# SOLVEDODE with Driac Delta Function

#### lvleph

Studying for my PhD qualifying exam in PDE and came across a problem that I am unsure how to solve.

Let $$\displaystyle t\in \mathbb{R}$$ find a general solution to
$$\displaystyle y'' - y = \delta(x)$$.

So what I have so far is
$$\displaystyle y'' - y = 0 \quad \forall t \ne \xi$$
Thus, the general solution would be of the form
$$\displaystyle y = A e^t + B e^{-t}$$
We also know that the first derivative makes a negative one jump when $$\displaystyle t = \xi$$ so
$$\displaystyle A_1 e^t - B_1 e^{-t} - A_2 e^t + B_2 e^{-t} = -1$$
But this doesn't really seem to help me get to a final solution.
Any help would be greatly appreciated. Thank you in advance.

#### Ackbeet

MHF Hall of Honor
I think you have to take limits of the derivative as x approaches zero in order to get your condition on the derivative.

#### lvleph

You mean as $$\displaystyle t \to \xi$$? But I am still not sure what you mean.

#### 1005

y(t) is not
$$\displaystyle y = A e^t + B e^{-t}$$

It is
$$\displaystyle y = (A e^t + B e^{-t})u(t)$$
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):
$$\displaystyle y' = (A e^t + B e^{-t})\delta(t) + (Ae^t-Be^{-t})u(t)$$

We then integrate both sides of the equation from 0- to 0+:
$$\displaystyle \int_{0^-}^{0^+}y''(t)\, dt-\int_{0^-}^{0^+}y(t)\, dt=\int_{0^-}^{0^+}\delta(t)\, dt$$
which becomes:
$$\displaystyle y'(0+) - y'(0-) - \int_{0^-}^{0^+}y(t)\, dt=u(0+) - u(0-)$$
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-:
$$\displaystyle y'(0+) - y'(0-)=1$$
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:
$$\displaystyle A - B = 1$$
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.

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• lvleph

#### lvleph

The unit step is 1 at 0+ yet 0 at 0-:
$$\displaystyle y'(0+) - y'(0-)=1$$
This may be in the definition of the delta function, but I don't understand why this is.

#### 1005

This may be in the definition of the delta function, but I don't understand why this is.
Specifically, what do you not understand? How integration from 0- to 0+ yielded 1? You can think of it using the area property of the dirac delta. By its definition, it has a unit area of 1, so if you integrate over when it occurs (which is exactly at t = 0), you accrue 1 area.

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.

• lvleph

#### lvleph

I understood that the area had to be of unit one. I guess I just didn't understand
The unit step is 1 at 0+ yet 0 at 0-:
. 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.

#### 1005

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:
$$\displaystyle \int\limits_{-\infty}^{t} \delta(t) \, dt = u(t),$$

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• lvleph

#### Ackbeet

MHF Hall of Honor
If it ever matters (I don't think it does here), the unit step is usually defined to be 1/2 right at the origin.

#### CaptainBlack

MHF Hall of Fame
Would it not be easier to find the solution (general or just the PI) using the LT?

CB