Textbook 1.7 Problem #5
I'm trying to solve Problem #5 on page 65 in the textbook, which @mark said would be the basis for our Writing Assignment 1. By following the hint suggested in the problem and then applying Green's identity, along with the Dirichlet conditions, I've derived the equation
$$-\int_{\Omega} \nabla u \cdot \nabla u\,dV = \lambda\int_{\Omega} u^2\,dV.$$
Presumably the final step would be to explain that since both integrals are nonnegative, $\lambda$ must be negative for the equation to be true for a nontrivial solution. But I'm a little confused as to why the integrals must be nonnegative. Of course the integrands cannot be negative, as they are square functions, but I'm not sure why the integrals cannot be negative either. Does anyone have any thoughts?
Comments
I'm pretty sure a positive integrand can never give a negative integral...that would be geometrically nonsensical.
@dualrey It depends on the domain, I would think. The integral of $x^2$, which is always positive, is $\frac{x^3}{3}$, which is negative for negative values of $x$.
@AbS
It depends on the domain, I would think. The integral of $x^2$, which is always positive, is $x^3/3$, which is negative for negative values of $x$.
That's an indefinite integral, though. This question deals with a definite integral, which (I guess) is why @dualrey wrote "geometrically nonsensical".
@mark and @Abs That's correct. For example, if we were to integrate $x^2$ from $-1$ to $0$,
$$\int_{-1}^{0}x^2dx=\left.\frac{1}{3}x^3\right|_{-1}^{0}=\frac{1}{3}(0)^3-\frac{1}{3}(-1)^3=-\left(-\frac{1}{3}\right)=\frac{1}{3}$$
or from $-2$ to $-1$:
$$\int_{-2}^{-1}x^2dx=\left.\frac{1}{3}x^3\right|_{-2}^{-1}=\frac{1}{3}(-1)^3-\frac{1}{3}(-2)^3=\frac{1}{3}\left(-1-(-8)\right)=\frac{7}{3}.$$
These are positive, as we would expect for an area above the $x$-axis. In any case, I think I have solved this conundrum. Let's start with your original equation,
$$-\int_{\Omega}\nabla u\cdot\nabla u dV=\lambda\int_{\Omega}u^2dV.$$
Since these are intregrated over the same domain, we'll add one to another:
$$\int_{\Omega}\left[\lambda u^2+\nabla u\cdot\nabla u\right] dV=0.$$
Now if the integrand is zero for an arbitrary domain, that can only mean it is zero too. So,
$$\lambda u^2+\nabla u\cdot\nabla u=0$$
$$\Rightarrow \lambda=\frac{-\nabla u\cdot\nabla u}{u^2}.$$
Since we've already established that both $\nabla u\cdot\nabla u$ and $u^2$ are positive, we must conclude that $\lambda$ is negative.
Thanks for the great explanation, @dualrey !
@dualrey wrote:
$$\lambda=\frac{-\nabla u\cdot\nabla u}{u^2}.$$
That can't be correct as the right side is a non-constant function while the left side is constant.
@mark I don't know what you mean. I thought I had proved that the right side $\textit{is}$ a constant, unless there is some external reason that $\nabla u\cdot\nabla u$ and $u^2$ cannot be proportional, in which case we would have a contradiction.
@dualrey
I thought I had proved that the right side is a constant
Hm... I guess we could always check an example. The easiest one I can think of would be the eigenfunction $u(x)=\sin(x)$ over $[0,\pi]$. Keeping in mind that, in one dimension, the Laplacian is just the second derivative and the gradient is just the first derivative.
@mark Ah, sweet example. I now see where my "proof" went wrong. I was rather cavalier about dividing by $u^2$, as if it would never be zero, but our boundary conditions say it must be zero somewhere! Any function where my ratio doesn't blow up can't be a solution to the problem, hence the contradiction. We should be good though, having established that the integrals are positive anyway, right?
@dualrey I think so. If we know that definite integrals of nonnegative integrands are nonnegative, than it follows that for the original equation to be true, $\lambda$ must be negative for both sides to have the same sign.