How Does the Dirac Delta Function Solve the Differential Equation?

[Summer95]

Homework Statement

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 $f(t)=\delta(t-t_{0})$.

Homework Equations

Convolution (Boas)
Laplace Transforms (Boas)

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?

 

[Ray Vickson]

Homework Statement

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 $f(t)=\delta(t-t_{0})$.

Homework Equations

Convolution (Boas)
Laplace Transforms (Boas)

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?

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

 

[Summer95]

Is f(t)f(t)f(t) supposed to be zero for t<t0t<t0t < t_0 and t>t0+1nt>t0+1nt > t_0+\frac{1}{n}?

Yes! Sorry I forgot to specify that.