Showing that the function is decreasing

Question

$y$ is a solution to the diff-equation: $\frac{dy}{dt} = \frac{d}{dx}\frac{dy}{dx}$

for $0\le x\le\pi$ and $t\ge 0$.

It also fulfills $y(0,t)=y(\pi,t)=0$ for $t\ge 0$.

I want to show that $\int_{0}^{\pi} (y(x,t))^{2} dx\,\,\,is\, a\, decreasing\, function\,of\,\,t. $

My attempt:

I assume that our function can be written as $y(x,t)=h(x)u(t)$. Then it satisfies the diff equation $u'(t)h(x)=u(t)h''(x)$.

So $\int_{0}^{\pi}(y(x,t))^{2} dx=(u(t))^{2}\int_{0}^{\pi}(h(x))^2 dx$

From the diff equation I get that $h(x)=\frac{u(t)}{u'(t)}h''(x)$.¨

So now we have $(u(t))^{2}\int_{0}^{\pi} (\frac{u(t)}{u'(t)})^{2} (h''(x))^{2} dx$.

Using integration by parts, integrating $(h''(x))^{2}$ yields me $-(\frac{(u(t))^{2}}{u'(t)})^{2}\int_{0}^{\pi} h''(x)h'(x) dx$.

And I am stuck here ... anyone has any idea if I can continue from here , or if this proves that it is decreasing?

Answer 1

The assumption that $y(x,t)=u(x)\,v(t)$ is not correct. This is an antzat to look for particular solutions. Multiply the equation by $y$: $$ y\,\frac{\partial y}{\partial t}=\frac12\,\frac{\partial}{\partial t}\,(y^2)=y\,\frac{\partial^2 y}{\partial x^2}. $$ Integrate with respect to $x$ on $[0,\pi]$, and using integration by parts we get $$\begin{align} \frac12\,\frac{\partial}{\partial t}\int_0^\pi y(x,t)^2\,dx&=\int_0^\pi y(x,t)\,\frac{\partial^2 y}{\partial x^2}(x,t)\,dx\\ &=y(x,t)\,\frac{\partial y}{\partial x}(x,t)\Bigr|_{x=0}^{x=\pi}-\int_0^\pi \Bigl(\frac{\partial y}{\partial x}(x,t)\Bigr)^2\,dx\\ &\le0. \end{align}$$