Suppose that $\varphi(x)\leq M$ for all $x\in\mathbb R$, where $M$ is a positive constant. Show that the solution $u$ to the corresponding Cauchy problem
$$
\begin{align}
u_t &= u_{xx}, \text{ over } x\in{\mathbb R}, t>0 \\
u(x,0) &= \varphi(x)
\end{align}
$$
satisfies $u(x,t)\leq M$ for all $x\in\mathbb R$ and $t>0$.
Lets start with $$u(x,t)=\int\limits_\mathbb{R} \phi(y)K(x-y,t)dy$$
Since the kernel will always be positive definite (except at $t=0$) we can safely use $\phi(x)\leq M$ to say$$u(x,t)\leq\int\limits_\mathbb{R} MK(x-y,t)dy$$
Well $M$ is constant by assumption giving $$u(x,t)\leq M\int\limits_\mathbb{R}K(x-y,t)dy$$ We know from the definition of it being a kernel that $\int_{-\infty }^{\infty }K(x) \, dx=1$. Giving
$$u(x,t)\leq M$$
What this tells us is that if there is no added heat and a maximum temperature then after any amount of time we will never find a point with higher temperature than that maximum. I believe you could show something similar for a minimum temperature under insulation.
Well, I think you've got some relevant stuff written down but you've also got some irrelevant stuff written down. I like this line a lot:
$$u(x,t)\leq\int\limits_\mathbb{R} MG(x-y,t)dy.$$
Now, clearly you can bring the $M$ out, right? So, what's
$$\int\limits_\mathbb{R} G(x-y,t)dy?$$
I don't know what your first line is all about. We already know $u(x,0)=\varphi(x)$.
The first line was to give a reference on the second line where we state what the integral of our kernel should be. I will edit a line in to make it more clear though.
Let me see if I understand your reasoning in the first part of your argument:
Because we have the property that
$$\lim_{t\rightarrow 0} \int_{\mathbb{R}} G(x-y,t) \varphi(y) \; dy = \varphi(x),$$
you're arguing that essentially, though we haven't used this language in class yet, that in the limit as $t \rightarrow 0$, $G(x-y,t)$ must go to $\delta(x-y)$? However, I do not know if this actually implies that without the limit, the integral still evaluates to $1$. I'm also curious if the integral going to $\varphi(x)$ in the limit truly implies that $G(x-y,t\rightarrow 0)$ is the delta function. I know that if you evaluate the integral directly you can show your claim to be true, but it would be really nice if there was an easier way to show it.
My understanding was that the kernel did not necessarily have area one and so I was specifying conditions to make it one. Turns out by definition of it being a kernel the area is one.
Looking back through one of Mark's mathematica notebooks (Heat Kernel) it is clearly written out that $\int_{-\infty }^{\infty }K_{\alpha }(x) \, dx=1$.I edited my initial post to reflect this.