I'm interested in understanding the following system of linear ODE's. Consider $\mu \leq 0$ and the system \begin{cases} \frac{d}{dt}u(t) = \mu u, \quad \quad \quad \quad \quad \quad \quad u(0)=1,\\ \frac{\partial}{\partial t} v(s,t) = \delta(t-s)u + \mu v \quad v(s,0)=0 \: \forall s \end{cases} I know the solutions to be $u(t) = e^{\mu t}$ and $$v(s,t) = \begin{cases} e^{\mu t} \quad \text{if } t\geq s \geq 0 \\ 0 \quad \quad \text{otherwise} \end{cases}$$
I'm a bit confused as to what it means to be a solution to the second ode in our system above. The Dirac delta distribution is throwing me off. What I expect the solution of the second system be is the following; $v$ is a solution to the second ODE if for all $\phi \in C^\infty_c(\mathbb{R}^2)$ that $$\int_\mathbb{R} v (s,t)\frac{\partial}{\partial t} \phi(s,t) = u(s)\phi(s) + \mu \int_\mathbb{R} \phi (s,t)v(s,t).$$
This is merely an educated guess from my limited knowledge of distribution theory. When I go to verify this I get a minus sign on the left hand side, but also I'm not even sure if this is even correct.
Many Thanks
You have made an error and missed a minus sign. The distributional derivative of $f$ is defined by $$ \int_{-\infty}^{\infty} f'(x) \, \phi(x) \, dx = - \int_{-\infty}^{\infty} f(x) \, \phi'(x) \, dx $$ so it should be $$ -\int_\mathbb{R} v (s,t)\frac{\partial}{\partial t} \phi(s,t) = u(s)\phi(s) + \mu \int_\mathbb{R} \phi (s,t)v(s,t). $$
Practically, if $f$ is piecewise smooth, $f'(x) = \delta(x-a)$ means that $f$ has a step discontinuity at $x=a$ with $f(a+) - f(a-) = 1,$ where $f(a\pm) = \lim_{x\to a\pm} f(x).$