2
$\begingroup$

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 $v$?

$\endgroup$
4
  • $\begingroup$ 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
0
$\begingroup$

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 $f_{xx} = f_{yy}$

Therefore

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.