Question

Solve the following Cauchy problem for the nonhomogeneous wave equation.

u_tt - u_xx = 1 on -infinity < x < infinity, t > 0.
u(x,0) = x^2
u_t(x,0) = 1.

Answer 1

Solve the following Cauchy problem for the nonhomogeneous wave equation.

$$\begin{array}{l} u_{tt}(x, t) - u_{xx}(x, t) = 1, -\infty < x < +\infty, t > 0, \\ u(x, 0) = x^2, \\ u_t(x, 0) = 1. \end{array}$$

Solution:

For Cauchy problem

$$\begin{array}{l} u_{tt}(x, t) - a^2 u_{xx}(x, t) = f(x, t), -\infty < x < +\infty, t > 0, \\ u(x, 0) = \varphi(x), \\ u_t(x, 0) = \psi(x) \end{array}$$

we have the next solution

$$u(x, t) = \frac{\varphi(x + at) + \varphi(x - at)}{2} + \frac{1}{2a} \int_{x - at}^{x + at} \psi(z) dz + \frac{1}{2a} \int_{0}^{t} \int_{x - a(t - \tau)}^{x + a(t - \tau)} f(z, \tau) dz \, d\tau.$$

In our case

$$\begin{array}{l} a = 1, \\ f(x, t) = 1, \\ \varphi(x) = x^2, \\ \psi(x) = 1. \end{array}$$

Thus

1) $\frac{\varphi(x + at) + \varphi(x - at)}{2} = \frac{(x + t)^2 + (x - t)^2}{2} = x^2 + t^2$

2) $\frac{1}{2a} \int_{x - at}^{x + at} \psi(z) dz = \frac{1}{2} \int_{x - t}^{x + t} dx = \frac{1}{2} \big(x + t - (x - t)\big) = t$

3) $\frac{1}{2a} \int_{0}^{t} \int_{x - a(t - \tau)}^{x + a(t - \tau)} f(z, \tau) dz \, d\tau = \frac{1}{2} \int_{0}^{t} \int_{x - (t - \tau)}^{x + (t - \tau)} dz \, d\tau = \frac{1}{2} \int_{0}^{t} (x + t - \tau - x + t - \tau) d\tau =$

$$= \int_{0}^{t} (t - \tau) d\tau = \left. (t \tau - \frac{\tau^2}{2}) \right|_{0}^{t} = \frac{t^2}{2}$$

Thus we have

$$u(x, t) = x^2 + t^2 + t + \frac{t^2}{2} = x^2 + \frac{3}{2} t^2 + t.$$

Answer:

$$u(x, t) = x^2 + \frac{3}{2} t^2 + t.$$