Like a bipolar transistor, a JFET consists of three layers of P and N silicon. However, these layers are connected in a different way. The picture below shows the inside of an N-channel JFET.

The two P layers are connected to each other and form the Gate. The Source and the Drain are both connected to the N layer. That's why this is called an N-channel JFET.

Let's connect an N-channel JFET to some voltage sources:

While discussing the inside of a diode, we saw that the depletion zone between a PN junction increases if the reverse voltage increases. So if V_GS becomes more negative, the depletion zone between the P layers and the N layer will become thicker, narrowing down the channel between the Source and the Drain. The voltage at which the channel is closed, is called the pinch-off voltage.

Now let's see what happens if V_GS remains constant and V_DS increases:

At first I_D increases as well, just like the D-S channel were a resistor. However, increasing V_DS also makes V_GD more negative, narrowing down the channel. When V_GD has reached a certain voltage, the pinch-off voltage, I_D cannot increase any further. The FET has now become saturated.

So if V_GD is between 0 and the pinch-off voltage, the S-D channel looks like a resistor; its resistance can be controlled by V_GS. If V_GD becoms less than the pinch-off voltage, the S-D channel acts like a current source that can be controlled by V_GS.

We see that both V_GS and V_GD have a "pinch-off value". Since JFETs are symmetrical, the pinch-off voltages are the same. You may even swap the Drain and the Source!