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

[mathmari]

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)

 

[S.G. Janssens]

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).

 

[mathmari]

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)

 

[S.G. Janssens]

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.

 

[mathmari]

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)

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)

 

[mathmari]

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)