Chain Rule/Second Partial Derivative Problem

Question

If you're given the arbitrary function:

$f(u(x,y),v(x,y))$

Solve for functions $u$ and $v$ that satisfy the following equation (for any function $f$).

$ \frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u \ \partial v} $

I attempted to solve this by first solving for $f_{xx}$ and $f_{yy}$ then subtracting them to see if something cancels.

$ f_{xx} = f_u u_{xx} + f_v v_{xx} + f_{uu} u_x^2 + f_{vv} v_x^2 + 2f_{uv} u_x v_x $

$ f_{yy} = f_u u_{yy} + f_v v_{yy} + f_{uu} u_y^2 + f_{vv} v_y^2 + 2f_{uv} u_y v_y $

Subtracting these results in a long nonsense equation that doesn't have much meaning.

$ f_{xx} - f_{yy} = f_u (u_{xx} - u_{yy}) + f_v(v_{xx} - v_{yy}) + f_{uu} (u_x^2 - u_y^2) + f_{vv} (v_x^2 - v_y^2) + 2f_{uv} (u_x v_x - u_y v_y) $

How would I go about solving this? Is this a partial differential equation problem?

Answer 1

Choosing $f(u,v)=u,v,u^2,v^2,uv$ successively implies that all coefficients of $f$ derivatives in your equation vanish. That is, we obtain an overdetermined system \begin{align} u_{xx}-u_{yy}&=0,\\ v_{xx}-v_{yy}&=0,\\ u_x^2-u_y^2&=0,\\ v_x^2-v_y^2&=0,\\ 2(u_xv_x-u_yv_y)&=1. \end{align} Let us focus on the first and third equations: \begin{align} u_{xx}=u_{yy},\\ u_x=\pm u_y, \end{align} for some sign $\pm$. Differentiating the second equation implies $u_{xx}=\pm u_{yy}$. If $+$ is chosen, then this is consistent with the first equation, i.e. we are left with $u_x=u_y$, and thus $$ u(x,y)=U(x+y) $$ for an arbitrary function $U$. But if $-$ is chosen, then the first equation implies $u_{xx}=u_{yy}=0$, and we get $$ u(x,y)=a+bx+cy+dxy $$ for some constants $a,b,c,d$. To summarize, the general solution of the "$u$ equations" is $$ u(x,y)=U(x+y),\qquad a+bx+cy+dxy. $$ Similarly, $$ v(x,y)=V(x+y),\qquad e+fx+gy+hxy. $$ We are thus left with solving the fifth equation $$ u_xv_x-u_yv_y=1/2. $$ Up to the natural $u\leftrightarrow v$ exchange symmetry, we have three possible combinations of solutions:

Case 1: $u=U,v=V$:

Here, the remaining equation reduces to $0=1/2$, so this combination yields no solutions.

Case 2: $u=a+bx+cy+dxy,v=V$.

In this case, the remaining equation becomes $$ V'(x+y)(b-c+dy+dx)=1/2. $$ This gives $V(z)=C+\frac{1}{2d}\ln|b-c+dz|$ if $d\neq 0$, or $V(z)=C+\frac{z}{2(b-c)}$ otherwise.

Case 3: $u=a+bx+cy+dxy,v=e+fx+gy+hxy$.

The remaining equation becomes $$ (b+dy)(f+hy)-(c+dx)(g+hx)=1/2. $$ Differentiating this equation in $x$ and $y$ implies $d=h=0$, so we are left with $$ bf-cg=1/2. $$ This determines $u,v$.