s t
c
o
= ⇔
2
.
Jadi
x k
s o
e s
t s
x u
−
= ,
~ .
Jadi L
= ,
t x
u
-1
} ,
~ {
s x
u
= ⇔
, t
x u
L
-1
⎪⎭ ⎪
⎬ ⎫
⎪⎩ ⎪
⎨ ⎧
− x
k s
o
e s
t
o
t t
x u
= ⇔
, L
-1
⎪⎭ ⎪
⎬ ⎫
⎪⎩ ⎪
⎨ ⎧
− x
k s
e s
1
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
= ⇔
kt x
erfc t
t x
u
o
2 ,
.
Jadi ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ =
kt x
erfc t
t x
u
o
2 ,
.
D. Pemrograman Komputer Persamaan Konduksi Panas Dimensi Satu
Berikut ini adalah pemrograman Maple untuk persamaan konduksi panas dimensi satu.
Interval setengah tak terbatas pada kasus parabolik restart;withPDEtools:
p1:=D2yx+yx=fx;
:= p1
= +
D
2
y x y x
f x
p2:=yx=Axexpm[1]x+Bxexpm[2]x;
:= p2
= y x
+ A x e
m 1
x
B x e
m 2
x
s1:=DAxexpm[1]x+DBxexpm[2]x=0;
:= s1
= +
D A x e
m 1
x
D B x e
m 2
x
s2:=m[1]DAxexpm[1]x+m[2]DBxexpm[2] x=fx;
:= s2
= +
m
1
D A x e
m 1
x
m
2
D B x e
m 2
x
f x
sol1:=solve{s1,s2},{DAx,DBx};
:= sol1
⎧ ⎩
⎪⎪ ⎪⎪⎪⎨
⎫ ⎭
⎪⎪ ⎪⎪⎪⎬
, =
D A x f x
e
m 1
x
− + m
2
m
1
= D B x
− f x
e
m 2
x
− + m
2
m
1
sol2:=sol1[1];
:= sol2
= D A x
f x e
m 1
x
− + m
2
m
1
sol3:=sol1[2];
:= sol3
= D B x
− f x
e
m 2
x
− + m
2
m
1
solu1:=dsolve{sol2},Ax;
:= solu1
= A x
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
e
−m 1
x
f x − +
m
2
m
1
x _C1
solu2:=dsolve{sol3},Bx;
:= solu2
= B x
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
− f x e
−m 2
x
− + m
2
m
1
x _C1
solu3:=subs_C1=_C2,solu2;
:= solu3
= B x
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
− f x e
−m 2
x
− + m
2
m
1
x _C2
pers1:=subssolu1,solu3,p2;
:= pers1
= y x
+ ⎛
⎝ ⎜⎜
⎜⎜ ⎜⎜⎜
⎞
⎠ ⎟⎟
⎟⎟ ⎟⎟⎟
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
e
−m 1
x
f x − +
m
2
m
1
x _C1 e
m 1
x
⎛
⎝ ⎜⎜
⎜⎜ ⎜⎜⎜
⎞
⎠ ⎟⎟
⎟⎟ ⎟⎟⎟
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
− f x e
−m 2
x
− + m
2
m
1
x _C2 e
m 2
x
simplifypers1;
y x e
m 1
x
d ⌠
⌡ ⎮
⎮ ⎮
e
−m 1
x
f x x
e
m 1
x
_C1 m
2
e
m 1
x
_C1 m
1
e
m 2
x
d ⌠
⌡ ⎮
⎮ ⎮
f x e
−m 2
x
x −
+ −
⎛ ⎝
⎜⎜ ⎜
=
e
m 2
x
_C2 m
2
e
m 2
x
_C2 m
1
− +
− + m
2
m
1
⎞ ⎠
⎟⎟ ⎟
restart;withinttrans: p:=diffux,s,x2-skux,s=0;
:= p
= −
⎛ ⎝
⎜⎜⎜ ⎞
⎠ ⎟⎟⎟
∂ ∂
2
x
2
u , x s
s u ,
x s k
p1:=dsolvep,ux,s;
:= p1
= u ,
x s +
_F1 s e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
_F2 s e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
p2:=subs_F1s=Ax,_F2s=Bx,p1;
:= p2
= u ,
x s +
A x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
B x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
s1:=DAxexpsqrtskx+DBxexp- sqrtskx=0;
:= s1
= +
D A x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
D B x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
s2:=sqrtskDAxexpsqrtskx- sqrtskDBxexp-sqrtskx=-fxk;
:= s2
= −
s k
D A x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
s k
D B x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
− f x
k
s3:=solve{s1,s2},{DAx,DBx};
:= s3
⎧
⎩ ⎪⎪
⎪⎪ ⎪⎪
⎪⎪⎪ ⎨
⎫
⎭ ⎪⎪
⎪⎪ ⎪⎪
⎪⎪⎪ ⎬
, =
D A x −
1 2
f x s
k e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
k =
D B x 1
2 f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
s k
k
sol1:=s3[1];
:= sol1
= D A x
− 1
2 f x
s k
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
k
sol2:=s3[2];
:= sol2
= D B x
1 2
f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
s k
k
so1:=dsolve{sol1},Ax;
:= so1
= A x
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ −
1 2
f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
s k
k x
_C1
so2:=dsolve{sol2},Bx;
:= so2
= B x
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ 1
2 f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
s k
k x
_C1
so3:=subs_C1=_C2,so2;
:= so3
= B x
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ 1
2 f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
s k
k x
_C2
pers1:=subsso1,so3,p2;
pers1 u ,
x s =
:=
+ ⎛
⎝ ⎜⎜
⎜⎜ ⎜⎜
⎜⎜ ⎜⎜
⎞
⎠ ⎟⎟
⎟⎟ ⎟⎟
⎟⎟ ⎟⎟
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ −
1 2
f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
s k
k x
_C1 e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
⎛
⎝ ⎜⎜
⎜⎜ ⎜⎜
⎜⎜ ⎜⎜
⎞
⎠ ⎟⎟
⎟⎟ ⎟⎟
⎟⎟ ⎟⎟
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ 1
2 f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
s k
k x
_C2 e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
pers2:=simplifypers1;
pers2 u ,
x s 1
2 e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
d ⌠
⌡ ⎮
⎮ ⎮
⎮ f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
x 2 e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
_C1 s
k k
− +
⎛ ⎝
⎜⎜ ⎜⎜⎜
= :=
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
d ⌠
⌡ ⎮
⎮ ⎮
⎮ f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
x 2 e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
_C2 s
k k
+ +
s k
k ⎞
⎠ ⎟⎟
⎟⎟⎟ ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟
per1:=subs_C1=0,pers2;
per1 u ,
x s =
:=
1 2
− +
+ e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
d ⌠
⌡ ⎮
⎮ ⎮
⎮ f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
d ⌠
⌡ ⎮
⎮ ⎮
⎮ f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
x 2 e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
_C2 s
k k
s k
k
invlaplaceper1,x,s;
invlaplace , ,
u , x s
x s 1
2 ⎛
⎝ ⎜⎜
⎜⎜⎜ ⎞
⎠ ⎟⎟
⎟⎟⎟ invlaplace
, , e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
d ⌠
⌡ ⎮
⎮ ⎮
⎮ f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
x x s −
⎛ ⎝
⎜⎜ ⎜⎜⎜
= ⎛
⎝ ⎜⎜
⎜⎜⎜ ⎞
⎠ ⎟⎟
⎟⎟⎟ invlaplace
, , e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
d ⌠
⌡ ⎮
⎮ ⎮
⎮ f x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
x x s 2 _C2
s k
k ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ Dirac
− s
s k k
+ +
⎞ ⎠
⎟⎟ ⎟⎟⎟
s k
k ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟
Interval terbatas pada kasus parabolik restart;withPDEtools:
D2yx+yx=fx;
= +
D
2
y x y x
f x
p:=Axcoshx+Bxsinhx;
:= p
+ A x
cosh x B x
sinh x
p1:=DAxcoshx+DBxsinhx=0;
:= p1
= +
D A x cosh x
D B x sinh x
p2:=DAxsinhx+DBxcoshx=fx;
:= p2
= +
D A x sinh x
D B x cosh x
f x
p3:=solve{p1,p2},{DAx,DBx};
:= p3
⎧ ⎩
⎪⎪ ⎪⎪⎨
⎫ ⎭
⎪⎪ ⎪⎪⎬
, =
D A x −
1 2
f x −
e
x 2
1 e
x
= D B x
1 2
+ e
x 2
1 f x
e
x
s1:=p3[1];
:= s1
= D A x
− 1
2 f x
− e
x 2
1 e
x
s2:=p3[2];
:= s2
= D B x
1 2
+ e
x 2
1 f x
e
x
sol1:=dsolve{s1},Ax;
:= sol1
= A x
+ d
⌠ ⌡
⎮ ⎮
⎮ ⎮
1 2
f x − +
e
x
e
−x
x _C1
sol2:=dsolve{s2},Bx;
:= sol2
= B x
+ d
⌠ ⌡
⎮ ⎮
⎮ ⎮
1 2
f x +
e
x
e
−x
x _C1
sol3:=subs_C1=_C2,sol2;
:= sol3
= B x
+ d
⌠ ⌡
⎮ ⎮
⎮ ⎮
1 2
f x +
e
x
e
−x
x _C2
pers1:=subssol1,sol3,p;
pers1 := +
⎛ ⎝
⎜⎜ ⎜⎜
⎞ ⎠
⎟⎟ ⎟⎟
+ d
⌠ ⌡
⎮ ⎮
⎮ ⎮
1 2
f x − +
e
x
e
−x
x _C1
cosh x ⎛
⎝ ⎜⎜
⎜⎜ ⎞
⎠ ⎟⎟
⎟⎟ +
d ⌠
⌡ ⎮
⎮ ⎮
⎮ 1
2 f x
+ e
x
e
−x
x _C2
sinh x
pers2:=simplifypers1;
pers2 1
2 cosh x
d ⌠
⌡ ⎮
⎮ f x
− + e
x
e
−x
x cosh x _C1
1 2
sinh x d
⌠ ⌡
⎮ ⎮
f x +
e
x
e
−x
x +
+ :=
sinh x _C2 +
restart;withPDEtools:withinttrans: p:=diffux,s,x2-skux,s=-fxk;
:= p
= −
⎛ ⎝
⎜⎜⎜ ⎞
⎠ ⎟⎟⎟
∂ ∂
2
x
2
u , x s
s u ,
x s k
− f x
k
p1:=dsolvep,ux,s;
p1 u ,
x s d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
− 1
2 e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
f x s k
x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
d ⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
1 2
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
f x s k
x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
+ =
:=
_F1 s e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
_F2 s e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
+ +
pp1:=subs_F1=Ax,_F2=Bx,p1;
pp1 u ,
x s d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
− 1
2 e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
f x s k
x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
d ⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
1 2
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
f x s k
x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
+ =
:=
A x s e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
B x s e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
+ +
p2:=DAxcoshsqrtskx+DBxsinhsqrt skx=0;
:= p2
= +
D A x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ cosh
s k
x D B x
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
sinh s
k x
p3:=sqrtskDAxsinhsqrtskx+sqrtsk DBxcoshsqrtskx=-fxk;
:= p3
= +
s k
D A x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ sinh
s k
x s
k D B x
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
cosh s
k x
− f x
k
s1:=solve{p2,p3},{DAx,DBx};
:= s1
⎧
⎩ ⎪⎪
⎪⎪ ⎪⎪
⎪⎪ ⎪⎪
⎨ ⎫
⎭ ⎪⎪
⎪⎪ ⎪⎪
⎪⎪ ⎪⎪
⎬ ,
= D A x
1 2
f x ⎛
⎝ ⎜⎜
⎜ ⎞
⎠ ⎟⎟
⎟ −
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x 2
1 s
k k e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
= D B x
− 1
2 f x
⎛ ⎝
⎜⎜ ⎜
⎞ ⎠
⎟⎟ ⎟
+ ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x 2
1 s
k k e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
s2:=s1[1];
:= s2
= D A x
1 2
f x ⎛
⎝ ⎜⎜
⎜ ⎞
⎠ ⎟⎟
⎟ −
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x 2
1 s
k k e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
s3:=s1[2];
:= s3
= D B x
− 1
2 f x
⎛ ⎝
⎜⎜ ⎜
⎞ ⎠
⎟⎟ ⎟
+ ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x 2
1 s
k k e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
sol1:=dsolves2,Ax;
:= sol1
= A x
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ −
1 2
f x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ −
+ e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
s k
k x
_C1
sol2:=dsolves3,Bx;
:= sol2
= B x
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ −
1 2
f x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ +
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
s k
k x
_C1
sol3:=subs_C1=_C2,sol2;
:= sol3
= B x
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ −
1 2
f x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ +
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
s k
k x
_C2
solu1:=subssol1,sol3,pp1;
solu1 u ,
x s d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
− 1
2 e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
f x s k
x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
d ⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
1 2
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
f x s k
x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
+ =
:= ⎛
⎝ ⎜⎜
⎜⎜ ⎜⎜
⎜⎜ ⎜⎜
⎞
⎠ ⎟⎟
⎟⎟ ⎟⎟
⎟⎟ ⎟⎟
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ −
1 2
f x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ −
+ e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
s k
k x
_C1 s e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
+
⎛
⎝ ⎜⎜
⎜⎜ ⎜⎜
⎜⎜ ⎜⎜
⎞
⎠ ⎟⎟
⎟⎟ ⎟⎟
⎟⎟ ⎟⎟
+ d
⌠
⌡ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ ⎮
⎮ −
1 2
f x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ +
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
s k
k x
_C2 s e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
+
solu2:=simplifysolu1;
solu2 u ,
x s 1
2 d
⌠ ⌡
⎮ ⎮
⎮ ⎮
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
f x x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
s s k s
k s −
⎛ ⎝
⎜⎜ ⎜⎜⎜
= :=
d ⌠
⌡ ⎮
⎮ ⎮
⎮ e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
f x x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
s s k s
k s +
s k e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
⎛ ⎝
⎜⎜ ⎜⎜⎜
⎞ ⎠
⎟⎟ ⎟⎟⎟
d ⌠
⌡ ⎮
⎮ ⎮
⎮ f x
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− +
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
x s
−
2 s k e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
_C1 s s s
k s k s
+
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
s k ⎛
⎝ ⎜⎜
⎜⎜⎜ ⎞
⎠ ⎟⎟
⎟⎟⎟ d
⌠ ⌡
⎮ ⎮
⎮ ⎮
f x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ +
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
x s
−
2 e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
s k _C2 s
s s k s
k s +
s k s s
k s k s
⎞ ⎠
⎟⎟ ⎟⎟⎟
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
solu3:=invlaplacesolu2,x,s;
solu3 invlaplace
, , u ,
x s x s
1 2
⎛ ⎝
⎜⎜ ⎜⎜⎜
= :=
s s k s
k s ⎛
⎝ ⎜⎜
⎜⎜⎜ ⎞
⎠ ⎟⎟
⎟⎟⎟ invlaplace
, , d
⌠ ⌡
⎮ ⎮
⎮ ⎮
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
f x x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
x s −
s s k s
k s ⎛
⎝ ⎜⎜
⎜⎜⎜ ⎞
⎠ ⎟⎟
⎟⎟⎟ invlaplace
, , d
⌠ ⌡
⎮ ⎮
⎮ ⎮
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
f x x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
x s +
s k ⎛
⎝ ⎜⎜
⎜⎜⎜ ⎞
⎠ ⎟⎟
⎟⎟⎟ d
⌠ ⌡
⎮ ⎮
⎮ ⎮
f x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ −
+ e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
x s
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
Dirac +
s s k
k −
2 s k _C1 s
s s k s
k s ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ Dirac
+ s
s k k
+
s k ⎛
⎝ ⎜⎜
⎜⎜⎜ ⎞
⎠ ⎟⎟
⎟⎟⎟ d
⌠ ⌡
⎮ ⎮
⎮ ⎮
f x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ +
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
x s
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
Dirac − +
s s k
k −
2 s k _C2 s
s s k s
k s ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ Dirac
− + s
s k k
+ s k
s s k s
k s ⎞
⎠ ⎟⎟
⎟⎟⎟ ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟
simplifysolu3;
invlaplace , ,
u , x s
x s 1
2 s s
k s k s
⎛ ⎝
⎜⎜ ⎜⎜⎜
⎞ ⎠
⎟⎟ ⎟⎟⎟
invlaplace , ,
d ⌠
⌡ ⎮
⎮ ⎮
⎮ e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
f x x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
x s ⎛
⎝ ⎜⎜
⎜⎜⎜ −
=
s s k s
k s ⎛
⎝ ⎜⎜
⎜⎜⎜ ⎞
⎠ ⎟⎟
⎟⎟⎟ invlaplace
, , d
⌠ ⌡
⎮ ⎮
⎮ ⎮
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k x k
f x x e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s k x
k
x s −
s k ⎛
⎝ ⎜⎜
⎜⎜⎜ ⎞
⎠ ⎟⎟
⎟⎟⎟ d
⌠ ⌡
⎮ ⎮
⎮ ⎮
f x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ −
+ e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
x s
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
Dirac +
s k s k
k +
2 s k _C1 s
s s k s
k s ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ Dirac
+ s k
s k k
−
s k ⎛
⎝ ⎜⎜
⎜⎜⎜ ⎞
⎠ ⎟⎟
⎟⎟⎟ d
⌠ ⌡
⎮ ⎮
⎮ ⎮
f x ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ +
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
s k
x
e
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
− s
k x
x s
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
Dirac − +
s k s k
k +
2 s k _C2 s
s s k s
k s ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟ Dirac
− + s k
s k k
− s k
s s k s
k s ⎞
⎠ ⎟⎟
⎟⎟⎟ ⎛
⎝ ⎜⎜
⎞ ⎠
⎟⎟
Contoh 2.
Gambarlah u pada contoh 1 sebagai sebuah permukaan dalam ruang yang mempuyai temperatur
dan persamaan konduksi panasnya
C t
o o
27 =
2 2
2 x
u t
u ∂
∂ =
∂ ∂
untuk 2
≤ ≤ t
dengan panjang L
x ≤
≤ ,
2 ,
= L
. Penyelesaian:
withplots:
Warning, the name changecoords has been redefined
u1:=ux,t=gammaerfx2sqrtkt;
:= u1
= u ,
x t γ
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
erf 1
2 x
k t
u2:=subsgamma=27,k=2,u1;
:= u2
= u ,
x t 27
⎛ ⎝
⎜⎜⎜ ⎞
⎠ ⎟⎟⎟
erf 1
4 x 2
t
plot3drhsu2,x=0..0.4,t=0..2,style=hidden,orienta tion=[25,30],title=Gambar 2;
Contoh 3.
Gambarlah u pada contoh 1 sebagai sebuah permukaan dalam ruang yang mempuyai temperatur
dan persamaan konduksi panasnya
C t
o o
35 =
2 2
8 x
u t
u ∂
∂ =
∂ ∂
untuk
8 ,
≤ ≤ t
dengan panjang L
x ≤
≤ ,
1 =
L
. Penyelesaian:
withplots:
Warning, the name changecoords has been redefined
p1:=ux,t=gammaerfx2sqrtkt;
:= p1
= u ,
x t γ
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
erf 1
2 x
k t
p2:=subsgamma=35,k=8,p1;
:= p2
= u ,
x t 35
⎛ ⎝
⎜⎜⎜ ⎞
⎠ ⎟⎟⎟
erf 1
16 x 8
t
plot3drhsp2,x=0..1,t=0..0.8,style=hidden,orienta tion=[35,40],title=Gambar 3;
Contoh 4.
Gambarlah u pada contoh 1 sebagai sebuah permukaan dalam ruang yang mempuyai temperatur
dan pada persamaan konduksi
panas
C t
o o
37 =
C t
o o
38 =
2 2
8 x
u t
u ∂
∂ =
∂ ∂
untuk
8 ,
≤ ≤ t
dengan panjang L
x ≤
≤ ,
1 =
L
. Penyelesaian:
withplots:
Warning, the name changecoords has been redefined
p:=ux,t=gammaerfx2sqrtkt;
:= p
= u ,
x t γ
⎛ ⎝
⎜⎜ ⎞
⎠ ⎟⎟
erf 1
2 x
k t
p1:=subsgamma=37,k=8,p;
:= p1
= u ,
x t 37
⎛ ⎝
⎜⎜⎜ ⎞
⎠ ⎟⎟⎟
erf 1
16 x 8
t
p1:=subsgamma=38,k=8,p;
:= p1
= u ,
x t 38
⎛ ⎝
⎜⎜⎜ ⎞
⎠ ⎟⎟⎟
erf 1
16 x 8
t
plot3d{rhsp1,rhsp2},x=0..1,t=0..0.8,style=hidd en,orientation=[35,40],axes=FRAMED,title=Gambar
4;
BAB V PENUTUP
A. Simpulan
Pemodelan persamaan konduksi panas dimensi satu adalah , dimana k adalah konstan. Bentuk transformasi Laplace dari masalah
nilai batas pada persamaan konduksi panas dimensi satu adalah.
tt t
ku u
=
1. Interval setengah tak terbatas pada kasus parabolik
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
+ ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ −
=
∫ ∫
− −
−
dx e
x f
e k
s k
dx e
x f
e k
s k
e c
s x
u
x k
s x
k s
x k
s x
k s
x k
s
2 1
2 1
, ~
2
dengan syarat batas:
~ ,
~ ,
~ s
g dx
s u
d s
u =
+
β α
. 2.
Interval terbatas pada kasus parabolik
x k
s dx
e e
x f
k s
k c
s x
u
x k
s x
k s
cosh 2
1 ,
~
1
⎟ ⎟
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎜ ⎜
⎝ ⎛
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
+ −
− =
∫
−
x k
s dx
e e
x f
k s
k c
x k
s x
k s
sinh 2
1
2
⎟ ⎟
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎜ ⎜
⎝ ⎛
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
+ −
+
∫
−
,
dengan syarat batas:
~ ,
~ ,
~
1
s g
dx s
u d
s u
= +
β α
, dan
53
~ ,
~ ,
~
2
s g
dx s
l u
d s
l u
= +
β α
. Sedangkan penyelesaian bentuk transformasi Laplace dari masalah
nilai batas pada persamaan konduksi panas dimensi satu adalah. 1.
Interval setengah tak terbatas pada kasus parabolik
= ,
t x
u
L
-1
⎪ ⎪
⎩ ⎪⎪
⎨ ⎧
+ ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ −
∫
− −
dx e
x f
e k
s k
e c
x k
s x
k s
x k
s
2 1
2
⎪ ⎪
⎭ ⎪⎪
⎬ ⎫
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
∫
−
dx e
x f
e k
s k
x k
s x
k s
2 1
2. Interval terbatas pada kasus parabolik
= ,
t x
u
L
-1
⎪ ⎪
⎩ ⎪⎪
⎨ ⎧
⎟ ⎟
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎜ ⎜
⎝ ⎛
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
+ −
−
∫
−
x k
s dx
e e
x f
k s
k c
x k
s x
k s
cosh 2
1
1
⎪ ⎪
⎭ ⎪⎪
⎬ ⎫
⎟ ⎟
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎜ ⎜
⎝ ⎛
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
+ −
+
∫
−
x k
s dx
e e
x f
k s
k c
x k
s x
k s
sinh 2
1
2
B. Saran
Pada penulisan skripsi ini, permasalahan hanya dibatasi penyelesaian masalah nilai batas pada persamaan diferensial parsial linear
orde dua dengan kasus parabolik pada persamaan konduksi panas dimensi satu