Unformatted text preview: ESE 502 Mathematics of Modern Engineering II
Lecture 6
”Wave equation continued. D’Alembert’s method”.
◦ We have solved the wave equation with specified boundary/initial conditions using the method of separation of variables.
◦ It is useful to notice that the obtained solution of the wave euqation can
be written in a closed form as well (that is, without an infinite series sign)
Indeed: If we assume for simplicity that g = 0, then the solution is written
as
∞
X
nπ
u(t, x) =
an cos λn t sin
x
L
n=1
or
u(t, x) = ∞
X an cos n=1 cnπ
nπ
t sin
x.
L
L Using the formulas for the sine of a sum and a difference of two variables,
we obtain that
u(t, x) = ∞
X 1
nπ
nπ
an [sin
(x − ct) + sin
(x + ct)] =
2
L
L
n=1 ∞ ∞ nπ
nπ
1X
1X
an sin
an sin
(x − ct) +
(x + ct)
2 n=1
L
2 n=1
L
where an are the Fourier coefficients of the function f .
When we define by f ? the odd periodic extension of the function f on IR
with the period 2L, then we can rewrite the solution u(t, x) as
1
1
u(t, x) = f ? (x − ct) + f ? (x + ct).
2
2 (1) Example 1. Let f (x) = Kx2 (1 − x), 0 < x < 1 where K = .01 and
g(x) = 0, c2 = 1. Find the solution of the wave equation specified by the
given conditions.
1 Solution: The initial deflection function satisfies the conditions needed to
allow the representation through Fourier series so that the solution can be
written in the form
u(t, x) = ∞
X an cos n=1 where
2
an =
L L Z f (x) sin
0 cnπ
nπ
t sin
x
L
L nπ
xdx, n = 1, 2, ...
L Alternatively, we can use the formula (1) as well.
Basically, we have to find only the coefficients an .
Z
an = 2 1 Kx2 (1 − x) sin nπxdx = 0 −Kx2 (1 − x) cos nπx 1 K
2[
0 +
nπ
nπ 1 Z 2K (2x − 3x2 ) sin nπx 1
1
[
0 −
nπ
nπ
nπ (2x − 3x2 ) cos nπxdx] = 0 Z 1 (2 − 6x) sin nπxdx] =
0 (2 − 6x) cos nπx 1
6
2K
[−

−
−
0
(nπ)2
nπ
nπ
−
− Z 1 cos nπxdx] =
0 2K 4 cos nπ + 2
[=
(nπ)2
nπ 4K
[1 + 2 cos nπ], n = 1, 2, ...
(nπ)3 Another way of solving the wave equation: change of variables
◦ Change of variables method is a common approach when trying to solve
a particular problem. The goal is that by introducing the new variables the
given problem becomes the one more easier to solve in the new variables.
After the transfromed problem is solved in new variables, one uses the
inverse change of variables to express the solution in terms of the original
variables.
2 Assume that we have to solve the wave equation
utt = c2 uxx . (2) We introduce new variables (τ, y) defined as
τ = x + ct and y = x − ct.
To rewrite the equation (2) in new variables (τ, y), we first use the chain
rule to find that:
ux = uτ τx + uy yx = uτ + uy ;
uxx = (uτ + uy )x = uτ τ τx + uτ y yx + uτ y τx + uyy yx =
uτ τ + 2uτ y + uyy
where we assume that the partial derivatives of u are continuous functions
thus the mixed partial derivatives coincide.
ut = uτ τt + uy yt = c(uτ − uy );
utt = c(uτ − uy )t == .... = c2 (uτ τ − 2uτ y + uyy ).
In new variables the equation (2) becomes (we substitute the expressions
for utt and uxx found into (2)):
uτ y = 0 (3) which is much easier to solve. In fact, the equation (3) can be solved by
simple integration in variables τ and y (in any order) . We obtain that
u(τ, y) = G(τ ) + H(y)
or
u(t, x) = G(x + ct) + H(x − ct) (4) where G and H are arbitrary differentiable functions.
◦ The function u given in (4) describes the general solution of the wave
equation (2).
◦ To find the unique solution of a particular wave equation, we need to use
the additional boundary/initial conditions.
◦ We assume here that those conditions are specified as in the previous
lecture and, for simplicity, we assume that g = 0.
3 Since u(0, x) = f (x) we obtain that
f (x) = G(x) + H(x). (5) On another hand, differentiating the function u(t, x) in t, we obtain that
ut (t, x) = cG0 (x + ct) − cH 0 (x − ct)
so that the condition ut (0, x) = g(x) = 0 yields
G0 (x) − H 0 (x) = 0.
Formally, integrating the last equation in x gives us that
G(x) − H(x) = K = const.
Together with the equation (5) this leads to
1
1
G(x) = f (x) + K
2
2
and 1
1
H(x) = f (x) − K.
2
2
Formally, using instead of x the expressions x + ct and x − ct we have that
1
1
G(x + ct) = f (x + ct) + K
2
2
and 1
1
H(x − ct) = f (x − ct) − K.
2
2
Adding G(x + ct) and H(x − ct), we obtain that
1
u(t, x) = [f (x + ct) + f (x − ct)].
2 (6) However, the last expression is merely a formal form of a function satisfying
the wave equation. We haven’t fully used the conditions on u that should
be satisfied: we used the initial conditions so far but not the boundary
conditions.
◦ Using the boundary conditions, will lead to the function f that should
be:
 an odd function;
4  must have the period 2L;
Indeed: The conditions u(t, 0) = 0 yields that
1
[f (ct) + f (−ct)] = 0
2
so that
f (−ct) = −f (ct)
meaning that f must be an odd function as a function defined on IR.
The conditions u(t, L) = 0 yields that
1
[f (ct + L) + f (−ct + L)] = 0
2
so that
f (ct + L) = −f (−ct + L) = f (ct − L)
since f is an odd function. But the relation f (ct + L) = f (ct − L) means
that f must be a periodic function with period 2L defined as a function
on IR.
In summary, we established that the solution of the wave equation with
the specified boundary/initial conditions is given in the form
1
u(t, x) = [f ? (x + ct) + f ? (x − ct)]
2 (7) wheer f ? is the odd periodic extension of the function f on IR. This
solution coinsides exactly with the solution u(t, x) we obtained earlier using
the method of separation of variables combined with the Fourier series
techniques.
Remark 1. The most important aspect in the change of variables idea
we have demonstrated was to choose the change of variables in a way that
the new PDE becomes a more easier one that can be solved using simple
integration techniques. For a particular class of 2nd order PDEs, there
is a procedure that suggests how that suitable change of variables can be
found. This idea is due originally to D’Alembert so that the corresponding
solution method in called D’Alembert’s method.
D’Alembert’s method
5 Assume that one has to solve a second order PDE of the form
Autt + 2Butx + Cuxx = F (t, x, u, ut , ux ) (8) where A, B, and C are the coefficients (they could be some real numbers
but also some functions depending on (t, x)).
Step 1: Consider what is called the characteristic equation of (8) given by
A(x0 )2 − 2B(x0 ) + C = 0 (9) where x = x(t) is seen as the function depending on t so that x0 is the
derivative of x in t.
Step 2: Solve the equation (9) implicitely as an ODE. Since the eqution is
quadratic in x0 , there will be two solutions and three possible cases: two
different real solutions, or one real solution only (of the multiplicity 2), or
two complexconjugate solutions.
Step 3: Since any implicit solution of the equation (9) can be written in
a form Φ(t, x) = const, the corresponding change of variables to be made
will be the following:
Case I (two real roots called also the hyperbolic case):
τ = Φ(t, x) and y = Ψ(t, x).
In this case the change of variables reduces the equation (8) to the equation
uτ y = F1 ;
Case II( one real root calle also the parabolic case):
τ = t and y = Φ(t, x);
In this case the change of variables reduces the equation (8) to the equation
uτ τ = F2 ;
Case III (two complexconjugate roots called also elliptic case):
1
1
τ = [Φ(t, x) + Ψ(t, x)] and y = [Φ(t, x) − Ψ(t, x)].
2
2i
In this case the change of variables reduces the equation (8) to the equation
uτ τ + uyy = F3 ;
6 The functions Fi , i = 1, 2, 3 are some functions of the similar form as the
function F .
Remark 2. Change of variables in case III does not really simplify the
original equation as the transformed PDE is of the same kind as the given
PDE. In this case one has to use other means in trying to find a better
change of variables as suggested by the D’Alembert’s method.
Remark 3. The hyperbolic case is equivalent to the condition: AC −B 2 <
0;
The parabolic case is equivalent to the condition: AC − B 2 = 0; and
The elliptic case is equivalent to the condition: AC − B 2 > 0.
Example 2. Find the general solution of the PDE utt − 16uxx = 0 using
the D’Alembert’s method.
Example 3. Do the same as in Example 2 for the PDE tutx − xuxx = 0. 7...
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 Fall '09
 Partial differential equation, wave equation, uτ τ