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I am trying to solve the following PDE:

\begin{equation} \partial_t u = \partial_x^2u + u - f(x,t) \end{equation}

with the initial condition: $u(x,t=0) = 0$. ($f(x,t)$ is known).

The first thing to think about is to separate the $u$ into $u = u_h + u_{nh}$, where $u_h$ solves the homogeneous version,and $u_{nh}$ solves the nonhomogeneous one. However, this does not work, as we will return to the initial problem for the latter.

So, what is the method to solve this kind of nonhomogeneous PDE?

AD Le
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Dec 08 '21 at 10:14
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    Write the equation as $$(\partial_t-\partial_x^2-\operatorname{id})u(x,t)=f(x,t)$$ Solve the specific case (i.e find the Green's function) $$(\partial_t-\partial_x^2-\operatorname{id})u(x,t)=\delta(t-\tau)\delta(x-\xi)$$ And then write the general case with $f$ as a convolution. I cannot predict how easy finding the Green's function will be, but this is typically the best way to do things. – K.defaoite Dec 08 '21 at 10:36
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    Consider the general equation $$(\partial_t-\partial_x^2-b\partial_x-c\operatorname{id})u=0$$ $u$ solves the above iff $v$ solves the standard heat equation $(\partial_t-\partial_x^2)v=0$ where $$v(x,t)=\exp\left(\frac{b}{2}x+\left(-c+\frac{b^2}{4}\right)t\right)u(x,t)$$ So this will give you the homogeneous solution anyway. – K.defaoite Dec 08 '21 at 10:59
  • The situation with of solving PDE is very different to that of ODE. In order to 'solve' a PDE you must specify a region and boundary/initial conditions. Even once you have done this only in the absolute simplest cases will you actually be able to 'solve' the PDE. – JackT Dec 09 '21 at 02:18
  • If you want to solve the PDE in all of $x\in \mathbb R$, $t>0$ then you will need assumptions on the growth of $u$ otherwise it is unlikely that solutions are unique. – JackT Dec 09 '21 at 02:19

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