Chap
.
12
Partial Differential Equations
(
PDEs
)
Sec
.
12
.
1
Basic Concepts
A PDE has a much greater variety of solutions than an ODE. Indeed, whereas the solution of an ODE of
second order contains 2 arbitrary
constants
, the solution of a PDE of second order generally contains 2
arbitrary
functions
. You determine the latter from intial conditions (from given initial functions, for instance,
given initial displacement and velocity of a mechanical system).
This section introduces basic concepts, lists the most important PDEs to be considered, and shows that
some PDEs can be solved as ODEs.
Problem Set 12
.
1
.
Page 537
1
.
PDE solvable as ODE
.
x
does not occur explicitly. As an ODE it would be
u
uu
U
16
u
±
0 with a
general solution
u
±
A
cos 4
y
U
B
sin 4
y
involving arbitrary constants
A
and
B
. Since you are dealing with
a PDE and look for a solution
u
±
u
u
x
,
y
U
, you can have arbitrary
functions A
±
A
u
x
U
and
B
±
B
u
x
U
, so
that
u
u
x
,
y
U
±
A
u
x
U
cos 4
y
U
B
u
x
U
sin 4
y
.
5
.
PDE solvable as a linear ODE
.
u
u
U
u
±
e
xy
can be solved by (4) in Sec. 1.5 with
x
±
y
and the
present
x
as a parameter, that is,
u
±
e
u
y
U
e
y
e
xy
dy
U
c
u
x
U
±
c
u
x
U
e
u
y
U
e
u
y
e
u
x
U
1
U
y
x
U
1
±
c
u
x
U
e
u
y
U
e
xy
x
U
1
.
11
.
Substitution
u
x
±
v
gives from
u
xy
±
u
x
the PDE
u
y
±
v
solvable as an ODE. Separate variables and
integrate with respect to
y
to get
v
±
c
1
u
x
U
e
y
and integrate, this time with respect to
x
, to get the answer
u
±
U
c
1
u
x
U
e
y
dx
U
c
2
u
y
U
±
e
y
U
c
1
u
x
U
dx
U
c
2
u
y
U
,
which on p. A33 in Appendix 2 of the book is denoted by
c
u
x
U
e
y
U
h
u
y
U
. This is the same, except for
notation, as you can see by noting the dependence of the two summands on the variables
x
and
y
.
27
.
Boundary value problem
. Verify the solution. Observing the chain rule, you obtain by differentiation
u
x
±
2
ax
x
2
U
y
2
and by another differentiation, with the product rule applied to 2
ax
and 1/
u
x
2
U
y
2
U
u
xx
±
2
a
x
2
U
y
2
U
u
1
±
2
ax
±
2
x
u
x
2
U
y
2
U
2
.
(A)
Similarly, with
y
instead of
x
,
u
yy
±
2
a
x
2
U
y
2
u
4
ay
2
u
x
2
U
y
2
U
2
.
(B)
Taking the common denominator
u
x
2
U
y
2
U
2
, you obtain in (A) the numerator
2
a
u
x
2
U
y
2
U
u
4
ax
2
±
u
2
ax
2
U
2
ay
2
and in (B)
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View Full DocumentChap. 12
Partial Differential Equations (PDEs)
119
2
a
u
x
2
u
y
2
U
u
4
ay
2
U
u
2
ay
2
u
2
ax
2
.
Addition of the two expressions on the right gives 0 and completes the verification.
Now determine
a
and
b
in
u
u
x
,
y
U
U
a
ln
u
x
2
u
y
2
U
u
b
from the boundary conditions. For
x
2
u
y
2
U
1 you have ln1
U
0, so that
b
U
110 from the first boundary condition. From this and the
second boundary condition 0
U
a
ln 100
u
b
you obtain
a
ln 100
U
u
110. Hence
a
U
u
110/ln 100, in
agreement with the answer on p. A33.
Sec
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 Spring '12
 M.lee
 Partial Differential Equations, Fourier Series, Partial differential equation

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