Suppose $X_t$ is a brownian motion with $X_0 \sim u_0$. What is the probability density of $X_t$? (heat equation) 
Suppose $u_0(x) = 2x$ for $0 \leq x \leq 1$ and $u_0(x)=0$ otherwise. Suppose $X_t$ is a brownian motion with $X_0 \sim u_0$. What is the probability density of $X_t$?

Since $X_t$ is a brownian motion, the density should satisfy the heat equation
$$ \partial_t u = \frac{1}{2} \partial_x^2 u .$$
The first part of the question gives us initial conditions for the heat equation $u_0(x) = 2x$ for $0 \leq x \leq 1$ and $u_0(x)=0$ otherwise.
I think the problem boils down to solving the heat equation for this initial condition. How would this be done?
 A: If $(B_t^x)_{t \geq 0}$ is a Brownian motion started at $x \in \mathbb{R}^d$, then the density of $B_t^x$ is given by
$$p_t^x(y) = \frac{1}{(2\pi t)^{d/2}} \exp \left(- \frac{|y-x|^2}{2t} \right).$$
Using the Dirac measure we can rewrite this as follows:
$$p_t^x(y) = \frac{1}{(2\pi t)^{d/2}} \int \exp \left(- \frac{|y-z|^2}{2t} \right) \delta_x(dz) $$
Note that $\delta_x$ is the initial distribution of the Brownian motion $(B_t^x)_{t \geq 0}$. If we replace $\delta_x$ by some general distribution, say $\mu$, we get
$$p_t^{\mu}(y) = \frac{1}{(2\pi t)^{d/2}} \int \exp \left(- \frac{|y-z|^2}{2t} \right) \, \mu(dz).$$
Roughly this means that we mix the densities $(p_t^x)_{x \in \mathbb{R}^d}$ according to our given initial distribution. In particular, if $\mu(dz) = u_0(z) \, dz$ for some density $u_0$, we have
$$p_t^{\mu}(y) = \frac{1}{(2\pi t)^{d/2}} \int \exp \left(- \frac{|y-z|^2}{2t} \right) u_0(z) \, dz.$$
A straight-forward calculation shows that this function does indeed satisfy the heat equation. Moreover, one can show that $p_t^{\mu}(y) \to u_0(y)$ as $t \to 0$. Note that the density $(p_t^{\mu})$ is the convolution of the initial distribution and the heat kernel.
This is the probabilistic approach. The more popular approach to solve the heat equation uses Fourier methods (i.e. take the Fourier transform of the heat equation, do some calculations and then invert the Fourier transform to get the solution $u$). You will find this in (almost) any book on PDEs.
