What is a Partial . . .
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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Title Page
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18MAB201T
Transforms and Boundary Value
Problems
Unit-3
PR
Prepared by
Dr. S. ATHITHAN
Assistant Professor
Department of of Mathematics
Faculty of Engineering and Technology
SRM I NSTITUTE OF S CIENCE AND T ECHNOLOGY
Kattankulathur-603203, Chengalpattu District.
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1.
What is a Partial Differential Equation?
What is a Partial . . .
∂y 2
=0
R
∂2φ
D
+
(1.2)
ED
B
∂x2
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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∂2φ
.S
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You’ve probably all seen an ordinary differential equation (ODE); for example the pendulum equation,
d2 θ
g
(1.1)
+ sin θ = 0,
2
dt
L
describes the angle, θ, a pendulum makes with the vertical as a function of time, t. Here
g and L are constants (the acceleration due to gravity and length of the pendulum respectively), t is the independent variable and θ is the dependent variable. This is an ODE
because there is only one independent variable, here t which represents time.
A partial differential equation (PDE) relates the partial derivatives of a function of two or
more independent variables together. For example, Laplace’s equation for φ(x, y),
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arises in many places in mathematics and physics. For simplicity, we will use subscript
notation for partial derivatives, so this equation can also be written φxx + φyy = 0.
We say a function is a solution to a PDE if it satisfy the equation and any side conditions
given. Mathematicians are often interested in if a solution exists and when it is unique.
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2.
Classifying PDE’s: Hyperbolic, Parabolic, Elliptic
What is a Partial . . .
Classify the following PDE’s
Classifying PDE’s: . . .
1. 3uxx − 4uxy − 3uy = 0.
Ans.: B 2 − 4AC = 16 > 0. ∴ The PDE is hyperbolic ∀ x, y.
2. y 2 uxx − 2xyuxy + x2 uyy − 3ux = 0.
Ans.: B 2 − 4AC = 0 ∀ x & y. ∴ Parabolic in this case.
A
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2
R
.S
4. x uxx + (1 − y )uyy = 0.
Ans.: B 2 − 4AC = 4x2 (1 − y 2 ) > 0 ∀x, x 6= 0 and y > 1, y < −1. ∴
The PDE is hyperbolic in this case.
B 2 − 4AC = 4x2 (1 − y 2 ) < 0 ∀x, x 6= 0 and −1 < y < 1. ∴ The PDE
is elliptic.
If x = 0 ∀y or ∀x, y = ±1, then B 2 − 4AC = 0. ∴ The PDE is parabolic in
this case.
Solution to the Heat . . .
Solved . . .
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D
The Plucked String as . . .
Heat Flow (Heat . . .
2
3. x uxx + 2xyuxy + (1 + y )uyy − 2ux = 0.
Ans.: B 2 − 4AC = −4x2 < 0 for x < 0 or x > 0. ∴ Elliptic in this case.
If x = 0, then B 2 − 4AC = 0. ∴ Parabolic in this case.
2
Solution of 1-D Wave . . .
Uniqueness of the . . .
N
2
Examples of Wave . . .
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Example : 1
Classify the follow differential equations as PDE’s hyperbolic, parabolic, elliptic.
What is a Partial . . .
Classifying PDE’s: . . .
1. The diffusion equation for u(x, t):
Examples of Wave . . .
Solution of 1-D Wave . . .
ut = c2 uxx
The Plucked String as . . .
Uniqueness of the . . .
2. The wave equation for w(x, t):
N
Heat Flow (Heat . . .
2
.A
TH
IT
H
A
wtt = c wxx
.S
3. The thin film equation for u(x, t):
Solved . . .
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ut = −(uuxxx )x
Solution to the Heat . . .
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4. The forced harmonic oscillator for y(t):
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ytt + ω 2 y = F cos(Ωt)
5. The Poisson Equation for the electric potential φ(x, y, z):
φxx + φyy + φzz = 4πρ(x, y, z)
where ρ(x, y, z) is a known charge density.
6. Burger’s equation for u(x, t):
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ut + uux = νuxx
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3.
Examples of Wave Equations in Different Situations
What is a Partial . . .
As we have seen before the ”classical” one-dimensional wave equation has the form:
utt = c2 uxx ,
(3.1)
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where u = u(x, t) can be thought of as the vertical displacement of the vibration of a
string.
The string can be fixed at both ends, or just at one end, or we can think of an ”infinite”
string, that is not bound at any end. Each will yield different boundary conditions for the
well-posed wave equation. We can also consider the case where the string is ”pushed”
with an external force h(x, t), or where we take under consideration the friction coefficient from the air that the string displaces. These two equations will be called ”forced”
and ”damped” respectively. In the ”forced” case, the wave equation is:
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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Y
D
utt = c2 uxx + h(x, t),
Classifying PDE’s: . . .
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where an example of the acting force is the gravitational force. In the ”damped” case the
equation will look like:
utt + kut = c2 uxx ,
where k can be the friction coefficient.
If we have more than one spatial dimension (a membrane for example), the wave equation
will look a bit different. In the case of the vibrating membrane we have two spatial
variables and the wave equation will look like:
utt = c2 (uxx + uyy ).
For n−dimensions (whatever THAT means...) the n-wave equation will be:
utt = c2 (ux1 x1 + ux2 x2 + · · · + uxn xn ).
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The initial conditions for the one-dimensional wave equation will be:
What is a Partial . . .
(i) u(x, 0) = f (x)
Classifying PDE’s: . . .
(ii) ut (x, 0) = g(x).
Examples of Wave . . .
Solution of 1-D Wave . . .
For the finite string the boundary conditions will be:
The Plucked String as . . .
Uniqueness of the . . .
(i) u(0, t) = A(t)
Heat Flow (Heat . . .
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(ii) u(L, t) = B(t).
.A
TH
IT
So, altogether we have 4 conditions which are repeated below together.
.S
(i) u(x, 0) = f (x)
Solved . . .
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(ii) ut (x, 0) = g(x)
Solution to the Heat . . .
ED
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(iv) u(L, t) = B(t).
B
(iii) u(0, t) = A(t)
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4.
Solution of 1-D Wave Equation (Dirichlet Problem)
What is a Partial . . .
If we tie the string at both ends we can have the following boundary conditions:
Classifying PDE’s: . . .
Examples of Wave . . .
u(0, t) = A(t), u(L, t) = B(t),
Solution of 1-D Wave . . .
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where A, B are C 1 piecewise functions. For example, we can have a sinusoidal function
at one end and a Heaviside function at the other.
When the boundary values A and B are 0 we obtain the Dirichlet Problem for the wave
equation:
utt = c2 uxx ,
.A
TH
0 < x < L, t > 0
t>0
.S
u(0, t) = 0, u(L, t) = 0,
0<x<L
Uniqueness of the . . .
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
DE
BC
IC.
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u(x, 0) = f (x), ut (x, 0) = g(x)
The Plucked String as . . .
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As you have seen in Lecture 5 for the diffusion equation, the method of separating the
variables is a very convenient way to obtain solutions for PDEs. In the case of the Dirichlet
Problem we will quickly review the method.
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Theorem 4.1. A solution of the problem:
utt = c2 uxx ,
0 < x < L, t > 0
u(0, t) = 0, u(L, t) = 0,
u(x, 0) = f (x), ut (x, 0) = g(x)
DE
t>0
BC
0<x<L
IC.
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is given by:
u(x, t) =
N X
L
n=1
nπc
Ān sin(
nπct
L
) + Bn cos(
nπx
) sin(
),
L
L
nπct
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where: f (x) =
N
X
Bn sin(
n=1
nπx
L
N
X
), and g(x) =
Ān sin(
nπx
L
n=1
). The coeffi-
Classifying PDE’s: . . .
cients Ān and Bn are given by:
Z
Z
nπx
2 L
nπx
2 L
f (x) sin(
), Ān =
g(x) sin(
).
Bn =
L 0
L
L 0
L
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
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Proof. We will use the method of separation of variables, namely think of the solution
u(x, t) as a product of a function that depends only on the variable t and of a function
that depends only on the variable x.
Let
u(x, t) = X(x)T (t)
and substitute in the equation
2
D
R
.S
utt = c uxx ,
Uniqueness of the . . .
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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we obtain:
What is a Partial . . .
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or
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T̈ (t)
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B
X(x)T̈ (t) = c2 Ẍ(x)T (t),
c2 T (t)
=
Ẍ(x)
X(x)
,
Page 7 of 34
thus the equality is one of functions of different variables, so both quotients have to be
constant.
Say
T̈ (t)
Ẍ(x)
=
= k = ±p2 ,
2
c T (t)
X(x)
then we can solve each ordinary differential equation separately. We have the following
three cases: k = −p2 , k = p2 , and k = p = 0.
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Case 1: When the constant is −p2 , then the solutions for
What is a Partial . . .
Ẍ(x)
Classifying PDE’s: . . .
2
= −p ,
X(x)
Examples of Wave . . .
Solution of 1-D Wave . . .
are:
The Plucked String as . . .
X(x) = c1 sin(px) + c2 cos(px),
Uniqueness of the . . .
and the solutions for
Heat Flow (Heat . . .
T̈ (t)
A
IT
H
c2 T (t)
N
= −p2 ,
.A
TH
are:
.S
T (t) = d1 sin(pct) + d2 cos(pct).
Solved . . .
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Then
Solution to the Heat . . .
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B
u(x, t) = [d1 sin(pct) + d2 cos(pct)][c1 sin(px) + c2 cos(px)].
2
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Case 2: When the constant is p , then the solutions for
Ẍ(x)
X(x)
= p2 ,
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are:
px
X(x) = c1 e
+ c2 e
−px
,
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and the solutions for
T̈ (t)
c2 T (t)
= p2 ,
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are ±cp, are
T (t) = d1 epct + d2 e−pct .
What is a Partial . . .
Classifying PDE’s: . . .
Then
pct
u(x, t) = (d1 e
−pct
+ d2 e
px
)(c1 e
+ c2 e
−px
Examples of Wave . . .
).
Solution of 1-D Wave . . .
Case 3: When the constant is 0, then the equations become
The Plucked String as . . .
Uniqueness of the . . .
Ẍ(x) = T̈ (t) = 0,
N
Heat Flow (Heat . . .
Solution to the Heat . . .
H
A
and X(x) = c1 x + c2 , and T (t) = d1 t + d2 . Then
IT
Solved . . .
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TH
u(x, t) = (d1 t + d2 )(c1 x + c2 ).
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To confirm which is most suitable solution, we may need the help of the actual nature of
the solutions. To check that we may proceed with boundary conditions. Let’s take a look
at the boundary conditions: u(0, t) = 0, u(L, t) = 0.
The only solution for u(x, t) that can satisfy them is
PR
u(x, t) = (d1 sin(pct) + d2 cos(pct))(c1 sin(px) + c2 cos(px)),
and the boundary conditions translate into:
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(d1 sin(pct) + d2 cos(pct))(c1 sin(0) + c2 cos(0))
=
0
(d1 sin(pct) + d2 cos(pct))(c1 sin(pL) + c2 cos(pL))
=
0,
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∀t > 0,
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namely:
c2
=
0
c1 sin(pL)
=
0.
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From the last condition we obtain p =
πn
L
, and
What is a Partial . . .
∞ X
πn
πn
πn
u(x, t) =
ct) + d2n cos(
ct) cn sin(
x).
d1n sin(
L
L
L
n=1
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Now, the conditions left to check are the initial conditions:
Uniqueness of the . . .
f (x) =
Bn sin(
=
g(x) =
IT
nπx
.A
TH
ut (x, 0)
N
X
Heat Flow (Heat . . .
),
H
n=1
L
N
=
nπx
A
u(x, 0)
N
X
Ān sin(
L
).
D
) + Bn cos(
Y
nπc
nπct
L
nπx
) sin(
),
L
L
nπct
Solved . . .
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A
R
n=1
Ān sin(
B
N X
L
ED
Then u(x, t) =
R
.S
n=1
Solution to the Heat . . .
PR
Remark 4.0.1. In the more general case for the Dirichlet Problem, when initial conditions (IC) change to more general homogeneous conditions: k1 u(x, 0)+k2 ux (x, 0) =
0, we can solve the problem in the same manner, using separation of variables.
Exercise: 4.1. Check that Case 1 is the only one that verifies the boundary conditions in
the proof above.
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Exercise: 4.2. Check that the solution found above verifies the initial conditions.
Close
Exercise: 4.3. A string of length π is held fixed at both endpoints. Its initial position
is f (x) = sin(x) and its initial velocity is g(x) = cos x. Assuming that c = 1,
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find the position of the string u(x, t) for every x ∈ [0, π] and for every t > 0. Find
an approximate value for u(x, t) by adding several terms of the series. Animate the
approximation and draw a 3D plot.
What is a Partial . . .
Classifying PDE’s: . . .
Examples of Wave . . .
Exercise: 4.4. Solve the following problem for the string equation:
Solution of 1-D Wave . . .
PDE
utt = uxx ,
The Plucked String as . . .
Uniqueness of the . . .
N
for every x ∈ [0, π] .
ut (x, 0) = 0,
A
u(x, 0) = sin(x),
ux (π, t) = 0 for every t > 0;
H
IC
u(0, t) = 0,
IT
BC
Solution to the Heat . . .
Solved . . .
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Notice the change in the boundary conditions. This will lead to different eigenvalues and
eigenfunctions.
Use Maple to animate the solution you found, to draw a 3D plot, and to check that the
solution satisfies the conditions of the problem.
Heat Flow (Heat . . .
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5.
The Plucked String as an Example for 1-D Wave Equation
The plucked string refers to the initial condition for the Dirichlet problem , where f (x)
looks like a ”plucked string”, namely:
What is a Partial . . .
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
utt = c2 uxx ,
0 < x < L, t > 0
t>0
u(x, 0) = f (x), ut (x, 0) = 0
0<x<L
A
H
IT
.A
TH
Solution to the Heat . . .
IC,
Solved . . .
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B
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x0 ≤ x ≤ L
PR
u
0

 ,
x
0
˙
f (x) =
u0

,

x0 − L
Heat Flow (Heat . . .
BC
Y
D
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0 ≤ x ≤ x0
EP
A
 u0

 x,
x0
f (x) =
x−L

u0
,
x0 − L
.S
where
We have that:
N
u(0, t) = 0, u(L, t) = 0,
DE
0 ≤ x ≤ x0
x0 ≤ x ≤ L,
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N
and f (0) = f (L) = 0, thus:
Z x0
Z L
2 L
u0
nπx
2 L
u0
nπx
Bn =
(
)
cos(
)dx + (
)
cos(
)dx
L nπ 0 x0
L
L nπ x0 x0 − L
L
2 L 2 u0
nπx0
u0
nπx0
=
(
) (
sin(
)−
sin(
))
L nπ
x0
L
x0 − L
L
2L2 u0
1
nπx0
=
).
sin(
2
2
π x0 (L − x0 ) n
L
A
H
IT
n=1
1
n2
sin(
nπx0
L
.A
TH
π 2 x0 (L − x0 )
) cos(
.S
2L2 u0
nπct
L
) sin(
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Heat Flow (Heat . . .
Solved . . .
nπx
L
).
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Musical Instruments (Vibrating Strings)
EP
A
5.1.
ED
B
Y
D
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u(x, t) =
Classifying PDE’s: . . .
Solution to the Heat . . .
Now we can write the formal solution to the plucked string equation:
∞
X
What is a Partial . . .
PR
Many instruments produce sound by making strings vibrate; such are the harp, the piano,
the harpsichord, the guitar, the violin, and others. Strings are kept fixed at the endpoints,
but they way the instruments are played create different initial conditions. In instruments
like the guitar, the string is plucked; this produces an initial perturbation with no initial
velocity. In the piano, on the other hand, the string is hit, which creates an initial velocity
but no initial perturbation from the initial position.
The oscillations of the string are described by
u(x, t) =
∞
X
n=1
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(An cos ωn t + Bn sin ωn t) sin pn t,
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where
π
ωn = cpn = cn .
L
L
The sound we hear is thus a combination of the main harmonic sounds (eigenfunctions)
pn = n
π
and
un (x, t) = (An cos ωn t + Bn sin ωn t) sin pn t.
D
R
M u20 L2 c2
πnx0
Y
n2 π 2 x20 (L − x0 )2
sin2
L
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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−2
PR
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A
The energy decreases as n , so only the main tone u1 and a few other harmonics are
audible.
On the other hand, if we hit the string with a flat hammer of length 2δ with center at x0
and producing an initial velocity v0 , the energy of the nth harmonic is
En =
Classifying PDE’s: . . .
.
ED
B
En =
.S
.A
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IT
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A
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The contribution of each particular harmonic is measured by its energy, which turns out
to be equal to:
ω2 M
2
),
En = n (A2n + Bn
4
where M = DL is the total mass of the string (recall that D was the density).
For the plucked string (Section 7.5), the energy is given by:
What is a Partial . . .
4M V02
n2
π 2 sin2
πnx0
L
sin2
πnδ
L
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−2
and the energy again decreases as n . However, if the hammer is sufficiently narrow,
letting δ tend to zero (the blade of a knife), we get the model of a string getting an impulse
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concentrated at a point x0 . The corresponding energy is:
En =
v02
sin2
πnx0
What is a Partial . . .
.
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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Title Page
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B
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H
A
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L
L
Thus, for a very narrow hammer the energies of all harmonics are of the same order and the
generated sound will be saturated with harmonics. This can be checked experimentally,
by hitting a string with the blade of a knife. The sound will have a metallic quality.
Not all harmonics are desirable. The first ones, u2 up to u6 , sound well together with the
main harmonic u1 . However, the 7th and the first harmonics sounding together produce
a sense of dissonance.
There are several ways to try to “kill” those harmonics by percussion (as in the piano).
πnx0
a). The position of the hammer. The presence of the factor sin
shows that by
L
choosing the center x0 of the hammer at the node of the undesired harmonic we may make
it disappear (make the corresponding An and Bn be equal to zero). In modern pianos
the position of the hammer is chosen near the nodes of the 7th and the 8th harmonics, to
“kill” them.
b). The shape of the hammer. In modern pianos the hammers are not flat, but rather round.
One can model this situation by choosing the initial velocity to be, say, a parabola on the
interval [x0 − δ, x0 + δ], instead of a horizontal line. Older pianos, which had flatter
and narrower hammers, produced a more piercing, shrilled sound.
c). The rigidity of the hammer. If instead of being rigid the hammer is softer. In this case
the motion is not described by its initial position and velocity, but rather by a short-time
acting force, which varies in time.
Exercise: 5.1. Find the solution u(x, t) of the string equation if l = π, c = 1, when
both endpoints are fixed, the initial velocity is zero, and u(x, t) is the following piecewise
linear function:
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6.
Uniqueness of the Solution to the 1-D Wave Equation
What is a Partial . . .
Theorem 6.1. Let u1 (x, t) and u2 (x, t) be two solutions of the problem:
2
utt = c uxx ,
0 < x < L, t > 0
u(0, t) = A(t), u(L, t) = B(t),
u(x, 0) = f (x), ut (x, 0) = g(x)
t>0
0 < x < L.
Classifying PDE’s: . . .
DE
BC
IC,
1
IT
H
A
N
where A, B, f, g are C piecewise continuous. Then u1 (x, t) = u2 (x, t) for all points
in the domain.
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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A
0
PR
We will prove that H(t) = 0 first. Differentiating with respect to t we obtain:
Z L
2
c 2vx vxt + 2vt vtt dx
Ḣ(t) =
0
=
2c
2
Z
[vx vxt + vt vxx ] dx
δ
L
(vx vt )dx = 2c2 [vx (x, t)vt (x, t)]0
δx
0
= 2c2 (vx (L, t)vt (L, t) − vx (0, t)vt (0, t)) = 0.
=
2c2
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L
0
Z L
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Proof. Let v(x, t) = u1 (x, t)−u2 (x, t), then v satisfies the wave equation with initial
conditions: u(x, 0) = ut (x, 0) = 0, and boundary conditions u(0, t) = u(L, t) =
0. Our goal is to prove that v(x, t) = 0 ∀x, t.
In order to accomplish this, define:
Z L
2
H(t) =
c vx (x, t)2 + vt (x, t)2 dx.
Examples of Wave . . .
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Since Ḣ(t) = 0, H(t) is constant, and as H(0) = 0, we conclude that H(t) = 0.
Z L
Then vt (x, t) = 0, and v(x, t) = v(x, t) − v(x, 0) =
vt (x, t)dt = 0.
0
Solution of 1-D Wave . . .
L
E(t) =
Classifying PDE’s: . . .
Examples of Wave . . .
Remark 6.0.1. The energy integral of the string at time t is:
Z
What is a Partial . . .
The Plucked String as . . .
T0 ux (x, t)2 + Dut (x, t)2 dx,
Uniqueness of the . . .
0
Heat Flow (Heat . . .
.S
.A
TH
IT
H
A
N
where D is the mass per unit length and T0 is the constant tension when the string is
straight. We can see that the energy is proportional to H if we construct H using u
instead of v. So the uniqueness proof comes from the conservation of energy for the
unforced string.
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B
A damped string of length 1 has equation
2
Solved . . .
Y
D
R
Example : 2
Solution to the Heat . . .
PR
EP
A
utt = c uxx − γut ,
where γ is a small damping coefficient. Find the solution u(x, t) assuming that
both endpoints are fixed, the initial condition is x(1 − x) and the initial velocity
is zero.
To see the nature of the by visualization, we can plot and animate the solution for
1
1
the case when c = and γ = . But that is the scope our course. So whoever
4
5
interested can do it later with the help of teacher.
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Example : 3
Find the solution u(x, y, t) of a square membrane with side 1 fixed on the boundary, if the initial position is u(x, y, 0) = (x − x2 )(y − y 2 ) and the initial
velocity is zero.
Same way as said in the previous problem, we can animate several eigenfunctions
un,m (x, y, t), say, u1,1 , u1,2 , u3,5 , assuming that c = 1.
What is a Partial . . .
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Heat Flow (Heat . . .
A
N
Example : 4
.S
.A
TH
IT
H
Solve the string equation utt = c2 uxx for L = 1, with the boundary conditions
u(0, t) = 0 and u(1, t) = 1, with zero initial velocity, assuming that the initial
position is
D
R
(a) u(x, 0) = sin x (easier),
Solved . . .
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Y
2
Solution to the Heat . . .
ED
B
(b) u(x, 0) = x (harder).
JJ
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EP
A
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Hint: You cannot use the superposition principle, since the boundary condition
at x = 1 is not homogeneous. Try a change of coordinates first, v(x, t) =
u(x, t) + h(x), where h(x) is a suitable (easy) function that would guarantee
that v also satisfies the string equation, now with homogeneous boundary conditions.
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Example : 5
A
N
Solve the string equation utt = c2 uxx for L = 1, with the boundary conditions
u(0, t) = 0 and ux (1, t) + u(1, t) = 0. This corresponds to the case when
the left end is fixed and the right end is attached to an elastic hinge. The initial
2
conditions are u(x, 0) = x − x2 and ut (x, 0) = x.
3
Note. This exercise is hard! The eigenvalues pn will be solutions of a transcendental equation.
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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A
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ED
B
Y
D
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.S
.A
TH
IT
H
We will see Heat Equation now. After that we will have few problems worked out here.
Then you can go through the books and solve related problems and discuss your doubts
with the teacher.
What is a Partial . . .
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7.
Heat Flow (Heat Equation or Diffusion Equation)
What is a Partial . . .
A classical example of the application of ordinary differential equations is Newton’s Law
of Cooling which, basically, answers the question “How does a cup of coffee cool?” Newton hypothesized that the rate at which the temperature, U (t), changes is proportional to
b,
the difference with the ambient temperature, which we call U
(7.1)
Solution of 1-D Wave . . .
The Plucked String as . . .
.A
TH
IT
H
Here κ is a positive rate constant (with units of inverse time) that measures how fast heat is
lost from the coffee cup to the ambient environment. If we specify the initial temperature,
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B
Y
we can solve for the evolution of the temperature,
Solved . . .
(7.2)
R
.S
U (0) = U0 ,
Heat Flow (Heat . . .
Solution to the Heat . . .
A
dt
Examples of Wave . . .
Uniqueness of the . . .
b ).
= −κ(U − U
N
dU
Classifying PDE’s: . . .
(7.3)
EP
A
R
b + (U0 − U
b )e−κt .
U (t) = U
PR
If we graph the temperature as a function of time, we see that it decays exponentially to
b , at a rate governed by κ.
the ambient temperature, U
When we derived Newton’s Law of cooling we made several assumptions – most importantly that the temperature in the coffee cup did not vary with location. If we account for
the variation of temperature with location, we can derive a PDE called the heat equation
or, more generally, the diffusion equation. If the temperature, U (x, t) is a function of a
single spatial variable, x, we will show that it satisfies the diffusion equation,
Ut = c2 Uxx ,
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where c is a constant known as the thermal diffusivity. In higher dimensions, the equation
can be written
Ut = c2 ∇2 U,
(7.5)
2
What is a Partial . . .
Classifying PDE’s: . . .
Examples of Wave . . .
where ∇ is the Laplacian.
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Solution to the Heat . . .
Solved . . .
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D
R
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.A
TH
IT
H
A
N
Heat Flow (Heat . . .
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8.
Solution to the Heat Equation (Diffusion Equation)
What is a Partial . . .
We can summarize the last section by restating a well-posed problem for the diffusion
equation on the interval 0 < x < L with Dirichlet boundary conditions,
T HE D IRICHLET P ROBLEM FOR THE D IFFUSION E QUATION
(N ON - HOMOGENEOUS B OUNDARY C ONDITIONS )
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Ut = c2 Uxx
N
t>0
A
U (L, t) = U1
H
U (0, t) = U0
0 < x < L, t > 0
0 < x < L.
Heat Flow (Heat . . .
BC
Solution to the Heat . . .
IC
Solved . . .
.A
TH
IT
U (x, 0) = f (x)
DE
.S
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ED
B
Y
D
R
Solving the general problem will have to wait, but we can find some specific solutions to
the problem using the ideas of Separation of Variables. For the moment, we will restrict
ourselves to homogeneous boundary conditions,
JJ
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T HE D IRICHLET P ROBLEM FOR THE D IFFUSION E QUATION
(H OMOGENEOUS B OUNDARY C ONDITIONS )
Ut = c2 Uxx
U (0, t) = 0 U (L, t) = 0
U (x, 0) = f (x)
0 < x < L, t > 0
DE
t>0
BC
0 < x < L,
IC
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If you want, you can skip the derivation for the moment and jump ahead to Exercise 1,
if you don’t mind the solution appearing deus ex machina ( a fancy term for “out of thin
air”).
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8.1.
A Solution to the Homogeneous Dirichlet Problem
What is a Partial . . .
Let us look for solutions to the homogeneous Dirichlet problem of the form
U (x, t) = X(x)T (t)
Classifying PDE’s: . . .
(8.1)
Examples of Wave . . .
Solution of 1-D Wave . . .
we find from the differential equation (DE) that
The Plucked String as . . .
2
XTt = c Xxx T
N
(8.2)
H
=
Solved . . .
IT
.A
TH
c2 T
Xxx
= k = ±p2 ,
X
(8.3)
.S
Tt
B
ED
R
=
Title Page
X(x)
Now, we have to solve each ordinary differential equation separately. We have the following three cases: k = −p2 , k = p2 , and k = p = 0.
Case 1: When the constant is −p2 , then the solutions for
Ẍ(x)
X(x)
JJ
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I
2
= k = ±p ,
EP
A
c2 T (t)
Ẍ(x)
PR
Ṫ (t)
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R
where p is to be determined.
We may write the above equation as
Heat Flow (Heat . . .
Solution to the Heat . . .
A
and dividing by XT we find
Uniqueness of the . . .
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= −p2 ,
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are:
X(x) = c1 sin(px) + c2 cos(px),
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and the solutions for
Ṫ (t)
c2 T (t)
What is a Partial . . .
= −p2 ,
Classifying PDE’s: . . .
are:
Examples of Wave . . .
2
2
−p c t
T (t) = d1 e
Solution of 1-D Wave . . .
.
The Plucked String as . . .
Then
2
u(x, t) = [d1 e−p
2
c t
][c1 sin(px) + c2 cos(px)].
Ẍ(x)
.A
TH
IT
H
A
N
Case 2: When the constant is p2 , then the solutions for
2
=p ,
.S
X(x)
D
R
are:
Uniqueness of the . . .
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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ED
B
Y
X(x) = c1 epx + c2 e−px ,
EP
A
Ṫ (t)
JJ
II
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R
and the solutions for
PR
c2 T (t)
= p2 ,
are ±cp, are
Page 24 of 34
2
2
p c t
T (t) = d1 e
Then
u(x, t) = d1 ep
2
2
c t
.
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(c1 epx + c2 e−px ).
Case 3: When the constant is 0, then the equations become
Ẍ(x) = Ṫ (t) = 0,
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and X(x) = c1 x + c2 , and T (t) = d1 . Then
What is a Partial . . .
u(x, t) = d1 (c1 x + c2 ).
Classifying PDE’s: . . .
To confirm which is most suitable solution, we may need the help of the actual nature of
the solutions. To check that we may proceed with boundary conditions. Let’s take a look
at the boundary conditions: u(0, t) = 0, u(L, t) = 0.
The only solution for u(x, t) that can satisfy them is
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Heat Flow (Heat . . .
(c1 sin(px) + c2 cos(px)).
H
IT
u(x, t) = d1 e
N
2
A
2
−p c t
Solution to the Heat . . .
Solved . . .
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Further, we can apply the boundary conditions we will proceed to get the solution.
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9.
Solved Problems/Examples
What is a Partial . . .
Classifying PDE’s: . . .
Example : 6
Examples of Wave . . .
Write one dimensional wave equation and its possible solutions.
Solution of 1-D Wave . . .
The Plucked String as . . .
Hints/Solution:
∂2u
One dimensional wave equation is given by
∂t2
=c
2
2∂ u
∂x2
(or) utt = c2 uxx
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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ED
B
Y
D
R
.S
.A
TH
IT
H
A
N
Its possible solutions are given by
(i) u(x, t) = (c1 epx + c2 e−px )(c3 epct + c4 e−pct ),
(ii) u(x, t) = (c5 cos px + c6 sin px)(c7 cos pct + c8 sin pct) and
(iii) u(x, t) = (c9 x + c10 )(c11 t + c12 ).
Uniqueness of the . . .
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Example : 7
PR
Write one dimensional heat equation and its possible solutions.
Page 26 of 34
Hints/Solution:
One dimensional heat equation is given by
∂u
∂t
2
= c2
∂ u
∂x2
Its possible solutions are given by
2 2
(i) u(x, t) = (c1 epx + c2 e−px )ec p t ,
2 2
(ii) u(x, t) = (c3 cospx + c4 sin px)e−c p t and
(iii) u(x, t) = (c5 x + c6 ).
(or) ut = c2 uxx
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Example : 8
In one dimensional wave equation
2
∂2u
= c2
∂t2
∂2u
∂x2
What is a Partial . . .
(or) utt = c2 uxx what does
Classifying PDE’s: . . .
Examples of Wave . . .
c stands for?.
Solution of 1-D Wave . . .
The Plucked String as . . .
Hints/Solution:
2
2
In one dimensional wave equation the value of c is stands for c =
m
, where T is the
.A
TH
IT
H
A
N
tension and m is mass per unit length.
T
2
∂x2
2
(or) ut = c uxx what does
EP
A
PR
Hints/Solution:
Solved . . .
JJ
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c stands for?.
Solution to the Heat . . .
Title Page
R
D
u
Y
= c
2
B
∂t
2∂
ED
In one dimensional heat equation
∂u
Heat Flow (Heat . . .
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.S
Example : 9
Uniqueness of the . . .
k
, where k is
ρh
thermal conductivity, ρ is the density of the material and h is the specific heat of the
material.
In one dimensional heat equation the value of c2 is stands for c2 =
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Example : 10
The ends of a rod of length 20 cm are maintained at the temperature 10◦ C and
20◦ C respectively until steady state prevails. Determine the steady state temperature of the rod.
What is a Partial . . .
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
Uniqueness of the . . .
R
.S
.A
TH
IT
H
A
N
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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ED
B
Y
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JJ
II
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R
PR
EP
A
Here, T0 = 10◦ C, Tl = 20◦ C and l = 20.
1
∴ (1) becomes u(x) = x + 10.
2
The Plucked String as . . .
D
Hints/Solution:
1-D heat equation is ut = c2 uxx .
In steady state, uxx = 0 =⇒ u(x) = ax + b.
Applying given conditions, we get
u(l) − u(0)
Tl − T0
=
b = u(0) = T0 and a =
l
l
Tl − T0
=⇒ u(x) =
x + T0 − − − (1)
l
Page 28 of 34
Example : 11
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When the ends of a rod of length 30 cm are maintained at the temperature 20◦ C
and 80◦ C respectively until steady state prevails. Determine the steady state
temperature of the rod.
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1-D heat equation is ut = c2 uxx .
In steady state, uxx = 0 =⇒ u(x) = ax + b.
Applying given conditions, we get
Tl − T0
u(l) − u(0)
b = u(0) = T0 and a =
=
l
l
Tl − T0
x + T0 − − − (1)
=⇒ u(x) =
l
◦
What is a Partial . . .
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
◦
Heat Flow (Heat . . .
.S
.A
TH
IT
H
A
N
Here, T0 = 20 C, Tl = 80 C and l = 30.
∴ (1) becomes u(x) = 2x + 20.
Solved . . .
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Example : 12
Solution to the Heat . . .
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II
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EP
A
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ED
B
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A tightly stretched string with fixed end points x = 0 and x = l is initially at
rest in its equilibrium position. If it is set vibrating giving each point a velocity
3x(l − x), find the displacement.
Hints/Solution:
One dimensional wave equation is given by
(1)
The boundary conditions becomes,
(i) u(0, t) = 0, t ≥ 0,
(ii) u(l, t) = 0 t ≥ 0,
(iii) u(x, 0) = 0, 0 ≤ x ≤ l
Page 29 of 34
∂2u
∂t2
= c2
∂2u
∂x2
(or) utt = c2 uxx − − −
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(iv)
∂u
∂t
(x, 0) = 3x(l − x), 0 ≤ x ≤ l
What is a Partial . . .
The suitable solution is given by
u(x, t) = (c1 cos px + c2 sin px)(c3 cos pct + c4 sin pct) − − − (2).
Applying boundary
conditions
(BC’s) (i) and (ii) in (2), weget
nπct
nπct
nπx
c3 cos
+ c4 sin
− − − (3).
u(x, t) = c2 sin
l
l
l
Classifying PDE’s: . . .
Applying boundary
conditions
(BC) (iii) in (3), we get
nπx
nπct
nπx
nπct
u(x, t) = c3 sin
c4 sin
= cn sin
sin
− − − (4).
l
l
l
l
Heat Flow (Heat . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
N
Uniqueness of the . . .
H
A
Solution to the Heat . . .
.A
TH
IT
Solved . . .
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B
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R
.S
Since cn is an arbitrary constant, and n is any integer and since (1) is homogeneous and
linear, the most general solution of (1) becomes,
∞
X
nπx
nπct
u(x, t) =
cn sin
sin
− − − (5).
l
l
n=1
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EP
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Using BC (iv), we have
∞
X
∂u
nπa
nπx
(x, 0) =
cn
sin
= 3x(l − x), 0 ≤ x ≤ l
∂t
l
l
n=1
Expanding 3x(l − x) as Fourier series in (0, l), we have
∞
X
nπx
, then
3x(l − x) =
bn sin
l
n=1

Zl
0
nπc
2
nπx
12l2
b n = cn
=
3x(l−x) sin
dx = 3 3 [1−(−1)n ] =
24l2

l
l
l
n π
0
n3 π 3
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if n = even
if n = odd
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=⇒
cn =

0
if n = even
24l3
if n = odd
cn4 π 4
∞
X
24l3
nπct
nπx
sin
u(x, t) =
sin
4
4
cn π
l
l
n=1,3,5,...
∞
3
X
24l
(2r − 1)πx
(2r − 1)πct
i.e. u(x, t) =
sin
sin
4
4
c(2r − 1) π
l
l
r=1
.A
TH
IT
H
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N

.S
Example : 13
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
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Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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A rod of length l through which heat flows is insulated at its sides. The ends are
kept at zero temperature.
If the initial temperature at the interior points of the bar
πx
3
is given by k sin
, find the temperature distribution in the bar after time
l
t.
What is a Partial . . .
Hints/Solution:
One dimensional heat equation is given by
Page 31 of 34
∂u
∂t
=c
The boundary conditions becomes,
(i) u(0, t) = 0, t ≥ 0,
(ii) u(l, t) = 0 t ≥ 0,
πx
(iii) u(x, 0) = k sin3
, 0≤x≤l
l
2
2∂ u
∂x2
(or) ut = c2 uxx − − − (1)
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The suitable solution is given by
2 2
u(x, t) = (c1 cos px + c2 sin px)e−c p t − − − (2).
Applying boundary
conditions
(BC’s) (i) and (ii) in (2), weh get 2 2 i
nπx h −c2 n22π2 t i
nπx −c2 n 2π t
l
l
u(x, t) = c2 sin
e
= cn sin
− − − (3).
e
l
l
.A
TH
.S
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Heat Flow (Heat . . .
Solution to the Heat . . .
Solved . . .
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Using BC (iii), we have
∞
X
πx
nπx
3
k sin
=
, then
bn sin
l
l
n=1
∞
X
k
πx
3πx
nπx
sin
− sin
=
bn sin
,
4
l
l
l
n=1
IT
H
A
N
Since cn is an arbitrary constant, and n is any integer and since (1) is homogeneous and
linear, the most general solution of (1) becomes,
∞
X
nπx h −c2 n22π2 t i
l
e
− − − (4).
u(x, t) =
cn sin
l
n=1
What is a Partial . . .
PR
EP
A
Equating like co-efficients and solving for bn we get
πx −c2 π22 t
3πx −c2 322π2 t
3k
k
l
l
u(x, t) =
sin
e
− sin
e
4
l
4
l
Page 32 of 34
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Example : 14
What is a Partial . . .
The ends A & B of a rod of length ‘`’ units long have their temperature kept
at 0◦ C & 120◦ C until steady state condition prevails. The temperature of the
end B is suddenly reduced to 0◦ C and that of A is maintained to 0◦ C. Find the
temperature distribution in the rod after time t.
Classifying PDE’s: . . .
Examples of Wave . . .
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
∂u
2∂
2
Heat Flow (Heat . . .
u
N
Hints/Solution:
2
B
Y
D
R
.S
.A
TH
IT
H
A
=c
(or) ut = c uxx − − − (1)
∂t
∂x2
In steady state, uxx = 0 =⇒ u(x) = ax + b. Applying given conditions, we get
u(l) − u(0)
Tl − T0
Tl − T0
=
=⇒ u(x) =
x+
b = u(0) = T0 and a =
l
l
l
T0 − − − (1)
One dimensional heat equation is given by
120
x.
EP
A
l
PR
The boundary conditions becomes,
(i) u(0, t) = 0, t ≥ 0,
(ii) u(l, t) = 0 t ≥ 0, 120x
(iii) u(x, 0) =
, 0≤x≤l
l
R
ED
Here, T0 = 0◦ C, Tl = 120◦ C and length = l. ∴ (1) becomes u(x) =
Solution to the Heat . . .
Solved . . .
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Page 33 of 34
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The suitable solution is given by
2 2
u(x, t) = (c1 cos px + c2 sin px)e−c p t − − − (2).
Applying boundary
conditions
(BC’s) (i) and (ii) in (2), weh get 2 2 i
nπx h −c2 n22π2 t i
nπx −c2 n 2π t
l
l
u(x, t) = c2 sin
e
= cn sin
e
− − − (3).
l
l
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sin
`
What is a Partial . . .
Classifying PDE’s: . . .
Examples of Wave . . .
e
Solution of 1-D Wave . . .
The Plucked String as . . .
Uniqueness of the . . .
Solution to the Heat . . .
Solved . . .
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Heat Flow (Heat . . .
B
(−1)
nπx
2 2 2
− c n 2π t
`
ED
nπ
n+1
as Fourier series in (0, l), we have
JJ
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n=1
240
l
EP
A
Ans.: u(x, t) =
∞
X
120x
PR
Using BC (iii) and expanding
Page 34 of 34
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