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Suppose that $u=u(x,y)$ and $v=v(x,y)$ have continuous second partial derivatives. If for each $f$, we have $$\frac{\partial^2f}{\partial u^2}+\frac{\partial^2f}{\partial v^2}=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2},$$ prove that the Jacobian $\begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{pmatrix}$ is constant.

By using the chain rule, $$\frac{\partial^2f}{\partial x^2}=\left(\frac{\partial u}{\partial x}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}+\left(\frac{\partial v}{\partial x}\right)^2\frac{\partial^2f}{\partial v^2}\\ \frac{\partial^2f}{\partial y^2}=\left(\frac{\partial u}{\partial y}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial y}\frac{\partial v}{\partial y}+\left(\frac{\partial v}{\partial y}\right)^2\frac{\partial^2f}{\partial v^2}$$ and we see that the Jacobian is orthogonal. But how to prove that it is constant?

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    $\begingroup$ How am I supposed to understand the constraint on $f$? Is it that if $g(x,y) = f(u(x,y), v(x,y))$, then $\partial_{uu} f + \partial_{vv} f = \partial_{xx} g + \partial_{yy} g$? $\endgroup$
    – Charles Hudgins
    Jun 22 at 17:33
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    $\begingroup$ I believe that you took your chain rule incorrectly (you forgot the product rule in there). Writing $g(x,y) = f(u(x,y), v(x,y))$, the second-order partial derivatives can be computed as $$\begin{align*} g_{xx} &= \frac{\partial}{\partial x}\left[f_u u_x + f_v v_x\right]\\ &= f_{uu}u_x^2 + f_{uv}u_xv_x + f_u u_{xx} + f_{vu}u_xv_x + f_{vv}v_x^2 + f_v v_{xx}\\ &= f_{uu}u_x^2 + f_u u_{xx} + 2f_{uv}u_xv_x + f_{vv}v_x^2 + f_v v_{xx} \end{align*}$$ and, similarly, $$g_{yy} = f_{uu}u_y^2 + f_u u_{yy} + 2f_{uv}u_yv_y + f_{vv}v_y^2 + f_v v_{yy}.$$ $\endgroup$
    – peabody
    2 days ago
  • $\begingroup$ @Charles I think so. $\endgroup$
    – Saunders
    2 days ago

1 Answer 1

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Bad calculation keeps me away from solving this.

As is pointed out by @peabody, we have $$\frac{\partial^2f}{\partial x^2}=\left(\frac{\partial u}{\partial x}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial x}\frac{\partial v}{\partial x}+\left(\frac{\partial v}{\partial x}\right)^2\frac{\partial^2f}{\partial v^2}+\frac{\partial^2u}{\partial x^2}\frac{\partial f}{\partial u}+\frac{\partial^2v}{\partial x^2}\frac{\partial f}{\partial v}\\ \frac{\partial^2f}{\partial y^2}=\left(\frac{\partial u}{\partial y}\right)^2\frac{\partial^2f}{\partial u^2}+2\frac{\partial u}{\partial y}\frac{\partial v}{\partial y}+\left(\frac{\partial v}{\partial y}\right)^2\frac{\partial^2f}{\partial v^2}+\frac{\partial^2u}{\partial y^2}\frac{\partial f}{\partial u}+\frac{\partial^2v}{\partial y^2}\frac{\partial f}{\partial v}$$ rather than I wrote before.

Using the given condition (since $f$ is arbitrary) we get $$ \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0\\ \left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2=1 $$

From the second identity $$ \frac{\partial^2u}{\partial x^2}\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}+\frac{\partial^2u}{\partial x\partial y}\left(\frac{\partial u}{\partial y}\right)^2=0\\ \frac{\partial^2u}{\partial x\partial y}\left(\frac{\partial u}{\partial x}\right)^2+\frac{\partial^2u}{\partial y^2}\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}=0 $$

Adding up we get $$\frac{\partial^2u}{\partial x\partial y}=0.$$

Let $\frac{\partial u}{\partial x}=A(x)$, $\frac{\partial u}{\partial y}=B(y)$. Put this in $\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2=1$ we get that both are constant.

Similarly $\frac{\partial v}{\partial x}$ and $\frac{\partial v}{\partial y}$ are constant. Hence $\begin{pmatrix}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{pmatrix}$ is constant.

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    $\begingroup$ How do you get $\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial y^2} = 0$? $\endgroup$
    – Charles Hudgins
    2 days ago
  • $\begingroup$ @Charles I think I got something wrong. Edited. $\endgroup$
    – Saunders
    2 days ago

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