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I thought about your problem and realized that there is no coordinate parametrization of the catenoid with the properties that you want.

Here is my argument: First, note that, in the given $uv$-parametrization, the first and second fundamental forms of the immersion are $$ I= \cosh(u)^2 (\mathrm{d}u^2 + \mathrm{d}v^2) \quad\text{and}\quad I\!I = \mathrm{d}u^2- \mathrm{d}v^2. $$ We want to find two (possibly local) foliations by geodesics, say $\mathcal{F}_i$ for $i=1,2$, such that the tangent vectors to the leaves of one foliation are conjugate to the tangent vectors to the leaves of the other foliation. What this means is that, if $X$ is a vector field tangent to the leaves of $\mathcal{F}_1$ and $Y$ is a vecor field tangent to the leaves of $\mathcal{F}_2$, then $I\!I(X,Y)=0$.

Of course, we can assume, without loss of generality, that $X$ and $Y$ are unit vector fields, i.e., $I(X,X)=I(Y,Y)=1$.

Now, given any unit tangent vector field to the surface, say, $Z$, the condition that the flow lines of $Z$ be geodesics is that $Z^\flat$, the dual $1$-form, be closed.

Thus, we require that $X$ and $Y$ be unit vector fields so that $I(X,X)=I(Y,Y)=1$ and that $X^\flat$ and $Y^\flat$ be closed, and we also require that $I\!I(X,Y) = 0$, since this is the conjugate property. If $$ X^\flat = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v), $$ for some angle function $\alpha$, then the equations $I(Y,Y)=1$ and $I\!I(X,Y) = 0$ force $$ \pm Y^\flat = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v). $$ Note that $\cos(2\alpha)\not=0$, otherwise, $X\pm Y = 0$, and the two foliations will not be transverse, so they can't be the level sets of the desired coordinate functions.

So we need that the following $1$-forms be closed: $$ \xi = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v) \quad\text{and}\quad \eta = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v). $$ If we take the exterior derivatives of these 1-forms and set them equal to zero, we find that $\alpha$ must satisfy the equation $$ \mathrm{d}\alpha = \frac{\tanh u}{\cos 2\alpha}\, \bigl(\mathrm{d}v - \sin 2\alpha\, \mathrm{d}u\bigr). $$ But now applying the exterior derivative to both sides of this relation, we see that $$ 0 = \frac{1-2\cosh^2u}{\cosh^2u\,\cos 2\alpha}\,\mathrm{d}u\wedge\mathrm{d}v. $$ Since $1-2\cosh^2u$ is never zero, we have a contradiction, so $\alpha$ cannot exist.

Remark: For comparison, it might be helpful to look at the case of the helicoid, parametrized by $(u\cos v, u\sin v, v)$. Here we have $$ I= \mathrm{d}u^2 + (u^2+1)\mathrm{d}v^2 \quad\text{and}\quad I\!I = \frac{2 du\,dv}{\sqrt{u^2+1}}. $$ now, reasoning as above, we seek an angle $\alpha$ such that the $1$-forms $$ \xi = \cos\alpha\,\mathrm{d}u +\sin\alpha\,\sqrt{u^2+1}\,\mathrm{d}v \quad\text{and}\quad \eta = \cos\alpha\,\mathrm{d}u -\sin\alpha\,\sqrt{u^2+1}\,\mathrm{d}v $$ are closed. The equation $d(\xi+\eta)=2\,\mathrm{d}(\cos\alpha\,\mathrm{d}u)=0$ implies that $\alpha$ must be a function of $u$ and the equation $d(\xi-\eta)=2\,\mathrm{d}(\sin\alpha\,\sqrt{u^2+1}\,\mathrm{d}v)=0$ then implies that $\sin\alpha\,\sqrt{u^2+1} = c$ for some constant $c\not=0$. Thus, $$ \xi = \sqrt{\frac{u^2+1-c^2}{u^2+1}}\,\mathrm{d}u +c\,\mathrm{d}v \quad\text{and}\quad \eta = \sqrt{\frac{u^2+1-c^2}{u^2+1}}\,\mathrm{d}u -c\,\mathrm{d}v, $$ defined on the range where $u^2+1-c^2>0$ (which is everywhere if $|c|<1$). The dual vector fields are $$ X_\pm = \sqrt{\frac{u^2+1-c^2}{u^2+1}}\, \frac{\partial}{\partial u} \pm \frac{c}{\sqrt{u^2+1}}\,\frac{\partial}{\partial v}, $$ so we choose $x$ and $y$ so that $\mathrm{d}x$ and $\mathrm{d}y$ each annihilate one of $X_\pm$, say $$ \mathrm{d}x = \frac{c\,\mathrm{d}u}{\sqrt{u^2+1-c^2}}+\mathrm{d}v \quad\text{and}\quad \mathrm{d}y = \frac{c\,\mathrm{d}u}{\sqrt{u^2+1-c^2}}-\mathrm{d}v, $$ which gives a 1-parameter family of solutions to the problem.

I thought about your problem and realized that there is no coordinate parametrization of the catenoid with the properties that you want.

Here is my argument: First, note that, in the given $uv$-parametrization, the first and second fundamental forms of the immersion are $$ I= \cosh(u)^2 (\mathrm{d}u^2 + \mathrm{d}v^2) \quad\text{and}\quad I\!I = \mathrm{d}u^2- \mathrm{d}v^2. $$ We want to find two (possibly local) foliations by geodesics, say $\mathcal{F}_i$ for $i=1,2$, such that the tangent vectors to the leaves of one foliation are conjugate to the tangent vectors to the leaves of the other foliation. What this means is that, if $X$ is a vector field tangent to the leaves of $\mathcal{F}_1$ and $Y$ is a vecor field tangent to the leaves of $\mathcal{F}_2$, then $I\!I(X,Y)=0$.

Of course, we can assume, without loss of generality, that $X$ and $Y$ are unit vector fields, i.e., $I(X,X)=I(Y,Y)=1$.

Now, given any unit tangent vector field to the surface, say, $Z$, the condition that the flow lines of $Z$ be geodesics is that $Z^\flat$, the dual $1$-form, be closed.

Thus, we require that $X$ and $Y$ be unit vector fields so that $I(X,X)=I(Y,Y)=1$ and that $X^\flat$ and $Y^\flat$ be closed, and we also require that $I\!I(X,Y) = 0$, since this is the conjugate property. If $$ X^\flat = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v), $$ for some angle function $\alpha$, then the equations $I(Y,Y)=1$ and $I\!I(X,Y) = 0$ force $$ \pm Y^\flat = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v). $$ Note that $\cos(2\alpha)\not=0$, otherwise, $X\pm Y = 0$, and the two foliations will not be transverse, so they can't be the level sets of the desired coordinate functions.

So we need that the following $1$-forms be closed: $$ \xi = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v) \quad\text{and}\quad \eta = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v). $$ If we take the exterior derivatives of these 1-forms and set them equal to zero, we find that $\alpha$ must satisfy the equation $$ \mathrm{d}\alpha = \frac{\tanh u}{\cos 2\alpha}\, \bigl(\mathrm{d}v - \sin 2\alpha\, \mathrm{d}u\bigr). $$ But now applying the exterior derivative to both sides of this relation, we see that $$ 0 = \frac{1-2\cosh^2u}{\cosh^2u\,\cos 2\alpha}\,\mathrm{d}u\wedge\mathrm{d}v. $$ Since $1-2\cosh^2u$ is never zero, we have a contradiction, so $\alpha$ cannot exist.

Remark: For comparison, it might be helpful to look at the case of the helicoid, parametrized by $(u\cos v, u\sin v, v)$. Here we have $$ I= \mathrm{d}u^2 + (u^2{+}1)\mathrm{d}v^2 \quad\text{and}\quad I\!I = \frac{2 du\,dv}{\sqrt{u^2{+}1}}. $$ now, reasoning as above, we seek an angle $\alpha$ such that the $1$-forms $$ \xi = \cos\alpha\,\mathrm{d}u +\sin\alpha\,\sqrt{u^2{+}1}\,\mathrm{d}v \quad\text{and}\quad \eta = \cos\alpha\,\mathrm{d}u -\sin\alpha\,\sqrt{u^2{+}1}\,\mathrm{d}v $$ are closed. The equation $\mathrm{d}(\xi+\eta)=2\,\mathrm{d}(\cos\alpha\,\mathrm{d}u)=0$ implies that $\alpha$ must be a function of $u$ and the equation $\mathrm{d}(\xi-\eta)=2\,\mathrm{d}(\sin\alpha\,\sqrt{u^2{+}1}\,\mathrm{d}v)=0$ then implies that $\sin\alpha\,\sqrt{u^2{+}1} = c$ for some constant $c\not=0$. Thus, $$ \xi = \sqrt{\frac{u^2{+}1{-}c^2}{u^2{+}1}}\,\mathrm{d}u +c\,\mathrm{d}v \quad\text{and}\quad \eta = \sqrt{\frac{u^2{+}1{-}c^2}{u^2{+}1}}\,\mathrm{d}u -c\,\mathrm{d}v, $$ defined on the range where $u^2{+}1{-}c^2>0$ (which is everywhere if $|c|<1$). The dual vector fields are $$ X_\pm = \sqrt{\frac{u^2{+}1{-}c^2}{u^2{+}1}}\, \frac{\partial}{\partial u} \pm \frac{c}{u^2{+}1}\,\frac{\partial}{\partial v}, $$ so we choose $x$ and $y$ so that $\mathrm{d}x$ and $\mathrm{d}y$ each annihilate one of $X_\pm$, say $$ \mathrm{d}x = \mathrm{d}v +\frac{c\,\mathrm{d}u}{\sqrt{(u^2{+}1{-}c^2)(u^2{+}1)}} \quad\text{and}\quad \mathrm{d}y = \mathrm{d}v -\frac{c\,\mathrm{d}u}{\sqrt{(u^2{+}1{-}c^2)(u^2{+}1)}}, $$ which gives a 1-parameter family of solutions to the problem.

I thought about your problem and realized that there is no coordinate parametrization of the catenoid with the properties that you want.

Here is my argument: First, note that, in the given $uv$-parametrization, the first and second fundamental forms of the immersion are $$ I= \cosh(u)^2 (\mathrm{d}u^2 + \mathrm{d}v^2) \quad\text{and}\quad I\!I = \mathrm{d}u^2- \mathrm{d}v^2. $$ We want to find two (possibly local) foliations by geodesics, say $\mathcal{F}_i$ for $i=1,2$, such that the tangent vectors to the leaves of one foliation are conjugate to the tangent vectors to the leaves of the other foliation. What this means is that, if $X$ is a vector field tangent to the leaves of $\mathcal{F}_1$ and $Y$ is a vecor field tangent to the leaves of $\mathcal{F}_2$, then $I\!I(X,Y)=0$.

Of course, we can assume, without loss of generality, that $X$ and $Y$ are unit vector fields, i.e., $I(X,X)=I(Y,Y)=1$.

Now, given any unit tangent vector field to the surface, say, $Z$, the condition that the flow lines of $Z$ be geodesics is that $Z^\flat$, the dual $1$-form, be closed.

Thus, we require that $X$ and $Y$ be unit vector fields so that $I(X,X)=I(Y,Y)=1$ and that $X^\flat$ and $Y^\flat$ be closed, and we also require that $I\!I(X,Y) = 0$, since this is the conjugate property. If $$ X^\flat = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v), $$ for some angle function $\alpha$, then the equations $I(Y,Y)=1$ and $I\!I(X,Y) = 0$ force $$ \pm Y^\flat = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v). $$ Note that $\cos(2\alpha)\not=0$, otherwise, $X\pm Y = 0$, and the two foliations will not be transverse, so they can't be the level sets of the desired coordinate functions.

So we need that the following $1$-forms be closed: $$ \xi = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v) \quad\text{and}\quad \eta = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v). $$ If we take the exterior derivatives of these 1-forms and set them equal to zero, we find that $\alpha$ must satisfy the equation $$ \mathrm{d}\alpha = \frac{\tanh u}{\cos 2\alpha}\, \bigl(\mathrm{d}v - \sin 2\alpha\, \mathrm{d}u\bigr). $$ But now applying the exterior derivative to both sides of this relation, we see that $$ 0 = \frac{1-2\cosh^2u}{\cosh^2u\,\cos 2\alpha}\,\mathrm{d}u\wedge\mathrm{d}v. $$ Since $1-2\cosh^2u$ is never zero, we have a contradiction, so $\alpha$ cannot exist.

I thought about your problem and realized that there is no coordinate parametrization of the catenoid with the properties that you want.

Here is my argument: First, note that, in the given $uv$-parametrization, the first and second fundamental forms of the immersion are $$ I= \cosh(u)^2 (\mathrm{d}u^2 + \mathrm{d}v^2) \quad\text{and}\quad I\!I = \mathrm{d}u^2- \mathrm{d}v^2. $$ We want to find two (possibly local) foliations by geodesics, say $\mathcal{F}_i$ for $i=1,2$, such that the tangent vectors to the leaves of one foliation are conjugate to the tangent vectors to the leaves of the other foliation. What this means is that, if $X$ is a vector field tangent to the leaves of $\mathcal{F}_1$ and $Y$ is a vecor field tangent to the leaves of $\mathcal{F}_2$, then $I\!I(X,Y)=0$.

Of course, we can assume, without loss of generality, that $X$ and $Y$ are unit vector fields, i.e., $I(X,X)=I(Y,Y)=1$.

Now, given any unit tangent vector field to the surface, say, $Z$, the condition that the flow lines of $Z$ be geodesics is that $Z^\flat$, the dual $1$-form, be closed.

Thus, we require that $X$ and $Y$ be unit vector fields so that $I(X,X)=I(Y,Y)=1$ and that $X^\flat$ and $Y^\flat$ be closed, and we also require that $I\!I(X,Y) = 0$, since this is the conjugate property. If $$ X^\flat = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v), $$ for some angle function $\alpha$, then the equations $I(Y,Y)=1$ and $I\!I(X,Y) = 0$ force $$ \pm Y^\flat = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v). $$ Note that $\cos(2\alpha)\not=0$, otherwise, $X\pm Y = 0$, and the two foliations will not be transverse, so they can't be the level sets of the desired coordinate functions.

So we need that the following $1$-forms be closed: $$ \xi = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v) \quad\text{and}\quad \eta = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v). $$ If we take the exterior derivatives of these 1-forms and set them equal to zero, we find that $\alpha$ must satisfy the equation $$ \mathrm{d}\alpha = \frac{\tanh u}{\cos 2\alpha}\, \bigl(\mathrm{d}v - \sin 2\alpha\, \mathrm{d}u\bigr). $$ But now applying the exterior derivative to both sides of this relation, we see that $$ 0 = \frac{1-2\cosh^2u}{\cosh^2u\,\cos 2\alpha}\,\mathrm{d}u\wedge\mathrm{d}v. $$ Since $1-2\cosh^2u$ is never zero, we have a contradiction, so $\alpha$ cannot exist.

Remark: For comparison, it might be helpful to look at the case of the helicoid, parametrized by $(u\cos v, u\sin v, v)$. Here we have $$ I= \mathrm{d}u^2 + (u^2+1)\mathrm{d}v^2 \quad\text{and}\quad I\!I = \frac{2 du\,dv}{\sqrt{u^2+1}}. $$ now, reasoning as above, we seek an angle $\alpha$ such that the $1$-forms $$ \xi = \cos\alpha\,\mathrm{d}u +\sin\alpha\,\sqrt{u^2+1}\,\mathrm{d}v \quad\text{and}\quad \eta = \cos\alpha\,\mathrm{d}u -\sin\alpha\,\sqrt{u^2+1}\,\mathrm{d}v $$ are closed. The equation $d(\xi+\eta)=2\,\mathrm{d}(\cos\alpha\,\mathrm{d}u)=0$ implies that $\alpha$ must be a function of $u$ and the equation $d(\xi-\eta)=2\,\mathrm{d}(\sin\alpha\,\sqrt{u^2+1}\,\mathrm{d}v)=0$ then implies that $\sin\alpha\,\sqrt{u^2+1} = c$ for some constant $c\not=0$. Thus, $$ \xi = \sqrt{\frac{u^2+1-c^2}{u^2+1}}\,\mathrm{d}u +c\,\mathrm{d}v \quad\text{and}\quad \eta = \sqrt{\frac{u^2+1-c^2}{u^2+1}}\,\mathrm{d}u -c\,\mathrm{d}v, $$ defined on the range where $u^2+1-c^2>0$ (which is everywhere if $|c|<1$). The dual vector fields are $$ X_\pm = \sqrt{\frac{u^2+1-c^2}{u^2+1}}\, \frac{\partial}{\partial u} \pm \frac{c}{\sqrt{u^2+1}}\,\frac{\partial}{\partial v}, $$ so we choose $x$ and $y$ so that $\mathrm{d}x$ and $\mathrm{d}y$ each annihilate one of $X_\pm$, say $$ \mathrm{d}x = \frac{c\,\mathrm{d}u}{\sqrt{u^2+1-c^2}}+\mathrm{d}v \quad\text{and}\quad \mathrm{d}y = \frac{c\,\mathrm{d}u}{\sqrt{u^2+1-c^2}}-\mathrm{d}v, $$ which gives a 1-parameter family of solutions to the problem.

I thought about your problem and realized that there is no coordinate parametrization of the catenoid with the properties that you want.

Here is my argument: First, note that, in the given $uv$-parametrization, the first and second fundamental forms of the immersion are $$ I= \cosh(u)^2 (\mathrm{d}u^2 + \mathrm{d}v^2) \quad\text{and}\quad I\!I = \mathrm{d}u^2- \mathrm{d}v^2. $$ For local $x$ and $y$ coordinates of the kind that you want, we can assume that $|\nabla x|^2 = |\nabla y|^2 = 1$, since this will make the level curves of $x$ and $y$ geodesics, and we also require that $I\!I(\nabla x,\nabla y) = 0$, since this is the conjugate property. In particular, it follows that there must be an angle function $\alpha$ such that $\cos 2\alpha \not=0$ and so that $$ \mathrm{d}x = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v) \quad\text{and}\quad \mathrm{d}y = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v). $$ If we take the exterior derivatives of these equations and use the fact that $\mathrm{d}^2=0$, we find that $\alpha$ must satisfy the equation $$ \mathrm{d}\alpha = \frac{\tanh u}{\cos 2\alpha}\, \bigl(\mathrm{d}v - \sin 2\alpha\, \mathrm{d}u\bigr). $$ But now applying the exterior derivative to this relation, we see that $$ 0 = \frac{1-2\cosh^2u}{\cosh^2u\,\cos 2\alpha}\,\mathrm{d}u\wedge\mathrm{d}v, $$ but $1-2\cosh^2u$ is never zero, so we have a contradiction.

I thought about your problem and realized that there is no coordinate parametrization of the catenoid with the properties that you want.

Here is my argument: First, note that, in the given $uv$-parametrization, the first and second fundamental forms of the immersion are $$ I= \cosh(u)^2 (\mathrm{d}u^2 + \mathrm{d}v^2) \quad\text{and}\quad I\!I = \mathrm{d}u^2- \mathrm{d}v^2. $$ We want to find two (possibly local) foliations by geodesics, say $\mathcal{F}_i$ for $i=1,2$, such that the tangent vectors to the leaves of one foliation are conjugate to the tangent vectors to the leaves of the other foliation. What this means is that, if $X$ is a vector field tangent to the leaves of $\mathcal{F}_1$ and $Y$ is a vecor field tangent to the leaves of $\mathcal{F}_2$, then $I\!I(X,Y)=0$.

Of course, we can assume, without loss of generality, that $X$ and $Y$ are unit vector fields, i.e., $I(X,X)=I(Y,Y)=1$.

Now, given any unit tangent vector field to the surface, say, $Z$, the condition that the flow lines of $Z$ be geodesics is that $Z^\flat$, the dual $1$-form, be closed.

Thus, we require that $X$ and $Y$ be unit vector fields so that $I(X,X)=I(Y,Y)=1$ and that $X^\flat$ and $Y^\flat$ be closed, and we also require that $I\!I(X,Y) = 0$, since this is the conjugate property. If $$ X^\flat = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v), $$ for some angle function $\alpha$, then the equations $I(Y,Y)=1$ and $I\!I(X,Y) = 0$ force $$ \pm Y^\flat = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v). $$ Note that $\cos(2\alpha)\not=0$, otherwise, $X\pm Y = 0$, and the two foliations will not be transverse, so they can't be the level sets of the desired coordinate functions.

So we need that the following $1$-forms be closed: $$ \xi = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v) \quad\text{and}\quad \eta = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v). $$ If we take the exterior derivatives of these 1-forms and set them equal to zero, we find that $\alpha$ must satisfy the equation $$ \mathrm{d}\alpha = \frac{\tanh u}{\cos 2\alpha}\, \bigl(\mathrm{d}v - \sin 2\alpha\, \mathrm{d}u\bigr). $$ But now applying the exterior derivative to both sides of this relation, we see that $$ 0 = \frac{1-2\cosh^2u}{\cosh^2u\,\cos 2\alpha}\,\mathrm{d}u\wedge\mathrm{d}v. $$ Since $1-2\cosh^2u$ is never zero, we have a contradiction, so $\alpha$ cannot exist.