# Initial data and heat equation

We assume all solutions to be bounded here!

Let $$y_{+},y_{-} \in C_c^{\infty}$$ be two positive functions.

If we then consider the heat equation

$$\partial_t u(t,x) = \Delta u(t,x)$$ for two different initial conditions

$$u_1(0,x) = y_+$$ and $$u_2(0,x)=y_+-y_-$$

then it is clear that $$u_1(t,x) \ge u_2(t,x)$$ for all $$t,x.$$

Now, let us consider the nonlinear heat equation

$$\partial_t u(t,x) = \Delta u(t,x)-\vert u(t,x) \vert^2$$

and assume that there exist two smooth solutions $$u_1,u_2$$ with initial data

$$u_1(0,x) = y_+$$ and $$u_2(0,x)=y_+-y_-$$

respectively.

Does this imply $$u_1(t,x) \ge u_2(t,x)$$?

This sounds plausible, as positive initial data stay positive and negative ones stay negative under evolution. Moreover, at the interface of positive and negative parts, the nonlinearity is small. Still, the nonlinearity makes it hard to say how this linear combination of initial data gets mixed up in the time evolution.

My thoughts

By Duhamel's formula

$$u_1(t,x) = e^{t\Delta}u_1(x) - \int_0^t e^{(t-s) \Delta}\vert u_1(s,x) \vert^2 \ ds.$$

$$u_2(t,x) = e^{t\Delta}u_2(x) - \int_0^t e^{(t-s) \Delta}\vert u_2(s,x) \vert^2 \ ds.$$

Subtracting the two from each other yields

$$u_1(t,x)-u_2(t,x) = e^{t\Delta}y_{-}(x) -\int_0^t e^{(t-s) \Delta}\left(\vert u_1(s,x) \vert^2-\vert u_2(s,x) \vert^2\right) \ ds.$$

Indeed the term $$e^{t\Delta}y_{-}(x)$$ is positive. Not sure how to conclude anything from the second term though. But it seems like for small $$t$$ this term is close to zero.

Observation

If instead we take initial conditions $$u_1(0,x) = y(x)$$ and $$u_2(0,x)= \widetilde{y}(x)$$ with $$0\le \widetilde{y} \le y$$ we seem to get $$u_1(t,x) \ge u_2(t,x)$$ by the above formula.

• The sentence in line 6 which begins with "then it is clear..." is only true under some additional conditions on the growth of $u_j$ at infinity. Commented Apr 13, 2019 at 0:36
• @AlexandreEremenko thanks, I added that the assumption that all solutions shall be bounded.
– user121558
Commented Apr 13, 2019 at 0:39

Yes, this is a simple comparison principle: writing $$u_1^2-u_2^2=(u_1+u_2)(u_1-u_2)$$ shows that the difference $$v(t,x):=u_1(t,x)-u_2(t,x)$$ solves the PDE $$\partial_t v=\Delta v-c(t,x)v$$ with bounded coefficient $$c(t,x):=u_1(t,x)+u_2(t,x)$$ and initial datum $$v(0,x)=y_-\geq 0.$$ The standard linear maximum principle then guarantees that $$v$$ remains nonnegative over time, hence $$u_1(t,x)\geq u_2(t,x)$$ for all $$t\geq 0$$ and all $$x$$.
NB: for more general nonlinearities $$f(u)$$ instead of $$u^2$$ one just writes $$f(u_1)-f(u_2)=\frac{f(u_1)-f(u_2)}{u_1-u_2}(u_1-u_2)=c(t,x) (u_1-u_2)$$, where the quotient is very naturally interpreted as $$f'(u_1)=f'(u_1)$$ whenever $$u_1(t,x)=u_2(t,x)$$. Then this "frozen coefficient" $$c(t,x)$$ is bounded as soon as $$f$$ is smooth enough and the solutions $$u_1,u_2$$ are bounded, and this puts us in position of applying the linear maximum principle for the difference $$v=u_1-u_2$$ (with bounded zeroth-order coefficient).