- #1

evinda

Gold Member

MHB

- 3,836

- 0

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)

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)