After C has been charge up (ie. S1 closed for a long time). The Voltage across C is same as E, effectively you have an open circuit at C. So initially, when S2 is closed, there cannot be any current through R1 because potential difference across it is E-V_C = 0, but as C discharges through R2, V_C drops and now E-V_C >0 and so E will charge up C again via R1 (so current through R1 becomes non-zero). And so as C discharges through R2, E through R1 constantly charges C up again. I believe the net effect would be that C is just a spectator in this process.
Wouldn't the current running through R1 also run through R2 because there is an open circuit at C? And the process you described seems to run like you said in that last sentence...that C would remain a spectator. In that case, when S2 is closed, can C be ignored?
Found the actual graph (the attachment). The horizontal asymptote is above 0, I1 starts at 0A and I2 starts at a positive value current. I was thinking both I1 and I2 would be under the horizontal asymptote; why is I2 above it?