Non-Homogeneous Wave Equation

Before learning about non-homogeneous wave equation, let’s recall what wave equation is. As we know, the wave equation is a second-order linear PDE that represents some medium competent in transferring waves. The wave equation solution provides us with all possible waves that can reproduce. Generally, it comprises a second-order derivative with respect to time, which derives from F = ma or something comparable, and a second derivative with respect to the position, which derives from F = -kx or similar ones, i.e. inertia and elasticity. For a vibrating elastic string, the one-dimensional wave equation is as follows:

u_tt = c² u_xx

Learn more about wave equations here.

In this article, you will learn one of the special types of wave equations called non-homogeneous wave equations and the easiest method of finding the solution to such equations.

Non-homogeneous Wave Equation in One Dimension

The non-homogeneous or inhomogeneous wave equation in 1D is given by:

u_tt(x, t) – c² u_xx(x, t) = s(x, t)

with initial conditions

u(x, 0) = f(x)

u_t(x, 0) = g(x)

Here, s(x, t) is the given function, i.e. the source function.

If the solution of the non-homogeneous wave equation exists, then it will be a unique solution. Also, we can find the solution to such equations using the d’Alembert formula. However, it is possible to determine the solution to a given non-homogeneous wave equation by Green’s theorem.

How to Solve Non-homogeneous Wave Equation?

Let’s learn how to solve the non-homogeneous wave equation (Cauchy problem) with the help of a solved example here.

Example:

Solve the following non-homogeneous wave equation (Cauchy problem).

u_tt – u_xx = t⁷; -∞ < x < ∞ and t > 0

With initial conditions

u(x, 0) = 2x + sin x

u_t(x, 0) = 0; -∞ < x < ∞

Solution:

Given:

u_tt – u_xx = t⁷; -∞ < x < ∞ and t > 0

With initial conditions

u(x, 0) = 2x + sin x

u_t(x, 0) = 0; -∞ < x < ∞

By comparing with the standard form, we have;

s(x, t) = t⁷

That means s(x, t) depends only on one variable, i.e. t.

v_tt = t⁷

Now, it is easy to find the solution set v(x, t) for the given Cauchy problem.

Also, we know that v(x, t) again depends only on t and v_xx = 0.

Let us integrate v_tt = t⁷ to get v(x, t).

∫ v_tt dt = ∫t⁷ dt

v(x, t) = (⅛) ∫t⁸ dt

= t⁹/(8.9)

= t⁹/72

Consider u(x, t) = v(x, t) + w(x, t)

Here, w(x, t) is the homogeneous problem’s solution.

Now,

w_tt − w_xx = 0, −∞ < x < ∞, t > 0

And

w(x, 0) = u(x, 0) − v(x, 0) = 2x + sin x,

wt(x, 0) = u(x, 0) − v(x, 0) = 0, −∞ < x < ∞

By d’Alembert’s formula, we have;

w(x, t) = (1/2) [2(x + t) + sin(x + t) + 2(x − t) + sin(x − t)]

= 2x + sin x cost.

Therefore, the original solution of the given problem is:

u(x, t) = 2x + sin x cost + (t⁹/72)

Practice Problems

1. Find the solution to the initial value problem:
   
   u_tt – 3² u_xx = e^x – e^-x; -∞ < x < ∞ and t > 0

With initial conditions

u(x, 0) = x

u_t(x, 0) = sin x; -∞ < x < ∞

2. Determine the solution of the following equation.
   
   u_xx – u_tt = 1
   
   u(x, 0) = sin x
   
   u_t(x, o) = x
3. Solve the following wave equation.
   
   u_tt = c² u_xx; x < 0

With initial conditions

u(x, 0) = cos x; x < 0

u_t(x, 0) = 0; x < 0

u(0, t) = e^-t; t > 0

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