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Given the following PDE: $$\partial_t\Psi(x,t)=\partial_{xx}\Psi(x,t)+k\partial_x\Psi(x,t)+g(x,t)-\beta\Psi(x,t)=0$$ where: $$g(x,t)=g_0\left(\frac{a_0}{2}+\sum_{k=1}^\infty\left[a_k\sin(kt)+b_kcos(kt)\right]\right)$$ with: $\Psi(0,t)=0,\Psi(L,t)=0$ and $\Psi(0,x)=\Psi_0$, $\beta$ constant $(\beta\gt\gt1)$ and $k$ constant, is it possible to find analytically a solution of this equation? Thanks in advance.

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1 Answer 1

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Let us write the operator, with the notation $D_x=-i\partial_x$, $$ L=D_x^2-ik D_x+\beta,\quad\text{the equation is}\quad\partial_t\Psi+L\Psi=g. $$ We start with noticing $ \partial_t-ik D_x+\beta=e^{iktD_x}e^{-t\beta}\partial _t e^{t\beta}e^{-iktD_x} $ so that with $\Psi=e^{iktD_x}e^{-t\beta} u$, we get the new equation, using the commutation properties with $L$, $$ \partial_t e^{iktD_x}e^{-t\beta} u+Le^{iktD_x}e^{-t\beta} u=g\Longleftrightarrow e^{-iktD_x}e^{t\beta}\partial_t e^{iktD_x}e^{-t\beta} u+Lu=e^{-iktD_x}e^{t\beta}g $$ i.e. $$ \partial_t u-\beta u+ikD_x u+D_x^2 u-ikD_x u+\beta u=\partial_t u+D_x^2 u=e^{-iktD_x}e^{t\beta}g $$ which is the heat equation: we get for positive times $t$ $$ u=e^{-tD_x^2}u(0)+\int_0^t e^{-(t-s)D_x^2}e^{-istD_x}e^{s\beta}g(s) ds. $$

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