Question

Suppose f is a function of one variable that has a continuous second derivative. Show that for any constants a and b, the function u(x,y) = f(ax + by) is a solution of the PDE uxxuyy −u2xy = 0.

Answer 1

Answer:

We can find the solution using chain rule,

For instance, if u(x,y) = f(ax+by), then

$u_{x}(x,y) = a f^{'} (ax+by)$  which represents derivative of x with respect to x

and then derivative of $u_{x}(x,y)$ with respect to y is

$u_{xy}(x,y) = (ab)^2 (f^{''} (ax+by))^2$,  

Now, the derivative of $u_{x}(x,y)$ with respect to x, which is the second derivative, which is

$u_{xx}(x,y) = a^2 f^{''} (ax+by)$

and the derivative $u_{y}(x,y)$ and   $u_{yy}(x,y)$ are

$u_{y}(x,y) = b f^{'} (ax+by)$,

$u_{yy}(x,y) = b^2 f^{''} (ax+by)$

Finally, the solution of PDE is

$u_{xx}(x,y) u_{yy}(x,y) - u_{xy} ^2 (x,y)$

$= (a^2 f^{''} (ax+by)) (b^2 f^{''} (ax+by)) - (ab)^2 (f^{''} (ax+by))^2$

$= (ab)^2 (f^{''} (ax+by))^2) - (ab)^2 (f^{''} (ax+by))^2$

= 0,

As the PDE is equal to 0, it means the function u(x,y) = f(ax+by) is the solution of the given PDE.