DTC Introductory Maths Module 2015: Partial Differential Equations
Partial Differential Equations Worksheet Solutions
Dr. David Robert Grimes
http://users.ox.ac.uk/~donc0074/
E-mail: [email protected]
Direct Integration
1. Using direct integration of otherwise, find solutions to the following PDEs.
(i) utt = 6xe−t when u(x, 0) = x2 and ut (x, 0) = x ;
ut = −6xe−t + g(x)
g(x) = 7x
ut = −6e−t + 7x
u = 6e−t + 7tx + f (x)
f (x) = x2 − 6x
u(x, t) = x2 + 6x(e−t − 1) + 7tx
(ii) uxy = sin(x) cos(y) when u(π, y) = sin(y) and ux (x, π2 ) = 2x ;
ux = sin x sin y + g(x)
g(x) = 2x − sin x
ux = 2x − sin x(1 − sin y)
2
u = x + cos x(1 − sin y) + f (y)
f (y) = 1 − π 2
u(x, y) = x2 + cos x(1 − sin y) + 1 − π 2
2
(iii) uxx = t2 (2 + 4x2 )ex + 8 when u(0, t) = t2 and ux (0, t) = 3t ;
2
ux = 2xt2 ex + 8x + g(x)
g(x) = 3t
ux = 2xt2 e
u = t2 e
x2
x2
+ 8x + 3t
+ 4x2 + 3tx + f (x)
f (x) = 0
u(x, y) = t2 e
x2
+ 4x2 + 3tx
DTC Introductory Maths Module 2015: Partial Differential Equations
Factorising Solutions
2. Investigate whether the following PDEs can be factorised. Where possible, write the
general solution in terms of λ for the given form of K.
(i) utt = 10uxx where K < 0 ;
T 00 + λ2 T = 0
λ2
X=0
10
λx
λx
u = (A cos(λt) + B sin(λt)) C cos( √ ) + D sin( √ )
10
10
X 00 +
(ii) ut = 2tuxx where K > 0 ;
T 0 − 2tλ2 T = 0
X 00 − λ2 X = 0
2 2
u = eλ t
Aeλx + Be−λx
(iii) ut = x2 tux where K < 0 ;
T 0 + tλ2 T = 0
x2 X 00 + λ2 X = 0
2
−λ2 t2
λ
u=A e 2
ex
(iv) ut = uxx − 10 where K = 0; - This PDE is not separable!
Separation of Variables
3. Using separation of variables or otherwise, find a solution to the following PDE:
∂u
∂2u
=
∂t
∂x2
subject to the conditions
1
u(0, t) = u( , t) = 0
3
u(x, 0) = 10 sin (3πx)
∂u
∂2u
=
∂t
∂x2
DTC Introductory Maths Module 2015: Partial Differential Equations
subject to the conditions
T 0 = KT, X 00 = KX
BC : K = −λ2
X = A cos λx + B sin λx
BC : A = 0, λ = 3nπ
X = B sin(3nπx)
2
T = Ae−λ t = Ae−λ
2 π2 t
U = A∗ sin(3nπx)e−9n
2t
IC : A = 10, n = 1
U = 10 sin(3πx)e−9π
2t
4. Using separation of variables or otherwise, find a solution to the following PDE:
∂2u
1 ∂2u
=
∂t2
4 ∂x2
subject to the conditions
u(0, t) = u(1, t) = 0
u(x, 0) = 20 sin(nπx), ut (x, 0) = 0
X 00 = KX, 4T 00 − KT = 0
BC : K = −λ2
X = A cos λx + B sin λx
BC : A = 0, λ = nπ
X = B sin(nπx)
nπt
nπt
T = C cos
+ D sin
2
2
nπt
nπt
T 0 = nπ D cos
− C sin
2
2
0
IC : T (0) = 0, D = 0
nπt
T = C cos
2
nπt
∗
U = A sin(nπx) cos
2
IC : A = 20
nπt
u(x, t) = 20 sin(nπx) cos
2
DTC Introductory Maths Module 2015: Partial Differential Equations
5. Using separation of variables or otherwise, find a solution to the following PDE:
t
∂u
∂2u
+2 2 =0
∂t
∂x
subject to the conditions
π
u(0, t) = 10t2 , u( , t) = 15t2
2
u(x, 0) = 0
tT 0 − KT = 0, 2X 00 + KX = 0
BC : K = λ2
λx
λx
X = A cos √ + B sin √
2
2
0
2
tT − λ T = 0
(1)
λ2
T = Ct
λx
λx
∗
∗
λ2
A cos √ + B sin √
U =t
2
2
√
2
∗
BC : u(0, t) = 10t , A = 10, λ = 2
(2)
U = t2 (10 cos x + B ∗ sin x)
(3)
BC : B ∗ = 15t2 , B = 15
u(x, t) = t2 (10 cos x + 15∗ sin x)
6. Advanced: A drug is injected into a blood vessel of length L. The vessel is clamped
at both ends so that there is no flux of drug out of the vessel. The initial concentration
of the drug is given by f (x) = 10e−kx . Using the 1D diffusion equation,
∂u
∂2u
= D 2,
∂t
∂x
(a) Write down the general solution for u(x, t), and denote it ū.
General solution : Use B.C to show K = −λ2 - then..
ū = e
−Dn2 π 2 t
L2
cos
nπx
L
(b) After you obtain the general solution, you will have an undetermined series constant
An . For this solution, Fourier series* tells us the solution will be in the form of
u(x, t) = A0 +
∞
X
An ū
n=1
If the Fourier coefficients are given by
Z
1 L
A0 =
f (x)dx
L 0
Z
2 L
nπx
f (x) cos
dx
An =
L 0
L
DTC Introductory Maths Module 2015: Partial Differential Equations
calculate the specific solution for this system.
10 1 − e−kL
kL
20kL(1 + e−kL )
An =
k 2 L2 + n 2 π 2
∞
X 20kL(1 + e−kL ) −Dn2 π2 t
10 nπx
e L2 cos
u(x, t) =
1 − e−kL +
kL
k 2 L2 + n2 π 2
L
A0 =
n=1
(c) How would you expect this system to evolve in time?
As there is no flux in or out, the system should decay to a constant concentration
along the vessel with time!