# Limit of Integral: Let u, A(x) be Functions

• MHB
• mathmari
In summary, the conversation discusses a partial differential equation and integrals with functions $u(x,t)$ and $A(x)$. The mean value theorem is suggested as a possible approach to solving the problem, and it is applied to the integrals to show that the limit as $\epsilon \rightarrow 0$ is equal to $A(u(a,t))$. The conversation also explores the use of integration by parts and choosing a specific function $\rho (t)$ in the proof, ultimately leading to the conclusion that $\rho (t)=0$ as $t\to \infty$.
mathmari
Gold Member
MHB
Hey!

Let $u(x,t), A(x)$ be functions, for which holds the following:

We have the pde $u_t+a(u)u_x=0$. Let $A'(u)=a(u)$ then the pde can be written as $u_t+A(u)_x=0$. We have the following integrals $$\int_{a-\epsilon}^au\cdot \left (\frac{x-a}{\epsilon}+1\right )\, dx+\int_a^budx+\int_b^{b+\epsilon}u\cdot \left (1-\frac{x-b}{\epsilon}\right )\, dx$$ and $$\frac{1}{\epsilon}\int_{a-\epsilon}^aA(u)\, dx-\frac{1}{\epsilon}\int_b^{b+\epsilon}A(u) \, dx$$ What do we get if we take the limit $\epsilon\rightarrow 0$ ? From the boundary th elimit goes to $0$ but the integrands go to $\infty$ or not? Or is the limit maybe something like the definition of the derivative? (Wondering)

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mathmari said:
What do we get if we take the limit $\epsilon\rightarrow 0$ ? From the boundary th elimit goes to $0$ but the integrands go to $\infty$ or not? Or is the limit maybe something like the definition of the derivative? (Wondering)

The hint is in the title of your previous thread: You may try to apply the mean value theorem (in integral form).

Krylov said:
The hint is in the title of your previous thread: You may try to apply the mean value theorem (in integral form).

Do you mean to write each integral as follows $$\int_a^bf(t)\, dt=f(c)(b-a)$$ ? (Wondering)

mathmari said:
Do you mean to write each integral as follows $$\int_a^bf(t)\, dt=f(c)(b-a)$$ ? (Wondering)

Yes, so if we let $t$ be fixed throughout, then follows for arbitrary $\epsilon > 0$ (with $a$ for the upper limit in the mean value integral and $a - \epsilon$ for the lower limit) that
$\frac{1}{\epsilon}\int_{a-\epsilon}^aA(u(x,t))\, dx = \frac{1}{\epsilon} A(u(\hat{x},t)),$
for some $\hat{x} \in [a-\epsilon,a]$. Now, if $u$ and $A$ are continuous, then $x \mapsto A(u(x,t))$ is continuous for fixed $t$, so by letting $\epsilon \downarrow 0$ we find that the right-hand side tends to $A(u(a,t))$.

For the other integrals, you may need to do a bit of manipulation (such as: make a simple change-of-variables by translation) to get them in the proper form for the mean value theorem, but the idea is the same.

Krylov said:
Yes, so if we let $t$ be fixed throughout, then follows for arbitrary $\epsilon > 0$ (with $a$ for the upper limit in the mean value integral and $a - \epsilon$ for the lower limit) that
$\frac{1}{\epsilon}\int_{a-\epsilon}^aA(u(x,t))\, dx = \frac{1}{\epsilon} A(u(\hat{x},t)),$
for some $\hat{x} \in [a-\epsilon,a]$. Now, if $u$ and $A$ are continuous, then $x \mapsto A(u(x,t))$ is continuous for fixed $t$, so by letting $\epsilon \downarrow 0$ we find that the right-hand side tends to $A(u(a,t))$.

Ah I understand! (Smile)

Krylov said:
For the other integrals, you may need to do a bit of manipulation (such as: make a simple change-of-variables by translation) to get them in the proper form for the mean value theorem, but the idea is the same.

For the other ones, for example the first one, can we not do the following?

We have that \begin{align*}\int_{a-\epsilon}^au\cdot \left (\frac{x-a}{\epsilon}+1\right )\, dx&=u(x_1, t)\cdot \left (\frac{x_1-a}{\epsilon}+1\right )\cdot (a-(a-\epsilon))=u(x_1, t)\cdot \left (\frac{x_1-a}{\epsilon}+1\right )\cdot \epsilon\\ & =u(x_1, t)\cdot \left (x_1-a+\epsilon\right )\end{align*} for some $x_1\in [a-\epsilon, a]$.
Then as $\epsilon\to 0$ we have that $x_1\to a$. So the integral goes to $0$ as $\epsilon\to 0$.

(Wondering)
I am trying to understand a proof where the above is used. Then there is the following:
$$\int_0^{\infty}\rho'(t)\left [\int_a^bu\, dx\right ]\, dt+\int_0^{\infty}\rho (t)\left [A(u(a,t))-A(u(b,t))\right ]\, dt=0$$ Since $\rho (t)$ is arbitrary we have $$-\frac{d}{dt}\left [\int_a^bu\, dx\right ]+A(u(a,t))-A(u(b,t))=0$$

How do get the last relation knowing that $\rho (t)$ is arbitrary? Do we take a specific formula for that function or what do we have to do in this step? (Wondering)

Using integration by parts at the first integral we get the following:

$$\int_0^{\infty}\rho '(t)\left (\int_a^bu\, dx\right )\, dt=\left [\rho (t)\left (\int_a^bu\, dx\right )\right ]_{t=0}^{\infty}-\int_0^{\infty}\rho (t)\frac{d}{dt}\left (\int_a^bu\, dx\right )\, dt$$

So, we get the following:

$$\left [\rho (t)\left (\int_a^bu\, dx\right )\right ]_{t=0}^{\infty}-\int_0^{\infty}\rho (t)\frac{d}{dt}\left (\int_a^bu\, dx\right )\, dt+\int_0^{\infty}\rho (t)\left [A(u(a,t))-A(u(b,t))\right ]\, dt=0\\ \Rightarrow \left [\rho (t)\left (\int_a^bu\, dx\right )\right ]_{t=0}^{\infty}+\int_0^{\infty}\rho (t)\left [-\frac{d}{dt}\left (\int_a^bu\, dx\right )+A(u(a,t))-A(u(b,t))\right ]\, dt=0$$

Right?

Do we maybe have to choose a specific function $\rho (t)$ such that $\left [\rho (t)\left (\int_a^bu\, dx\right )\right ]_{t=0}^{\infty}=0$ ?

(Wondering) To say something about the function $\rho (t)$ : At the beginning of the proof we had the function $\psi (x,t)$ which is any test function defined on the half-plane. Then we chose $\psi (x,t)=\phi (x)\cdot \rho (t)$.

It is given for a test function $\psi (x,t)$ that it is a $C^{\infty}$ function in the $xt$ plane that is zero outside a bounded set.

This means that $\rho (t)=0$ as $t\to \infty$, right? But what about at $t=0$, i.e. how can we conclude that $\rho (0)\cdot \int_a^bu(x,0)\, dx=0$ ? (Wondering)

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## What is a limit of integral?

A limit of integral refers to the value that a definite integral approaches as its limits of integration approach a certain value. It can also refer to the maximum or minimum value of a definite integral over a given interval.

## How is a limit of integral calculated?

A limit of integral is typically calculated using the fundamental theorem of calculus, which states that the limit of a definite integral can be evaluated by finding the antiderivative of the integrand and evaluating it at the limits of integration.

## What is the relationship between u and A(x) in a limit of integral?

The function u is typically used as a substitution variable in a limit of integral, while A(x) represents the integrand itself. Together, they are used to calculate the limit of the integral.

## What are some common applications of limits of integral?

Limits of integral are commonly used in mathematical analysis, physics, and engineering to calculate areas, volumes, and other quantities in real-world problems. They are also used in optimization problems to find the maximum or minimum value of a function.

## What are some key properties of limits of integral?

Some important properties of limits of integral include linearity, the mean value theorem for integrals, and the substitution rule. These properties allow for easier calculation and manipulation of limits of integral in various contexts.

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