0
5
10
-5
0
5
10
-5
0
5
10
-5
0
5
10
-5
0
5
10
Finally, we plotted the figures that was including two kinds of expression
We used the parametric equation to draw this illustration
-5
0
5
10-5
0
5
10
-5
0
5
10
-5
0
5
10
We made the express in terms of x and y to find this
• The end• Thanks for comment
