Been given the following problem:
Consider the PDE $$u_t = u_{xx} + u, 0 < x < 1$$ $$u(0, t) = u(1, t) = 0, t > 0$$ $$u(x, 0) = f(x), 0 < x < 1$$ (a) Use the method of separation of variables to obtain a solution for $v(x,t)$, where $ u(x,t) = e^{rt}v(x, t)$ , choosing $r$ such that the above problem reduces to the heat equation, and then, obtain the solution for $u(x, t)$
How would I find $r$ to reduce the problem to a heat equation as asked? Cannot seem to find this information anywhere
With
$u_t = u_{xx} + \beta u, \tag 1$
and
$u(x, t) = e^{rt} v(x, t), \tag 2$
we have
$u_t = re^{rt}v + e^{rt}v_t \tag 3$
and
$u_{xx} = e^{rt}v_{xx}; \tag 4$
then (1) becomes
$re^{rt}v + e^{rt}v_t = e^{rt}v_{xx} + \beta e^{rt} v; \tag 5$
we may cancel the factor $e^{rt}$:
$rv + v_t = v_{xx} + \beta v, \tag 6$
and re-arrange things slightly to obtain
$v_t = v_{xx} + (\beta - r) v; \tag 7$
if we now choose
$r = \beta, \tag 8$
we find that (7) yields
$v_t = v_{xx}, \tag 9$
the usual heat equation.
We conform to the requirement of the present question but setting
$r = 1, \tag{10}$
whence
$u(x, t) = e^t v(x, t), \tag{11}$
or
$v(x, t) = e^{-t} u(x, t). \tag{12}$