Before digital processing got fast and cheap enough to do convolutions, various ways were developed to do it in analog electronics. If you want to convolve two arbitrary signals, then you're out of luck unless you are willing to make a lot of compromises and/or spend a lot of money. Historically, analog convolutions were limited to convolving one real time signal by a pre-determined fixed signal, called the "filter kernel". Either way, some storage is required for each signal, but with one signal fixed it can be implemented by a "permanent" memory, which allows for a lot more possibilities than doing it on the fly.
You still have the problem of storing some portion of the live signal, since some interval of that needs to be multiplied by the kernel as the signal passes by. Systems have been developed that do this with delay lines, traveling electron beams, bucket-brigage charges on a CCD, and accoustic waves. There are probably others that I am not aware of or forgot about.
Once you can somehow store a snapshot of the live signal wide enough to match the filter kernel, you will have to then multiply it by that kernel and sum up the products. In delay line systems, this would be done with "taps" at regular intervals. The signal at each tap would be multiplied by a fixed gain (the filter kernel value at that tap), then all these resulting signals summed. CCDs had split pickups over each charge bucket so that the gain for each bucket was set by where the split was located. This would be set when the chip was made, so there were CCD filter chips with certain pre-determined filters. The most common use was for a sync filter, which is a low pass filter with a sharp frequency cutoff. Surface accoustic wave devices had the signal propagate accross the chip accoustically, which is much slower than light so a large enough time snapshot would be on the chip at any one time. Like with CCD, the pickups were arranged on the chip with pre-determined gains. These parts were typically used for IF and RF notch filters at a well tuned frequency.
Since others have provided answers, I'll try to answer with an intuitive approach.
The original equation is:
\$ y(n) = \sum\limits_{m=-\infty}^\infty u(m)(.5)^{n-m} u(n-m) \$
Now, due the to the presence of \$u(m)\$, the terms for \$m < 0 \$ are zero. Thus, the sum is unchanged by changing the lower limit from \$- \infty \$ to zero. This change of lower limit makes the presence of \$u(m)\$ redundant and so it can be removed.
\$ y(n) = \sum\limits_{m=-\infty}^\infty u(m)(.5)^{n-m} u(n-m) = \sum\limits_{m=0}^\infty (.5)^{n-m} u(n-m) \$
Similarly, due to the presence of \$u(n-m)\$, the terms for \$m > n \$ are zero. Thus, the sum is unchanged by changing the upper limit from \$\infty \$ to \$m=n\$. This change of upper limit makes the presence of \$u(n - m)\$ redundant and so it can be removed.
\$ y(n) = \sum\limits_{m=0}^\infty (.5)^{n-m} u(n-m) = \sum\limits_{m=0}^{m=n} (.5)^{n-m} \$
Best Answer
In addition there should be somewhere mentioned the integration period, it's from time=0 to current time t.
$$\int_0^t{x(\tau)\cdot h(t-\tau)d\tau }$$
Tau is the formal variable for integration process, t is the running physical time. The integral should be calculated completely from blank board for every t. That's quite a job for a signal processor. Fortunately there are accelerating algorithms developed.