What are the subsequent inputs to the DES key schedule after round 1? I've just been studying the DES crypto algorithm as presented by Christof Paar in his book entitled, Understanding Cryptography as well as his lecture. There is a diagram on page 68 of the text where the key schedule is shown and I understand the following information about it already:


A 64 bit "key" is originally input
That 64 bit key has 8 bits removed from it during the Permuted Choice 1 permutation, making it now a 56 bit key.
The 56 bits output from PC-1 are now split into two subsections, $C_0$ and $D_0$ , each of which are 28 bits.
$C_0$ and $D_0$ each have their respective 28 bits rotated left once if the round is 1,2,9, or 16, otherwise their respective bits are rotated left by 2 bits.
** After the above occurs, the permuted 56 bits gets processed by the Permuted Choice 2 permutation and the output is a 48 bit round key. **


I highlighted item #5 here because this is where my confusion lies and the precursor to my question: Once we have a round key, $k_1$ for example, it is obviously then placed into the $f$ function as input along with the $R_i$. However, after this round completes, the round key $K_i$ is 48 bits but the key schedule transforms appear to take a 56 bit input. Can someone please explain how we get from a round key such as $k_2$ (the subkey for round 2 of the algorithm) to round key $k_3$? Is the original key re-used again for each round transformation or is the previous rounds key used?
As you already correctly observed, to get $k_1$, you rotate $C_0$ and $D_0$ left to obtain $C_1$ and $D_1$ and apply PC-2 to $C_1$ and $D_1$. 

To get $k_2$, you simply rotate $C_1$ and $D_1$ left to obtain $C_2$ and $D_2$ and apply PC-2. To get $k_3$, you rotate $C_2$ and $D_2$ left to obtain $C_3$ and $D_3$ and apply PC-2. This continues until $k_{16}$. Also see Figure 3 in the Wikipedia article on DES for a graphical illustration. The round keys $k_1, \dots, k_{16}$ are then used in the round function of the respective round.