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Let $c \in \mathbb R, d \in \mathbb R ^n$ konstant, $u_0\in C_b(\mathbb R ^n), f\in C^2_1((0,\infty)\times\mathbb R ^n), \text{supp}(f), \text{supp}(u_0)$ compact.

Using the approach $v(t,x):= \exp(-x\cdot d)u(t,x)$ solve the PDE:

$$\begin{cases}\partial_t u - \Delta u + cu + d\cdot \nabla u = f &\text{in }(0,\infty)\times\mathbb R ^n \\ u = u_0 & \text{in } \{0\}\times\mathbb R ^n\end{cases}$$

MY ATTEMPT: we have a parabolic 2nd order linear PDE, so the point of the substitution of the function $u$ must be to get rid of the terms $cu$ and $d\cdot \nabla u$ and use the fundamental solution of the heat equation $(\partial_t - \Delta)v=g$ for some $g$ to get the solution of the initial equation. However, setting $u(t,x)=\exp(x\cdot d)v(t,x)$ and taking the terms putting it into the equation yields:

$$(\partial_t-\Delta)v(t,x) = \exp(-x\cdot d)((\partial _t -\Delta) u + 2 d\cdot \nabla u + |d|^2u)$$ which is not quite the initial equation.

So the substitution must be somehow modified to yield the transformation the PDE that will allow us to use $\partial_t u - \Delta u + cu + d\cdot \nabla u = f$, but unfortunately I don't see how. I hoped maybe someone could give me a clue.

Thank you in advance!

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    Have you tried multiplying also a term $e^{-\gamma t}$? With your substitution you probably get rid only of the gradient term. – 2017-01-29
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    Thank you, apparently some parameters were required to cancel out redundant terms, setting $v(t,x):=\exp(\alpha t + \beta x\cdot d)u(t,x)$ cancels the redundant terms out with $\beta = -\frac 12, \alpha =c + \frac 14 |d|^2$. – 2017-01-29

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