I am trying to teach myself PDEs, and I am working on the following problem from Strauss:
(a) If $u_{t}-ku_{xx}=f$, $v_{t}-kv_{xx}=g$, $f \leq g$, and $u\leq v$ at $x=0$, $x = l$, and $t = 0$, prove that $u \leq v$ for $0 \leq x \leq l$, $0 \leq t < \infty$.
(b) If $v_{t}-v_{xx}\geq \sin x$ for $0 \leq x \leq \pi$, $0 < t < \infty$, and if $v(0,t)\geq 0$, $v(\pi,t)\geq 0$ and $v(x,0) \geq \sin x$, use Part (a) to show that $v(x,t) \geq (1-e^{-t})\sin x$.
Now, I am having trouble with both Parts (a) and (b), but more with Part (b).
This is what I've done for each of them, and my thoughts:
(a) Let $w=v-u$, $h=g-f$. Then, by linearity, $w(x,t)$ is a solution to the equation $w_{t}-kw_{xx}=h$, with $w(x,t) \geq 0$ at $x=0$, $x = l$, and $t=0$, and $h \geq 0$ at $x=0$, $x=l$, $t=0$.
From the strong version of the Minimum Principle, the minimum can only be assumed on the bottom or on the lateral sides.
On the lateral sides, either $x=0$ or $x=l$, which implies that $w(x,t) \geq 0$ and $h \geq 0$.
On the bottom, $t = 0$, which implies that $w(x,t) \geq 0$, and $h \geq 0$.
Since on $ 0 \leq x \leq l$, and on $0 \leq t < \infty$, $w(x,t) \geq 0$, we must have that $v(x,t) \geq u(x,t)$ on $0 \leq x \leq l$.
(b) Basically, when $x = 0$ and $x = \pi$, $\sin x = 0$, so $v \geq 0$. When $t = 0$, $(1-e^{-t}) = 0$, so $v \geq (0 \cdot \sin x)$ implies that $v \geq 0$ in this case as well. Since $-1 \leq \sin x \leq 1$, having $v \geq 0$ takes care of the strongest condition. So, I can see why $v(x,t) \geq (1-e^{-t})\sin x$, but I am not sure mathematically HOW to show that it is, especially using Part (a).
(a) looks fine, assuming "the strong version of the maximum principle" indeed applies here (don't have a book nearby, so did not check).
(b) let $u(x,t)=(1-e^{-t})\sin x$. You want to show $u\le v$. Calculate $$u_t-u_{xx} = e^{-t}\sin x + (1-e^{-t})\sin x = \sin x$$ So, the inequality $u_t-u_{xx} \le v_t-v_{xx}$ holds by assumption. Since $u\le v$ holds at $t=0$, $x=0$, and at $x=\pi$, part (a) applies.