Boundary Value Problems
u
x
x
+
u
y
y
=
0
{\displaystyle \displaystyle u_{xx}+u_{yy}=0}
Disc of radius c For a disk with a radius of "c", let the polar coordinates be
0
<
r
<
c
{\displaystyle 0<r<c}
−
π
<
θ
<
π
{\displaystyle -\pi <\theta <\pi }
r
2
u
r
r
+
r
u
r
+
u
θ
θ
=
0
{\displaystyle r^{2}u_{rr}+ru_{r}+u_{\theta \theta }=0}
u
(
c
,
θ
)
=
f
(
θ
)
{\displaystyle u(c,\theta )=f(\theta )}
u
(
r
,
π
)
=
u
(
r
,
−
π
)
{\displaystyle u(r,\pi )=u(r,-\pi )}
u
θ
(
r
,
π
)
=
u
θ
(
r
,
−
π
)
{\displaystyle u_{\theta }(r,\pi )=u_{\theta }(r,-\pi )}
u
(
r
,
θ
)
=
R
(
r
)
Θ
(
θ
)
{\displaystyle \displaystyle u(r,\theta )=R(r)\Theta (\theta )}
r
2
R
″
Θ
+
r
R
′
Θ
+
R
Θ
″
=
0
{\displaystyle \displaystyle r^{2}R''\Theta +rR'\Theta +R\Theta ''=0}
r
2
R
″
+
r
R
′
R
+
Θ
″
Θ
=
0
{\displaystyle \displaystyle {\frac {r^{2}R''+rR'}{R}}+{\frac {\Theta ''}{\Theta }}=0}
r
2
R
″
+
r
R
′
R
=
−
Θ
″
Θ
=
Constant
{\displaystyle \displaystyle {\frac {r^{2}R''+rR'}{R}}=-{\frac {\Theta ''}{\Theta }}={\mbox{Constant}}}
The constant may be greater than , equal to or less than zero.
λ
2
>
0
{\displaystyle \displaystyle \lambda ^{2}>0}
−
Θ
″
Θ
=
λ
2
{\displaystyle \displaystyle -{\frac {\Theta ''}{\Theta }}=\lambda ^{2}}
Θ
″
+
λ
2
Θ
=
0
{\displaystyle \displaystyle \Theta ''+\lambda ^{2}\Theta =0}
Θ
(
θ
)
=
A
c
o
s
(
λ
θ
)
+
B
s
i
n
(
λ
θ
)
{\displaystyle \displaystyle \Theta (\theta )=Acos(\lambda \theta )+Bsin(\lambda \theta )}
u
(
r
,
π
)
=
u
(
r
,
−
π
)
{\displaystyle \displaystyle u(r,\pi )=u(r,-\pi )}
Θ
(
π
)
=
Θ
(
−
π
)
{\displaystyle \displaystyle \Theta (\pi )=\Theta (-\pi )}
A
c
o
s
(
λ
π
)
+
B
s
i
n
(
λ
π
)
=
A
c
o
s
(
λ
−
π
)
+
B
s
i
n
(
λ
−
π
)
{\displaystyle \displaystyle Acos(\lambda \pi )+Bsin(\lambda \pi )=Acos(\lambda -\pi )+Bsin(\lambda -\pi )}
B
s
i
n
(
λ
π
)
=
−
B
s
i
n
(
λ
π
)
{\displaystyle \displaystyle Bsin(\lambda \pi )=-Bsin(\lambda \pi )}
2
B
s
i
n
(
λ
π
)
=
0
{\displaystyle \displaystyle 2Bsin(\lambda \pi )=0}
B
=
0
{\displaystyle \displaystyle B=0}
s
i
n
(
λ
π
)
=
0
{\displaystyle \displaystyle sin(\lambda \pi )=0}
The continuity of the derivative provides a second condition:
u
θ
(
r
,
π
)
=
u
θ
(
r
,
−
π
)
{\displaystyle \displaystyle u_{\theta }(r,\pi )=u_{\theta }(r,-\pi )}
Θ
θ
(
π
)
=
Θ
θ
(
−
π
)
{\displaystyle \displaystyle \Theta _{\theta }(\pi )=\Theta _{\theta }(-\pi )}
−
A
λ
s
i
n
(
λ
π
)
+
λ
B
c
o
s
(
λ
π
)
=
−
A
λ
s
i
n
(
λ
−
π
)
+
B
λ
c
o
s
(
λ
−
π
)
{\displaystyle \displaystyle -A\lambda sin(\lambda \pi )+\lambda Bcos(\lambda \pi )=-A\lambda sin(\lambda -\pi )+B\lambda cos(\lambda -\pi )}
−
A
λ
s
i
n
(
λ
π
)
=
A
λ
s
i
n
(
λ
π
)
{\displaystyle \displaystyle -A\lambda sin(\lambda \pi )=A\lambda sin(\lambda \pi )}
2
A
λ
s
i
n
(
λ
π
)
=
0
{\displaystyle \displaystyle 2A\lambda sin(\lambda \pi )=0}
A
=
0
{\displaystyle \displaystyle A=0}
s
i
n
(
λ
π
)
=
0
{\displaystyle \displaystyle sin(\lambda \pi )=0}
s
i
n
(
λ
π
)
=
0
{\displaystyle \displaystyle sin(\lambda \pi )=0}
s
i
n
(
λ
π
)
=
0
{\displaystyle \displaystyle sin(\lambda \pi )=0}
λ
=
n
{\displaystyle \displaystyle \lambda =n}
s
i
n
(
n
π
)
=
0
,
n
=
1
,
2
,
.
.
.
{\displaystyle \displaystyle sin(n\pi )=0,{\mbox{ }}n=1,2,...}
Example of Potential equation on semi-annulus. [ edit | edit source ] %%(php)
<?php
echo "Hello, World!";
?>
%%
Boundary Value Problems
