The given PDE is
4uxx + 5uxy + uyy + ux + uy = 2 (1)
Re-arrange (1) so that it can be compared with the standard
4uxx + 5uxy + uyy = 2 − (ux + uy) (2)
A PDE of the form
Auxx + Buxy + Cuyy = Φ(x, y, u, ux, uy)
can be classified as hyperbolic, parabolic or elliptical, based on the sign of
the discriminant (Sahil, 2014)
∆ = B 2A
2C B = B
2 − 4AC
Comparing (2) with the standard form, we get A = 4, B = 5 and C = 1.
Therefore,
∆ = (5)2 − 4(4)(1) = 9
Since ∆ > 0, the PDE in (1) is hyperbolic.
b)
According to Sahil (2014), “In order to obtain the canonical form of (1) we
1
first need to transform the independent variables x and y to new independent
variables ξ and η through change of variables”
η = η(x, y), ξ = ξ(x, y)
According to Sahil (2014), “Both ξ and η are twice continuously differentiable
and the Jacobian does not vanish in the region of consideration”, that is
J =
∂(ξ, η)
∂(x, y)
6= 0
Define, w(ξ, η) = u(x(ξ, η), y(ξ, η)), then u(x, y) = w(ξ(x, y), η(x, y)). Therefore, by chain rule
ux = wξξx + wηηx
uy = wξξy + wηηy
uxx = wξξξ
2
x + 2wξηξxηx + wηηη
2
x + wξξxx + wηηxx
uyy = wξξξ
2
y + 2wξηξyηy + wηηη
2
y + wξξyy + wηηyy
uxy = wξξξxξy + wξη(ξxηy + ξyηx) + wηηηxηy + wξξxy + wηηxy
Substituting the derivatives in (2) we get
awξξ + bwξη + cwηη = ζ(ξ, η, w, wξ, wη)
where,
a = 4ξ
2
x + 5ξxξy + ξ
2
y
b = 8ξxηx + 5(ξxηy + ξyηx) + 2ξyηy
c = 4η
2
x + 5ηxηy + η
2
y
If a and c become equal to 0, through a transformation, then the PDE is
reduced into a much simpler form called canonical form. To obtain this
transformation we require
a = 4ξ
2
x + 5ξxξy + ξ
2
y = 0
c = 4η
2
x + 5ηxηy + η
2
y = 0
2
Divide a by ξ
2
y
and c by η
2
y
. Therefore,
4

ξx
ξy
2

• 5 
  ξx
  ξy
  
• 1 = 0 (3)
  4
  
  ηx
  ηy
  2
• 5 
  ηx
  ηy
  
• 1 = 0 (4)
  Along the ξ direction: dξ = ξxdx + ξydy = 0, which gives dy
  dx
  = −
  ξx
  ξy
  .
  Similarly we can obtain: dy
  dx
  = −
  ηx
  ηy
  . Substitute in (3) and (4)
  4
  
  dy
  dx
  2
  − 5
  
  dy
  dx
  
• 1 = 0 (5)
  Equation (5) is called characteristic polynomial of (2) and its roots are
  dy

dx

5 ±
p
5
2 − 4(4)(1)
8
= 1,
1
4
The characteristic curves or simply characteristics of the equation are the
curves along which ξ and η remain constant. These are obtained by solving
for dy
dx
in the above equation. Therefore
dy
dx
= 1 =⇒ y = x + c1 &
dy

dx

1
4
=⇒ y =
1
4
x + c2
where, c1 and c2 are constants of integration. This implies that y − x = c1,
and y −
1
4
x = c2, are the lines along which ξ and η constant. Hence
ξ = y − x & η = y −
1
4
x
are the characteristics of equation (1).
To find the canonical form of the equation, we focus on the equation
awξξ + bwξη + cwηη = ζ(ξ, η, w, wξ, wη)
3
Since a = c = 0 for the transformation obtained, we get
bwξη = ζ(ξ, η, w, wξ, wη)
b can be obtained using formula (which can be verified easily)
b =
−∆
A
= −
9
4
Since ξ and η are linear in x and y, the second derivatives of ξ and η in
x and y will vanish, which leaves the right hand side as ζ(ξ, η, w, wξ, wη) =
2 − (ux + uy), or more correctly
ζ = 2−(wξξx+wηηx+wξξy+wηηy) = 2−(wξ(−1)+wη(−1/4)+wξ+wη) = 2−

wη
3
4

Therefore,
−
9
4
wξη = 2 −

wη
3
4

∴ wξη −
wη
3
= −
8
9
(6)
is the canonical form of (1).
c)
The general solution to the PDE in (1) can be obtained by solving the canonical form of the equation as given in (6)
wξη −
wη
3
= −
8
9
Integrate w.r.t ξ, therefore
wη −
w
3
= −
8
9
ξ + g(η)
[Integrating factor µ = e
−η/3
]
∴ w(ξ, η) = e
η/3
Z
e
−η/3

−
8
9
ξ + g(η)

∂η + C
u(x, y) = e
(y−0.25x)/3
Z
e
−(y−0.25x)/3

−
8
9
(y − x) + g(y − 0.25x)

(∂y − 0.25∂x) + C
4
is the general solution to (1).
Answer
a)
The given PDE is:
∂u
∂t = h
2
∂
2u
∂
2x
− ku (1)
5
Let us define a new variable v, such that:
u(x, t) = e
−ktv(x, t)
Take partial derivatives of both sides:
∂
2
∂x2

[u(x, t)]

= ∂
2
∂x2

[e
−ktv(x, t)]

∂
2u
∂x2
= e
−kt ∂
2
v
∂x2
(2)
Also,
∂
∂t[u(x, t)] = ∂
∂t[e
−ktv(x, t)]
∂u
∂t = e
−kt ∂v
∂t + v
∂
∂t(e
−kt)
∂u
∂t = e
−kt ∂v
∂t − kve−kt
= e
−kt ∂v
∂t − ku (3)
Put (2) and (3) in (1):
e
−kt ∂v
∂t − ku = h
2

e
−kt ∂v
∂x
− ku
[Add ku to both sides]
e
−kt ∂v
∂t = h
2

e
−kt ∂v
∂x
[Divide throughout by e
−kt]
∴
∂v
∂t = h
2
∂
2
v
∂x2
(4)
b)
Boundary conditions for u(x, t):
u(0, t) = u(1, t) = 0
6
Initial conditions for u(x, t):
u(x, 0) = x(1 − x)
Replace u by ve−kt. Therefore, the boundary conditions become:
v(0, t)e
−kt = v(1, t)e
−kt = 0
e
−kt cannot be zero, since it’s an exponential term. Therefore, the boundary
conditions in terms of v are:
v(0, t) = v(1, t) = 0 (5)
For the initial condition we have:
v(x, 0)e
−k·0 = x(1 − x)
∴ v(x, 0) = x(1 − x) (6)
c)
Equations (4), (5) and (6) define a boundary value problem in v(x, t), presented together below:
∂v
∂t = h
2
∂
2
v
∂x2
v(0, t) = v(1, t) = 0
v(x, 0) = x(1 − x)
We seek solution in the form:
v(x, t) = X(x) T(t)
7
Substituting it in the PDE we get:
∂
∂t(XT) = h
2
∂
2
∂x2
(XT)
=⇒ X
∂T
∂t = h
2T
∂
2X
∂x2
[Separating variables]
∴
1
h
2T
dT

dt

1
X
d
2X
dx
2
(6)
Partial derivatives have been changed to ordinary in (6) because X and T
are functions of single variable.
Inspecting (6) we find that the left side of the equation is function of t and
the right side is function of x. And since the two terms are equal, they must
be equal to a constant (say −λ).
1
h
2T
dT

dt

1
X
d
2X
dx
2
= −λ (7)
The boundary conditions become:
v(0, t) = X(0) T(t) = 0
v(1, t) = X(1) T(t) = 0
T(t) = 0 will result in a trivial solution, so ignoring that possibility the
boundary conditions are given as
X(0) = X(1) = 0
Solving for X(x):
From (7) we have:
d
2X
dx
2

• λX = 0
  with the boundary conditions
  X(0) = X(1) = 0
  8
  This is a Strum-Liouville problem.
  λ can be positive, negative or zero. We will use method of elimination to
  decide what sign should λ bear.
  i) λ < 0: Let, λ = −k
  2
  , which implies the solution to ODE is
  X = C1e
  kx + C2e
  −kx
  The BC: X(0) gives C1 + C2 = 0 or C1 = −C2. The second boundary
  condition: X(1) = 0, gives C1(e
  2k − 1) = 0, which implies C1 = 0 and
  therefore C2 = 0. This leads to a trivial solution, so we discard the possibility
  of λ < 0
  ii) λ = 0: The solution obtained is
  X = Ax + B
  The first BC gives X(0) = B = 0 and the second one gives X(1) = A = 0.
  This too leads to a trivial solution and we discard this possibility also.
  Thus, we have established that λ should be positive. Let, λ = k
  2
  , therefore
  d
  2X
  dx
  2
• λX = 0
  Solution to this ODE is given by
  X = C1 cos √
  λx + C2 sin √
  λx
  X(0) = 0 =⇒ 0 = C1, therefore
  X = C2 sin √
  λ
  X(1) = 0 =⇒ 0 = C2 sin √
  λx
  C2 cannot be 0 as it will lead to trivial solution. Therefore
  sin √
  λ = 0
  which implies
  p
  λn = nπ or λn = n
  2π
  2 n = 1, 2, 3, . . .
  Note: n 6= 0, or it will lead to λ = 0, the possibility of whic