# The Dirac Delta Function

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1. Jun 4, 2016

### Summer95

1. The problem statement, all variables and given/known data
Differential equation: $Ay''+By'+Cy=f(t)$ with $y_{0}=y'_{0}=0$

Write the solution as a convolution ($a \neq b$). Let $f(t)= n$ for $t_{0} < t < t_{0}+\frac{1}{n}$. Find y and then let $n \rightarrow \infty$.

Then solve the differential equation with $f(t)=\delta(t-t_{0})$.

2. Relevant equations

Convolution (Boas)
Laplace Transforms (Boas)

3. The attempt at a solution

So when I go through the first part with $f(t)= n$ for $t_{0} < t < t_{0}+\frac{1}{n}$ and do convolution I get $y=\frac{1}{A(b-a)}\int_0^t(e^{-a(t-\tau)}-e^{-b(t-\tau)})f(\tau)d\tau$ which has different cases depending on t:

0 if $t<t_{0}$

$\frac{n}{A(b-a)}(\frac{1}{a}(1-e^{a(t_{0}-t)})-\frac{1}{b}(1-e^{b(t_{0}-t)}))$ if $t_{0}<t<t_{0}+\frac{1}{n}$

$\frac{n}{A(b-a)}(\frac{1}{a}(e^{-a(t-t_{0}-\frac{1}{n})}-e^{a(t_{0}-t)})-\frac{1}{b}(e^{-b(t-t_{0}-\frac{1}{n})}-e^{b(t_{0}-t)}))$ if $t>t_{0}+\frac{1}{n}$

I don't understand what happens as $n\rightarrow \infty$. I know it should become

$\frac{1}{A(b-a)}(e^{a(t_{0}-t)}-e^{b(t_{0}-t)})$ for $t>t_{0}$ because that is what I get when I use $f(t)=\delta(t-t_{0})$ from the beginning. But what happens to all the extra terms? And the n out front that goes to infinity?

2. Jun 4, 2016

### Ray Vickson

Is $f(t)$ supposed to be zero for $t < t_0$ and $t > t_0+\frac{1}{n}$?

3. Jun 4, 2016

### Summer95

Yes! Sorry I forgot to specify that.