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Let $f\in C^0(\mathbb R^n)\cap L^\infty(\mathbb R^n)$ and $\alpha\in\mathbb R$. How can you find a solution $u$ of $$\begin{cases}\frac{\partial u}{\partial t}(x,t)-\Delta u(x,t)&=&-\alpha\cdot u(x,t) &\text{ in }]0,\infty[\times\mathbb R^n\\ \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space u&=&f&\text{ on }\mathbb R^n\end{cases}$$ ?

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Construct the solution $u$ of the usual heat equation

$$\frac{\partial}{\partial t} u = \Delta u, \quad \quad u(0,x) = f(x)$$

in your favorite manner. Now consider $w(t,x) = e^{-\alpha t}u(t,x)$. Then by the product rule

$$\frac{\partial}{\partial t} w(t,x) = e^{-\alpha t}\frac{\partial}{\partial t}u(t,x) - \alpha e^{\alpha t}u(t,x)$$ $$ = e^{-\alpha t}\Delta u(t,x) - \alpha w(t,x) = \Delta (e^{-\alpha t}u(t,x)) - \alpha w(t,x)$$ $$= \Delta w(t,x) - \alpha w(t,x)$$


$$w(0,x) = e^{-\alpha \cdot 0}u(0,x) = u(0,x) = f(x)$$

so that $w$ is the desired solution. For the construction of solutions to the usual heat equation, it would be easier if you imposed some decay conditions on $f$, say, $f \in L^2$. In that case, one way of doing it is by using the Fourier transform. Consider the Laplacian $\Delta: H^2 \to L^2$. Conjugating by the Fourier Transform yields

$$(F\Delta F^{-1})u(x) = -\|x\|^2u(x)$$

and thus in Fourier space, the heat equation reads

$$\frac{\partial}{\partial t}\hat{u}(t) = -\|\xi\|^2\hat{u}(t)$$ which yields $$\hat{u}(t) = e^{-\|\xi\|^2t}\hat{u}(0) = e^{-\|\xi\|^2t}\hat{f}$$

Transforming back yields

$$u(t) = (F^{-1}e^{-\|\xi\|^2t}F)f$$

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