if $f(x,y)=\phi(x-cy)+\phi(x+cy)$ then $f_{22}=c^2f_{11}.$

Question

Any hint how to solve this:

Show that if $f(x,y)=\phi(x-cy)+\phi(x+cy)$ then $f_{22}=c^2f_{11}.$

I don't know how to proceed with this .Kindly help with some hint how to solve this...

Answer 1

The task is to show that a $C^2$ function $\phi$ can be used to build a solution to the wave equation $$\partial_{yy} f(x,y) = c^2 \partial_{xx} f(x,y)\\ f(x,y) = \phi(x-cy) + \phi(x+cy)$$ Not sure why you write $f_{11}$ (like matrix indexing).
To prove this, just differentiate using the chain rule: $$\begin{align*} \partial_{yy} f(x,y) & = \partial_y ( -c\phi'(x-cy) + c\phi'(x+cy) ) \\ & = (-c)^2 \phi''(x-cy) + c^2 \phi''(x+cy) \\ & = c^2 (\phi''(x-cy) + \phi''(x+cy)) \\ &= c^2 \partial_x (\phi'(x-cy) + \phi'(x+cy)) \\ & = c^2 \partial_{xx} (\phi(x-cy) + \phi(x+cy)) \\ & =c^2 \partial_{xx} f(x,y) \end{align*}$$

Answer 2

$$\frac{\partial^2 f}{\partial y^2}=\frac{\partial }{\partial y}\left(-c\phi'(x-cy)+c\phi'(x+cy)\right)=c^2\phi''(x-cy)+c^2\phi''(x+cy)$$

Now compute $\dfrac{\partial^2 f}{\partial x^2}$ and check if the equality holds.