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MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #2 VERSION B
- Consider the general solution of the wave equation for a vibrating string with fixed ends:
u(x, t) =
∑^ ∞
n=
sin
( (^) πnx L
(bn cos (λnt) + b∗ n sin (λnt))
where λn = πcn L
Suppose that at initial time we have initial position f (x) = 0 (a flat string). At what later time t will the string be completely flat again?
Solution: The bn vanish because it is initially flat, and at time t = 0, we have
0 = u(x, 0) =
∑^ ∞
n=
sin
( (^) πnx L
bn cos (λnt)
The function sin x has zeros at all multiples of π. These sin (λnt) functions have frequencies λn = πcn L which are multiples of the first one
λ 1 = πc L so they go back to zero when the first one does, which is at
λ 1 t = π
or t =
L
c
- Use separation of variables to find the general solution of the heat equation
∂u ∂t = c^2
∂^2 u ∂x^2
(^2) u ∂y^2
in a rectangular plate with edges of lengths a and b, and with top, bottom and right edges insulated, while the left edge is kept at 0o. Hint: recall the solutions of the harmonic oscillator type equation
−
h
d^2 h dt^2 = λ
are
h(t) =
A cos
λt
λt
if λ > 0 A + Bt if λ = 0 A cosh
|λ|t
|λ|t
if λ < 0 Date: November 21, 2000. 1
where A and B can be chosen to be any constants. Also recall the identities about cosh and sinh: d dx cosh x = sinh x d dx sinh x = cosh x sinh 0 = 0 cosh 0 = 1
and the fact that cosh x is positive for every value of x.
Solution: The equations to solve are
∂u ∂t = c^2
∂^2 u ∂x^2
∂^2 u ∂y^2
u = 0 at x = 0 & ∂u ∂x = 0 at x = a ∂u ∂y = 0 at^ y^ = 0 &^ y^ =^ b
The equations for the insulated sides use Fourier’s law to ensure that there is no heat flux through the boundary. Separating variables:
u(x, y, t) = E(x)F (y)G(t)
(we just split u into a product of functions, one of each variable) we find the heat equation becomes EF G′^ = c^2 (E′′F G + EF ′′G)
which, once we divide by EF G, is just
G′ G = c^2
E′′
E
F ′′
F
The left side involves only t, and the right side only x and y. So there is a constant γ so that G′ G = −γ = c^2
E′′
E
+ F^
′′ F
Therefore G(t) = e−γt
while
− γ c^2
E′′
E
F ′′
F
or
−
E′′
E
γ c^2
F ′′
F
The left side depends only on x, and the right only on y, so that there is some constant β so that
−
E′′
E
= β = γ c^2
F ′′
F
so that √ βa = π
for some integer m, and we have E(x) oscillating, with
E(x) = sin
π a
x
In terms of m and n, everything looks like
E(x) = sin
π a
F (y) = cos
( (^) πny b
G(t) = e−γmnt
where
γmn = c^2 π^2
m + (^12)
a^2
n^2 b^2
So
u(x, y, t) = exp (−γmnt) sin
π a
x
cos
( (^) πmy b
This is the same function if we replace n by −n, so we can suppose n = 0, 1 , 2 ,.... It just changes sign if we replace
m +
2 → −m^ −^
2 = (−^1 −^ m) +
so we can suppose that m = 0, 1 , 2 ,... as well. The general solution of the heat equation with three sides insulated and the fourth at 0o^ is
u(x, y, t) =
∑^ ∞
m=
∑^ ∞
n=
Amn exp (−γmnt) sin
π
x
cos
( (^) πmy b
- Recall that the steady state of the heat equation in a disk of radius R with edge temperature f (θ) is
u(r, θ) = a 0 +
∑^ ∞
n=
( (^) r R
)n (an cos (nθ) + bn sin (nθ)).
Find the total heat
Q =
u(x, y) dx dy =
u(r, θ) r dr dθ
where the integral is carried out over the whole disk.
0 –0.5 – 1 0.5 x
–0. 0 0. 1
y
1
Figure 1. A round peak
Solution:
Q =
u(r, θ) r dr dθ
∫ R
0
r dr
∫ (^2) π
0
u(r, θ) dθ
= 2πa 0
∫ R
0
r dr +
∑^ ∞
n=
∫ R
0
( (^) r R
)n r dr
∫ (^2) π
0
(an cos (nθ) + bn sin (nθ)) dθ
= πR^2 a 0 +
∑^ ∞
n=
Rn+ n + 2
Rn^ (an · 0 + bn · 0)
= 2πR^2 a 0
which is just the area of the disk multiplied by the average value a 0 of the edge temperature f (θ).
- Looking at the heat equation in polar coordinates ∂u ∂t = c^2
∂^2 u ∂r^2
r
∂u ∂r
r^2
∂^2 u ∂θ^2
and recalling first and second derivatives from calculus, explain mathematically why a round peak in the graph of u (as in figure 1) representing a blob of heat will be pushed downward (will cool).
Solution: The first derivative vanishes at a maximum, so ∂u ∂r
The second derivatives in r and in θ will be negative, so the right hand side of the wave equation is negative. Therefore the left hand side, the rate of change of temperature, must be negative too. So it will cool.
