Find the form of the wave equation (2.1.1) with respect to new axes \(x^{\prime}\) and \(y^{\prime}\), which are rotated an angle \(\theta\) with respect to the \(x y\) axes. Hint: \(x^{\prime}=x \cos \theta+y \sin \theta, y^{\prime}=-x \sin \theta+\) \(y \cos \theta\).

Solve the above equations for \(x\) and \(y\): \(x = x^{\prime} \cos \theta - y^{\prime} \sin \theta\) \(y = x^{\prime} \sin \theta + y^{\prime} \cos \theta\)

Apply the chain rule to find the partial derivatives: \(\frac{\partial u}{\partial x^{\prime}} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial x^{\prime}} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial x^{\prime}}\) \(\frac{\partial u}{\partial y^{\prime}} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial y^{\prime}} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial y^{\prime}}\) Now, find the second partial derivatives: \(\frac{\partial^2 u}{\partial x^{\prime 2}} = \frac{\partial}{\partial x^{\prime}} \left( \frac{\partial u}{\partial x} \frac{\partial x}{\partial x^{\prime}} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial x^{\prime}} \right)\) \(\frac{\partial^2 u}{\partial y^{\prime 2}} = \frac{\partial}{\partial y^{\prime}} \left( \frac{\partial u}{\partial x} \frac{\partial x}{\partial y^{\prime}} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial y^{\prime}} \right)\)

Replace the second derivatives with respect to \(x\) and \(y\) in the wave equation with the ones found in step 2: \(\frac{\partial^2 u}{\partial x^{\prime 2}} + \frac{\partial^2 u}{\partial y^{\prime 2}} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2}\) Now, you have the form of the wave equation (2.1.1) with respect to the new axes \(x'\) and \(y'\).