Help with Dirac Delta Function Problem

[danai_pa]

i can't solve there problem, please help me

1) delta(y^2-a^2) = 1/absolute 2a[delta(y-a)+delta(y+a)]


2) f(y)delta(y-a) = f(a)delta(y-a)

 

[dextercioby]

Well,the first is really easy,you just need to apply the theory.

For the second,integrate both sides wrt "y" on an interval containing the point y=a.

Daniel.

 

[thepatientmental]

I'm sorry to hear that you're having trouble with the Dirac Delta Function problem. Let's take a look at both of the equations you provided and see if we can break them down step by step to help you solve them.

1) delta(y^2-a^2) = 1/absolute 2a[delta(y-a)+delta(y+a)]

First, let's simplify the expression on the right side. We can see that it is a sum of two delta functions, one with an argument of (y-a) and the other with an argument of (y+a). We can apply the sifting property of the delta function, which states that delta(ax) = 1/absolute a * delta(x), to each of these delta functions.

So, we have 1/absolute 2a * [1/absolute (y-a) * delta(y-a) + 1/absolute (y+a) * delta(y+a)]

Now, we can simplify this further by noticing that 1/absolute (y-a) is the same as 1/absolute (a-y), and similarly for 1/absolute (y+a). This allows us to combine the two fractions and we end up with 1/absolute 2a * 1/absolute (a-y) * [delta(y-a) + delta(y+a)].

Finally, we can simplify this even further by noticing that 1/absolute (a-y) is the same as 1/absolute (y-a), so we can combine these two terms and we end up with 1/absolute 2a * [delta(y-a) + delta(y+a)].

2) f(y)delta(y-a) = f(a)delta(y-a)

For this equation, we can use the scaling property of the delta function, which states that delta(ax) = 1/absolute a * delta(x). In this case, we have delta(y-a), so we can apply this property and we end up with 1/absolute (y-a) * delta(y-a).

Now, we can use the definition of the delta function, which states that the integral of the delta function over any interval containing 0 is equal to 1. In this case, we can integrate the expression 1/absolute (y-a) * delta(y-a) from -infinity to +infinity, and we should get a value of 1.

So, we have f(y)