2332 Tutorial Sheet
1
Questions
1. Prove uniqueness for solutions of the Klein-Gordon or Helmholz equation
4φ = m2 φ
(1)
on a region D and with Dirichlet or Neumann boundary conditions on δD.
The way to do this is to again define an energy term
Z
E=
dV [(∇φ)2 + m2 φ2 ]
(2)
V
and, if there are two solutions φ1 and φ2 , let φ = φ1 − φ2 ; this will satisfy the pde
with zero boundary conditions. As for the Laplace equation case, we take one of the
∇’s outside to give a boundary term
Z
Z
2
2 2
E=
dV [(∇φ) + m φ ] =
dV [∇(φ∇φ) + m2 φ2 − φ4φ]
(3)
V
V
where we have taken away a φ4φ since, from our identities, for φ a scalar and F a
vector field
∇(φF) = ∇φ · F + φ(∇ · F)
(4)
Now the last two terms cancel because of the equation; if we hadn’t added the m2 φ2
term to the energy this bit would not have worked. Hence
Z
Z
E=
dV [∇(φ∇φ) =
φ(∇φ) · dS = 0
(5)
V
δV
where we have used the Gauss theorem and we get zero because of the boundary
conditions. Now E can only be zero if φ = 0.
1
Sinéad Ryan, ryan@maths.tcd.ie
1
2. Prove uniqueness for solutions to the heat equation
4u = k
∂u
∂t
(6)
on a region D × [0, ∞) and with Dirichlet or Neumann boundary conditions on
δD × [0, ∞), initial condition u(x, 0) = f (x) on D at time t = 0 and decay condition
u(x, t) → 0 exponential fast as t goes to infinity, k is a constant.
Again, say we have two solutions, u1 and u2 , then the difference u = u1 − u2 satisfies
the equation with zero boundary conditions. Consider
Z ∞Z
E=
dV [(∇u)]
(7)
0
V
and again taking one of the ∇ outside
Z ∞ Z
E =
dt
dV [(∇u)]
Z0 ∞ ZV
=
dt
dV [∇(u∇u) − u4u]
Z0 ∞ ZV
∂
dV [u u]
dt
=
∂t
V
0
(8)
Where again we have used the Gauss theorem and the boundary conditions. The only
difference is that getting rid of the Laplace term has left us with a time derivative
term, however, this can be rewritten as a total derivative
Z ∞ Z
∂
E =
dt
dV [u u]
∂t
0Z
VZ
∞
∂
1
dt[ u2 ]
dV
=
2 ZV
∂t
0
1
∞
=
dV u2 0 = 0
(9)
2 V
so again we have ∇u = 0 which means u is constant in space and substituting into
the heat equation tells us it is also constant in time, and, by the initial condition,
that constant is zero.
2
3. The function φ(x, y) is harmonic in the square 0 ≤ x ≤ π, 0 ≤ y ≤ π. On three sides
φ is zero and on the lower side
φ(x, 0) = cos x.
(10)
Determine φ(x, y) within the square. Let φ(x, y) = X(x)Y (y) and substitute in to
get
X 00 Y + Y 00 X = 0
(11)
and moving stuff around, this gives
X 00
Y 00
=−
X
Y
(12)
and so each side depends on a different independent variable, so they must be equal
to a constant
X 00 = EX
Y 00 = EY
(13)
so there are three classes of solutions E = 0
X = Ax + B
(14)
X = Aekx + Be−kx
(15)
X = A sin kx + B cos kx
(16)
or E = k 2
or E = −k 2
with similar solutions for Y but, because of the extra minus, with the positive and
negative E solutions the other way around. Now, we try to match the boundary
conditions. At x = 0 and x = π we have zero φ, hence, for Y non-trivial we must
have X(0) = X(π) = 0, as in the notes, it is not possible to match these boundary
conditions for the E positive and E = 0 solutions, this leaves the E = −k 2 , X(0) = 0
gives B = 0 and X(π) = 0 give
sin kπ = 0
(17)
and hence k = n a positive natural number, negative numbers give the same solutions
and n = 0 is just the E = 0 solution which was dealt with seperately.
Now, our general solution, satisfying the x boundary conditions and the differential
equation, is
∞
X
φ(x, y) =
sin nx Cn eny + Dn e−ny
(18)
n=1
and putting φ(x, π) = 0 gives
Cn enπ = −Dn e−nπ
3
(19)
Now, to satisfy the y = 0 condition
cos x =
∞
X
An sin nx
(20)
n=1
where An = Cn + Dn and 0 < x < π. We want to calculate this as a Fourier series,
so we have to extend the boundary function as an odd period function: we need a
f (x) such that f (x) = cos x for x ∈ (0, π), f (−x) = −f (x) and f (x + 2π) = f (x).
Hence, we define
cos x
x ∈ (0, π)
f (x) =
(21)
− cos x x ∈ (−π, 0)
and f (x + 2π) = f (x) and we expect to be able to find An such that
f (x) =
∞
X
An sin nx
(22)
n=1
because it is a sine series for an odd function, and if we do that, we will satisfy
the boundary condition because f (x) reduces to the boundary condition along the
boundary. By the formula from Fourier analysis
Z
Z
1 π
2 π
An =
f (x) sin nxdx =
cos x sin nxdx
(23)
π −π
π 0
Now
2
π
Z
0
π
Z
1 π
cos x sin nxdx =
(sin (n + 1)x − sin (n − 1)x) dx
π 0
π
1
1
1
=
cos (n − 1)x −
cos (n + 1)x
π n − 1
n +1
0
2 1 + (−1)n 1 + (−1)n
=
−
π
n+1
n−1
(24)
so
4
−1
for n even and zero for n odd. Now, solving for Cn and Dn we get
X 4
1
1
ny
−ny
e +
e
φ(x, y) =
sin nx
n2 − 1
1 − e2πn
1 − e−2πn
An = −
where the sum is over even values of n.
4
n2
(25)
(26)