To a regular function the solution is a number of system (set?) of numbers. These crazy things have solutions that are actually functions, or sets of functions! That is to say that when you finally get to the point that you have to solve one of these things you may be able to find a non unique solution. These related functions are controlled by constants and referred to as classes. The constants serve the purpose of scaling values.
Pretty nutty, to be sure, but they are also very useful when you have to figure out a function.
This class is focused on first and second order Ordinary DEs, (1st & 2nd ODEs) in opposition to Partial DEs (PDEs). The order of a DE is the highest derivative of the equation. PDEs are derived from partial derivatives and thus, beyond me currently =)
A second important measure of a DE is its linearity, which is a measure of whether a function is linear or not, we don’t care about partial or how close to linear it is.
This function is a first order linear DE
Second order linear DE
First order non-linear DE
Second order non-linear DE
You can see from the pattern that we are looking at the powers of the function. If it is only to the second power we consider that a linear DE. Otherwise life gets messy and we aren’t playing that game yet.
Things are just getting started so I don’t have a clue how this quarter will end, but so far so good. We are currently dealing with the Existence and Uniqueness Thm.
(a) is continuous within (R) AND
For those of you who don’t already understand what they are saying, ill break it down into English, for mathematicians of course.
- Restate the function in normal form, this may require some algebraic kung-fu and or bending of reality. Don’t be afraid, just do it.
- Check to see if the function is continuous (you can do this with the tricks you picked up in earlier calculus courses, such as a smooth check).
- Take the partial differentiation of the function in terms of y and check to make sure THAT function is continuous and differentiable.
- If both of those are smooth functions then a solution exists and it is going to be a unique solution.
You may be wondering the same thing that I was about this, What on earth is this useful for? One example that I have found quite relevant is to think about it like this: We are dealing with problems where the rate of change is related to the function itself. Examples include temperature change, population growth and mixtures. This mixture example is one that we are discussing currently and I think illustrates things in the most explicit terms, I’ll be paraphrasing as best I can because I am not my instructor, nor are my notes detailed enough for an exact quote, I will also jazz it up so you care more.
Let’s say we work at a salinity testing plant. We have been approached by a new contractor who claims to have a new salinity testing device and we need to test it. With a couple meetings with your testing team you come up with the following experiment setup:
We have a cylindrical tank of brine solution (salt water for you non pirates). This tank has some 300.0 gal. of water contained within it, and two pipes, one at the top, an inlet from another tank allowing 3.0 gal./min of 2.0 lbs/gal brine solution, and an outlet from the tank into a salinity measuring device which allows the solution to come out at 3.0 gal./min. The salinity of the tank will be changed four times for four tests so that value is going to be left out at the moment. A stirring device will keep the water at perfect solution, so the salinity in the machine is instantaneously mixed and the output will be instantaneously pumped out.
Knowing these values alone allows us to take some time before we get into the test to find out what our expected values, which is where our DEs come into play. The salinity of our tank is able to be found with a function of A, the amount of salt in our tank and we can simply divide that value by the tank’s volume (300 gal.). First we should find out what the rate of change of our tank is.
This is saying that because of the different in pumping the volume of our tank will stay the same. You can of course integrate this all you wish but the value of our tank is going to be be a constant of 300gal. How about we ignore the quantity and pay attention to the salt itself being pumped in/out? That does make a significant difference, Because the Amount (A(t)) of salt in the tank will be changing over time. It will start at some initial content, but with time it will increase or decrease depending upon the amount of salt being added by the input pump.
With these three steps we now have the ability to calculate the change in salt content within our tank. Our answer is stated in terms of our derivative and the function itself, which is again what our problem was discussing previously. From here you simply need to know what the initial quantity of salt in the tank is.
To be sure, you can take a look at some input conditions that are intuitive. If you had started with a 2 lbs./gal solution your A(t) would be … (2.0 lbs./gal * 300 gal) … 600 lbs and if you plug that into the equation it says it would not change the salinity…
This follows our intuition and holds true to our understanding. You can use this to build off and find a solution to the problem if you are so inclined. I am going to get back to life and all that mess so I hope this gets some of your motors rolling.
Oh, and enjoy this video.