Fourier Series: Solutions of One Dimensional Wave Equation

SOLUTIONS OF ONE DIMENSIONAL WAVE EQUATION

Equation of a vibrating string :‒ One dimensional wave equation: Consider an elastic string tightly stretched between two points O and A. Let O be the origin and OA as x‒axis. On giving a small displacement to the string, perpendicular to its length (parallel to the y‒axis). Let y be the displacement at the point P (x, y) at any time. The wave equation is

Example 1: Obtain the solution of one dimensional wave equation:

Solution:

One dimensional wave equation is

Solve this by using method of separation of variables.

Let

 y = XT              ………….(2)

where X is a function of x alone and T is a function of t alone.

  differentiate (2) partially, we get

The nature of the solutions of (6) & (7) depends on the value of k.

Case (i) Let k be positive say k = p²

(6) ⇒ X" ‒ p²X = 0

(7) ⇒ T" ‒ c²p²T = 0

Now, we get two ordinary diff. equations.

The auxiliary equations are

m² ‒ p² = 0

m = p, ‒p

 So, X = A₁e^px + A₂e^‒px

m² ‒ c²p² = 0

m = cp, −cp

T = A₃ e^cpt + A₄ e^‒cpt

(2) ⇒ y = XT

 y = (A₁e^px + A₂e^‒px )( A₃ e^cpt + A₄ e^‒cpt )

Case (ii) Let k be negative say k = −p²

(6) & (7) = X" + p²X = 0 and T" + c²p²T = 0

The auxiliary equations are

m² + p² = 0

m = ± pi

So, X = (A₅ cos px + A₆ sin px),

m² + c²p² = 0

m = ± cpi

So, T = (A₇ cos cpt + A₈ sin cpt),

 (2) ⇒ y = XT

 Y(x,t) = (A₅ cos px + A₆ sin px) (A₇ cos cpt + A₈ sin cpt)

Case (iii) Let k = 0

 (6) & (7) ⇒ X'' = 0 & T'' = 0

The auxiliary equations are

m² = 0,

m = 0,0

X = A₉x + A₁₀,

m² = 0

m = 0, 0

T = A₁₁t + A₁₂

 equation (2) ⇒ y(x,t) = (A₉x + A₁₀) (A₁₁t + A₁₂)

Thus the various possible solutions of the wave equations are

y(x, t) = (A₁e^Px + A₂e^‒px) (A₃ e^cpt + A₄ e^‒cpt)       ……..(8)

y(x, t) = (A₅ cos px + A₆ sin px) (A₇ cos cpt + A₈ sin cpt)       ……..(9)

 y(x, t) =  (A₉x+A₁₀) (A₁₁t + A₁₂)       ……..(10)

Out of the three mathematically possible solutions derived, we have to choose the solution which is consistent with the physical nature of the problem and the given boundary conditions. In the case of vibration of an elastic string, y (x, t) representing the displacement of the string at any point x, y must be periodic in t. Hence solution (9), which consists of periodic functions in t is the suitable solution of the problems on vibration of strings. The arbitrary constants in the suitable solution are found out by using the boundary conditions of the problem.

In problems, we directly assume that (9) is the suitable solution of vibration of string problems.

The suitable solution (9) which is periodic in 't' is incidentally periodic in 'x' also.

Example 2: Derive D' Alembert's solution of the wave equation:

Solution: Consider the one dimensional wave equation

 Let D = ∂/∂t and D' = ∂/∂x

Hence, (1) can be written as (D² ‒ c²D'²)y = 0

The auxiliary equation is

 m² ‒ c² = 0

The general solution of the wave equation is

  y = f(x + ct) + g(x − ct)         …….(2)

where f and g are arbitrary functions

Suppose, initially y(x, 0) = ϕ(x) and ∂y/∂t(x, 0) = 0

From (2),  y(x, 0) = f(x) + g(x) = ϕ(x)

i.e.,  ϕ(x) = f(x) + g(x)        …………(3)

From (2), ∂y/∂t (x,t) = c [f ' (x + ct) − g' (x ‒ ct)]

 ∂y/∂t (x,0) = 0

 c [f '(x) − g '(x)] = 0

 f '(x) − g '(x) = 0

Integrating, we get_f(x) − g(x) = k          ………(4)

From (3) & (4), we get

 f(x) = (ϕ(x) + k) / 2

and g (x) = (ϕ(x) ‒ k) / 2

 (2) can be written as

 y (x, t) = 1/2 [ϕ (x + ct) + k] + 1⁄2 [ϕ (x − ct) − k]

= 1/2 [ ϕ(x − ct) + ϕ(x+ct) ]

This solution is called D' Alemberts solution of the one dimensional wave equation.

Example 3: State the assumptions made in the derivation of one dimensional wave equation.

Solution:

 The wave equation is 

This gives y (x, t), the transverse vibration of a string stretched to a constant tension T. In deriving this equation, we make the following assumptions.

1. The motion takes place entirely in one plane i.e., xy plane.

2. We consider only transverse vibrations, the horizontal displacement of the particles of the string is negligible.

3. The tension T is constant at all times and at all points of the deflected string.

4. T is considered to be so large compared with the weight of the string and hence the force of gravity is negligible.

5. The effect of friction is negligible.

6. The string is perfectly flexible, i.e., it can transmit tension but not bending or shearing forces.

7. The slope of the deflection curve at all points and at all instants is so small that sina can be replaced by a, where a is the inclination of the tangent to the deflection curve,

Example 4: Derive one dimensional wave equation :

Solution: Consider a tightly stretched elastic string of length l with its end points fixed. Let the string be released from rest and allowed to vibrate.

The problem is to determine the deflection y (x, t) at any point x and at any time t > 0.

For deriving the p.d.e. we make the following assumption.

See Example 3.

To obtain the differential equation

Take one end of the string as origin O and take X‒axis along the string.

We assume that the motion takes place entirely in the XY plane.

Consider the forces acting on a small portion PQ of the string where P is (x, y) and Q is (x + Δx, y + Δy)

By assumption, the string does not offer resistance to bending and shearing hence the tension at each point of the string is tangential to the curve of the string.

Let T₁ and T₂ be the tensions at the points P and Q.

Let ᴪ and ᴪ+Δᴪ be the angles made by the tangents at P and Q respectively with the X axis.

Let m be the mass per unit length of the string

Since, there is no motion in the horizontal portion of the string the tension must be constant.

 T₁ cos ᴪ = T₂ cos (ᴪ + Δᴪ) which is a constant.

Since ᴪ is small, cos ᴪ and cos (ᴪ+Δᴪ) are approximately equal to 1.

Thus, T₁ = T₂ = T, a constant.

The vertical component of the force acting on the element PQ is Tsin(ᴪ + Δᴪ) ‒ Tsinᴪ = T

T(ᴪ + Δᴪ) ‒ Tᴪ

= TΔᴪ

      ['. sin ᴪ = ᴪ where ᴪ is small]

The acceleration of the element in the y direction is ∂²y/∂t².

Hence, by Newton's second law of motion,

 mΔS ∂²y/∂t² = T Δ Ψ where m is the mass per unit length of the string as ΔS is the length of the element PQ.

Now, dᴪ / dS represents the curvature of the curve at P.

= ∂²y/∂x²

[∂y/∂x is very small and (∂y/∂x)² is negligible]

using this in (1), we get ∂²y/∂t² = T/m . ∂²y/∂x²

∂²y/∂x² = a² ∂²y/∂x²

where a² = T/m

This partial differential equation is known as the one dimensional wave equation. It is a homogeneous of second order.

Example 5a: The p.d.e. of a vibrating string is  what is a²?

Solution:

a² = T/m = Tension / mass per unit length of the string

Example 5b: Explain why a² (instead of a) is used in the p.d.e. of the vibrating string .

Solution:

 a² = T/m = Tension / mass per unit length of the string = + ve

As T/m is positive, it is denoted by a² (and not by a).