dirac delta

2523 days ago by chrisphan

var('t') html("<h1>Dirac delta “function”</h1>Let $$d_\\tau(t) = \\left\{\\begin{matrix} 0, & t \leq -\\tau \cr \\frac{1}{2\\tau}, & -\\tau < t < \\tau \cr 0, & t \geq \\tau \end{matrix}. \\right.$$ If this is an external forcing function, its<br /><b>impulse</b> (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 $\\tau = 2$:") plot((1/4)*(heaviside(t+2) - heaviside(t-2)), -3, 3, rgbcolor=(1,0,0)) 
       

Dirac delta “function”

Let
d_\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, its
impulse (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”

Let
d_\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, its
impulse (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:

                                
html("Now, the <b>Dirac delta</b> “function” (it's not really a function)<br />is thought of as a function with impuse 1 that is concentrated<br />completely at one moment in time, at time $t=0$. It can be thought<br />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:") a = animate([plot((tau/2)*(heaviside(t + 1/tau) - heaviside(t - 1/tau)), -1, 1, rgbcolor=(1,0,0)) for tau in [2, 3, 4, 5, 6, 10, 20, 30, 40, 80, 100]], ymin=0, ymax=20) show(a) 
       
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:
html("<h1>Example 1</h1>Consider the IVP $$3y''+ y' + 2y = \delta(t-4), \; \; y(0)=0, \; \; y'(0) = 0.$$You can think of this a model for a spring-mass system beginning at rest and equilibrium,<br />with an external force of impulse 1 concentrated at time $t = 4$. The solution to the IVP is $$y = u_4(t)\\frac{2}{\sqrt{23}}e^{-(t-4)/6}\sin\left(\\frac{\sqrt{23}}{6} (t-4) \\right).$$ Here is a plot of $y$:") def h(t): return (2/sqrt(23))*exp(-t/6)*sin(sqrt(23)*t/6) plot(lambda t: heaviside(t-4)*h(t-4), 0, 30) 
       

Example 1

Consider the IVP
3y''+ y' + 2y = \delta(t-4), \; \; y(0)=0, \; \; y'(0) = 0.
You can think of this a model for a spring-mass 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^{-(t-4)/6}\sin\left(\frac{\sqrt{23}}{6} (t-4) \right).
Here is a plot of y:

                                
                            

Example 1

Consider the IVP
3y''+ y' + 2y = \delta(t-4), \; \; y(0)=0, \; \; y'(0) = 0.
You can think of this a model for a spring-mass 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^{-(t-4)/6}\sin\left(\frac{\sqrt{23}}{6} (t-4) \right).
Here is a plot of y:

                                
html("<h1>Example 2</h1>Consider the IVP $$y'' + y = \delta(t- \pi/4)\sin(t), \;\; y(0)=0, \;\; y'(0)= 0.$$The solution is $$y = \\frac{\sqrt{2}}{2} u_{\pi/4}(t) \sin(t - \pi/4).$$Here is a plot of $y$:") plot(lambda t:(sqrt(2)/2)*heaviside(t-pi/4)*sin(t-pi/4), 0, 10) 
       

Example 2

Consider the IVP
y'' + 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 2

Consider the IVP
y'' + 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: