# Thread: ODE with Driac Delta Function

1. ## ODE with Driac Delta Function

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.

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

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

4. 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.

5. Originally Posted by 1005
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.

6. Originally Posted by lvleph
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.

7. 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.

8. 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:
$\displaystyle \int\limits_{-\infty}^{t} \delta(t) \, dt = u(t),$

9. 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.

10. Would it not be easier to find the solution (general or just the PI) using the LT?

CB

11. I would think that the simplest way to find a particular solution is to use "variation of parameters".

Since we know that $\displaystyle e^x$ and $\displaystyle e^{-x}$ are independent solutions to the homogeneous equation, try a solution of the form $\displaystyle y(x)= u(x)e^x+ v(x)e^{-x}$ where "u" and "v" are functions to be determined.

$\displaystyle y'(x)= u'(x)e^x+ u(x)e^x+ v'(x)e^{-x}- v(x)e^{-x}$

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 $\displaystyle u'e^x+ v'e^{-x}= 0$. That leaves $\displaystyle y'= u(x)e^x- v(x)e^{-x}$ and so

$\displaystyle y"= u'(x)e^x+ u(x)e^x- v'(x)e^{-x}+ v(x)e^{-x}$
and
$\displaystyle y"- y= u'(x)e^x- v'(x)e^{-x}= \delta(x)$

We now have the two equations $\displaystyle u'(x)e^x+ v'(x)e^{-x}= 0$ and $\displaystyle u (x)e^x- v'(x)e^{-x}= \delta(x)$ which are easy to solve for u'(x) and v'(x).

Adding the two equations, $\displaystyle 2u'(x)e^x= \delta(x)$ so that $\displaystyle u'(x)= \frac{1}{2}e^{-x}\delta(x)$.

Subtracting the two equations, $\displaystyle 2v'(x)e^{-x}= -\delta(x)$ so that $\displaystyle v'(x)= -\frac{1}{2}e^x\delta(x)$.

Now, integrate by using the basic "definition" of the $\displaystyle \delta$ function: $\displaystyle \int_a^b f(x)\delta(x)dx= f(0)$ if $\displaystyle a\le 0\le b$, 0 otherwise.

$\displaystyle u(x)= \frac{1}{2}\int_{-\infty}^x e^{-x}\delta(x)dx= \{\begin{array}{cc}0 & x< 0 \\ \frac{1}{2} & x\ge 0\end{array}$

$\displaystyle v(x)= -\frac{1}{2}\int_{-\infty}^x e^x\delta(x)dx= \{\begin{array}{cc}0 & x< 0 \\ -\frac{1}{2} & x\ge 0\end{array}$

12. Writing $\displaystyle Y(s)= \mathcal {L} \{y(x)\}$ , the DE written in terms of Laplace Transform is...

$\displaystyle \displaystyle Y(s) (s^{2}-1) = s\ y(0) + y^{'} (0) + 1$ (1)

... so that is...

$\displaystyle \displaystyle y(x)= \mathcal {L}^{-1} \{ \frac{s\ y(0) + y^{'}(0)}{s^{2}-1} + \frac{1}{s^{2}-1} \} = y(0)\ \cosh x + \{1+y^{'}(0)\}\ \sinh x$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

13. Originally Posted by HallsofIvy
I would think that the simplest way to find a particular solution is to use "variation of parameters".
This is essentially what I was attempting.

14. Originally Posted by HallsofIvy

Now, integrate by using the basic "definition" of the $\displaystyle \delta$ function: $\displaystyle \int_a^b f(x)\delta(x)dx= f(0)$ if $\displaystyle a\le 0\le b$, 0 otherwise.

$\displaystyle u(x)= \frac{1}{2}\int_{-\infty}^x e^{-x}\delta(x)dx= \{\begin{array}{cc}0 & x< 0 \\ \frac{1}{2} & x\ge 0\end{array}$

$\displaystyle v(x)= -\frac{1}{2}\int_{-\infty}^x e^x\delta(x)dx= \{\begin{array}{cc}0 & x< 0 \\ -\frac{1}{2} & x\ge 0\end{array}$
I was just looking over this, and I believe we should be integrating over $\displaystyle \mathbb{R}$ so that we have
$\displaystyle u(x)= \frac{1}{2}\int_{-\infty}^{+\infty} e^{-x}\delta(x)dx= \frac{1}{2}\left[\int_{-\infty}^x e^{-x}\delta(x)dx + \int_x^{+\infty} e^{-x}\delta(x)dx\right]= \frac{1}{2}$
$\displaystyle u(x)= -\frac{1}{2}\int_{-\infty}^{+\infty} e^{x}\delta(x)dx= -\frac{1}{2}\left[\int_{-\infty}^x e^{x}\delta(x)dx + \int_x^{+\infty} e^{x}\delta(x)dx\right]= -\frac{1}{2}$
Or am I missing something?