let $f(x, y)$ function, and $f_{xx}, f_{xy}, f_{yx}, f_{yy}$ are continuous in $\mathbb{R}^2$, $f_{xx}\equiv f_{yy}$. $z(u, v)$ is defined as $z(u, v)=f(u+v, u-v)$, I need to show that $\frac{\partial z}{\partial u}$ is not dependent on $v$ and deduce that exist functions $h_1(t)$, $h_2(t)$ so that $z(u, v)=h_1(u) + h_2(v)$.

Because $f_{xx}, f_{xy}, f_{yx}, f_{yy}$ are continuous in $\mathbb{R}^2$ $\Longrightarrow$ $f_x, f_y$ are differentiable $\Longrightarrow$ $f$ is differentiable. let $x(u, v)=u+v$ and $y(u, v)=u-v$. So according to the chain rule $$\frac{\partial z}{\partial u} = f_x\cdot x_u + f_y\cdot y_u,\space\space \frac{\partial z}{\partial v} = f_x\cdot x_v + f_y\cdot y_v \Longrightarrow \frac{\partial z}{\partial u} = f_x + f_y,\space\space \frac{\partial z}{\partial v} = f_x - f_y$$ but I don't know how to progress from here, how can I show that $\frac{\partial z}{\partial u}$ doesn't depend on $v$?

  • $\begingroup$ You want to show that $f_x + f_y$ is constant along lines $(u+v, u-v)$ for fixed $u$. Show that the gradient of $f_x + f_y$ is orthogonal to those lines, using the hypotheses given in the first sentence. $\endgroup$ – Joe Jun 23 '20 at 14:21
  • $\begingroup$ I'm not sure I understand. the gradient is $\nabla(f_x+f_y)(u+v, u-v)=((f_{xx}+f_{yx})(u+v, u-v), (f_{xy}+f_{yy})(u+v, u-v))$ and the inner product is $\langle \nabla(f_x+f_y)(u+v, u-v), (u+v, u-v) \rangle=(f_{xx}+f_{yx})(u+v, u-v)\cdot (u+v) + (f_{xy}+f_{yy})(u+v, u-v)(u-v)=(f_{xx}+f_{yx})(u+v, u-v)\cdot u + (f_{xy}+f_{yy})(u+v, u-v)\cdot u$ which isn't 0 so the vectors aren't orthogonal $\endgroup$ – CforLinux Jun 23 '20 at 14:55
  • $\begingroup$ Do you know what continuity of the second partials implies about $f_{xy}$ and $f_{yx}$? Use that and $f_{xx}=f_{yy}$ to show that the gradient is orthogonal to $\langle 1, -1 \rangle$ $\endgroup$ – Joe Jun 23 '20 at 15:47
  • $\begingroup$ @Joe so if I understand correctly, let $p_0=(u, u)$ the gradient is $\nabla(f_x+f_y)(u,u)=((f_{xx}+f_{yx})(u,u)+(f_{xy}+f_{yy})(u,u))$, the inner product is $\langle \nabla(f_x+f_y)(u,u), (1, -1) \rangle = (f_{xx}+f_{yx})(u,u) - (f_{xy}+f_{yy})(u,u) = f_{xx}(u,u) - f_{yy}(u,u) + f_{yx}(u,u) - f_{xy}(u,u) = 0$ hence any vector $(v, -v)$ is orthogonal to the gradient, thus $\frac{\partial z}{\partial u}((u, u) + v(1,-1))$ doesn't depend on v $\endgroup$ – CforLinux Jun 23 '20 at 18:48

I'm not sure that I follow the notation in your last comment, but I think you've got it:

$$\frac{\partial z}{\partial u} = \frac{\partial }{\partial u} f(u+v, u-v) = f_x(u+v,u-v) + f_y(u+v,u-v)$$

In the basis of $x$ and $y$, the direction of increasing $v$ is $\langle 1, -1 \rangle$, so $\frac{\partial }{\partial v} \left(\frac{\partial z(u,v)}{\partial u}\right) = \left(\nabla_{x,y} \left[ f_x(u+v,u-v) + f_y(u+v,u-v) \right] \right)\cdot \langle 1, -1 \rangle$.

$$\nabla_{x,y} \left[ f_x(u+v,u-v) + f_y(u+v,u-v) \right] = \langle f_{xx}(u+v,u-v) + f_{yx}(u+v,u-v), f_{xy}(u+v,u-v) + f_{yy}(u+v,u-v) \rangle$$

Since the second derivatives are continuous, by Clairaut's theorem $f_{xy} = f_{yx}$. Also, we are told that $f_{xx} = f_{yy}$


$$\frac{\partial }{\partial v} \left(\frac{\partial z(u,v)}{\partial u}\right) = \left(f_{xx}(u+v,u-v) + f_{yx}(u+v,u-v)\right) - \left(f_{xy}(u+v,u-v) + f_{yy}(u+v,u-v)\right) = 0$$


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