Question

If f and g are arbitrary function of their respective arguments, show that u= f( x -vt+i alpha y ) is a solution of (d^2u/dx^2 ) + (d^2u/dy^2) = (1/c^2) (d^2u/dt^2), where (alpha)^2 = 1-(v^2/c^2)?

Answer 1

Consider $u=f(x-vt+i\alpha y).$

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

$${\partial u \over \partial x}=f'(x-vt+i\alpha y)$$ $${\partial^2 u \over \partial x^2 }=f''(x-vt+i\alpha y)$$

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

$${\partial u \over \partial y}=f'(x-vt+i\alpha y)\cdot i\alpha$$ $${\partial^2 u \over \partial y^2 }=f''(x-vt+i\alpha y)\cdot (i\alpha)^2$$

Then

$${\partial^2 u \over \partial y^2 }=-\alpha^2{\partial^2 u \over \partial x^2 }$$

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

$${\partial u \over \partial t}=f'(x-vt+i\alpha y)\cdot (-v)$$ $${\partial^2u \over \partial t^2 }=f''(x-vt+i\alpha y)\cdot (-v)^2$$

Then

$${\partial^2u \over \partial t^2 }=v^2 {\partial^2u \over \partial x^2 }$$

Substitute into the given equation

$${\partial^2u \over \partial x^2 }-\alpha^2{\partial^2u \over \partial x^2 }={v^2 \over c^2}\cdot {\partial^2u \over \partial x^2 }$$

Take

$$\alpha^2=1-{v^2 \over c^2}$$

$${\partial^2u \over \partial x^2 }-\bigg(1-{v^2 \over c^2}\bigg){\partial^2u \over \partial x^2 }={v^2 \over c^2}\cdot {\partial^2u \over \partial x^2 }$$

$${v^2 \over c^2}\cdot {\partial^2u \over \partial x^2 }={v^2 \over c^2}\cdot {\partial^2u \over \partial x^2 }$$

$${\partial^2u \over \partial x^2 }= {\partial^2u \over \partial x^2 },\ True$$

Therefore, $u=f(x-vt+i\alpha y)$ satisfies the given differential equation and hence the solution of the equation.