格林函数习题解答

10.05.1D Green#2
Find the Green function G = G(x, t) for the equation en that y (0) = 0 and y (1) = 0. d2 y − y = f (x) givdx2
Solution:
3
d2 y Ly = 2 − y = 0 dx y (0) = 0 y (1) = 0
′
y1 = x y2 = 1
As Green function is continuous, so A(t)y2 (t) = B (t)Y1 (t) ⇒ One the other hand,
′ ′ A(t)y2 (t) − B (t)y1 (t) = −
A(t) = B (t)t
1 1 = − = −1 p(t) 1
= 0 (β is a non-zero constant. )
x=0
Solution:
d2 y dy +β = f (x) dx2 dx α(x) d2 y dy + α(x)β = α(x)f (x) dx2 dx dα = βα dx ⇒ ⇒ α(x) = eβx d βx dy (e ) = eβx f (x) dx dx d βx dy (e )=0 Ly = dx dx y (0) = 0 dy dx tion for Ly is y = C1 e−βx + C2 , As the boundary condition is y (0) = 0, y ′ (0) = 0, then G(x, t) = 0, C1 e−βx + C2 , 0 x<t
⇒ 0 − B (t) = −1 B (t) = 1 A(t) = t
1
{ ⇒ G(x, t) =
x, t,
0 x<t t<x 1
10.05.1D Green#1
Find the Green function G = G(x, t) for the equation given that y (0) = 0 and dy dx dy d2 y +β = f (x) dx2 dx
⇒
t < x < +∞
As the Green function G(x, t) should be continuous at point t, then, C1 e−βt + C2 = 0 ⇒ C2 = −C1 e−βt
2
On the other hand, the differential of Green function G′ (x, t) should satisfy G′ (x, t)|t = − 1 1 = − βt p(t) e
⇒ −C1 βe−βt = −e−βt ⇒ C1 = 1 , β G(x, t) = C2 = −C1 e−βt = − 0, 0 e−βt β x<t
⇒
⇒
e−βx − e−βt , t < x < +∞ β −βx e − e−βt , 0 t<x β G(x, t) = 0, x < t < +∞ ∫ +∞ ⇒ y (x) = − G(x, t)eβt f (t)dt
Assign 7, mathematical Method First name: LIANG Last name : ZHANG Oct-25-2012
10.5.1
Find the Green function for this operator and boundary condition: { d2 y y (0) = 0, Ly (x) ≡ 2 , and y ′ (1) = 0 dx Where we assume that the ends of the interval are at x = 0 and x = 1.
0
t<x 0
t<x 1
x<t
4
Ay2 (t)y1 (x), G(x, t) = Ay1 (t)y2 (x),
t<x
⇒ G(x, t) =
sint −( tan1 − cost)sinx,
−sint( sinx − cosx), tant1 sinx −( tant1 − cosx)sint, ⇒ G(x, t) = −sinx( sint − cost), tan1 ∫ 1 y (x) = − G(x, t)f (t)dt
The general solution for Ly = 0 is y = C1 sin x + C2 cos x, According to the boundary condition, we have y1 = sinx y2 = sin x − cos x tan1 cos t sin t ′ ′ ( t) − y 1 y2 = sin t × ( + sin t) − cos t × ( − cos t) = 1 W [y1 , y2 ]|t = y1 (t)y2 tan 1 tan 1 p(t) = 1 ⇒ A=− 1 = −1 W [y1 , y2 ]p(t) 0 x<t 1 0 x<t 1
0
∫ =−
0
x
e−βx − e−βt βt e f (t)dt β
Comment: It is apparent that G(x, t) ̸= G(t, x). As the operator ′ L′ is second order differential, so if the two boundary conditions focus on one side x = a, ∂G then the Green function for a < x < t is definitely determined, and = 0. ∂t For the interval t < x < b, the two boundary conditions also focus at one side, 1 ′ ′ (t) − p( (t) = y1 that is, y2 (t) = y1 (t), y2 t) . We can get the Green function for ∂G t < x < b and ̸= 0. In short, the Green function won’t be symmetrical if ∂t the two boundary conditions focus on one side.
Solution:
First, ∵ ∴ Ly (x) ≡ d2 y dx2 the general solution is y = kx + b y (0) = 0 ⇒ y (1) = 0 ⇒ Let’s Green function to be, { G(x, t) = B (t)y1 , A(t)y2 , t<x 1 0 x<t