# Is heat equation considered as LTI system?

The heat equation

$$\frac{\partial u}{\partial t} = \alpha \nabla^2 u + f(x,t) \tag 1$$

Here, $$u$$ represents the temperature field, $$\alpha$$ is the thermal diffusivity of the material, $$\nabla^2$$ is the Laplacian operator, and $$f(x,t)$$ represents the sink/source term. I guess that it would depends on what kind of sink/source function would be to be able to consider it linear and time invariant.

My physical intuition is that it should be linear similar to how heat source can add up, but maybe of we put heat source as a function of the temperature itself it might break linearity? Not sure about time invariant too.

• I think if $u(t)$ is measured in Kelvin and if $f(x,t)$ has no dependence on $u(t)$, it's an LTI system. Commented Mar 24 at 16:12

An equation alone does not describe a system. You can define a linear system from it if you consider the heat equation as defining a system $$h$$ where $$u(\mathbf x, t) = h(f(\mathbf x, t)) \tag a$$ and the dynamics of $$u(\mathbf x, t)$$ conform to your (1), which I repeat here: $$\frac{\partial}{\partial t} u(\mathbf x, t) = \alpha \nabla^2 u(\mathbf x, t) + f(\mathbf x,t) \tag b$$
The important thing that I've done here is to define $$f(\mathbf x,t)$$ as an input to your system -- this means that you don't take its dependency on $$\mathbf x$$ or $$t$$ into account when determining if $$h$$ is linear.
This leaves with the operations on $$u$$. $$\frac {\partial}{\partial t}$$ is a linear operation, and so is $$\nabla^2$$ -- so the system is linear.
Note that what the is not a finite-state system. $$u$$ is the system state, but even if it's describing a finite-sized chunk of stuff, the states are continuous. This means that even if you did find a closed-form Laplace-domain transfer function from $$f$$ to $$\mathbf u$$ it would not be a ratio of polynomials in $$s$$.