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On PDE Evans, 2nd edition, page 46, it says the following:

LEMMA (Integral of fundamental solution). For each time $t > 0$, \begin{align} \int_{\mathbb{R}^n} \Phi(x,t) \, dx = 1 \end{align}

Note that \begin{align} \Phi(x,t) := \begin{cases} \frac{1}{(4\pi t)^\frac{n}{2}} e^{-\frac{|x|^2}{4t}} \, \, (x \in \mathbb{R}^n, t > 0) \\ 0 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (x \in \mathbb{R}^n, t > 0) \end{cases} \end{align}

The following is Evan's proof, as printed exactly in the textbook:

Proof. We calculate \begin{align} \int_{\mathbb{R}^n} \Phi(x,t) \, dx &= \frac{1}{(4\pi t)^\frac{n}{2}} \int_{\mathbb{R}^n} e^{-\frac{|x|^2}{4t}} \, dx \\ &= \frac{1}{\pi^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-|z|^2}\, dx \\ &= \frac{1}{\pi^{\frac{n}{2}}} \prod_{i=1}^{n} \int_{-\infty}^{\infty} e^{-z_i^2} \, dz_i \\ &= 1 \end{align}

What I do not understand from the above proof are from step 1 to step 2, where u-substitution is employed. When I tried this (as in, let $z = \frac{x}{(4t)^{\frac{1}{2}}}$), I got: \begin{align} |z|^2 = \frac{|x|^2}{4t} \text{ and } dz = \frac{1}{(4t)^{\frac{1}{2}}} dx \end{align}

which means

\begin{align} \int_{\mathbb{R}^n} \Phi(x,t) \, dx &= \frac{1}{(4\pi t)^\frac{n}{2}} \int_{\mathbb{R}^n} e^{-\frac{|x|^2}{4t}} \, dx \\ &= \frac{1}{(\pi)^{\frac{n}{2}}(4t)^\frac{n-1}{2}} \int_{\mathbb{R}^n} e^{-\frac{|x|^2}{4t}} \frac{1}{(4t)^\frac{1}{2}} \, dx \\ &= \frac{1}{(\pi)^{\frac{n}{2}}(4t)^\frac{n-1}{2}} \int_{\mathbb{R}^n} e^{-|z|^2} \, dz \\ &\not= \frac{1}{(\pi)^{\frac{n}{2}}} \int_{\mathbb{R}^n} e^{-|z|^2} \, dz \\ &= 1 \end{align}

I do not understand why my proof does not match the proof printed in the Evans textbook.

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    $\begingroup$ You're dividing each of the $n$ coordinates by $\sqrt{4t}$, so the determinant of the transformation is $\dfrac{1}{(\sqrt{4t})^n}$. $\endgroup$ – Daniel Fischer Jan 18 '14 at 0:14

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