### Author Topic: Take a correct branch  (Read 1261 times)

#### Jingxuan Zhang

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##### Take a correct branch
« on: February 27, 2018, 10:12:26 AM »
In chap 5.2 repeatedly I find "we need to take a correct branch". For instance, for the Schrodinger equation with $t\gtrless 0$, what in fact is the reason of taking $\sqrt{i}=\pm e^{i\pi/4}$?

#### Victor Ivrii

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##### Re: Take a correct branch
« Reply #1 on: February 27, 2018, 04:09:03 PM »
If we solve a crossbreed of Schr\"odinger and Heat

u_t =au_{xx}
\tag{1}

with $a\in \mathbb{C}$ with the positive  real part (which is well-posed in the direction of positive $t$) we get a formula with kernel $\frac{1}{\sqrt {4\pi a t}} e^{-x^2/4at}$. This square root is positive for real $a>0$.

Let us look, what happens if $a=bi+\varepsilon$, $\varepsilon \to +0$ and $b>0$. We get
$\frac{1}{\sqrt {4\pi (bi+\varepsilon) t}} e^{-x^2/4(bi+\varepsilon)t}$.  But if $\varepsilon \to +0$ then this equation (1) becomes Schr\"odinger equation

u_t =ibu_{xx}
\tag{2}

exponent factor tends to $e^{-x^2/4b i t}=e^{ix^2/4b t}$, but since $bi +\varepsilon \to bi$ from the right, then its argument tends to $\pi/2$ from below, and argument of the square root tends to ${\pi/4}$, so we get

\frac{1}{\sqrt {4\pi bt}} e^{-i\pi/4}e^{ix^2/4b t}
\tag{3}

Recall that $b>0$, $t>0$.

The same is true for $b<0, t<0$. Indeed, changing $t\mapsto -t$, $b\mapsto -b$ preserves equation (2).

So (3) holds for $bt>0$. But taking complex conjugation and $b\to -b$ also preserves (2). This operation with (3) brings

\frac{1}{\sqrt {4\pi |bt|}} e^{i\pi/4}e^{ix^2/4b t}.
\tag{4}

So, correct formula (I suspect I skipped on the lecture) for convolution kernel is

\frac{1}{\sqrt {4\pi |bt|}} e^{\mp i\pi/4}e^{ix^2/4b t}\qquad \text{for} \ \ bt \gtrless 0.
\tag{5}

#### Jingxuan Zhang

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##### Re: Take a correct branch
« Reply #2 on: February 27, 2018, 08:14:49 PM »
Indeed, changing $t\mapsto -t$, $b\mapsto -b$ preserves equation (2).

So (3) holds for $bt>0$. But taking complex conjugation and $b\mapsto -b$ also preserves (2). This operation with (3) brings

\frac{1}{\sqrt {4\pi |bt|}} e^{i\pi/4}e^{ix^2/4b t}.
\tag{4}

Is this because (heuristically) since $u$ is real-valued
$$u_t=ibu_{xx}=\overline{i(-b)u_{xx}} \implies u = \frac{1}{\sqrt {4\pi bt}} e^{i (-\pi/4+x^2/4b t)} = \overline{\frac{1}{\sqrt {4\pi (-b)t}} e^{i (-\pi/4+x^2/4(-b) t)}}=\frac{1}{\sqrt {4\pi (-b)t}} e^{i (\pi/4+x^2/4bt)}?$$
I hesitated a lot before posting. This $-b$ under square root is really poignant. How would you then justify your "taking complex conjugation and $b\mapsto -b$ also preserves (2)"?