What is the Inequality for the Heat Equation?

[mmmboh]

So I multiplied the heat equation by 2u, and put the substitution into the heat equation, and get 2uu_t-2uu_xx=(u²)_t=2(uu_x)_x+2(u_x)².

I`m not sure where to go from there, I can integrate with respect to t, then I would have a u<sup>2</sup> under the integral on the left side, but them I`m not sure where to go.

I also tried using the fact that the solution to the heat equation is

And I said that the integral was less than
\int_{-infinity}^{infinity} \ f(y)dy

and then I squared both sides, and said that the integral squared was less than the square of the integral. Then I integrated both sides with respect to x, but the problem is now, I have a double integral on the right side :S.

 

[hunt_mat]

I think the idea is to derive a differential equation for the integral of u^2 w.r.t x from 0 to 1.

 

[mmmboh]

How do I do that?

 

[hunt_mat]

To start with, use the hint and integrate over x from 0 to 1 and what do you get?

 

[hunt_mat]

I think the idea is to show that:

<br /> \int_{0}^{1}u^{2}(t,x)dx<br />

is a decreasing function and so it will always be less than or equal to it's initial value, and so you get the inequality you are asked to prove.

 

[mmmboh]

<br /> \int_{0}^{1} \ (u^{2})_{t}dx = 2uu_{x}(1,t)-2uu_{x}(0,t)-2<br /> \int_{0}^{1} \ (u_{x})^{2}dx<br />

This is what I get.

 

[hunt_mat]

Riiiight, I got the following equation:

<br /> \frac{\partial}{\partial t}u^{2}=2\frac{\partial}{\partial x}\left( u\frac{\partial u}{\partial x}\right) -2\left(\frac{\partial u}{\partial x}\right)^{2}<br />

Integrate this w.r.t. x from 0 to 1 and what do you get?

 

[mmmboh]

I fixed my equation, any way that`s what I get, but I don`t see what to do with it.

 

[hunt_mat]

Do as I said and integrate to obtain:

<br /> \frac{\partial}{\partial t}\int_{0}^{1}u^{2}dx=2\left[ u(t,x)\frac{\partial u}{\partial x}(t,x)\right]_{0}^{1}-2\int_{0}^{1}\left(\frac{\partial u}{\partial x}\right)^{2}dx<br />

Now the stuff in the square brackets vanishes (why?). So what can't we say about:

<br /> \int_{0}^{1}u^{2}dx<br />

 

[mmmboh]

? My equation is the same as yours, I just used subscript notation. If the stuff in the square brackets vanishes, then as you change t, <br /> <br /> \int_{0}^{1}u^{2}dx<br /> <br /> decreases, so it will be smaller than <br /> <br /> \int_{0}^{1}f(x)^{2}dx<br /> <br />

since this is when t=0. Is that right?

Edit: The term in the square brackets vanishes because u(1,t)=u(0,t)=0.

So what would the uniqueness theorem be then? It`s always decreasing with respect to t, and u(x,t)=f(x) only once? I`m not sure.

 

[hunt_mat]

We're told that in the question, u(t,0)=u(t,1)=0.

Finally, you have the solution, but formally, i would say that:

<br /> \frac{\partial}{\partial t}\int_{0}^{1}u^{2}dx=-2\int_{0}^{1}\left(\frac{\partial u}{\partial x}\right)^{2}dx\leqslant 0<br />

as the quantity in the integrand in postive, so the integral will be positive which shows that the quantity we spoke about is clearly decreasing.

 

[mmmboh]

Thank you. You have been a great help. What do you think of my uniqueness theorem though? It does not seem like much of a theorem.

Edit: I guess it means that since the integral is decreasing, the solution to the heat equation with a boundary condition is unique...

 

[mmmboh]

I edited my post.

 

[hunt_mat]

The usual way for uniqueness theorems is to assume that there are two solutions u_1 and u_2 and examine the quantity w=u_1-u_2 and show that the only possible case is w=0.