I want to solve the problem
$$ \frac {\partial^2u}{\partial t^2} + \alpha \frac {\partial u} {\partial t} = f(t)$$
$$u(0)=u'(0)=0$$
$$f(t)=exp(-\beta t)$$
Using distributions (Green functions).
So i did
$$ \frac {\partial^2G}{\partial t^2} + \alpha \frac {\partial G} {\partial t} = \delta (t-s)$$ $$G(0)=G'(0)=0$$
When I solved this as I would usually do, by solving the equation for $t \gt s$ and $t \lt s$ and then giving conditions for $G$ (continuous as it is in 1D, with a jump in the derivative and satisfying the homogeneous bondary conditions) I end up with $G=0$, and that is nonsense. So, how would you do this?
If you solve this equations directly, you can show that it actually has a solution. What is my problem? Maybe the green function is not always continuous, but then I don't get why I always ask that condition in order to solve the PDE in 1D using distributions...
Thank you all so much!
First I'll respond to your concerns about generalities: the Green's function for a problem of order $n$ has a jump in the derivative of order $n-1$, and the derivatives of order less than or equal to $n-2$ are all continuous. So for a second order equation, the Green's function is continuous but not differentiable.
In this case you want to have $G$ such that $G''+\alpha G'=0$ except that at $s$ there is a jump in $G'$ of size $1$. Based on that equation, up until $s$, $G$ will simply remain zero. Then at $s$, $G$ stays the same but $G'$ jumps up to $1$. In other words, to handle $t>s$ you need to solve the problem
$$G''+\alpha G'=0,G(s)=0,G'(s)=1.$$
The solution to this problem is not zero, and this is what you need to find.
A nice tip that simplifies life a little bit: in an IVP for an autonomous equation, you do not need to introduce the $s$, because $G$ will always depend only on $t-s$. Instead you can simply put $\delta(t)$ on the right side and be done with it (assuming the initial conditions are at zero).