# ODE with Driac Delta Function

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• Jul 14th 2010, 03:18 PM
lvleph
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 $t\in \mathbb{R}$ find a general solution to
$y'' - y = \delta(x)$.

So what I have so far is
$y'' - y = 0 \quad \forall t \ne \xi$
Thus, the general solution would be of the form
$y = A e^t + B e^{-t}$
We also know that the first derivative makes a negative one jump when $t = \xi$ so
$A_1 e^t - B_1 e^{-t} - A_2 e^t + B_2 e^{-t} = -1$
But this doesn't really seem to help me get to a final solution.
Any help would be greatly appreciated. Thank you in advance.
• Jul 14th 2010, 03:46 PM
Ackbeet
I think you have to take limits of the derivative as x approaches zero in order to get your condition on the derivative.
• Jul 14th 2010, 03:50 PM
lvleph
You mean as $t \to \xi$? But I am still not sure what you mean.
• Jul 14th 2010, 03:57 PM
1005
y(t) is not
$y = A e^t + B e^{-t}$

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

We then integrate both sides of the equation from 0- to 0+:
$\int_{0^-}^{0^+}y''(t)\, dt-\int_{0^-}^{0^+}y(t)\, dt=\int_{0^-}^{0^+}\delta(t)\, dt$
which becomes:
$y'(0+) - y'(0-) - \int_{0^-}^{0^+}y(t)\, dt=u(0+) - u(0-)$
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-:
$y'(0+) - y'(0-)=1$
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:
$A - B = 1$
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.
• Jul 14th 2010, 04:16 PM
lvleph
Quote:

Originally Posted by 1005
The unit step is 1 at 0+ yet 0 at 0-:
$y'(0+) - y'(0-)=1$

This may be in the definition of the delta function, but I don't understand why this is.
• Jul 14th 2010, 04:21 PM
1005
Quote:

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.
• Jul 14th 2010, 04:25 PM
lvleph
I understood that the area had to be of unit one. I guess I just didn't understand
Quote:

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.
• Jul 14th 2010, 04:28 PM
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:
$\int\limits_{-\infty}^{t} \delta(t) \, dt = u(t),$
• Jul 15th 2010, 03:20 AM
Ackbeet
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.
• Jul 15th 2010, 03:34 AM
CaptainBlack
Would it not be easier to find the solution (general or just the PI) using the LT?

CB
• Jul 15th 2010, 03:58 AM
HallsofIvy
I would think that the simplest way to find a particular solution is to use "variation of parameters".

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

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

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

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

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

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

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

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

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

$v(x)= -\frac{1}{2}\int_{-\infty}^x e^x\delta(x)dx= \{\begin{array}{cc}0 & x< 0 \\ -\frac{1}{2} & x\ge 0\end{array}$
• Jul 15th 2010, 07:08 AM
chisigma
Writing $Y(s)= \mathcal {L} \{y(x)\}$ , the DE written in terms of Laplace Transform is...

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

... so that is...

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

Kind regards

$\chi$ $\sigma$
• Jul 15th 2010, 07:49 AM
lvleph
Quote:

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.
• Jul 20th 2010, 08:06 AM
lvleph
Quote:

Originally Posted by HallsofIvy

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

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

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

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