This answer uses the smoothness of the system so I'm not sure it's the most general.

The high school physics setup describes a system that is

Time reversible. If q(t) is a trajectory, then q(-t) is another solution. In this case a ball that starts from rest and slides down the hill.

But staying at rest at the top forever is another valid solution to the initial conditions (at top, v=0)

These are two different solutions for the same initial conditions, violating uniqueness. Now I invoke the smoothness of the slide to say the solution must be unique (this is a standard theorem depending on Lipschitz continuity, which we have). Since permanent rest at the top is a fine solution, the other trajectory must be something else (one that takes infinite time). This is in contrast with Norton's dome which has a cusp at the top.

Poor nephew!

If the top of the slide is parabolic we get an example where traversing

a finite distance takes infinite time.

Let's suppose we trace the parabola y = x^2. When we start at some (x0, y0) we must have kinetic energy = gravitational potential energy: 1/2*m*v0^2 = m*g*(-y0). Notice that the velocity is proportional to the square root of the vertical distance remaining.

What does it look like when we're halfway there? Now we're at position (x0/2, y0/4). We have used up 3/4 of our kinetic energy. Now our velocity (v1) is given by 1/2*m*v1^2 = mg(-y0/4). After substitution v_1=v_0/2. Ah.

The key is to notice that the remaining parabolic trajectory, starting from (x_0/2, y_0/4) is exactly a scaled copy of our original situation. We have climbed, but we haven't made any progress.

This answer uses the smoothness of the system so I'm not sure it's the most general.

The high school physics setup describes a system that is

Time reversible. If q(t) is a trajectory, then q(-t) is another solution. In this case a ball that starts from rest and slides down the hill.

But staying at rest at the top forever is another valid solution to the initial conditions (at top, v=0)

These are two different solutions for the same initial conditions, violating uniqueness. Now I invoke the smoothness of the slide to say the solution must be unique (this is a standard theorem depending on Lipschitz continuity, which we have). Since permanent rest at the top is a fine solution, the other trajectory must be something else (one that takes infinite time). This is in contrast with Norton's dome which has a cusp at the top.

Poor nephew!

If the top of the slide is parabolic we get an example where traversing

a finite distance takes infinite time.

Let's suppose we trace the parabola y = x^2. When we start at some (x0, y0) we must have kinetic energy = gravitational potential energy: 1/2*m*v0^2 = m*g*(-y0). Notice that the velocity is proportional to the square root of the vertical distance remaining.

What does it look like when we're halfway there? Now we're at position (x0/2, y0/4). We have used up 3/4 of our kinetic energy. Now our velocity (v1) is given by 1/2*m*v1^2 = mg(-y0/4). After substitution v_1=v_0/2. Ah.

The key is to notice that the remaining parabolic trajectory, starting from (x0/2, y0/4) is exactly a scaled copy of our original situation. We have climbed, but we haven't made any progress.

This answer uses the smoothness of the system so I'm not sure it's the most general.

The high school physics setup describes a system that is

Time reversible. If q(t) is a trajectory, then q(-t) is another solution. In this case a ball that starts from rest and slides down the hill.

But staying at rest at the top forever is another valid solution to the initial conditions (at top, v=0)

These are two different solutions for the same initial conditions, violating uniqueness. Now I invoke the smoothness of the slide to say the solution must be unique (this is a standard theorem depending on Lipschitz continuity, which we have). Since permanent rest at the top is a fine solution, the other trajectory must be something else (one that takes infinite time). This is in contrast with Norton's dome which has a cusp at the top.

Poor nephew!

This answer uses the smoothness of the system so I'm not sure it's the most general.

The high school physics setup describes a system that is

Time reversible. If q(t) is a trajectory, then q(-t) is another solution. In this case a ball that starts from rest and slides down the hill.

But staying at rest at the top forever is another valid solution to the initial conditions (at top, v=0)

These are two different solutions for the same initial conditions, violating uniqueness. Now I invoke the smoothness of the slide to say the solution must be unique (this is a standard theorem depending on Lipschitz continuity, which we have). Since permanent rest at the top is a fine solution, the other trajectory must be something else (one that takes infinite time). This is in contrast with Norton's dome which has a cusp at the top.

Poor nephew!

If the top of the slide is parabolic we get an example where traversing

a finite distance takes infinite time.

Let's suppose we trace the parabola y = x^2. When we start at some (x0, y0) we must have kinetic energy = gravitational potential energy: 1/2*m*v0^2 = m*g*(-y0). Notice that the velocity is proportional to the square root of the vertical distance remaining.

What does it look like when we're halfway there? Now we're at position (x0/2, y0/4). We have used up 3/4 of our kinetic energy. Now our velocity (v1) is given by 1/2*m*v1^2 = mg(-y0/4). After substitution v_1=v_0/2. Ah.

The key is to notice that the remaining parabolic trajectory, starting from (x_0/2, y_0/4) is exactly a scaled copy of our original situation. We have climbed, but we haven't made any progress.