# Steepest descent for Linearized KdV equation

I take steepest descent on Linearzied KdV equation, $$u_t+u_{xxx}=0$$ And by Fourier transform I know the phase is $$i(k^3+k\frac{x}{t})$$ I want to know asymptotic of the exponential integral $$\int^{\infty}_{-\infty}f(k)e^{it(k^3+k\frac{x}{t})}dk$$ when $$t\rightarrow\infty$$, so I fix $$\frac{x}{t}$$.

Now I know the path of steepest descent is such that imaginary part of phase is constant and pass through saddle points locally with the steepest angle. I get two saddle points (I consider $$\frac{x}{t}<0$$) $$-\lvert\frac{x}{3t}\rvert^{\frac{1}{2}}$$ and $$\lvert\frac{x}{3t}\rvert^{\frac{1}{2}}$$ and locally steepest angle is $$\frac{3\pi}{4}$$ and $$\frac{\pi}{4}$$ respectively.

According to the conditions of steepest descent path, I know the path is hyperbola, and I want to match saddle points, so I get the following graph

There are two questions I want to know:

1.In steepest descent we want the new contour $$\mathcal{C}$$ s.t. $$e^{t\phi}$$ vanishes ($$\phi$$ is the phase), so in my view, the graph would approach $$i$$ since increaseing veolocity on y-axis faster than x-axis. Then by asymptotic I get $$\phi\thicksim ik^3=i|k|^3e^{3i\theta}\rightarrow i|k|^3(i)=-|k|^3$$ and $$e^{it\phi}$$ vanishes, I'm not sure it's okay or not.

2.In steepest descent, I just know how to deform the new contour $$\mathcal{C}$$ and $$\int^{\infty}_{-\infty}f(k)e^{t\phi}dk=\int_{\mathcal{C}}f(k)e^{t\phi}dk ,$$ but I don't know how to prove it or give a explanation, books just give "since Cauchy theorem and the integrand is analytic, we can deform another contour and not affect the integral."

Cauchy theorem is from the same endpoints, the integrals of holomorphic functions are path-independent, and I have no idea how to match it with real line and the contour I deform. If possible, plz give me some details.

• If $g(k) = \operatorname {Im} (i (k^3 + a k))$, then $g(k) = \text {const}$ is a cubic curve, we would have to take a branch of $g(k) = g(k_L)$ going to $-i \infty$ and a branch of $g(k) = g(k_R)$ going to $-i \infty$. But this is not necessary, it is sufficient to consider a small neighborhood of $k_L$ and $K_R$, where we can apply Laplace's method, and show that the contribution from the rest of the contour is negligible. To get an idea of why we can move the contour off the real axis, consider $\left| \int_A^{A + i} e^{i t (k^3 + a k)} dk \right|$ for $A \to \infty$. Commented Jun 30, 2022 at 15:06