I am currently interested in solving the inhomogeneous heat equation for a two dimensional flow for a bounded domain. The PDE is given as:
$\frac{\partial u_1'}{\partial t}-\nu(\frac{\partial^2 u_1'}{\partial x_1^2}+\frac{\partial^2 u_1'}{\partial x_2^2})=f(x_1,x_2,t)$
with $f(x_1,x_2,t)=cos(\sqrt{a}x_1)cos(\sqrt{b}x_2)e^{-\nu(a+b)t}$.
I was thinking about solving this by using the Fundamental solution for two dimensions, given as:
$\Phi ={\frac {1}{\sqrt {(4\pi \nu t)^{n}}}}\exp \left(-{\frac {\mathbf {x} \cdot \mathbf {x} }{4 \nu t}}\right)$
The solution would then be given (atleast for the 1D case) as:
$u_1^{'p}=\int_0^t \int_{-\infty}^{\infty}\Phi(\vec{x}-\vec{y},t-s)f(\vec{y},s)d\vec{y}ds$
However since the inhomogeneous function is given as a rather "complicated" structure in terms of integration, I struggle quiete a lot there.
Do you have any ideas or experience in how to solve suche an inhomogeneous heat equation?
I would also appreciate any guidance in how to solve the inhomogeneous heat equation in genereal for a two dimensional flow, because I am not quite sure if this approach is the correct one.
I am very thankful for any advice
Best Answer
The PDE is linear so the general solution is the sum of a homogeneous and particular solution, $u = u_h + u_p$.
The particular solution satisfies
$$ \frac{\partial u_p}{\partial t} - \nu \left(\frac{\partial^2 u_p}{\partial x_1^2} + \frac{\partial^2 u_p}{\partial x_2^2}\right) = f $$
Making the ansatz $u_p(x_1,x_2,t) = T(t)\cos (\sqrt{a}x_1)\cos (\sqrt{b}x_2)$ and subsituting into the PDE, we get
$$T'(t)\cos (\sqrt{a}x_1)\cos (\sqrt{b}x_2) + \nu T(t)(a+b) \cos (\sqrt{a}x_1)\cos (\sqrt{b}x_2) = e^{-\nu(a+b) t}\cos (\sqrt{a}x_1)\cos (\sqrt{b}x_2), $$
and a solution is obtained when
$$T'(t) + \nu(a+b)T(t) = e^{-\nu(a+b) t}$$
Solving the ODE with an integrating factor we get $T(t) = (C+t)e^{-\nu (a+b)t}$, and the particular solution is
$$u_p = (C+t)e^{-\nu (a+b)t}\cos (\sqrt{a}x_1)\cos (\sqrt{b}x_2)$$
As you have not specified an initial condition, nothing can be said yet about the constant $C$ appearing in the solution.
It remains to solve the homogeneous part,
$$ \frac{\partial u_h}{\partial t} - \nu \left(\frac{\partial^2 u_h}{\partial x_1^2} + \frac{\partial^2 u_h}{\partial x_2^2}\right) = 0$$
As you want to solve on a "bounded domain", once you supply the initial and boundary conditions you can apply separation of variables seeking a solution of the form
$$u_h(x_1,x_2,t) = X_1(x_1)X_2(x_2)T(t)$$