How does one show that $\biggl(\dfrac{\partial}{\partial{t}}-\dfrac{\partial^{2}}{\partial x^{2}}\biggr)$ is a linear operator?

Question

How does one show that $\dfrac{\partial{u}}{\partial{t}}-\dfrac{\partial^{2}{u}}{\partial{x^{2}}}$ is a linear operator? At first, I was thinking that there might be some distributive property that I've forgotten from Calculus, i.e. $\biggl(\dfrac{\partial}{\partial{t}}-\dfrac{\partial^{2}}{\partial x^{2}}\biggr)u$ should somehow equate to $\dfrac{\partial{u}}{\partial{t}}-\dfrac{\partial^{2}{u}}{\partial{x^{2}}}$.

Answer 1

Let $$D=\partial_t-\partial_x^2$$ i.e. $$D: u \mapsto \partial_tu-\partial_x^2u$$ And $D$ is linear iff $\forall$ scalar $\alpha$ and vectors $u,v$ we have that $$D(u+\alpha v)=Du+\alpha Dv$$ So you just need show that $$\partial_t (u+\alpha v)-\partial_x^2(u+\alpha v)=\partial_t u -\partial_x^2 u+\alpha (\partial_t v - \partial_x^2 v)$$ Using the properties of the partial derivatives.