The adjoint operator of the operator L(u) = xE0, 1] dx2 with boundary condition: u(1)= u(0), and u'(1)=2 u'(0), is L(v)= - d'v with boundary conditions dx? v(1)=2 v(0), and v(0)= v'(1). Ob. L(v)= with boundary conditions
The adjoint operator of the operator L(u) = xE0, 1] dx2 with boundary condition: u(1)= u(0), and u'(1)=2 u'(0), is L(v)= - d'v with boundary conditions dx? v(1)=2 v(0), and v(0)= v'(1). Ob. L(v)= with boundary conditions
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![du
xE0, 1]
The adjoint operator of the operator L(u)
%3D
dx
with boundary conditions u(1)= u(0), and u'(1)=2 u'(0),
is
O a
d'v
L(v)=
dx?
with boundary conditions
v(1)=2 v(0), and v (0)= v'(1).
d?v
with boundary conditions
Ob.
L(v)=
dx2
v(0) = 2 v(1), and v (0)= v'(1).
L(v) = -
d'v
with boundary conditions
dx
2 v(1) = v(0), and v (0) = v'(1).
L+(v) =
with boundary conditions
v(1) = v(0), and 2 v'(0)= v'(1).
%3D
dfv
L+(v) =
dx
with boundary conditions
V(1) = v(0), and v (0)=2 v' (1).
L(V) = -
dx
with boundary conditions
v(1) = v(0), and v (0) = 2 v'(1).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b22b5bd-5b6c-46b5-91f9-e0569ad4ab63%2Fa7f506a8-df04-456f-8408-e56076420720%2Fr21idzr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:du
xE0, 1]
The adjoint operator of the operator L(u)
%3D
dx
with boundary conditions u(1)= u(0), and u'(1)=2 u'(0),
is
O a
d'v
L(v)=
dx?
with boundary conditions
v(1)=2 v(0), and v (0)= v'(1).
d?v
with boundary conditions
Ob.
L(v)=
dx2
v(0) = 2 v(1), and v (0)= v'(1).
L(v) = -
d'v
with boundary conditions
dx
2 v(1) = v(0), and v (0) = v'(1).
L+(v) =
with boundary conditions
v(1) = v(0), and 2 v'(0)= v'(1).
%3D
dfv
L+(v) =
dx
with boundary conditions
V(1) = v(0), and v (0)=2 v' (1).
L(V) = -
dx
with boundary conditions
v(1) = v(0), and v (0) = 2 v'(1).
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