MHB Show that the function is the solution

[evinda]

Hello! (Wave)

Theorem: Suppose that $\phi(x)$ is continuous and bounded in $\mathbb{R}^n$, then the solution of the problem

$$u_t=\Delta u \text{ in } (0,T) \times \mathbb{R}^n , \forall T>0 \\ u(0,x)=\phi(x), x \in \mathbb{R}^n$$

is given by the formula $u(t,x)=\int_{\mathbb{R}^n} \Gamma(t,x-\xi) \phi(\xi) d{\xi}$.

We have that $\Gamma(t-\tau,x- \xi)=\frac{1}{2^n [\pi(t-\tau)]^{\frac{n}{2}}} e^{-\frac{|x-\xi|^2}{4 (t-\tau)}}$.

We can see that $\Gamma_t-\Delta_x \Gamma=0$ and so we get that $u_t-\Delta u=0, \forall t>0$.

In order to show that $u(0,x)=\phi(x)$ we consider the difference $u(t,x)-\phi(x)$.

$u(t,x)-\phi(x)=\int_{\mathbb{R}^n} \Gamma(t,x-\xi) [\phi(\xi)-\phi(x)] d{\xi}$.

We set $\xi_i-x_i=2 \sqrt{t} \sigma_i, i=1, \dots, n$ and we get $u(t,x)-\phi(x)=\pi^{-\frac{n}{2}} \int_{\mathbb{R}^n} e^{-|\sigma|^2} [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] d{\sigma}$.

Let $Q_R=\{ |\sigma|<R\}$,

then $u(t,x)-\phi(x)=\pi^{-\frac{n}{2}} \int_{Q_R} e^{-|\sigma|^2} [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] d{\sigma}+\pi^{-\frac{n}{2}} \int_{\mathbb{R}^n \setminus{Q_R}} e^{-|\sigma|^2} [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] d{\sigma}(\star)$

From the definition of the generalized integral $\forall \epsilon>0$ there exists a $R>0$ such that $\pi^{-\frac{n}{2}} \left| \int_{\mathbb{R}^n \setminus{Q_R}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] d{\sigma} \right| \leq \frac{\epsilon}{2}$.

Since $\phi$ is continuous for a given $R>0$ there is a $\delta>0$ such that for $0<t<\delta$ it holds that

$\pi^{-\frac{n}{2}} \int_{Q_R} e^{-|\sigma|^2} |\phi(x+2 \sqrt{t}\sigma)-\phi(x)| d{\sigma} \leq \frac{\epsilon}{2}$

So from the $(\star)$ we will have that $\forall \epsilon>0$ there is a $\delta>0$ such that $|u(t,x)-\phi(x)| \leq \epsilon$ if $0<t< \delta$, i.e. $\lim_{t \to 0} u(t,x)=\phi(x)$.First of all, I have found that the definition of the generalized Riemann integral is the following:

View attachment 6357

In our case, at the integral where we use this definition , we have an open set, namely this one $\mathbb{R}^n \setminus{Q_R}$. Do we pick an arbitrary $[a,b] \in \mathbb{R}^n \setminus{Q_R}$ and consider a partition of it?

If so, then we have that for each $\epsilon>0$ there is a $R>0$ such that for every partition of $[a,b]$

$a=x_0 < x_1< x_2< \dots<x_n=b$

with tags $x_{k-1} \leq t_k \leq x_k, k=0,1, \dots,n$ we have $$\left| \int_{\mathbb{R}^n \setminus{Q_R}} e^{-|\sigma|^2} [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] d{\sigma}- \sum_{k=1}^n e^{-|\sigma|^2} [\phi(x+2 \sqrt{t} t_k)-\phi(x)] (x_k-x_{k-1}) \right|< \epsilon$$

whenever $x_k-x_{k-1}<R$.Do we pick now $x_k-x_{k-1} \to 0$ in order to get the inequality for the integral?

Also, in order to get the following inequality:$\pi^{-\frac{n}{2}} \int_{Q_R} e^{-|\sigma|^2} |\phi(x+2 \sqrt{t}\sigma)-\phi(x)| d{\sigma} \leq \frac{\epsilon}{2}$we use the continuity of $\phi$, so for each $\epsilon>0, \exists \delta'>0$ such that if $0< x+ 2\sqrt{t} \sigma-x=2 \sqrt{t} \sigma< \delta'$ we have that $|\phi(x+ 2 \sqrt{t} \sigma)-\phi(x)| \leq \epsilon$.

Then $\pi^{-\frac{n}{2}} \int_{Q_R} e^{-|\sigma|^2} |\phi(x+2 \sqrt{t} \sigma)-\phi(x)|d{\sigma} \leq \pi^{-\frac{n}{2}} \epsilon \int_{Q_R} e^{-|\sigma|^2} d{\sigma} \leq \pi^{-\frac{n}{2}} \epsilon \int_{\mathbb{R}^n} e^{-|\sigma|^2} d{\sigma}=\epsilon$.

Right? Do we set $\delta=\frac{\delta'^2}{4 \sigma^2}$ ?

Also does the first inequality hold for any $t$ and so also for $0<t< \delta$? (Thinking)

 

[I like Serena]

Hi evinda! (Smile)

I believe you've quoted the definition of a non-generalized Riemann integral.
However, it only applies to functions with a domain that is an interval in $\mathbb R$.

Shouldn't we look at a multiple integral, since we have a function with a domain in $\mathbb R^n$? (Wondering)

Or actually the Lesbesgue integral that applies to functions on any domain? (Wondering)
Note that your integral is of the form $\int f\,d\mu$.

 

[evinda]

Hi evinda! (Smile)

I believe you've quoted the definition of a non-generalized Riemann integral.
However, it only applies to functions with a domain that is an interval in $\mathbb R$.

Shouldn't we look at a multiple integral, since we have a function with a domain in $\mathbb R^n$? (Wondering)

Or actually the Lesbesgue integral that applies to functions on any domain? (Wondering)
Note that your integral is of the form $\int f\,d\mu$.

Ok... where can we find the $\epsilon, \delta$-proof of these integrals?

 

[I like Serena]

Let $Q_R=\{ |\sigma|<R\}$,

From the definition of the generalized integral $\forall \epsilon>0$ there exists a $R>0$ such that $\pi^{-\frac{n}{2}} \left| \int_{\mathbb{R}^n \setminus{Q_R}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] d{\sigma} \right| \leq \frac{\epsilon}{2}$.

How about we calculate it as:
$$\pi^{-\frac{n}{2}} \left| \int_{\mathbb{R}^n \setminus{Q_R}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] d{\sigma} \right|
= \pi^{-\frac{n}{2}} \left| \int_{r=R}^\infty\int_{\sigma\in\partial{Q_r}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] \,ds\,dr\right| \\
\le \pi^{-\frac{n}{2}} \left| \int_{R}^\infty e^{-r^2 }\cdot 2M \cdot w_nr^{n-1}\,dr\right|
$$
(Wondering)

Now it's a Riemann Integral. (Happy)

 

[evinda]

How about we calculate it as:
$$\pi^{-\frac{n}{2}} \left| \int_{\mathbb{R}^n \setminus{Q_R}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] d{\sigma} \right|
= \pi^{-\frac{n}{2}} \left| \int_{r=R}^\infty\int_{\sigma\in\partial{Q_r}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] \,ds\,dr\right| \\
\le \pi^{-\frac{n}{2}} \left| \int_{R}^\infty e^{-r^2 }\cdot 2M \cdot w_nr^{n-1}\,dr\right|
$$
(Wondering)

Now it's a Riemann Integral. (Happy)

Why does the equality you write hold? Namely this one:

$$\pi^{-\frac{n}{2}} \left| \int_{\mathbb{R}^n \setminus{Q_R}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] d{\sigma} \right|
= \pi^{-\frac{n}{2}} \left| \int_{r=R}^\infty\int_{\sigma\in\partial{Q_r}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] \,ds\,dr\right|$$

 

[I like Serena]

Why does the equality you write hold? Namely this one:

$$\pi^{-\frac{n}{2}} \left| \int_{\mathbb{R}^n \setminus{Q_R}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] d{\sigma} \right|
= \pi^{-\frac{n}{2}} \left| \int_{r=R}^\infty\int_{\sigma\in\partial{Q_r}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] \,ds\,dr\right|$$

Let's pick an example.
Suppose we pick the 1-sphere and integrate some $f(\sigma)$ from $R$ to $\infty$.

Then it looks like:
\begin{tikzpicture}[scale=0.7]
\draw[->] (0,0) -- (40:3) node

{R};
\draw[blue, ultra thick] circle (3);
\draw[gray, thin] circle (5) circle (6);
\path[blue, ultra thick] (50:5) -- node[below left] {$ds$} (70:5) -- node

{$dr$} (70:6) -- node[below] {$d\sigma$} (50:6) -- cycle;
\draw[blue, ultra thick] (50:5) arc (50:70:5) -- node

{$dr$} (70:6) arc (70:50:6) -- cycle;
\end{tikzpicture}
We have a "volume" element $d\sigma$ in a ball that is equal to a "surface" element $ds$ on the corresponding sphere times $dr$.
So:
$$\int_{\mathbb R^2 \setminus B_R} f(\sigma)\,d\sigma
= \int_R^\infty \int_0^{2\pi} f(\sigma)\,rd\phi\,dr
= \int_R^\infty \int_{\partial B_r} f(\sigma)\,ds\,dr
$$

Can we generalize it to $\mathbb R^n$ with an $(n-1)$-sphere? (Wondering)​