Question

If f and g are arbitrary functions of their respective arguments, show that u=f(x-vt+iay)+g(x-vt+iay) is a solution of d2u/dx2+d2u/dy2=1/c^2 .d2u/dt2,where a^2=1-(v^2/c^2)

Answer 1

Question #88320, Math, Differential Equations

1. If $f$ and $g$ are arbitrary functions of their respective arguments, show that $u = f(x - vt + iay) + g(x - vt + iay)$ is a solution of $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2}$, where $a^2 = 1 - \frac{v^2}{c^2}$.

Solution:

Consider $u = f(x - vt + iay) + g(x - vt + iay)$.

Differentiating $u$ partially with respect to $x$, we get

$$\frac{\partial u}{\partial x} = f'(x - vt + iay) + g'(x - vt + iay)$$ $$\frac{\partial^2 u}{\partial x^2} = f''(x - vt + iay) + g''(x - vt + iay)$$

Differentiating $u$ partially with respect to $y$, we get

$$\frac{\partial u}{\partial y} = f'(x - vt + iay) \cdot ia + g'(x - vt + iay) \cdot ia$$ $$\begin{array}{l} \frac{\partial^2 u}{\partial y^2} = f''(x - vt + iay) \cdot (-a^2) + g''(x - vt + iay) \cdot (-a^2) \\ = -a^2 \left(\frac{\partial^2 u}{\partial x^2}\right) \quad \text{(by using Equation 1)} \end{array}$$

Differentiating $u$ partially with respect to $t$, we get

$$\begin{array}{l} \frac{\partial u}{\partial t} = f'(x - vt + iay) \cdot (-v) + g'(x - vt + iay) \cdot (-v) \\ \frac{\partial^2 u}{\partial t^2} = f''(x - vt + iay) \cdot (v^2) + g''(x - vt + iay) \cdot (v^2) \\ = v^2 \left(\frac{\partial^2 u}{\partial x^2}\right) \quad \text{(by using Equation 1)} \end{array}$$

Taking $a^2 = 1 - \frac{v^2}{c^2}$, Equation (2) becomes,

$$\begin{array}{l} \frac{\partial^2 u}{\partial y^2} = -a^2 \left(\frac{\partial^2 u}{\partial x^2}\right) \\ = -\left(1 - \frac{v^2}{c^2}\right) \left(\frac{\partial^2 u}{\partial x^2}\right) \end{array}$$ $$\begin{array}{l} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{v^2}{c^2} \left(\frac{\partial^2 u}{\partial x^2}\right) \\ = \frac{1}{c^2} \left(v^2 \left(\frac{\partial^2 u}{\partial x^2}\right)\right) \end{array}$$ $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} \quad \text{(by using Equation 3)}$$

From Equation (4), $u = f(x - vt + iay) + g(x - vt + iay)$ satisfies the given differential equation and hence the solution of the equation.

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