Show that the function is the solution

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In summary: This equality holds because both integrals are equal, just written in different forms. The left side is an integral over the entire domain, while the right side is an integral over the boundary of the domain multiplied by a constant. This is a common technique in integration to simplify the calculation.
  • #1
evinda
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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)
 

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  • #2
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 $\int f\,d\mu$.
 
  • #3
I like Serena said:
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?
 
  • #4
evinda said:
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)
 
  • #5
I like Serena said:
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:

$$\pi^{-\frac{n}{2}} \left| \int_{\mathbb{R}^n \setminus{Q_R}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] d{\sigma} \right|
= \pi^{-\frac{n}{2}} \left| \int_{r=R}^\infty\int_{\sigma\in\partial{Q_r}} e^{-|\sigma|^2 } [\phi(x+2 \sqrt{t} \sigma)-\phi(x)] \,ds\,dr\right|$$
 
  • #6
evinda said:
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)​
 

FAQ: Show that the function is the solution

What does it mean to "show that the function is the solution"?

Showing that a function is the solution means proving that it satisfies a given equation or set of equations. This involves demonstrating that when the function is plugged into the equation, it results in a true statement.

How do you show that a function is the solution?

To show that a function is the solution, you can use a variety of methods such as substitution, algebraic manipulation, or calculus techniques. The specific method will depend on the equation and the function being evaluated.

Why is it important to show that a function is the solution?

Showing that a function is the solution is important because it provides evidence that the function is a valid solution to the given equation. This is crucial in mathematics and science, where accuracy and proof are essential.

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Any equation that involves one or more variables can potentially be solved by showing that a function is the solution. These can include linear equations, quadratic equations, polynomial equations, and more.

Are there any limitations to showing that a function is the solution?

While showing that a function is the solution is a powerful method for solving equations, it may not always be possible or practical. In some cases, it may be more efficient to use other techniques such as graphing or numerical methods to find a solution.

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