**Unformatted text preview: **MEC702 Wk 13 Lect 1 Partial Dierential Equations (PDEs)
Denition.
• A partial dierential equation (PDE) is an equation involving one or more
partial derivatives of an (unknown) function, call it u, that depends on
two or more variables, often time t and one or several variables in space. • Just as ODEs, we have that a PDE is linear if the function u and all its partial derivatives are of the rst degree. • But the denition of homogeneous PDE is dierent from homogeneous
ODE. A PDE is homogeneous if all the terms of the PDE contain either u or one of its partial derivatives. Otherwise, the PDE is nonhomogeneous. Example. Important PDEs.
∂2u
∂2u
= c2 2
2
∂t
∂x
2
∂u
∂
u
= c2 2
∂t
∂x
∂2u ∂2u
+ 2 =0
∂x2
∂y
∂2u ∂2u
+ 2 = f (x, y)
∂x2
∂y
2
∂2u
∂ u ∂2u
2
=
c
+
∂t2
∂x2
∂y 2
∂2u ∂2u ∂2u
+ 2 + 2 =0
∂x2
∂y
∂z One-dimensional wave eqn
One-dimensional heat eqn
Two-dimensional Laplace eqn
Two-dimensional Poisson eqn
Two-dimensional wave eqn
Three-dimensional Laplace eqn where c > 0 is a constant, t is the time variable, x, y , z are the Cartesian
coordinates, and dimension is the number of these coordinates in the equation. Example. Verify that the following functions are all solutions to the twodimensional Laplace equation.
1. u = x2 − y 2
2. u = ex cos y
1 3. u = ln(x2 + y 2 )
The above example shows that number of solutions that a PDE can have is
very large. To obtain a unique solution corresponding to a physical problem,
we make use of additional conditions which maybe
• Initial Conditions - when time t is one of the variables, the value of the
function u may be given for t = 0 • Boundary Conditions - for the region R where the function u is dened,
values of u may be given at the boundaries of R. Remark. A partial derivative ∂f
∂x can also be written as fx , hence ∂2f
∂x2 = fxx . Theorem. (Fundamental Theorem on Superposition) If u1 and u2 are solutions
of a homogeneous linear PDE in some region R, then
u = c1 u1 + c2 u2 with any constants c1 and c2 is also a solution of that PDE in the region R. Example. Solving the PDE uxx − u = 0 like an ODE. Find the solutions u of the PDE uxx − u = 0 depending on x and y . Solution. Since no y-derivatives occur, we can solve this PDE like the ODE
u00 − u = 0. For the ODE, we have the solution
u = Aex + Be−x with constant A and B .
For the PDE, we have the solution
u(x, y) = A(y)ex + B(y)e−x with arbitrary functions A(y) and B(y), thus obtaining a variety of solutions. Example. Solving the PDE uxy = ux Find solutions u = u(x, y) of this PDE. like an ODE. Solution. Setting ux = p, we have py = −p. We have
py
=
p
ln |p| =
p = −1
−y + ∂(x)
c(x)e−y and by integration with respect to x,
where u(x, y) = f (x)e−y + g(y) where f (x) and g(y) are arbitrary functions.
2 f (x) = ´ c(x)dx Modeling: Vibrating String, Wave Equation
The model of a vibrating elastic string (a violin string, for instance) consists of
the one-dimensional wave equation
∂2u
∂2u
= c2 2
2
∂t
∂x where c2 = Tρ (the term c2 indicates that the constant is positive).
Since the string is fastened at the ends x = 0 and x = L, we have the boundary conditions u(0, t) = 0
u(L, t) = 0 for all t ≥ 0. Furthermore, the form of the motion of the string will depend on its
initial deection (deection at time t = 0), call it f (x), and on its initial velocity
(velocity at t = 0), call it g(x). Thus, we have the two initial conditions
u(x, 0) = f (x)
ut (x, 0) = g(x) for all 0 ≤ x ≤ L. Solution by Separating Variables. Use of Fourier
Series.
We will now solve the one-dimensional wave equation.
Step 1: Two ODEs from the Wave Equation In the method of separating variables, or product method, we determine solutions of the wave equation of the form
u(x, t) = F (x) · G(t) which are product of two functions. This is a powerful general method that
has various applications in engineering mathematics. Dierentiating the two
functions,
∂2u
∂2u
00
00
and
∂t2 = F G
∂x2 = F G.
Inserting into the wave equation gives
F G00 = c2 F 00 G. 3 Dividing by c2 F G and simplifying gives
1 G00
F 00
=
.
c2 G
F The variables are now separated; the left-side depending only on t and the
right-side only on x. Hence both sides must be constant. Thus,
1 G00
F 00
=
= k.
c2 G
F Multiplying by the denominators gives immediately two ODEs
F 00 − kF = 0 and G00 − c2 kG = 0 where the separation constant k is still arbitrary.
Step 2: Satisfying the Boundary Conditions We now determine solutions F (x) and G(t) of the wave equation so that u =
F (x)G(t) satises the boundary conditions, that is,
u(0, t) = F (0)G(t) = 0
u(L, t) = F (L)G(t) = 0 for all t.
We now solve for F (x). If G ≡ 0, then u = F G = 0, which is of no interest.
Hence consider G 6= 0 and then by the boundary conditions,
F (0) = 0, F (L) = 0 . We must show that k must be negative.
• Assume that k = 0, the general solution of F 00 − kF = 0 is
F = ax + b, and from the conditions above, a = b = 0, so that F ≡ 0 and u = F G ≡ 0,
which is of no interest.
• Assume that positive k = µ2 , a general solution of F 00 − µ2 F = 0 is
F (x) = Aeµx + Be−µx , and from the conditions above, we obtain F ≡ 0 as before.
4 • Hence we are left with the possibility of choosing k to be negative, say
k = −p2 . Then
F 00 + p2 F = 0 and has a general solution
F (x) = A cos px + B sin px. From this and the boundary conditions, we have
F (0) = A=0 F (L) = B sin pL = 0. We must assume B 6= 0 otherwise F ≡ 0. Then we have sin pL = 0, thus
so that pL = nπ p= nπ
L . Setting B = 1, we thus obtain innitely many solutions F (x) = Fn (x),
where
Fn (x) = sin nπ
x
L for n = 1, 2, . . ..
We next solve for G(t). We have
k = −p2 = − then G00 + λ2 G = 0 where nπ 2
L
λn = cp = cnπ
L . We obtain the general solution of Gn (t) is
Gn (t) = Bn cos (λn t) + Bn∗ sin (λn t) Therefore the general solution of the one-dimensional wave equation
h nπ i
un (x, t) = [Bn cos (λn t) + Bn∗ sin (λn t)] sin
x
L where n = 1, 2, . . ..
• For each n = 1, 2, . . ., the functions un (x, t) are called the eigenfunctions, or characteristics functions. • The values λn = cnπ
L the vibrating string. are called eigenvalues, or characteristic values, of • The set [λ1 , λ2 , . . .] is called the spectrum.
5 Step 3: Solution of the Entire Problem. Fourier Series. By the Fundamental Theorem on Superposition, the complete general solution of the one-dimensional wave equation is
u(x, t) = ∞
X un (x, t) = n=1 where λn = ∞ h
nπ i
X
x
(Bn cos (λn t) + Bn∗ sin (λn t)) sin
L
n=1 which is an innite series. xnπ
L Applying Initial Conditions. • Applying the rst initial condition u(x, 0) = f (x) (initial displacement),
u(x, 0) = ∞
X Bn sin n=1 nπ
x = f (x).
L Hence we must choose the terms Bn so that u(x, 0) becomes the Fourier
sine series of f (x), thus
2
Bn =
L ˆ L f (x) sin nπx 0 L dx • Applying the second initial condition ut (x, 0) = g(x) (initial velocity),
"∞
# nπ o
Xn
∂u ∗
=
x
[−Bn λn sin (λn t) + Bn λn cos (λn t)] sin
∂t t=0
L
n=1 t=0 = ∞
X nπ
Bn∗ λn sin
x = g(x).
L
n=1 This gives a Fourier sine series of g(x). Then the Fourier coecients are
Bn∗ λn Since λn = cnπ
L 2
=
L ˆ L g(x) sin
0 nπ
x dx
L , we obtain by division
Bn∗ 2
=
cnπ ˆ L g(x) sin
0 6 nπ
x dx
L Establishing the solution. The general solution is purely a formal expression. For the sake of simplicity, we consider only the case when initial velocity
g(x) = 0. Then the terms Bn∗ = 0, thus
u(x, t) = P∞
n=1 Bn cos (λn t) sin nπx
L , λn = cnπ
L . Using the formula,
cos nπ
nπ 1 h nπ
nπ
i
sin
ct sin
x =
(x − ct) + sin
(x + ct) ,
L
L
2
L
L we have
u(x, t) =
= = ∞
X Bn cos (λn t) sin nπx n=1
∞
X L h nπ
nπ
i
1
Bn sin
(x − ct) + sin
(x + ct)
2 n=1
L
L
"∞
#
∞
n nπ
o X
n nπ
o
1 X
Bn sin
(x − ct) +
Bn sin
(x + ct) .
2 n=1
L
L
n=1 Notice the two Fourier sine series. Denote
f ∗ (x) = Bn ∞
X
n=1 then
u(x, t) = sin nπ
x ,
L 1 ∗
[f (x − ct) + f ∗ (x + ct)]
2 where f ∗ (x) is the odd extension of f (x) with the period 2L. 7 ...

View
Full Document

- Winter '16
- Alveen Chand
- Sin, Partial differential equation, wave equation, one-dimensional wave equation