Expansion of Integrals. 5


Method of the steepest descent

We consider $$I(k)= \int_L e^{k \phi(z)}f(z)\,dz \label{eq-2.5.1}$$ where now $L\subset \Omega$ is a contour from $z_0$ to $z_1$, $\Omega \subset \bC$ is a simple-connected domain and $\phi$, $f$ are holomorphic in $\Omega$. Recall from complex variables that in this framework $I(k)$ does not depend on the choice of $L$.

We are interested in the asymptotics of $I(k)$ as $k\to +\infty$.

Selecting a contour

Obviously we need to select a contour $L$ in such way that $\max_{z\in L} \Re\phi(z)$ was a small as possible. Consider a landscape of $\Re \phi(z)$. From Complex Variables we know that both $\Re \phi(z)$ and $\Im \phi(z)$ are harmonic functions. We also know that harmonic functions do not have local maxima or minima in the inner points, only saddle points.

Lemma 1.

1. $|\nabla \Re\phi |=|\nabla \Im \phi|=|\phi'|$;
2. $z_0$ is a stationary point of $\Re \phi$ if and only if $\phi'(z_0)=0$.
3. $\partial^\alpha \Re\phi (z_0)=0$ for all $\alpha:\,|\alpha|\le m-1$ with $m\ge 2$ if and only if $\phi' (z_0)=\ldots =\phi^{(m-1)}(z_0)=0$.

Proof. From Complex Variables.

If $\phi' (z_0)=\ldots =\phi^{(m-1)}(z_0)=0$ and $\phi^{(m)}(z_0)\ne 0$ we call $(m-1)$ multiplicity of the saddle. If $m=2$ then the saddle is simple.

Lemma 2. Consider lines along which $\Re \phi$ changes the fastest. Those are tangent to $\nabla \Re \phi$. Along these lines $\Im \phi=\const$.

Lemma 3. Let $\phi' (z_0)=\ldots =\phi^{(m-1)}(z_0)=0$ and $A:=\phi^{(m)}(z_0)\ne 0$. Then in the vicinity of $z_0$

1. $\{z:\,\Re \phi(z) =\Re \phi(z_0)\}$ consists of $2m$ "rays" (curved away from $z_0$) issued from $z_0$ in the directions $$e^{i(\arg A + \pi k +\pi/2)/m}\qquad k=0,1,\ldots, 2m-1 \label{eq-2.5.2}$$ dividing vicinity into $2m$ sectors in which $\Re \phi(z)\gtrless \Re\phi(z_0)$ alternatively. Here $\arg A$ is an argument of $A=|A|e^{i\arg A}$.
2. In each of $m$ sectors in which $\Re \phi(z)< \Re\phi(z_0)$ there is a single ray of the steepest descent issued from $z_0$ in the direction $$e^{i(\arg A + \pi k)/2m}\qquad k=1,3,\ldots, 2m-1 \label{eq-2.5.3}$$ and in each of $m$ sectors in which $\Re \phi(z)> \Re\phi(z_0)$ there is a single ray of the steepest ascent issued from $z_0$ in the direction $$e^{i(\arg A + \pi k +\pi/2)/m}\qquad k=0,2,\ldots, 2m-2. \label{eq-2.5.4}$$

Proof. Sufficient to consider a toy-model $\phi(z)=Az^m$ with $z_0=0$.

 $m=2$ $m=3$ Orange lines are $\{z:\,\Re \phi(z) =\Re \phi(z_0)\}$, blue lines are of the steepest descent and red lines of the steepest ascent

Lemma 4. One can select contour from $z_0$ to $z_1$ in such a way that it can be broken into several contours $L_1,\ldots ,L_K$ such that

1. In some contours $L_j$ equality $\Re\phi(z)=\max_{\zeta\in L} \Re\phi (\zeta)$ holds in the single point $z^*_j$ which is its end-point. Near this point $L_j$ is a line of the steepest descent.
2. In the remaining contours $\Re\phi(z)<\max_{\zeta\in L} \Re\phi (\zeta)$ everywhere.

Proof. This lemma looks intuitively obvious (but the rigorous proof is a bit tedious).

Calculations

Obviously we need to calculate only contributions of the contours of type (a). We consider just one contour $L$ of type (a) from $z^*$ to some point (does not matter).

Theorem 1. Let $L$ be a contour from $z^*$ to some point (does not matter) and $\Re\phi(z)<\Re\phi(z^*)$ in each point of this contour $z\ne z^*$. Let $\phi' (z^*)=\ldots =\phi^{(m-1)}(z^*)=0$ and $A:=\phi^{(m)}(z^*)\ne 0$, $m\ge 2$. Let $L$ be a contour of the steepest descent.

Then $$I(k)\sim e^{ik\phi(z^*)}\sum_{n\ge 0}\kappa_n k^{-(n+1)/m} \label{eq-2.5.5}$$ where $$\kappa_0= \Gamma ((m+1)/m) |f^{(m)}(z^*)/m!|^{-1/m}e^{i\theta}f(z^*) \label{eq-2.5.6}$$ where $e^{i\theta}$ is a direction of $L$ in $z^*$.

Remark 1. If $m=1$ (\ref{eq-2.5.5}) holds with $$\kappa_0= -(\phi'(z^*))^{-1}f(z^*). \label{eq-2.5.7}$$