1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

0.5 Green’s Functions 45

Second, the boundary condition atx=rbecomes

βu(r)+β′u′(r)=c 1

(

βu 1 (r)+β′u′(r)

)

+

∫r

l

[

u 1 (z)

(

βu 2 (r)+β′u′ 2 (r)

)

−u 2 (z)

(

βu 1 (r)+β′u′ 1 (r)

)]f(z)

W(z)dz=^0. (11)

Now, the boundary condition (6) onu 2 atx=reliminates one term of the
integrand, leaving

c 1

(

βu 1 (r)+β′u′ 1 (r)

)

−

∫r

l

u 2 (z)

(

βu 1 (r)+β′u′ 1 (r)

)f(z)

W(z)

dz= 0. (12)

The common factor ofβu 1 (r)+β′u′ 1 (r)can be canceled from both terms, and
we then find

c 1 =

∫r

l

u 2 (z)Wf((zz))dz. (13)

Now we have foundc 1 andc 2 so thatu(x)in Eq. (7) satisfies both boundary
conditions. If we use the values ofc 1 andc 2 as found, we have

u(x)=u 1 (x)

∫r

l

u 2 (z)

f(z)

W(z)dz

+

∫x

l

(

u 1 (z)u 2 (x)−u 2 (z)u 1 (x)

)f(z)

W(z)dz. (14)

The solution becomes more compact if we break the interval of integration at
xin the first integral, making it
∫r

l

u 2 (z)

f(z)

W(z)dz=

∫x

l

u 2 (z)

f(z)

W(z)dz+

∫r

x

u 2 (z)

f(z)

W(z)dz. (15)

When the integrals on the rangeltoxare combined, there is some cancella-
tion, and our solution becomes

u(x)=

∫x

l

u 1 (z)u 2 (x)f(z)

W(z)

dz+

∫r

x

u 1 (x)u 2 (z)f(z)

W(z)

dz. (16)

Finally, these two integrals can be combined into one. We first define the
Green’s functionfor the problem (1), (2), (3) as

G(x,z)=







u 1 (z)u 2 (x)

W(z) , l<z≤x,

u 1 (x)u 2 (z)

W(z)

, x≤z<r.

(17)