---
title: "Derive 1D wave equation and solve"
subject: "Applied Mathematics"
chapter: 3
key: "semester_4/applied_mathematics:wave-eq-derive-and-solve"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/wave-eq-derive-and-solve
site: BCT Engineering
---

# Derive 1D wave equation and solve

Applied Mathematics · Chapter 3: Partial Differential Equations

- Asked in: 2074 Bhadra, 5+5 marks; 2073 Magh, 10 marks; 2072 Asoj, 10 marks; 2070 Bhadra, 10 marks; 2068 Bhadra (asked 5 times)
- Video: [Derivation of one dimensional wave equation | VTU Mathematics | Module 3](https://www.youtube.com/watch?v=c2IeHJOhSAg) — Dr Dhananjaya M
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/wave-eq-derive-and-solve
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-3

## Question

Derive one dimensional wave equation (transverse vibration of a string) and solve it completely.

## Short answer

From transverse forces on a small string element, Newton's law gives ∂²u/∂t²=c²∂²u/∂x² with c²=T/ρ. Separation of variables with fixed ends yields u(x,t)=∑[A_n cos(nπct/l)+B_n sin(nπct/l)] sin(nπx/l); A_n,B_n from Fourier sine expansions of initial displacement and velocity.

## Answer

#### Part 1: Derivation of the one-dimensional wave equation

**Assumptions.** A perfectly flexible, uniform string of mass ρ per unit length is stretched to a constant tension T along the x-axis and fixed at x = 0 and x = l. It vibrates in the xy-plane with small displacements u(x, t); the slope is small, so cos θ ≈ 1 and sin θ ≈ tan θ. Gravity is neglected.

```text
 u
 |            Q ○──→ T (angle θ₂)
 |          ╱
 |   P ○───╱
 |  ↙ T (angle θ₁)
 |_________________________ x
      x        x+δx
```

Take a small element PQ between x and x + δx. The tensions T at the two ends make small angles θ₁ and θ₂ with the x-axis.

- Horizontal: T cos θ₂ − T cos θ₁ ≈ 0 (no horizontal motion).
- Vertical net force = T sin θ₂ − T sin θ₁ ≈ T (tan θ₂ − tan θ₁).

Since tan θ = ∂u/∂x:

net force = T [ (∂u/∂x)_{x+δx} − (∂u/∂x)_x ]

By Newton's second law, mass × acceleration = force. The mass of the element is ρ δx:

```text
ρ δx ∂²u/∂t² = T [ (∂u/∂x)_{x+δx} − (∂u/∂x)_x ]
∂²u/∂t² = (T/ρ) · [ (∂u/∂x)_{x+δx} − (∂u/∂x)_x ] / δx
```

Letting δx → 0 and writing c² = T/ρ:

**∂²u/∂t² = c² ∂²u/∂x²**

#### Part 2: Solution

**Boundary and initial conditions**

1. u(0, t) = 0 and u(l, t) = 0 for all t (fixed ends)
2. u(x, 0) = f(x) (initial shape)
3. ∂u/∂t at t = 0 equals g(x) (initial velocity)

**Step 1: separation of variables.** Put u = X(x) T(t):

X T″ = c² X″ T, so X″/X = T″/(c²T) = −λ² (constant, shown negative below).

```text
X″ + λ² X = 0
T″ + λ²c² T = 0
```

(A zero or positive separation constant gives only the trivial solution X = 0 once the ends are fixed, so the constant is −λ², a negative value.)

**Step 2: apply boundary conditions.** X = A cos λx + B sin λx.

- u(0, t) = 0 gives A = 0.
- u(l, t) = 0 gives B sin λl = 0, so λ = nπ/l (n = 1, 2, 3, …) since B ≠ 0.

Xₙ(x) = sin(nπx/l), and T(t) = C cos(λct) + D sin(λct).

**Step 3: general solution.** By superposition:

**u(x, t) = Σ_{n=1}^{∞} [ Bₙ cos(nπct/l) + Bₙ* sin(nπct/l) ] sin(nπx/l)**

**Step 4: initial conditions.**

- u(x, 0) = f(x) gives f(x) = Σ Bₙ sin(nπx/l), a Fourier sine series, so

  Bₙ = (2/l) ∫₀^l f(x) sin(nπx/l) dx

- ∂u/∂t at t = 0: g(x) = Σ Bₙ* (nπc/l) sin(nπx/l), so

  Bₙ* = (2/(nπc)) ∫₀^l g(x) sin(nπx/l) dx

**Special case (string released from rest, g = 0):** u(x, t) = Σ Bₙ cos(nπct/l) sin(nπx/l).

**Answer:** u(x, t) as in Step 3, with Bₙ and Bₙ* from Step 4. The n-th term is the n-th normal mode, with frequency nc/(2l).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The wave equation arises from
   A. mass * acceleration = net vertical tension force
   B. conservation of energy only
   C. Hooke's law of the string material
   D. heat balance
   Answer: A. rho dx u_tt = T[u_x(x+dx) - u_x(x)]. (Key point: Newton's law)

2. c^2 equals
   A. rho/T
   B. T/rho
   C. T rho
   D. T/l
   Answer: B. Tension over mass per unit length. (Key point: Constant)

3. The separation constant is -lambda^2 because
   A. it makes T(t) grow
   B. it is arbitrary
   C. otherwise only trivial solution once ends are fixed
   D. c becomes imaginary
   Answer: C. Non-negative constants force X = 0. (Key point: Separation constant)

4. Fixed ends give lambda =
   A. n pi c
   B. n pi
   C. l/(n pi)
   D. n pi/l
   Answer: D. sin lambda l = 0. (Key point: Coefficients)

## Related in Applied Mathematics

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