This is a solved example from Calculus A Complete Course by Robert Adams , section 12.5 . I think this example has been solved incorrectly :
Example 9: If $f(x, y)$ is harmonic, show that $f(x^2 - y^2, 2xy)$ is also harmonic.
Solution: Let $ u = x^2 - y^2 $ and $ v = 2xy $. If $ z = f(u, v) $, then
\begin{align*} \frac{\partial z}{\partial x} &= 2x f_1(u, v) + 2y f_2(u, v) \\ \frac{\partial z}{\partial y} &= -2y f_1(u, v) + 2x f_2(u, v) \end{align*}
Differentiating again,
\begin{align*} \frac{\partial^2 z}{\partial x^2} &= 2f_1(u, v) + 2x(2x f_{11}(u, v) + 2y f_{12}(u, v)) \\ &\phantom{=} + 2y(2x f_{21}(u, v) + 2y f_{22}(u, v)) \\ &= 2f_1(u, v) + 4x^2 f_{11}(u, v) + 8xy f_{12}(u, v) + 4y^2 f_{22}(u, v) \end{align*}
and
\begin{align*} \frac{\partial^2 z}{\partial y^2} &= -2f_1(u, v) - 2y(-2y f_{11}(u, v) + 2x f_{12}(u, v)) \\ &\phantom{=} + 2x(-2y f_{21}(u, v) + 2x f_{22}(u, v)) \\ &= -2f_1(u, v) + 4y^2 f_{11}(u, v) - 8xy f_{12}(u, v) + 4x^2 f_{22}(u, v) \end{align*}
Therefore, $$ \frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 4(x^2 + y^2) (f_{11}(u, v) + f_{22}(u, v)) = 0 $$ because f is harmonic. Thus, z = f(x^2 - y^2, 2xy) is a harmonic function of x and y .
My First Question:
I wonder why he is sure that $f_{12} =f_{21}$ at the end , because we know that the condition for $f_{12} =f_{21}$ is f and $f_1$ and $f_2$ and $f_{11}$ and $f_{12}$ and $f_{22}$should all be continuous. How does he conclude that $f(x^2−y^2,2xy)$ is continuous just based on f(x,y)? Because we don't have the modified function expression.
My Second Question:
Suppose $f(x^2−y^2,2xy)$ works in Laplace, but how we are sure that it is harmonic?