Solve the heat problem
$$
\begin{array}{cc}
u_t = u_{xx} & t\geq0, \, x\geq0 \\
u(x,0) = x^2 e^{-x} & u(0,t) = 0
\end{array}
$$
Comments: Note the $x$ domain: $\{x:x\geq0\}$. See if you can relate this problem to one on the full line where symmetry dictates that the boundary condition at zero will be maintained.
For this problem, I believe we have a boundary on the left at x=0 and heat diffusing only on the positive x. Notably $u(x,t)=0$, so the boundary at x=0 on the left is held a constant temperature of 0.
Let:
$$ \phi (x)=x^2 e^{-x}$$
Then we can use this formula:
$$u(x,t)=\int _{-\infty }^{\infty }\phi (y)G(x-y,t)dy$$
I think we can break this integral into two pieces for only the positive x satisfying the half line condition. $\phi(x)$ can be extended as an odd function given piecewise (apologies for the formatting):
$$\phi _{odd}= \left(
\begin{array}{cc} \ &
\begin{array}{cc}
\phi (x) & x>0 \\
0 & x=0 \\
-\phi (-x) & x>0 \\
\end{array}
\\
\end{array}
\right)$$
Breaking up the integral:
$$\int _0^{\infty }\phi (y) G(x-y,t)dy+\int _{-\infty }^0\phi (y)G(x-y,t)dy$$
Applying piece wise condtions:
$$\int _0^{\infty }\phi (y) G(x-y,t)dy-\int _0^{\infty }\phi (y)G(x+y,t)dy$$
Simplifying and substituting for $\phi(x)$ in terms of y we get as our solution:
$$u(x,t)=\int _0^{\infty }y^2e^{-y}[G(x-y,t)-G(x+y,t)]dy$$
I think I have to leave it at that since we don't have a kernel. As for symmetry, I loosely suspect that the boundary condition at zero is maintained in part because the function $\phi(x)$ is extended as an odd function? Any corrections would be appreciated.
EDIT: Since we have the heat equation I think we can use the heat kernel and then solve with terms containing the error function. Will update soon.
Pretty good, though, I'm not sure what you mean by "we don't have a kernel". What's wrong with the heat kernel?