Show that the most general solution of the wave equation (5.1.8) consisting of plane waves traveling in the direction of the unit vector \(\mathbf{n}\) is given by $$ p=f_{1}\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right)+f_{2}\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right), $$ where \(f_{1}\) and \(f_{2}\) are arbitrary functions.

First, we need to compute the first and second time-derivatives of \(p\). Since \(f_1\) and \(f_2\) are arbitrary functions and only depend on the arguments inside the parenthesis, we can write: $$ \frac{\partial p}{\partial t} = -c_{f} f_{1}'\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right) + c_{f} f_{2}'\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right), $$ $$ \frac{\partial^2 p}{\partial t^2} = c_{f}^2 f_{1}''\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right) + c_{f}^2 f_{2}''\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right), $$ where \(f_1'\) and \(f_2'\) are the first derivatives of \(f_1\) and \(f_2\), and \(f_1''\) and \(f_2''\) are the second derivatives of \(f_1\) and \(f_2\), respectively.

Next, we need to compute the Laplacian of pressure wave \(p\), which is given by the double derivative with respect to each spatial coordinate: $$ \nabla^2 p = \nabla \cdot \nabla p = \nabla \cdot\left(\mathbf{n} f_{1}'\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right)-\mathbf{n} f_{2}'\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right)\right). $$ Since \(\mathbf{n}\) is a constant vector, we can rewrite the expression as: $$ \nabla^2 p =\nabla \cdot \left[f_{1}'\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right) \mathbf{n} - f_{2}'\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right) \mathbf{n}\right]. $$ Now, using the property \(\nabla \cdot (f \mathbf{A}) = f \nabla \cdot \mathbf{A} + \mathbf{A} \cdot (\nabla f)\), where \(f\) is a scalar function and \(\mathbf{A}\) is a vector field, we have: $$ \nabla^2 p =\left[f_{1}'\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right) + f_{2}'\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right)\right] \nabla \cdot \mathbf{n} + \nabla\left[f_{1}'\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right) - f_{2}'\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right)\right] \cdot\mathbf{n}. $$ Since \(\nabla \cdot \mathbf{n} = 0\) as \(\mathbf{n}\) is a constant unit vector, we are left with: $$ \nabla^2 p= \nabla\left[f_{1}'\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right) - f_{2}'\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right)\right] \cdot\mathbf{n}. $$ Given that \(f_1'\) and \(f_2'\) only depend on the arguments inside the parenthesis, we can rewrite the last term as: $$ \nabla^2 p= \left[ f_{1}''\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right) - f_{2}''\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right)\right] \mathbf{n} \cdot\mathbf{n}. $$ Since \(\mathbf{n}\cdot\mathbf{n}=1\), we have: $$ \nabla^2 p= f_{1}''\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right) - f_{2}''\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right). $$

Finally, we want to check if the given solution \(p\) satisfies the wave equation: $$ \frac{\partial^2 p}{\partial t^2} = c_{f}^2\nabla^2 p, $$ Plugging in our expressions for \(\frac{\partial^2 p}{\partial t^2}\) and \(\nabla^2 p\), we observe that they are indeed equal: $$ c_{f}^2 f_{1}''\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right) + c_{f}^2 f_{2}''\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right) = c_{f}^2\left[f_{1}''\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right) - f_{2}''\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right)\right]. $$

Therefore, the given function: $$ p=f_{1}\left(\mathbf{n} \cdot \mathbf{r}-c_{f} t\right)+f_{2}\left(\mathbf{n} \cdot \mathbf{r}+c_{f} t\right), $$ does indeed satisfy the wave equation (5.1.8) and represents the most general solution consisting of plane waves traveling in the direction of the unit vector \(\mathbf{n}\), with \(f_1\) and \(f_2\) being arbitrary functions.