Dirac delta “function”Letd_\tau(t) = \left\{\begin{matrix} 0, & t \leq \tau \cr \frac{1}{2\tau}, & \tau < t < \tau \cr 0, & t \geq \tau \end{matrix}. \right. If this is an external forcing function, itsimpulse (a measure of the total amount of force exerted) is \int_{\infty}^{\infty} d_\tau(t) = \int_{\tau}^\tau \frac{1}{2\tau} dt = 1. Here is a plot of d_{\tau} when \tau = 2:
Dirac delta “function”Letd_\tau(t) = \left\{\begin{matrix} 0, & t \leq \tau \cr \frac{1}{2\tau}, & \tau < t < \tau \cr 0, & t \geq \tau \end{matrix}. \right. If this is an external forcing function, itsimpulse (a measure of the total amount of force exerted) is \int_{\infty}^{\infty} d_\tau(t) = \int_{\tau}^\tau \frac{1}{2\tau} dt = 1. Here is a plot of d_{\tau} when \tau = 2:

Now, the Dirac delta “function” (it's not really a function)
is thought of as a function with impuse 1 that is concentrated completely at one moment in time, at time t=0. It can be thought of as the limit of d_\tau as \tau \rightarrow 0: \delta(t) = \lim_{\tau \rightarrow 0} d_\tau(t). This animation illustrates the limiting process:
Now, the Dirac delta “function” (it's not really a function)
is thought of as a function with impuse 1 that is concentrated completely at one moment in time, at time t=0. It can be thought of as the limit of d_\tau as \tau \rightarrow 0: \delta(t) = \lim_{\tau \rightarrow 0} d_\tau(t). This animation illustrates the limiting process:

Example 1Consider the IVP3y''+ y' + 2y = \delta(t4), \; \; y(0)=0, \; \; y'(0) = 0. You can think of this a model for a springmass system beginning at rest and equilibrium,with an external force of impulse 1 concentrated at time t = 4. The solution to the IVP is y = u_4(t)\frac{2}{\sqrt{23}}e^{(t4)/6}\sin\left(\frac{\sqrt{23}}{6} (t4) \right). Here is a plot of y:
Example 1Consider the IVP3y''+ y' + 2y = \delta(t4), \; \; y(0)=0, \; \; y'(0) = 0. You can think of this a model for a springmass system beginning at rest and equilibrium,with an external force of impulse 1 concentrated at time t = 4. The solution to the IVP is y = u_4(t)\frac{2}{\sqrt{23}}e^{(t4)/6}\sin\left(\frac{\sqrt{23}}{6} (t4) \right). Here is a plot of y:

Example 2Consider the IVPy'' + y = \delta(t \pi/4)\sin(t), \;\; y(0)=0, \;\; y'(0)= 0. The solution is y = \frac{\sqrt{2}}{2} u_{\pi/4}(t) \sin(t  \pi/4). Here is a plot of y:
Example 2Consider the IVPy'' + y = \delta(t \pi/4)\sin(t), \;\; y(0)=0, \;\; y'(0)= 0. The solution is y = \frac{\sqrt{2}}{2} u_{\pi/4}(t) \sin(t  \pi/4). Here is a plot of y:

