MA 201: Partial Differential Equations Lecture - 7 · MA 201: Partial Diﬀerential Equations...

Second Order Linear PDEWave Equation

Method of Characteristics for Wave EquationInfinite String Problem

MA 201: Partial Differential EquationsLecture - 7

IIT Guwahati Dr Bhupen Deka, Department of Mathematics, IIT Guwahati, 2020



• A 2nd order PDE with two independent variables x and y is given by

F (x , y , u, ux , uy , uxy , uxx , uyy ) = 0. (1)






What is the linear form?







• The unknown function u = u(x , y) satisfies an equation:

Auxx + Buxy + Cuyy + Dux + Euy + Fu + G = 0. (2)

Coefficients and G are sufficiently smooth functions of x and y .










Facts:

• Introducing the differential operators X = ∂∂x

and Y = ∂∂y

, equation

(2) can be written as

(AX 2 + BXY + CY 2 + DX + EY + FI )u = G . (3)










Facts:


and Y = ∂∂y

, equation



• Classification of such PDEs is based on expression










Facts:


and Y = ∂∂y

, equation



• Classification of such PDEs is based on expression

Lu = Auxx + Buxy + Cuyy ,

which is called the Principal part of the equation.




• At a point (x , y), the above equation is said to be

Hyperbolic if B2(x , y)− 4A(x , y)C (x , y) > 0






Parabolic if B2(x , y)− 4A(x , y)C (x , y) = 0







Elliptic if B2(x , y) − 4A(x , y)C (x , y) < 0








• Each category relates to specific problems









1. An Elliptic Equation

uxx + uyy = 0.

Which is known ‘Laplace’s Equation’.










uxx + uyy = 0.


2. A Hyperbolic Equation

utt − uxx = 0.

Which is known as ‘Wave’ or ‘Vibrating String Equation’.










uxx + uyy = 0.


2. A Hyperbolic Equation

utt − uxx = 0.

Which is known as ‘Wave’ or ‘Vibrating String Equation’.

3. A parabolic Equation

ut = uxx .

Which is known as ‘Heat Equation’.




Vibration and Wave




Vibrating String ProblemConsider a stretched string of

length π with the ends fastenedto the ends x = 0 and x = π.






Suppose that the string is setto vibrate by displacing it fromits equilibrium position.







• Let u(x , t) denote thetransverse displacement attime t ≥ 0 of the point onthe string at position x .







• Let u(x , t) denote thetransverse displacement attime t ≥ 0 of the point onthe string at position x . Thatis, we assume that each pointof the string moves only inthe vertical direction.








• In particular, u(x , 0) denotesthe initial shape of the stringand ut(x , 0) denotes theinitial velocity.








• In particular, u(x , 0) denotesthe initial shape of the stringand ut(x , 0) denotes theinitial velocity.

• That is, the string is set tovibrate by supplying an initialvelocity ut(x , 0) to it from itsequilibrium position u(x , 0)and then releasing it.




Vibrating String Problem Contd..





• Under some physical assumptions, we arrive at the following equation

utt(x , t) = c2uxx (x , t), (4)

which governs the entire process.






utt(x , t) = c2uxx (x , t), (4)


• Here, c represents a physical quantity.






utt(x , t) = c2uxx (x , t), (4)


• Here, c represents a physical quantity.

• Equation (4) is also well-known as Wave Equation.




Solution for Wave Equation




Solution for Wave Equation• Consider following wave equation

utt(x , t) = c2uxx (x , t). (5)





utt(x , t) = c2uxx (x , t). (5)

• It is easy to recognize that above equation reduces to followingequations

( ∂

∂t+ c

∂

∂x

)( ∂

∂t− c

∂

∂x

)

u = 0





utt(x , t) = c2uxx (x , t). (5)


( ∂

∂t+ c

∂

∂x

)( ∂

∂t− c

∂

∂x

)

u = 0 or (6)

( ∂

∂t− c

∂

∂x

)( ∂

∂t+ c

∂

∂x

)

u = 0. (7)





utt(x , t) = c2uxx (x , t). (5)


( ∂

∂t+ c

∂

∂x

)( ∂

∂t− c

∂

∂x

)

u = 0 or (6)

( ∂

∂t− c

∂

∂x

)( ∂

∂t+ c

∂

∂x

)

u = 0. (7)

• So, we try to solve individual first order equations, that is

ut + cux = 0





utt(x , t) = c2uxx (x , t). (5)


( ∂

∂t+ c

∂

∂x

)( ∂

∂t− c

∂

∂x

)

u = 0 or (6)

( ∂

∂t− c

∂

∂x

)( ∂

∂t+ c

∂

∂x

)

u = 0. (7)

• So, we try to solve individual first order equations, that is

ut + cux = 0 and (8)

ut − cux = 0. (9)




Solution for Wave Equation




Solution for Wave Equation• Consider equation

ut + cux = 0, (10)





ut + cux = 0, (10)

whose characteristic equations are

dt

1=

dx

c=

du

0.





ut + cux = 0, (10)


dt

1=

dx

c=

du

0.

Considering 1st and 2nd pair, we obtain

x − ct = c1.





ut + cux = 0, (10)


dt

1=

dx

c=

du

0.


x − ct = c1.

Then, due to Lagrange theorem any arbitrary function ofφ = x − ct will be the solution to the PDE (10).





ut + cux = 0, (10)


dt

1=

dx

c=

du

0.


x − ct = c1.

Then, due to Lagrange theorem any arbitrary function ofφ = x − ct will be the solution to the PDE (10). We assume that

u = f (φ) = f (x − ct)

is a solution for the PDE (10),





ut + cux = 0, (10)


dt

1=

dx

c=

du

0.


x − ct = c1.


u = f (φ) = f (x − ct)

is a solution for the PDE (10), where f is an arbitrary function of φ.





ut + cux = 0, (10)


dt

1=

dx

c=

du

0.


x − ct = c1.


u = f (φ) = f (x − ct)


• Similarly, it is an easy exercise to verify that u = g(x + ct) is thesolution for the PDE ut − cux = 0,





ut + cux = 0, (10)


dt

1=

dx

c=

du

0.


x − ct = c1.


u = f (φ) = f (x − ct)


• Similarly, it is an easy exercise to verify that u = g(x + ct) is thesolution for the PDE ut − cux = 0, where g is an arbitrary functionof x + ct.




Solution for Wave Equation Contd..




Solution for Wave Equation Contd..• We are in a process of solving wave equation

utt(x , t)− c2uxx = 0. (11)





utt(x , t)− c2uxx = 0. (11)

• In this process, we have found that u = f (x − ct) is a solution tothe PDE

ut + cux = 0 (12)





utt(x , t)− c2uxx = 0. (11)


ut + cux = 0 (12)

• And u = g(x + ct) is a solution for the PDE

ut − cux = 0, (13)

where f is an arbitrary function of x − ct and g is an arbitraryfunction of x + ct.





utt(x , t)− c2uxx = 0. (11)


ut + cux = 0 (12)


ut − cux = 0, (13)


• Then, observe that u(x , t) = f (x − ct) + g(x + ct) is a solution tothe equation (11).





utt(x , t)− c2uxx = 0. (11)


ut + cux = 0 (12)


ut − cux = 0, (13)


• Then, observe that u(x , t) = f (x − ct) + g(x + ct) is a solution tothe equation (11). In fact, it is the general solution for the waveequation.





utt(x , t)− c2uxx = 0. (11)


ut + cux = 0 (12)


ut − cux = 0, (13)


• Then, observe that u(x , t) = f (x − ct) + g(x + ct) is a solution tothe equation (11). In fact, it is the general solution for the waveequation.

• The physical interpretations of the functions f and g are quiteinteresting.





For instance, consider

y(x , t) = sin(x − ct).






y(x , t) = sin(x − ct).

Then observe that






y(x , t) = sin(x − ct).

Then observe that

• At time t = t1, position of a object traveling with the wave is givenby

(x , y) = (x , sin(x − ct1)).






y(x , t) = sin(x − ct).

Then observe that


(x , y) = (x , sin(x − ct1)).

• Location of the object at time t = 0 and x = 0 is (0, 0).






y(x , t) = sin(x − ct).

Then observe that


(x , y) = (x , sin(x − ct1)).


• Location of the object at time t = t1 and x = ct1 is (ct1, 0).






y(x , t) = sin(x − ct).

Then observe that


(x , y) = (x , sin(x − ct1)).



• In time t1 the object has moved from 0 to ct1 horizontally.






y(x , t) = sin(x − ct).

Then observe that


(x , y) = (x , sin(x − ct1)).



• In time t1 the object has moved from 0 to ct1 horizontally. So, cdenotes the velocity of the wave propagation.





Thus, we observe that

• The function f (x − ct) defines a wave propagation moving at aconstant speed c in the positive direction of the x-axis.







• Similarly the function g(x + ct) defines a wave progressing in thenegative direction of the x-axis with constant velocity c .








• The solutionu(x , t) = f (x − ct) + g(x + ct)









to the wave equation

utt − c2uxx = 0









to the wave equation

utt − c2uxx = 0

is, therefore, the algebraic sum of these two traveling waves.




The D’Alembert Solution For Wave Equation





We consider the following initialvalue problem

utt − c2uxx = 0, x ∈ (−∞,∞), t > 0 (14)

Displacement: u(x , 0) = φ(x) (15)

Velocity: ut(x , 0) = ψ(x) .(16)






utt − c2uxx = 0, x ∈ (−∞,∞), t > 0 (14)



From the solution u(x , t) = f (x − ct) + g(x + ct) to the wave equation(14), we find that






utt − c2uxx = 0, x ∈ (−∞,∞), t > 0 (14)




f (x) + g(x) = φ(x),






utt − c2uxx = 0, x ∈ (−∞,∞), t > 0 (14)




f (x) + g(x) = φ(x), (17)

−cf ′(x) + c g ′(x) = ψ(x), (18)

for all values of x .




The D’Alembert solution of the wave equation (Contd...)




The D’Alembert solution of the wave equation (Contd...)Integrating the second equation with respect to x

−f (x) + g(x) =1

c

∫ x

0

ψ(τ)dτ + A. (19)

where A = g(0)− f (0).





−f (x) + g(x) =1

c

∫ x

0

ψ(τ)dτ + A. (19)

where A = g(0)− f (0). Again, we have f (x) + g(x) = φ(x).





−f (x) + g(x) =1

c

∫ x

0

ψ(τ)dτ + A. (19)

where A = g(0)− f (0). Again, we have f (x) + g(x) = φ(x). Whichleads to

f (x) =1

2

[

φ(x)−1

c

∫ x

0

ψ(τ)dτ

]

− A/2, (20)

g(x) =1

2

[

φ(x) +1

c

∫ x

0

ψ(τ)dτ

]

+ A/2, (21)





−f (x) + g(x) =1

c

∫ x

0

ψ(τ)dτ + A. (19)


f (x) =1

2

[

φ(x)−1

c

∫ x

0

ψ(τ)dτ

]

− A/2, (20)

g(x) =1

2

[

φ(x) +1

c

∫ x

0

ψ(τ)dτ

]

+ A/2, (21)

Substituting these expressions into u(x , t) = f (x − ct) + g(x + ct), wehave

u(x , t) =1

2

[

φ(x − ct) + φ(x + ct)]

+1

2c

∫ x+ct

x−ct

ψ(τ)dτ. (22)





−f (x) + g(x) =1

c

∫ x

0

ψ(τ)dτ + A. (19)


f (x) =1

2

[

φ(x)−1

c

∫ x

0

ψ(τ)dτ

]

− A/2, (20)

g(x) =1

2

[

φ(x) +1

c

∫ x

0

ψ(τ)dτ

]

+ A/2, (21)

Substituting these expressions into u(x , t) = f (x − ct) + g(x + ct), wehave

u(x , t) =1

2

[

φ(x − ct) + φ(x + ct)]

+1

2c

∫ x+ct

x−ct

ψ(τ)dτ. (22)

This is D’Alembert’s (1717-1783) solution of one-dimensional wave

equation, which he actually derived around 1746.




Non-Homogeneous Wave Equation: Duhamel’s Principle





Find u(x , t) such that

utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞)






utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (23)

with IC’s u(x , 0) = 0, ut(x , 0) = 0 ∀x ∈ (−∞,∞). (24)

Due to the presence of F in the right side of the equation (35),






utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (23)

with IC’s u(x , 0) = 0, ut(x , 0) = 0 ∀x ∈ (−∞,∞). (24)

Due to the presence of F in the right side of the equation (35), wedirectly can not apply D’Alembert’s (1717-1783) solution.






utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (23)

with IC’s u(x , 0) = 0, ut(x , 0) = 0 ∀x ∈ (−∞,∞). (24)

Due to the presence of F in the right side of the equation (35), wedirectly can not apply D’Alembert’s (1717-1783) solution. Rather, werelate the given IVP (23)-(24) with a homogeneous IVP






utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (23)

with IC’s u(x , 0) = 0, ut(x , 0) = 0 ∀x ∈ (−∞,∞). (24)

Due to the presence of F in the right side of the equation (35), wedirectly can not apply D’Alembert’s (1717-1783) solution. Rather, werelate the given IVP (23)-(24) with a homogeneous IVP by convertingF (x , t) into a initial function.






utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (23)

with IC’s u(x , 0) = 0, ut(x , 0) = 0 ∀x ∈ (−∞,∞). (24)


More precisely, we consider an user defined problem

vtt − c2vxx = 0, (x , t) ∈ (−∞,∞)× (0,∞) (25)

with IC’s v(x , 0) = 0, vt(x , 0) = F (x , ξ), ξ > 0. (26)






utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (23)

with IC’s u(x , 0) = 0, ut(x , 0) = 0 ∀x ∈ (−∞,∞). (24)


More precisely, we consider an user defined problem

vtt − c2vxx = 0, (x , t) ∈ (−∞,∞)× (0,∞) (25)


Then, Duhamel’s Principle yields the solution u from v




Duhamel’s Principle Contd..




Duhamel’s Principle Contd..Now, we have a user defined problem

vtt − c2vxx = 0, (x , t) ∈ (−∞,∞)× (0,∞) (27)


Note that the solution v of the above problem depends on x , t and ξ.





vtt − c2vxx = 0, (x , t) ∈ (−∞,∞)× (0,∞) (27)


Note that the solution v of the above problem depends on x , t and ξ.Thus, v = v(x , t, ξ).





vtt − c2vxx = 0, (x , t) ∈ (−∞,∞)× (0,∞) (27)


Note that the solution v of the above problem depends on x , t and ξ.Thus, v = v(x , t, ξ). Accordingly, we can modify the ICs as

v(x , 0, ξ) = 0, vt(x , 0, ξ) = F (x , ξ), ξ > 0.





vtt − c2vxx = 0, (x , t) ∈ (−∞,∞)× (0,∞) (27)



v(x , 0, ξ) = 0, vt(x , 0, ξ) = F (x , ξ), ξ > 0.

Due to D’Alembert’s, v = v(x , t, ξ) is given by

v(x , t, ξ) =1

2c

∫ x+ct

x−ct

F (τ, ξ)dτ. (29)





vtt − c2vxx = 0, (x , t) ∈ (−∞,∞)× (0,∞) (27)



v(x , 0, ξ) = 0, vt(x , 0, ξ) = F (x , ξ), ξ > 0.

Due to D’Alembert’s, v = v(x , t, ξ) is given by

v(x , t, ξ) =1

2c

∫ x+ct

x−ct

F (τ, ξ)dτ. (29)

Then the solution u is given by

u(x , t) =

∫ t

0

v(x , t − τ, τ)dτ




Leibniz Rule for Differentiation

d

dt

(

∫ b(t)

a(t)

G (x , t)dx)

=

∫ b(t)

a(t)

Gtdx + G (b(t), t)b′(t)

−G (a(t), t)a′(t). (30)





d

dt

(

∫ b(t)

a(t)

G (x , t)dx)

=

∫ b(t)

a(t)


−G (a(t), t)a′(t). (30)

Use Leibniz rule on

u(x , t) =

∫ t

0

v (x , t − τ, τ)dτ, v (x , 0, ξ) = 0, vt(x , 0, ξ) = F (x , ξ), ξ > 0,

to have

ut =

∫ t

0

vt(x , t − τ, τ)dτ + v(x , t − t, t)×dt

dt− v(x , t − 0, 0)× 0

=

∫ t

0

vt(x , t − τ, τ)dτ + v(x , 0, t) =

∫ t

0

vt(x , t − τ, τ)dτ.





d

dt

(

∫ b(t)

a(t)

G (x , t)dx)

=

∫ b(t)

a(t)


−G (a(t), t)a′(t). (30)

Use Leibniz rule on

u(x , t) =

∫ t

0

v (x , t − τ, τ)dτ, v (x , 0, ξ) = 0, vt(x , 0, ξ) = F (x , ξ), ξ > 0,

to have

ut =

∫ t

0

vt(x , t − τ, τ)dτ + v(x , t − t, t)×dt

dt− v(x , t − 0, 0)× 0

=

∫ t

0

vt(x , t − τ, τ)dτ + v(x , 0, t) =

∫ t

0

vt(x , t − τ, τ)dτ.

Simillarly

utt(x , t) =

∫ t

0

vtt (x , t − τ, τ)dτ + vt(x , 0, t) =

∫ t

0

vtt(x , t − τ, τ)dτ + F (x , t)

c2uxx(x , t) = c2

∫ t

0

vxx (x , t − τ, τ)dτ so that utt − c2uxx = F (x , t).




Completely Non-Homogeneous Wave Equation: Duhamel’sPrinciple






utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (31)

with IC’s u(x , 0) = φ(x), ut(x , 0) = ψ(x) ∀x ∈ (−∞,∞). (32)






utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (31)


Try to break the problem as (ρ, θ) problems as: Find θ such that

θtt(x , t)− c2θxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (33)

with IC’s θ(x , 0) = 0, θt(x , 0) = 0 ∀x ∈ (−∞,∞). (34)






utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (31)



θtt(x , t)− c2θxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (33)

with IC’s θ(x , 0) = 0, θt(x , 0) = 0 ∀x ∈ (−∞,∞). (34)

Find ρ such that

ρtt(x , t)− c2ρxx = 0, (x , t) ∈ (−∞,∞)× (0,∞) (35)

with IC’s ρ(x , 0) = φ(x), ρt(x , 0) = ψ(x) ∀x ∈ (−∞,∞). (36)






utt(x , t)− c2uxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (31)



θtt(x , t)− c2θxx = F (x , t), (x , t) ∈ (−∞,∞)× (0,∞) (33)

with IC’s θ(x , 0) = 0, θt(x , 0) = 0 ∀x ∈ (−∞,∞). (34)

Find ρ such that

ρtt(x , t)− c2ρxx = 0, (x , t) ∈ (−∞,∞)× (0,∞) (35)

with IC’s ρ(x , 0) = φ(x), ρt(x , 0) = ψ(x) ∀x ∈ (−∞,∞). (36)

Principle of Super Position tells that ρ+ θ solve the completelynon-homogeneous problem


Second Order Linear PDEWave EquationMethod of Characteristics for Wave EquationInfinite String Problem

MA 201: Partial Differential Equations Lecture - 7 · MA 201: Partial Diﬀerential Equations...

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Transcript of MA 201: Partial Differential Equations Lecture - 7 · MA 201: Partial Diﬀerential Equations...