# Thread: Heaviside or dirac delta integration

1. ## Heaviside or dirac delta integration

Hello,

I basically have this equation (and another very similar) to solve for my biomechanics course:

Where H(t) is the heaviside function. I can find the solution ε(t) as the sum of the general solution and the particular solution. The general solution is easy to get, and I can find the particular solution of type:

By replacing it into the first equation and determine P(t), however, this lead me to this equation:

The integral can be transformed by part into:

But I can't solve this, mostly because the integral is undefinite.

2. ## Re: Heaviside or dirac delta integration

Are you allowed to use Laplace transforms?

For the particular solution (slight change in notation so I don't have to type as much)

let $\beta = \dfrac{\sigma_0}{\eta}$

$e^\prime(t) + \dfrac{e}{\tau} = \beta H(t)$

take Laplace transforms

$sE(s)-e(0) + \dfrac{1}{\tau}E(s) = \dfrac{\beta}{s}$

$(s^2 + \dfrac s \tau)E(s) = \beta+e(0)$

$E(s) = \dfrac{\beta+e(0)}{s^2 + \frac s \tau} = \dfrac{\beta+e(0)}{s\left(s+\frac 1 \tau\right)}$

apply partial fractions

$E(s) = \dfrac{e(0)-\beta \tau }{s \tau +1}+\dfrac{\beta }{s}$

and invert the Laplace transform

$e(t) =\beta H(t) + \dfrac 1 \tau \left(e(0)-\beta \tau\right)e^{-t/\tau}$

and you would add this to the homogeneous solution to find the solution to the overall diff eq.

3. ## Re: Heaviside or dirac delta integration

Originally Posted by Kyraz
Hello,

I basically have this equation (and another very similar) to solve for my biomechanics course:

Where H(t) is the heaviside function. I can find the solution ε(t) as the sum of the general solution and the particular solution. The general solution is easy to get, and I can find the particular solution of type:

By replacing it into the first equation and determine P(t), however, this lead me to this equation:

The integral can be transformed by part into:

But I can't solve this, mostly because the integral is undefinite.
The fact that the integral "is indefinite" just means it can be written as $\displaystyle \int_0^t H(t) e^{t/\tau} dt+ C$ where C is an unknown constant. And, of course, since H(t) is 1 for all x greater than 0, $\displaystyle \int H(t)e^{t/\tau}t= \int_0^t e^{t/\tau} dt+ C= \tau e^{t/\tau}+ C$.

4. ## Re: Heaviside or dirac delta integration

Thank you very much for your two solutions Romsek and HallsofIvy (I realized some english mistakes in my previous post, sorry for that, I am used to write in french).
By continuing my calculation using the result of HallsofIvy in my particular solution, I finally have this for ε(t) =

However, the professor has the same solution, with a factor H(t).

Given that this model represents the elongation of a tendon, I can guess ε(t) should be =0 for t<0, hence the H(t) in the solution. I don't see from where this additional term comes given that neither my general solution, nor my particular solution introduce the H(t).

5. ## Re: Heaviside or dirac delta integration

Originally Posted by Kyraz
Thank you very much for your two solutions Romsek and HallsofIvy (I realized some english mistakes in my previous post, sorry for that, I am used to write in french).
By continuing my calculation using the result of HallsofIvy in my particular solution, I finally have this for ε(t) =

However, the professor has the same solution, with a factor H(t).

Given that this model represents the elongation of a tendon, I can guess ε(t) should be =0 for t<0, hence the H(t) in the solution. I don't see from where this additional term comes given that neither my general solution, nor my particular solution introduce the H(t).
Let $f(x)$ be an integrable function. Define $g(x) = H(x)\cdot f(x)$. It should be obvious that $\displaystyle \int g(x)dx = H(x)\int f(x)dx$. Think about Riemann sums or Lebesgue integrals (depending on what level you are at in your integration theory) to see why this would be true.

6. ## Re: Heaviside or dirac delta integration

Indeed, I saw Riemann sums and Lebesgue integrals some time ago, but I have never been extremely comfortable with the theory. Thank you again, I finally got the expected solution (but for that, I had to assume the integration constant of my general solution =0, because my initial condition was not enough to determine the two constants in my final solution).