I am trying to solve the following differential equation
$$\frac{dx}{dt} = x \delta(t-\tfrac{1}{2})$$
from $t=0$ to $t=1$ where $\delta$ denotes the typical Dirac function. If I separate variables and integrate, I get
$$ \int_{x_0}^{x_t} \frac{1}{x} dx = \int_0^t \delta (\tau-\tfrac{1}{2}) d\tau$$
leading to
$$ \ln (x_t) - \ln(x_0) = \begin{cases} 0, & t \leq 1/2 \\ 1, & t > 1/2 \end{cases} $$
On the other hand, if I rewrite the original equation as
$$ \int_{x_0}^{x_t} dx = \int_0^t x(\tau) \delta (\tau-\tfrac{1}{2}) d \tau$$
and simply apply the definition of the Dirac function, I get
$$ x_t - x_0 = \begin{cases} 0, & t \leq 1/2 \\ x\left(\tfrac{1}{2}\right), &t > 1/2 \end{cases} $$ Why do I get two different answers? Is either of these right or is the original equation ill-defined in some sense ... perhaps the Radon-Nikodym derivative does not exist since the LHS is wrt to Lebesgue and the RHS is wrt Dirac? Thanks in advance!
The second way is not very well defined, but there's not really a contradiction between them. The only problem is that $x$ is not really the same for $t < 1/2$ as it is for $t = 1/2$, it's tricky to define $x$ at this point, since there's a discontinuity in the function here.
If you look at the first equation, you have
$$ \ln (x_t) - \ln(x_0) = \begin{cases} 0, & t < 1/2 \\ 1, & t > 1/2 \end{cases} $$
Using logarithm rules to rewrite it,
$$ \ln (x_t/x_0) = \begin{cases} 0, & t < 1/2 \\ 1, & t > 1/2 \end{cases} $$ $$ x_t/x_0 = \begin{cases} e^0, & t < 1/2 \\ e^1, & t > 1/2 \end{cases} $$ $$ x_t = \begin{cases} x_0, & t < 1/2 \\ x_0 e, & t > 1/2 \end{cases} $$
And the second equation is
$$ x_t = \begin{cases} x_0, & t < 1/2 \\ x_0 + x(1/2), & t > 1/2 \end{cases} $$
So the definitions are the same for $t < 1/2$ and if you pretend $x(1/2) = x_0(e-1)$, then they are the same for $t > 1/2$ as well.