solve 1?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
[Second Order Equations] How do you solve 1? The second picture is for context
Transcribed Image Text:1.
2.
find u (x, t) for 0 < x < 1 and t> 0 which solves
UtUxx = 0,
u (x,0) = 1,
ux (1,t) = t,
find u (x, t) for 0 < x < 1 and t > 0 which solves
Ut - Uxx = et,
u (x,0) = 1,
ux (0, t) = 0.
u (1, t) = u (0,t) = 0.
![5.6 INHOMOGENEOUS BOUNDARY CONDITIONS
In this section we consider problems with sources given at the boundary. We
shall see that naive use of the separation of variables technique will not work.
Let's begin with the diffusion equation with sources at both endpoints.
u₁ = kuxx
u(0,t)=h(t)
0<x<l, t>0
u(l,t) = j(t)
u (x, 0) = 0.
A separated solution u = X(x)T(t) just will not fit the boundary conditions.
So we try a slightly different approach.
EXPANSION METHOD
We already know that for the corresponding homogeneous problem the correct
expansion is the Fourier sine series. For each t, we certainly can expand
u(x,t) = Σun(t) sin -
n=1
2
-76-
un(t) =
0=ut - kuxx =
for some coefficients un(t), because the completeness theorems guarantee
that any function in (0, 1) can be so expanded. The coefficients are necessarily
given by
nTX
1
Σ[
илх
u(x, t) sin dx.
1
(3)
You may object that each term in the series vanishes at both endpoints and
thereby violates the boundary conditions. The answer is that we simply do not
insist that the series converge at the endpoints but only inside the interval. In
fact, we are exactly in the situation of Theorems 3 and 4 but not of Theorem
2 of Section 5.4.
Now differentiating the series (2) term by term, we get
dun
dt
(1)
+kun (1)
(7) ²] sin
(2)
плх
So the PDE seems to require that dun/dt + kλnun = 0, so that un(t) =
Anet. There is no way for this to fit the boundary conditions. Our method
fails! What's the moral? It is that you can't differentiate term by term. See
Example 3 in Section 5.4 for the dangers of differentiation.
Let's start over again but avoid direct differentiation of the Fourier series.
The expansion (2) with the coefficients (3) must be valid, by the completeness
theorem 5.4.3, say, provided that u(x, t) is a continuous function. Clearly, the
initial condition requires that u,,(0) = 0. If the derivatives of u(x, t) are also
continuous, let's expand them, too. Thus
with
V₁ (1) =
du
at
with the coefficients
2
a²u
əx²
dx2 =
Wn(t)=
u(x, t) sin
n=1
vn (1) - kwn (1) =
V₁ (1) sin
ди ηπε
sin
at T
The last equality is valid since we can differentiate under an integral sign if
the new integrand is continuous (see Section A.3). We also expand
илх
1
n=1
2 a²u
=75%
əx²
By Green's second identity (5.3.3) the last expression equals
2
+ ²7 (ux
ηπ.Χ
1
Wn(1) sin
dx =
dx +
sin
плх
1
dun
dt
NAX
=76
So from (5) and (8) we deduce that u, (1) satisfies
е
dx.
7(77)²(x,1
Here come the boundary conditions. The sine factor vanishes at both ends.
The last term will involve the boundary conditions. Thus
w₁(t)=-λun(t)-2nal-2(-1)" j(t) + 2xl-²h(t),
where λ = (n/1). Now by (5) and (7) the PDE requires
η π.χ
ux sin
пл
nлx
Tucos T
(u₁-kux) sindx =
-fo=o.
0=0.
(4)
dun
= k{-λnun(1) - 2nлl-²[(-1)" j(t)-h(t)]}.
dt
(5)
(6)
(7)
1
(8)
(9)
This is just an ordinary differential equation, to be solved together with the
initial condition u,(0) = 0 from (1). The solution of (9) is
Un(t) = Ce-akt - 2n71-²ks e-hak(-s)[(-1)" j(s)- h(s)] ds. (10)
As a second case, let's solve the inhomogeneous wave problem
un-c²uxx = f(x, t)
u(0,t)=h(t)
u(x, 0) = p(x)
and, just as before,
Again we expand everything in the eigenfunctions of the corresponding ho-
mogeneous problem:
Therefore,
u(x, t) = [u,(t) sin
n=1
u₁(x, t) with coefficients un(t), uxx(x, t) with coefficients wn(t), f(x, t) with
coefficients fn(t), p(x) with coefficients on, and (x) with coefficients Vn.
Then
u(l,t)=k(t)
u,(x, 0) = (x).
Wn (1) =
2 2²u ηπχ
Un (1) ==
sin dx =
at² 1
=76
From the PDE we also have
Un(1) - C²w₁(1) = ²√ √ ²
2 a²u η π.χ
sin
ax²
= -λnun (1) +2nлl-²[h(t)-(-1)"k(t)].
d²un
d1²
with the initial conditions
илх
I
- dx
d²un
d1²
(un-c²uxx) sin dx = fn(t).
ηπχ
1
(11)
+c²λnun(t) = -2nлl-² [(-1)"k(t)-h(t)] + fn(t)
un (0) = Φη u' (0) = Vn.
The solution can be written explicitly (see Exercise 11).
(12)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c4bfc92-ad55-477d-ac4d-39185defff1d%2Ffca62a07-d1ca-4d2e-a52f-84ba86a6c3e2%2Fzbzhne_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5.6 INHOMOGENEOUS BOUNDARY CONDITIONS
In this section we consider problems with sources given at the boundary. We
shall see that naive use of the separation of variables technique will not work.
Let's begin with the diffusion equation with sources at both endpoints.
u₁ = kuxx
u(0,t)=h(t)
0<x<l, t>0
u(l,t) = j(t)
u (x, 0) = 0.
A separated solution u = X(x)T(t) just will not fit the boundary conditions.
So we try a slightly different approach.
EXPANSION METHOD
We already know that for the corresponding homogeneous problem the correct
expansion is the Fourier sine series. For each t, we certainly can expand
u(x,t) = Σun(t) sin -
n=1
2
-76-
un(t) =
0=ut - kuxx =
for some coefficients un(t), because the completeness theorems guarantee
that any function in (0, 1) can be so expanded. The coefficients are necessarily
given by
nTX
1
Σ[
илх
u(x, t) sin dx.
1
(3)
You may object that each term in the series vanishes at both endpoints and
thereby violates the boundary conditions. The answer is that we simply do not
insist that the series converge at the endpoints but only inside the interval. In
fact, we are exactly in the situation of Theorems 3 and 4 but not of Theorem
2 of Section 5.4.
Now differentiating the series (2) term by term, we get
dun
dt
(1)
+kun (1)
(7) ²] sin
(2)
плх
So the PDE seems to require that dun/dt + kλnun = 0, so that un(t) =
Anet. There is no way for this to fit the boundary conditions. Our method
fails! What's the moral? It is that you can't differentiate term by term. See
Example 3 in Section 5.4 for the dangers of differentiation.
Let's start over again but avoid direct differentiation of the Fourier series.
The expansion (2) with the coefficients (3) must be valid, by the completeness
theorem 5.4.3, say, provided that u(x, t) is a continuous function. Clearly, the
initial condition requires that u,,(0) = 0. If the derivatives of u(x, t) are also
continuous, let's expand them, too. Thus
with
V₁ (1) =
du
at
with the coefficients
2
a²u
əx²
dx2 =
Wn(t)=
u(x, t) sin
n=1
vn (1) - kwn (1) =
V₁ (1) sin
ди ηπε
sin
at T
The last equality is valid since we can differentiate under an integral sign if
the new integrand is continuous (see Section A.3). We also expand
илх
1
n=1
2 a²u
=75%
əx²
By Green's second identity (5.3.3) the last expression equals
2
+ ²7 (ux
ηπ.Χ
1
Wn(1) sin
dx =
dx +
sin
плх
1
dun
dt
NAX
=76
So from (5) and (8) we deduce that u, (1) satisfies
е
dx.
7(77)²(x,1
Here come the boundary conditions. The sine factor vanishes at both ends.
The last term will involve the boundary conditions. Thus
w₁(t)=-λun(t)-2nal-2(-1)" j(t) + 2xl-²h(t),
where λ = (n/1). Now by (5) and (7) the PDE requires
η π.χ
ux sin
пл
nлx
Tucos T
(u₁-kux) sindx =
-fo=o.
0=0.
(4)
dun
= k{-λnun(1) - 2nлl-²[(-1)" j(t)-h(t)]}.
dt
(5)
(6)
(7)
1
(8)
(9)
This is just an ordinary differential equation, to be solved together with the
initial condition u,(0) = 0 from (1). The solution of (9) is
Un(t) = Ce-akt - 2n71-²ks e-hak(-s)[(-1)" j(s)- h(s)] ds. (10)
As a second case, let's solve the inhomogeneous wave problem
un-c²uxx = f(x, t)
u(0,t)=h(t)
u(x, 0) = p(x)
and, just as before,
Again we expand everything in the eigenfunctions of the corresponding ho-
mogeneous problem:
Therefore,
u(x, t) = [u,(t) sin
n=1
u₁(x, t) with coefficients un(t), uxx(x, t) with coefficients wn(t), f(x, t) with
coefficients fn(t), p(x) with coefficients on, and (x) with coefficients Vn.
Then
u(l,t)=k(t)
u,(x, 0) = (x).
Wn (1) =
2 2²u ηπχ
Un (1) ==
sin dx =
at² 1
=76
From the PDE we also have
Un(1) - C²w₁(1) = ²√ √ ²
2 a²u η π.χ
sin
ax²
= -λnun (1) +2nлl-²[h(t)-(-1)"k(t)].
d²un
d1²
with the initial conditions
илх
I
- dx
d²un
d1²
(un-c²uxx) sin dx = fn(t).
ηπχ
1
(11)
+c²λnun(t) = -2nлl-² [(-1)"k(t)-h(t)] + fn(t)
un (0) = Φη u' (0) = Vn.
The solution can be written explicitly (see Exercise 11).
(12)
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