false

Class Notes
(839,469)

United States
(325,982)

Pennsylvania State University
(4,860)

Mathematics
(573)

MATH 250
(23)

Unknown
(6)

Lecture

Description

Second Order Linear Partial Differential Equations
Part IV
One▯dimensional undamped wave equation; D’Alembertsolution of the
wave equation; damped wave equation and the general wave equation; two▯
dimensional Laplace equation
The second type of second order linear partial differential equations in 2
independent variables is the one▯dimensional wave equation. Together with
the heat conduction equation, they are sometimes referred to as the
“evolution equations” because their solutions “evolve”, or change, with
passing time. The simplest instance of the one▯dimensional wave equation
problem can be illustrated by the equation that describes the standing wave
exhibited by the motion of a piece of undamped vibrating elastic string.
© 2008, 2012 Zachary S Tseng E▯4 ▯ 1 Undamped One-Dimensional Wave Equation:
Vibrations of an Elastic String
Consider a piece of thin flexible string of length L, of negligible weight.
Suppose the two ends of the string are firmly secured (“clamped”) at some
supports so they will not move. Assume the set▯up has no damping. Then,
the vertical displacement of the string, 0 < x < L, and at any time t > 0, is
given by the displacement function u(x,t). It satisfies the homogeneous one▯
dimensional undamped wave equation:
2
a u xxu tt
Where the constant coefficient a is given by the formula a = T/ρ, such that
a = horizontal propagation velocity of the wave motion, T = force of tension
exerted on the string, ρ = mass density (mass per unit length). It is subjected
to the homogeneous boundary conditions
u(0,t) = 0, and u(L,t) = 0, t > 0.
The two boundary conditions reflect that the two ends of the string are
clamped in fixed positions. Therefore, they are held motionless at all time.
The equation comes with 2 initial conditions, due to the fact that it contains
the second partial derivative term utt The two initial conditions are the
initial (vertical) displacement u(x,0), and the initial (vertical) velocity
ut(x,0), both are arbitrary functions of x alone. (Note that the string is
merely the medium for the wave, it does not itself move horizontally, it only
vibrates, vertically, in place. The resulting wave form, or the wave▯like
“shape” of the string, is what moves horizontally.)
© 2008, 2012 Zachary S Tseng E▯4 ▯ 2 Hence, what we have is the following initial▯boundary value problem:
(Wave equation) a u = u , 0 < x < L, t > 0 ,
xx tt
(Boundary conditions) u(0,t) = 0 , and u(L,t) = 0 ,
(Initial conditions) u(x,0) = f (x) , and utx,0) = g(x) .
We first let u(x,t) = X(x)T(t) and separate the wave equation into two
ordinary differential equations. Substituting u = X″T and u = XT″ into
xx tt
the wave equation, it becomes
2
a X″T = XT″.
© 2008, 2012 Zachary S Tseng E▯4 ▯ 3 Dividing both sides a X T :
X ′′ T ′′
= 2
X a T
A2 for the heat conduction equation, it is customary to consider the constant
a as a function of t and group it with the rest of t▯terms. Insert the constant
of separation and break apart the equation:
X ′′= T ′′=−λ
X a T
X ′′
=−λ → X″ = −λX → X″ + λX = 0,
X
T′′
2 =−λ 2 2
a T → T″ = −a λT → T″ + a λT = 0.
The boundary conditions also separate:
u(0,t) = 0 → X(0)T(t) = 0 → X(0) = 0 or T(t) = 0
u(L,t) = 0 → X(L)T(t) = 0 → X(L) = 0 or T(t) = 0
As usual, in order to obtain nontrivial solutions, we need to choose
X(0) = 0 and X(L) = 0 as the new boundary conditions. The result,
after separation of variables, is the following simultaneous system of
ordinary differential equations, with a set of boundary conditions:
X″ + λX = 0, X(0) = 0 and X(L) = 0,
2
T″ + a λT = 0 .
© 2008, 2012 Zachary S Tseng E▯4 ▯ 4 The next step is to solve the eigenvalue problem
X″ + λX = 0, X(0) = 0, X(L) = 0.
We have already solved this eigenvalue problem, recall. The solutions are
n π 2
Eigenvalues: λ = 2 , n = 1, 2, 3, …
L
nπ x
Eigenfunctions: X nsin , n = 1, 2, 3, …
L
Next, substitute the eigenvalues found above into the second equation to find
T(t). After putting eigenvalues λ into it, the equation of T becomes
2 2
2 n π
T ′+ a 2 T =0 .
L
It is a second order homogeneous linear equation with constant coefficients.
It’s characteristic have a pair of purely imaginarycomplex conjugate roots:
anπ
r = ± i.
L
Thus, the solutions are
anπt anπt
T nt) = A cns + B nin , n = 1, 2, 3, …
L L
Multiplying each pair of n and Tntogether and sum them up, we find the
general solution of the one▯dimensional wave equation, with both ends fixed,
to be
© 2008, 2012 Zachary S Tseng E▯4 ▯ 5 ∞
an t anπt nπ x
u x t ) =∑ An cos L +B nsin L sin L .
n=1
There are two sets of (infinitely many) arbitrary coefficients. We can solve
for them using the two initial conditions.
Set t = 0 and apply the first initial condition, the initial (vertical)
displacement of the string u(x,0) = f(x), we have
∞ nπx
u(x,0) =∑ (Ancos(0)+ B nin(0) sin )
n=1 L
∞
= A sin nπ x = f (x)
n=1 n L
Therefore, we see that the initial displacement f(x) needs to be a Fourier sine
series. Since f(x) can be an arbitrary function, this usually means that we
need to expand it into its odd periodic extension (of period 2L). The
coefficientsnA are then found by the relanion n = b , wnere b are the
corresponding Fourier sine coefficients of fhat is
L
A =b = 2 f (x)sinnπ x dx
n n L ∫ L .
0
Notice that the entire sequence of the coeffinients A are determined exactly
by the initial displacement. They are completely independent of the other
sequence of coefficients B , which are determined solely by the second
n
initial condition, the initial (vertical) velocityof the string. To find B , we
differentiate u(x,with respect to t and apply the initial velocity,
u (x,0) = g(x).
t
© 2008, 2012 Zachary S Tseng E▯4 ▯ 6 ∞ anπ an t an π an t n x
utx t ) =∑ − An sin + Bn cos sin
n 1 L L L L L
Set t = 0 and equate it with g(x):
∞ an π n x
utx ,0) ∑ Bn sin = g(x).
n=1 L L
We see that g(x) needs also be a Fourier sine series. Expand it into its odd
periodic extension (period 2L), if necessary. Once g(x) is written into a sine
series, the previous equation becomes
∞ anπ nπ x ∞ nπ x
u tx,0) =∑ B n sin = g(x)= ∑ bnsin
n=1 L L n=1 L
Compare the coefficients of theLlike sine terms, we see
anπ 2 nπ x
Bn =bn= ∫g(x)sin dx .
L L 0 L
Therefore,
L
B = L b = 2 g(x)sin nπ x dx
n anπ n anπ ∫ L .
0
As we have seen, half of the particular solution is determined by the initial
displacement, the other half by the initial velocity. The two halves are
determined independent of each other. Hence, if the initial displacement
f(x) = 0, then nll A = 0 and u(x,t) contains no sine▯terms of t. If the initial
velocity g(x) = 0, then all B = 0 and u(x,t) contains no cosine▯terms of t.
© 2008, 2012 Zachary S Tseng E▯4 ▯ 7 Let us take another look and summarize the result for these 2 easy special
cases, when either f(x) or g(x) is zero.
Special case I: Nonzero initial displacement, zero initial velocity: f(x) ≠ 0,
g(x) = 0.
Since g(x) = 0, then B = 0 for all n.
2 L nπ x
A n= ∫ f ( )sin dx, n = 1, 2, 3, …
L 0 L
Therefore,
∞
u( , ) = A cosanπt sinnπ x
∑ n L L .
n=1
© 2008, 2012 Zachary S Tseng E▯4 ▯ 8 The D’Alembert Solution
In 1746, Jean D’Alembert produced an alternate formof solution to the
wave equation. His solution takes on an especiallysimple form in the above
case of zero initial velocity.
Use the product formula sin(A)cos(B) = [sin(A − B) + sin(A + B)]/2, the
solution above can be rewritten as
∞
1 nπ (x at) nπ (x at)
u(x,t) = 2 ∑ A nn L +sin L
n 1
Therefore, the solution of the undamped one▯dimensional wave equation
with zero initial velocity can be alternatively expressed as
u(x,t) = [F(x − at) + F(x + at)]/.
Such that F(x) is the odd periodic extension (period 2L) of the initial
displacement f(x).
An interesting aspect of the D’Alembert solution is that it readily shows that
the starting waveform given by the initial displacement would keep its
general shape, but it would also split exactly into two halves. The two
halves of the wave form travel in the opposite directions at the same finite
speed of a. (Notice that the two halves of the wave form are being
translated/moved in the opposite direction at the rate of distance a per unit
time.)
© 2008, 2012 Zachary S Tseng E▯4 ▯ 9 Special case II: Zero initial displacement, nonzero initial velocity: f(x) = 0,
g(x) ≠ 0.
Since f(x) = 0, thnn A = 0 for all n.
L
2 nπ x
B n ∫ g(x)sin dx , n = 1, 2, 3, …
anπ 0 L
Therefore,
∞
u( , ) = B sinanπt sin nπ x
∑ n L L .
n=1
© 2008, 2012 Zachary S Tseng E▯4 ▯ 10 Example: Solve the one▯dimensional wave problem
9u xxu tt , 0 < x < 5, t > 0,
u(0,t) = 0, and u(5,t) = 0,
u(x,0) = 4sin(πx) − sin(2πx) − 3sin(5πx),
u (x,0) = 0.
t
2
First note thata = 9 (so a = 3), and L = 5.
The general solution is, therefore,
∞ 3nπt 3nπt nπ x
u(x,t) = ∑ A nos +B snn sin .
n 1 5 5 5
Since g(x) = 0, it must be that allnB = 0. We just need to find n . We
also see that u(x,) = f(x) is already in the form of a Fourier sine
series. Therefore, we just need to extract the corresponding Fourier
sine coefficients:
A 5 b 5 4,
A 10b =101,
A 25 =253,
A n b n 0, for all other n, n ≠ 5, 10, or 25.
Hence, the particular solution is
u(x,t) = 4cos(3πt)sin(πx) − cos(6πt)sin(2πx)
− 3cos(15πt)sin(5πx).
© 2008, 2012 Zachary S Tseng E▯4 ▯ 11 We can also solve the previous example using D’Alembert’s solution. The
problem has zero initial velocity and its initial displacement has already been
expanded into the required Fourier sine series, u(x,0) = 4sin(πx) − sin(2πx) −
3sin(5πx) = F(x). Therefore, the solution can also be found by using the
formula u(x, t) = [F(x − at) + F(x + at)]/2, where a = 3. Thus
u(x,t) = [ [ 4sin(π(x + 3t)) + 4sin(π(x − 3t)) ] − [sin(2π(x +
3t)) + sin(2π(x + 3t)) ] − [3sin(5π(x + 3t)) + 3sin(5π(x +
3t)) ] ]/2
Indeed, you could easily verify (do this as an exercise) that the solution
obtained this way is identical to our previous answer. Just apply the addition
formula of sine function ( sin(α ± β) = sin(α)cos(β) ± cos(α)sin(β) ) to each
term in the above solution and simplify.
© 2008, 2012 Zachary S Tseng E▯4 ▯ 12 Example: Solve the one▯dimensional wave problem
9u = u , 0 < x < 5, t > 0,
xx tt
u(0,t) = 0, and u(5,t) = 0,
u(x,0) = 0,
ut(x,0) = 4.
2
As in the previous exampla = 9 (so a = 3), and L = 5.
Therefore, the general solution remains
∞
u(x,t) = A cos 3nπt +B sin 3nπt sinnπ x
n 1 n 5 n 5 5 .
Now, f(x) = 0, consequently anl A = 0. We just need tn find B . The
initial velocity g(x) = 4 is a constant function. It is not an odd periodic
function. Therefore, we need to expand it into its odd periodic
extension (period T = 10), then equate it with u (x,0). In short:
L 5
B = 2 g x)sin n x dx = 2 4sin n x dx
n anπ ∫ L 3nπ ∫ 5
0 0
80
= 2 2 , n = odd
3n π
0 , n = even
Therefore,
∞ 80 3(2n −1)πt (2n −1)π x
u(x,t) =∑ sin sin
n=13(2n−1) 2π 2 5 5 .
© 2008, 2012 Zachary S Tseng E▯4 ▯ 13 The Structure of the Solutions of the Wave Equation
In addition to the fact that the constant a is the standing wave’s propagation
velocity, several other observations can be readily made from the solution of
the wave equation that give insights to the nature of the solution.
To reduce the clutter, let us look at the form of the solution when there is no
initial velocity (when g(x) = 0). The solution is
∞ anπt nπ x
u(x,t) = ∑ A nos sin
n 1 L L .
The sine terms are functions of x. They described the spatial wave patterns
(the wavy “shape” of the string that we could visually observe), called the
normal modes, or natural modes. The frequencies of those sine waves that
we could see, nπ/L, are called the spatial frequencies of the wave.
Meanwhile, the cosine terms are functions of t, they give the vertical
displacement of the string relative to its equilibrium position (which is just
the horizontal, or the x▯axis). They describe the up▯and▯down vibrating
motion of the string at each point of the string. These temporal frequencies
(the frequencies of functions of t; in this case, the cosines’) are the actual
frequencies of oscillating motion of vertical displacement. Since this is the
undamped wave equation, the frequencies of the cosine terms, anπ/L
(measured in radians per second), are called the natural frequencies of the
string. In a string instrument, they are the frequencies of the sound that we
could hear. The corresponding natural periods (= 2π/natural frequency) are
therefore 2L/an.
For n = 1, the observable spatial wave pattern is sin(πx/L). It is the string’s
first natural mode. The first natural frequency of oscillation, aπ/L, is called
the fundamental frequency of the string. It is also called, in acoustics, as the
first harmonic of the string.
© 2008, 2012 Zachary S Tseng E▯4 ▯ 14 For n = 2, the spatial wave pattern is sin(2πx/L) is the second natural mode.
The second natural frequency of oscillation, 2aπ/L, is also called the second
harmonic (or the first overtone) of the string. It is exactly twice of the
string’s fundamental frequency. Acoustically, it produces a tone that is one
octave higher than the first harmonic. For n = 3, the third natural frequency
is also called the third harmonic (or the second overtone), and so forth.
The motion of the string is the combination of all its natural modes, as
indicated by the general solution.
Lastly, notice that the “wavelike” behavior of the solution of the undamped
wave equation, quite unlike the solution of the heat conduction equation
discussed earlier, does not decrease in amplitude/intensity with time. It
never reaches a steady state. This is a consequence of the fact that the
undamped wave motion is a thermodynamically reversible process that
needs not obey the second law of Thermodynamics.
© 2008, 2012 Zachary S Tseng E▯4 ▯ 15 First natural mode (oscillates at the fundamental frequency / 1st harmonic):
Second natural mode (oscillates at the 2nd natural frequency / 2nd harmonic):
Third natural mode (oscillates at the 3rd natural frequency / 3rd harmonic):
© 2008, 2012 Zachary S Tseng E▯4 ▯ 16 Summary of Wave Equation: Vibrating String Problems
The vertical displacement of a vibrating string of length L, securely clamped
at both ends, of negligible weight and without damping, is described by the
homogeneous undamped wave equation initial▯boundaryvalue problem:
a u xxu tt, 0 < x < L, t > 0,
u(0,t) = 0, and u(L,t) = 0,
u(x,0) = f(x), and t (x,0) = g(x).
The general solution is
∞ anπt anπt nπ x
u(x,t) = ∑ A nos +B snn sin .
n 1 L L L
The particular solution can be found by the formulas:
L
A = 2 f (x)sinnπ x dx
n L ∫ L , and
0
2 L nπ x
B n= g(x)sin dx
anπ 0 L .
The solution waveform has a (horizontal) propagation velocity of a.
© 2008, 2012 Zachary S Tseng E▯4 ▯ 17 The General Wave Equation
The most general form of the one▯dimensional wave equation is:
2
a u +xx(x,t) = u + γu ttku. t
Where a = the propagation velocity of the wave,
γ = the damping constant
k = (external) restoration factor, such as when vibrations occur
in an elastic medium.
F(x, t) = arbitrary external forcing function (If F = 0 then the
equation is homogeneous, else it is nonhomogeneous.)
© 2008, 2012 Zachary S Tseng E▯4 ▯ 18 The Telegraph Equation
The most well▯known example of (a homogeneous version of) the general
wave equation is the telegraph equation. It describes the voltage u(x, t)
inside a piece of telegraph / transmission wire, whose electrical properties
per unit length are: resistance R, inductance L, capacitance C, and
conductance of leakage current G:
2
a u =xx + γutt ku. t
2
Where a = 1/LC, γ = G/C + R/L, and k = GR/CL.
© 2008, 2012 Zachary S Tseng E▯4 ▯ 19 Example: The One▯Dimensional Damped Wave Equation
2
a u xxu +ttu t, γ≠ 0.
Suppose boundary conditions remain as the same (both ends fixed): (0,t) = 0,
and u(L,t) = 0.
The equation can be separated as follow. First rewrite it as:
a X″T = XT″ + γXT′ ,
2
Divide both sides ba XT, and insert a constant of separation:
X ′′= T +γT ′= −λ
X a T .
Rewrite it into 2 equations:
X″ = −λX → X″ + λX = 0,
T″ + γT′ = − a λT → T″ + γT′ + a λT = 0.
The boundary conditions also are separated, as usual:
u(0,t) = 0 → X(0)T(t) = 0 → X(0) = 0 or T(t) = 0
u(L,t) = 0 → X(L)T(t) = 0 → X(L) = 0 or T(t) = 0
As before, setting T(t) = 0 would result in the constant zero solution
only. Therefore, we must choose the two (nontrivial) conditions in
terms of x: X(0) = 0, and X(L) = 0.
© 2008, 2012 Zachary S Tseng E▯4 ▯ 20 After separation of variables, we have the system
X″ + λX = 0, X(0) = 0 and X(L) = 0 ,
2
T″ + γT′ + α λT = 0 .
The next step is to find the eigenvalues and their corresponding
eigenfunctions of the boundary value problem
X″ + λX = 0, X(0) = 0 and X(L) = 0.
This is a familiar problem that we have encountered more than once
previously. The eigenvalues and eigenfunctions are, recall,
n π 2
Eigenvalues: λ = 2 ,

More
Less
Unlock Document

Related notes for MATH 250

Only pages 1,2,3,4 are available for preview. Some parts have been intentionally blurred.

Unlock DocumentJoin OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.