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Consider $u_t+\gamma u+ \Delta u=0$ with initial condition $u(0,x)=f(x)$, with f continous and bounded.

I know that the main idea to solve this is to insert $u(t,x)=e^{\alpha t} \omega(t,x)$, establish the equation $\omega$ must satisfy, reduce this equation to the classical heat equation by choosing $\alpha=-\gamma$. And since we know the equation to the heat eq, express the $u$ by this. However, there is some logic of solving a PDE this way I dont understand. Why are we allowed to assume that $u(t,x)=e^{\alpha t} \omega(t,x)$ satisfies the heat eq. with reaction term, and thus substitue it into the equation and set equal zero? Why are we allowed to assume the solution is "seperated" this way in two factors?

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Since $e^{\alpha t}$ is never zero, multiplying by $e^{\alpha t}$ is an invertible operation. That is, for every function $f(t, x)$ and every $\alpha$, there exists a unique function $\omega(t, x, \alpha)$ such that $f(t, x) = e^{\alpha t}\omega(t, x, \alpha)$. Since every function has this decomposition, then in particular our solution $u(t, x)$ also has that decomposition.

You might have noticed I wrote $\omega(t, x,\alpha)$ instead of $\omega(t,x)$. Indeed, $\omega$ will be a different function for different $\alpha$. However, because $\alpha$ is just a constant, it's conventional to leave out the $\alpha$ from the argument of $\omega$. Just like it's conventional to write $u(t, x)$ instead of the technically more correct $u(t, x, \gamma)$.

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  • $\begingroup$ Do you know some reference where I can see more about the phisics behind reaction term $\gamma u$? $\endgroup$ Commented Aug 4, 2023 at 3:27

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