第十三章 线性偏微分方程的通解

u0 = u0 (x, y ) 1. f (x, y ) = eax+by
L(a, b) = 0 1 1 eax+by = eax+by . L(Dx , Dy ) L(a, b)
L(a, b) = 0 L(Dx , Dy ) = bDx − aDy (bDx − aDy )u = eax+by . 13.2 Lagrange dy du dx = = ax+by , b −a e a dx + b dy = 0, adu + eax+by dy = 0.
u0 = =
13.3
11
u = x2 (y + 3x) + xφ(y + 3x) + ψ (y + 3x).
2
f (x, y )
= (Dx + Dy )(Dx − 2Dy + 2)u = 0.
u = φ(x − y ) + e−2x ψ (y + 2x). (Dx − αDy − β )2 z = 0 z = xeβx φ(y + αx) + eβx ψ (y + αx).
13.3
7ห้องสมุดไป่ตู้
13.3
= +
L(Dx , Dy )u = f (x, y ) 1 f (x, y ), L(Dx , Dy )
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13.1
2
13.1
L[u] = f,
u L f f ≡0
13.1
L ∂2u ∂2 2 2 − a ∇ u = f L ≡ 2 − a2 ∇2 ∂t2 ∂t ∂u ∂ − κ∇2 u = f L≡ − κ∇2 ∂t ∂t ∇2 u = f L ≡ ∇2
Poisson
u 1 u1 u2
2 Dx , Dx D y 2 Dy
2 2 L(Dx , Dy ) = G(Dx , Dx Dy , Dy ),
1 sin(ax + by ) 2 , D D , D2 ) G(Dx x y y 1 = sin(ax + by ), 2 G(−a , −ab, −b2 ) 1 cos(ax + by ) 2 , D D , D2 ) G(Dx x y y 1 = cos(ax + by ). G(−a2 , −ab, −b2 ) 3. f (x, y ) = eax+by g (x, y ) 1 eax+by g (x, y ) L(Dx , Dy ) = eax+by
= 0, A0 , A1 , · · · , An , B0 , · · · , M, N, P 1. L(Dx , Dy ) Dx , Dy
n n−1 n−2 2 n A0 Dx + A1 Dx Dy + A2 Dx Dy + · · · + An Dy u = 0.
L(Dx , Dy )
n
L(Dx , Dy ) = A0 (Dx − α1 Dy )(Dx − α2 Dy ) · · · (Dx − αn Dy ), α1 , α2 , · · · , αn u = φ(y + αx),
u = xn−1 φ1 (y + αx) + xn−2 φ2 (y + αx) + · · · +xφn−1 (y + αx) + φn (y + αx).
2
2 2 (Dx − 2Dx Dy + Dy )u = 0
u = xφ(x + y ) + ψ (x + y ).
2. L(Dx , Dy )
Dx , D y (Dx − αDy − β )z = 0. ( )
= 12
= x4 + 2x3 y,
1 1 x = x2 Dx 2 1 1 x = x3 2 Dx 6 1 1 4 x= x 3 Dx 24 2
∵ ∵ ∵
d x2 =x , dx 2 d2 x3 =x , dx2 6 d3 x4 =x . dx3 24
u = xφ(x + y ) + ψ (x + y ) + x4 + 2x3 y. 1/L(Dx , Dy ) 1 1 Dx = 2 1− 2 (Dx − Dy ) Dy Dy
f (x, y, z ) = 0 ∂f ∂f ∂f dx + dy + dz = 0. ∂x ∂y ∂z Dx z = − ( ) ∂f /∂x , ∂f /∂z Dy z = − ∂f /∂y . ∂f /∂z ( ) ( )
∂f ∂f ∂f −α + βz = 0. ∂x ∂y ∂z dx dy dz = = . 1 −α βz Lagrange y + αx = C, βx = ln z − ln C .
13.2
4
13.2
∂nu ∂nu ∂nu + A + · · · + A 1 n ∂xn ∂xn−1 ∂y ∂y n n−1 ∂ u ∂u ∂u +B0 n−1 + · · · + M +N + P u = 0, ∂x ∂x ∂y
A0
n n−1 n L(Dx , Dy )u ≡ A0 Dx + A1 Dx D y + · · · + An D y n−1 + B0 Dx + · · · + M Dx + N Dy + P u
∂2u ∂2u − a2 2 = 0 2 ∂x ∂y u = φ(y + αx)
u = φ1 (y + ax) + φ2 (y − ax).
13.2
α (Dx − αDy )2 u = 0,
5
u = xφ1 (y + αx) + φ2 (y + αx). α n (Dx − αDy )n u = 0,
L(Dx , Dy ) eax+by
= eax+by L(Dx + a, Dy + b) = eax+by g (x, y ).
13.3
9
4. 4
f (x, y ) = xm y n
2 (Dx
1/L(Dx , Dy ) − 2Dx Dy +
2 Dy )u
Dx , D y
= 12xy
u0 = = =
L[u] = f L[u] = 0 L[u1 ] = 0,
u
L[u] = f
L[u2 ] = 0,
c1 u1 + c2 u2 L[c1 u1 + c2 u2 ] = 0, c1 c2 2 u1 u2 L[u] = f L[u1 ] = f, u1 − u2 L[u1 − u2 ] = 0. L[u2 ] = f,
k Dx u = αk φ(k) (y + αx), k Dy u = φ(k) (y + αx), r s Dx Dy u = αr φ(r+s) (y + αx),
n
A0 αn + A1 αn−1 + · · · + An φ(n) (y + αx) = 0. ( auxiliary equation) A0 αn + A1 αn−1 + · · · + An = 0 α1 , α2 , · · · , αn u = φ1 (y + α1 x) + φ2 (y + α2 x) + · · · + φn (y + αn x), φi , i = 1, 2, · · · , n 1 ( ) (n a α 2 − a2 = 0 α = ±a )
−2
=
1 Dx 1−2 + ··· . 2 Dy Dy
u0 =
12 xy = 2xy 3 + y 4 . (Dx − Dy )2
x4 + 2x3 y − 2xy 3 − y 4 = (x − y )(x + y )3 = 2x(x + y )3 − (x + y )4
1
f (ax + by )
L(Dx , Dy ) Dx , Dy
1 g (x, y ). L(Dx + a, Dy + b)
Dx eax+by g (x, y ) = eax+by (Dx + a)g (x, y ), Dy eax+by g (x, y ) = eax+by (Dy + b)g (x, y ), L(Dx , Dy )eax+by g (x, y ) = eax+by L(Dx + a, Dy + b)g (x, y ). 1 g (x, y ) L(Dx + a, Dy + b) 1 g (x, y ) L(Dx + a, Dy + b)
( ) ( )
z = C eβx = eβx φ(y + αx). L(Dx , Dy ) Dx 3 Dy Dx ) ∂2u ∂2u ∂u ∂2u ∂u − −2 2 +2 +2 =0 2 ∂x ∂x∂y ∂y ∂x ∂y Dy L(Dx , Dy ) n (
13.2
6
2 2 (Dx − Dx Dy − 2Dy + 2Dx + 2Dy )u
n n−1 n L(Dx , Dy )u ≡ A0 Dx + A1 Dx D y + · · · + An D y n−1 + B0 Dx + · · · + M Dx + N Dy + P u
= f (x, y ), Dx ≡ ∂/∂x Dy ≡ ∂/∂y A0 , A1 , · · · , An , B0 , · · · , M, N, P x, y
5
v = (x + y )3 + φ(x + iy ) + ψ (x − iy ). L(a, b) = 0
(Dx − αDy )u = xr ψ (y + αx), Lagrange dx dy du = = r . 1 −α x ψ (y + αx) y + αx = c u= 1 1 xr+1 ψ (c) = xr+1 ψ (y + αx). r+1 r+1
(n)
r Dx g (ax + by ) = ar g (r) (ax + by ), s Dy g (ax + by ) = bs g (s) (ax + by ).
13.3
10
L(Dx , Dy )g (ax + by ) = L(a, b)g (n) (ax + by ). L(a, b) = 0 1 1 g (n) (ax + by ) = g (ax + by ). L(Dx , Dy ) L(a, b) ∂2v ∂2v + 2 = 12(x + y ) 2 ∂x ∂y 1 v0 = 12 12 (x + y )3 = (x + y )3 . (x + y ) = 2 2 + D2 Dx 1 + 12 · 3! y
3
u1
u2 L[u1 ] = f1 , L[u2 ] = f2 ,
13.1
c1 u1 + c2 u2 L[c1 u1 + c2 u2 ] = c1 f1 + c2 f2 .
3
A0
∂nu ∂nu ∂nu + A + · · · + A 1 n ∂xn ∂xn−1 ∂y ∂y n n−1 ∂ u ∂u ∂u +B0 n−1 + · · · + M +N + P u = f (x, y ), ∂x ∂x ∂y
1 1 xr ψ (y + αx) = xr+1 ψ (y + αx). Dx − αDy r+1
r! 1 xr ψ (y + αx) = xr+k ψ (y + αx). (Dx − αDy )k (r + k)! 6
2 2 )u = 6x + 2y − 6Dx Dy + 9Dy (Dx
(Dx − 3Dy )2 u = 6x + 2y. 3 xφ(y + 3x) + ψ (y + 3x) 1 (6x + 2y ) (Dx − 3Dy )2 2 (3x + y ) = x2 (y + 3x). (Dx − 3Dy )2
12 12 xy xy = 2 − 2D D + D 2 Dx ( D − Dy )2 x y x y 12 2 Dx 1− Dy Dx
−2
xy =
Dy 12 1+2 + · · · xy 2 Dx Dx
2 2 1 12 xy + x = 12 y 2 x + 3 x 2 Dx Dx Dx Dx 1 3 1 x y + x4 6 12
ax + by = c. adu + ec dy = 0.
1 1 u = − y ec = − y eax+by , a a
1 1 eax+by = − y eax+by . bDx − aDy a
13.3
( x c c)
8
2.
f (x, y ) = ei(ax+by) 1 1 ei(ax+by) = ei(ax+by) . L(Dx , Dy ) F (ia, ib) a b L(Dx , Dy ) 1 1 sin(ax + by ) = Im ei(ax+by) , L(Dx , Dy ) L(ia, ib) 1 1 cos(ax + by ) = Re ei(ax+by) . L(Dx , Dy ) L(ia, ib) L(Dx , Dy )