PDE with a condition

Question

Considering the heat equation, $$\frac{du}{dt}=\frac{d^2u}{dx^2}$$ if $$u(x,t)=t^{\alpha}\phi(\xi)$$ with $$\xi=x/\sqrt{t} \enspace then \enspace \phi \enspace satisfies \enspace \alpha\phi-(1/2)\xi\phi'=\phi"$$ show that $$\int_{-\infty}^{\infty} u(x,t) \, \mathrm{d}x =\int_{-\infty}^{\infty} t^{\alpha}\phi(\xi) \, \mathrm{d}x$$ is independent of t only if $$\alpha=-1/2$$ and show that if $$\alpha=-1/2, \enspace C-(1/2)\xi\phi=\phi'$$ with C an arbitrary constant.

Answer 1

Under certain reasonable regularity/integrability conditions (which I'm almost certain you're allowed to assume), we see $$\frac{d}{dt} \int_{\mathbb R} u(x,t) dx = \int_{\mathbb R} \frac{\partial u}{\partial t}(x,t) dx.$$ With the the given ansatz $u(x,t) = t^\alpha \phi(x/\sqrt t)$, we see \begin{align*} \frac{d}{dt} \int_\mathbb R u(x,t) dt &= \int_\mathbb R \alpha t^{\alpha-1}\phi\left( \tfrac{x}{\sqrt t}\right ) + t^\alpha \phi'\left( \tfrac{x}{\sqrt t}\right ) \cdot \tfrac{-x}{2t^{3/2}} dx \\ &= \int_\mathbb R \alpha t^{\alpha-1}\phi\left( \tfrac{x}{\sqrt t}\right ) - \tfrac 1 2 xt^{\alpha - 3/2} \phi'\left( \tfrac x {\sqrt t} \right) dx \\ &=\int_\mathbb R \alpha t^{\alpha-1}\phi\left( \tfrac{x}{\sqrt t}\right ) - \tfrac 1 2 xt^{\alpha - 1} \frac{d}{dx} \left( \phi\left( \tfrac x {\sqrt t} \right) \right)dx \end{align*} Now integrating by parts for the second term (and assuming the boundary term goes to zero, which another reasonable assumption), we see \begin{align*}\frac{d}{dt} \int_\mathbb R u(x,t) dt &= \int_\mathbb R \alpha t^{\alpha-1}\phi \left( \tfrac x {\sqrt t}\right) + \tfrac{1}2 t^{\alpha -1}\phi \left( \tfrac x {\sqrt t}\right) dx \\ &= \int_\mathbb R \left(\alpha + \tfrac 1 2 \right) t^{\alpha-1}\phi \left( \tfrac x {\sqrt t}\right) dx. \end{align*} Now $\int_\mathbb R u(x,t) dx$ is constant in $t$ iff this derivative is zero iff $\alpha = -1/2$ (again, this last line of reasoning holds if we know, for example, that $\phi$ is positive).

When $\alpha = -1/2$, the equation for $\phi$ becomes $$\phi'' + \tfrac 1 2 \xi \phi' +\tfrac 1 2 \phi = 0.$$ But $\xi \phi' + \phi = \frac d {d\xi} (\xi \phi) = (\xi \phi)'$ so we can re-write this as $$\phi'' + \tfrac 1 2 (\xi \phi)' = 0$$ whence integrating yields $$\phi' + \tfrac 1 2 \xi \phi = C$$ for some arbitrary constant $C$.