# PDEs notes and Fourier transforms ```Coefficients:
SUMMARY ON FOURIER TRANSFORMS
NB:
WE USE FOURIER SINE AND COSINE TRANSFORM METHODS TO SOLVE PDE’S:
(1)Wave Equation
(2)Heat Equation
(3)Laplace’s Equation
PARTIAL DIFFERENTIAL EQUATIONS (PDEs)
INTRODUCTION
A ‘curly dee’, ∂, is used to denote a differential coefficient in an expression
containing more than one variable.
If V =πr2h then
∂V means ‘the partial derivative of V with respect to r , with h
∂r
remaining constant’.
Thus,
∂V = (πh)πh)h) d (πh)r2) = (πh)πh)h)(πh)2r ) = 2πh)rh
∂r
dr
Similarly, ∂V means ‘the partial derivative of V with respect to h, with r remaining constant’.
∂h
Thus,
∂V = (ππr2) d (πh) = (ππr2)(π1) = πr2
∂h
dh
∂V and∂V are examples of first order partial derivatives, since n=1 when written in the
form∂nV
∂r
∂h
∂r n
A partial differential equation is an equation that contains one or more partial derivatives.
Examples
include:
(πi)
a ∂u + b ∂u = c
∂x
∂y
(πh)ii)
∂2u = 1 ∂2u
∂x2
c2 ∂t2
(known as the Wave equation)
(πiii)
∂2u = 1 ∂u
∂x2
c2 ∂t
(known as the Heat conduction equation)
(πiv)
∂2u
∂x2
(known as Laplace’s equation)
+ ∂2u = 0
∂y2
Equation (πi) is a first-order partial differential equation (PDE).
Equations (πii). (πiii) and (πiv) are second-order PDEs since the highest power of the
differential is 2.
PDEs occur in many areas of engineering and technology; electrostatics, heat conduction,
magnetism, wave motion, hydrodynamics and aerodynamics all use models that PDEs.
Such equations are difficult to solve, but techniques have been developed for the simpler
types.
In fact, for all but for the simplest cases, there are a number of numerical methods of
solutions of partial
differential equations available.
To be able to solve simple partial differential equations knowledge of the following is
required:
(πa) Partial integration,
(πb) First- and second-order partial differentiation.
(πc) The solution of ordinary differential equations.
(a)
Partial integration.
Integration is the reverse process of differentiation.
Example 1: Solve the differential equation ∂u = 5 cos x sin t.
∂t
This means that the equation above is integrated partially wrt t.
Then, 5 cos x term is considered as a constant.
u = ∫ 5 cos x sin t dt
= 5cos x ∫ sin t dt
= 5cos x . – cost + c
= - 5 cos x cos t + f(x)
Example 2: Solve the differential equation ∂2u = 6x2 cos 2y.
(πa)
∂x∂y
Firstly, the equation (πh)a) above is integrated partially wrt y. Then, 6x2 is a constant.
∂u = ∫ 6 x 2 cos 2y dy.
∂x
∂u = 6 x2
∂x
∫ cos 2y
dy.
∂u = 6 x2 . 1 sin 2y + f (x).
∂x
2
∂u = 6 x2 . sin 2y + f (πx).
∂x
2
∂u = 3 x2 . sin 2y + f (x).
∂x
(πh)b)
Lastly, the equation (πh)b) above is integrated partially wrt x. Hence, 3 sin 2y is a constant.
u = 3 sin 2y ∫ &iquest; &iquest; 2 + f(πh)x)] dx.
u = 3 sin 2y. x2+1 + x . f(πh)x) + g(πh)y).
2 +1
u = sin 2y + x .f(πh)x) + g(πh)y).
N.B : f(x)
called
and g(y) are functions that may be determined if extra information,
boundary conditions or initial conditions, are known.
Example 3.
Solve the differential equation ∂2u = 6x2(2y−1) given the boundary conditions that
at
∂x2
x =0, ∂u = sin 2y and u = cos y.
∂x
Hint: Integrate partially wrt x twice first and you apply boundary conditions.
Try it yourself
u = −cos(x+y) + 2x + cosx + y2 −1 + cosy
SOLUTION
∂u = (2y−1) 6
∂x
∫x
2
dx
∂u = (2y−1) 6 x3 + f(y),
∂x
3
∂u = 2 x3 (2y−1)
∂x
+ f(y),
where f (y) is an arbitrary function.
From the boundary conditions, when x =0,∂u = sin 2y
∂x
Hence, sin 2y =2(0)3(2y −1) + f (y)
[Obtained after substituting given boundary conditions]
from which, f (πh)y) = sin 2y.
Now ∂u = 2x3(2y −1) + sin 2y integrating partially again with respect to x gives:
∂x
u = ∫&iquest;&iquest;
u = 2 x4 (πh)2y – 1) + x sin 2y + F(πh)y)
4
From the boundary conditions, when x = 0, u = cos y, hence
cos y = (πh)0)4(πh)2y −1) + (πh)0)sin 2y + F(πh) y)
2
from which, F( y)=cos y.
Hence, the solution for the given boundary conditions is:
u = x4 (πh)2y−1)
2
+ x sin 2y + cos y
SECOND ORDER PDEs
For a function of a single real variable f(πh)x) , the central difference formula approximating the
second derivative is
Separating the variables
Let u(x, t)= X(x)T (t), where X(x) is a function of x only and T (t) is a function of t only, be a trial
solution to the wave equation
∂2u = 1 ∂2u
∂x2
c2 ∂t2 .
If the trial solution is simplified to u=XT, then
∂u = X’ T
and ∂2u = X’’ T
∂x
∂x2
Also ∂u = XT’ and
∂t
∂2u = XT’’
∂t2
Substituting into the partial differential equation ∂2u = 1 ∂2u gives:
∂x2 c2 ∂t2
X’’T = 1 XT’
c2
Separating the variables gives:
X’’ = 1 T’’
X
c2 T
Let μ= X’’ = 1 T’’ where μ is a constant.
X c2 T
Thus, since μ= X’’ (πh)a function of x only), it must be independent of t; and, since μ= 1 T ‘’
X
c2 T
(πh)a function of t only), it must be independent of x. If μ is independent of x and t, it can only be a
constant.
If μ= X’’ then X’’ = μX or X’’ − μX =0 and if μ= 1 T ‘’ then T’’ = c2μT or T’’ − c2μT =0.
X
c2 T
Please take note of two relevant exercises that appear on Worksheet Twelve number 2.
Example 4
At
At
At
At
At
A
B
C
D
E
3 + 4
A +3
B+2
3 +A
D +B
+B +D
+C +E
+G+F
+E + 4
+F +3
- 4A
- 4B
- 4C
- 4D
- 4E
= 0
-4A +B + 0C + D + 0 E + 0F = -7
=0
=0
=0
At F
E + C + H + 2 – 4F = 0 Please complete
SOLUTION OF THE WAVE EQUATION
a constant, say k.
=
Example 6.
A metal bar, insulated along its sides, is 1m long. It is initially at room temperature of 15◦C and
at time t =0, the ends are placed into ice at 0◦C. Find an expression for the temperature at a point
P at a distance x m from one end at any time t seconds after t =0.
Thus, the general solution is given by:
u(x, t) = {P cos px +Q sin px}e−p2c2t (πe-p squared. c
)
squared. t
u(0, t) = 0 thus 0 = P e−p2c2t
from which, P =0 and u(x, t)={Qsin px}e−p2c2t
(πe-p squared .c squared t)
(πe-p squared.c squared. t)
u(1, t)=0 thus 0={Q sin p}e−p2c2t
(πe-n squared . c squared t)
Also,
Since Q ≠ 0, sin p = 0 from which, p = nπ where n=1,2,3,. . .
∞
u(x, t)= ∑ &iquest;&iquest; Qn e-p2c2 t sin n𝜋x}
Hence,
(πe-p squared 𝜋.c squared. t)
n =1
The final initial condition given was that at t =0,u=15, i.e. u(x,0)= f (x)=15
∞
Hence,
15=∑ {Qn sin n𝜋x} where, from Fourier coefficients, Qn = 2 &times; mean value of 15
sin nπx from
n =1
x =0 to x =1,
1
i.e.
Qn = 2
∫ 15 sin nπx dx
0
1
1
−cos nπx
= 30
nπ
[
]
0
=−30 [cosnπ −cos0]
nπ
= 30 (π1− cosnπ)
nπ
= 0 (πwhen n is even) and
nπ
Hence, the required solution is:
60 (πwhen n is odd)
∞
u(x, t) =
∑
{Q n e-p2c2t sin n𝜋x}
(πe-p squared. c squared. t)
{sin nπx }e-n2𝜋2c2t
(πe-n squared 𝜋 squared c squared t)
n ( odd ) =1
∞
= 60
∑
n ( odd ) =1
Π
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