I think you have to take limits of the derivative as x approaches zero in order to get your condition on the derivative.
Studying for my PhD qualifying exam in PDE and came across a problem that I am unsure how to solve.
Let find a general solution to
So what I have so far is
Thus, the general solution would be of the form
We also know that the first derivative makes a negative one jump when so
But this doesn't really seem to help me get to a final solution.
Any help would be greatly appreciated. Thank you in advance.
y(t) is not
where u(t) is the unit step function. This is true, because a causal system excited at t = 0 by an impulse cannot have a response before t = 0.
Now, on to your question. We will need to know also the form of y'(t):
We then integrate both sides of the equation from 0- to 0+:
Now, it is obvious from looking at the form of y(t) that it has no time to develop area nor does it have any higher order singularity functions. Therefore, an integral from 0- to 0+ yields zero. The unit step is 1 at 0+ yet 0 at 0-:
Now, the impulse is zero at both 0- and 0+, and the u(t) is 1 at 0+ and 0 at 0-. Therefore, y'(0-) = 0. Further, y'(0+) = ae^0+ - be^0+ = a - b. So we can finally write:
To find the next equation, you need to integrate twice. We can do that one a little faster since we've gone through some of the reasoning:
integrating twice yields y(0+) - y(0-) + 0 = 0
then A + B = 0
edit: Also, an impulse is another name for the dirac delta. impulse integrated = u(t). u(t) integrated = ramp(t) = t u(t). That's how I said the double integration equals zero. It was a double integral of delta(t) = ramp(t), and ramp(0+) = ramp(0-) = 0.
To further understand the dirac delta, it's best to think of it as a limit of a rectangular function. If we had a function defined as:
0 if |t| > a/2
1/a if |t| <= a/2
That makes a rectangle from -a/2 to a/2 of height 1/a. The area is height times width: (a/2 - (-a/2) ) *(1/a) = (a) (1/a) = 1. So this rectangular pulse always has 1 area regardless of the value of a. We then take the limit as a -> 0. In this limit, we have a "function" that occurs exactly at t = 0 with an undefined height(infinity height, 1/a as a -> 0), yet it still has an area of 1. Integration, a numerical method to find area under a curve, therefore, always returns 1 if you integrate over the occurrence of the impulse.
I understood that the area had to be of unit one. I guess I just didn't understand. Defining it this way of course leads to a positive one step rather than a negative one step. It seems then it must be from the definition.The unit step is 1 at 0+ yet 0 at 0-:
You can think of the unit step as the integral of the dirac delta. If this were the case, the unit step will return 0 for all values before the impulse occurs. So from -infinity to 0- (right before zero), it returns 0. Then, right at t = 0, the integral accumulates 1 area, so the unit step starts returning 1 from there on out. Technically, the unit step is undefined at t = 0. However, we don't run into this problem, because we are evaluating it at 0+ (a little after 0). So now do you see how u(0+) = 1?
The way you should write it is like so:
I would think that the simplest way to find a particular solution is to use "variation of parameters".
Since we know that and are independent solutions to the homogeneous equation, try a solution of the form where "u" and "v" are functions to be determined.
Since, in fact, there are an infinite number of functions u and v that would work, we can simplify- and narrow the search by requiring that . That leaves and so
We now have the two equations and which are easy to solve for u'(x) and v'(x).
Adding the two equations, so that .
Subtracting the two equations, so that .
Now, integrate by using the basic "definition" of the function: if , 0 otherwise.