I am trying to solve the equation $$ \frac{\partial}{\partial\alpha}F(x; \alpha) = \lambda \frac{\partial^2}{\partial x^2}F(x; \alpha) \qquad (1)$$ with the condition $F(x; \alpha = 0) = \exp(\ln 2 \, e^{x})$. Since (1) is the diffusion equation, it seems that one can use the point source solution $$G(x; \alpha) = \frac{1}{(4 \pi \lambda \alpha)^{1/2}} \exp\left( - \frac{x^2}{4 \lambda \alpha}\right), \qquad (2)$$ which satisfies $$\frac{\partial}{\partial\alpha}G(x; \alpha) = \lambda \frac{\partial^2}{\partial x^2}G(x; \alpha)\, ; \qquad G(x; \alpha=0) = \delta (x),$$ to write the general solution \begin{align} F(x; \alpha) = \int^{\infty}_{-\infty} dx' \, F(x'; 0) G(x-x'; \alpha). \qquad (3) \end{align} However with $F(x;\alpha) = \exp(\ln 2 e^{x})$ and $G(x; \alpha)$ given in (2), the integrand of (3) diverges at $x \to \infty$. I'm guessing I'm violating some boundary condition assumption for (3) by using an initial condition that diverges at infinity, but I'm also not sure how else to construct a general solution to (1).
Is it possible to solve (1) with the initial condition $F(x;\alpha) = \exp(\ln 2 e^{x})$?