Problem text: Using Fourier transforms solve the linear Schrödinger equation $$i \frac{\partial u}{\partial t} + \frac{\partial^2 u }{\partial x^2} = 0,~~~u(x,0) = f(x)$$ Hint: Obtain the Green's function in closed form and use superposition. Recall $$\int_{-\infty}^\infty e^{iu^2}{\rm d}u = \sqrt{\pi}\,e^{i\pi/4}$$

I know how to solve this problem by a simple manipulation of the Fourier transform. The question is how can I use the given hint to solve the problem. I'm not aware of such a setting involving Green's function. Any help is much appreciated.

  • 1
    $\begingroup$ The Greens function here is the Heat kernel for $t \to i t$. The Greens function in this context refers to $F(x,y,t)$ such that $u(x,t) = \int F(x,y,t) f(y) {\rm d}y$ with $F(0,y,t) = \delta(x-y)$. $\endgroup$
    – Winther
    Dec 28, 2016 at 0:18

2 Answers 2


To solve it using the Green's function approach we first note that if we knew a function $G(x,y,t)$ satisfying

$$iG_t + G_{xx} = 0,~~~G(x,y,0) = \delta(x-y)$$

then we can write the solution for $u$ as

$$u(x,t) = \int_{-\infty}^\infty G(x,y,t)f(y){\rm d}y$$

It's simple to check that this form satisfy both the PDE and the initial condition. Taking the Fourier transform of the PDE for $G$ with respect to $x$ using the convention $\hat{f}(k) \equiv \int_{-\infty}^\infty f(x) e^{ikx}{\rm d}x$ we get

$$i\hat{G}_t - k^2\hat{G} = 0,~~~\hat{G}(k,y,0) = \int_{-\infty}^\infty \delta(x-y) e^{ikx}{\rm d}x = e^{iky}$$

Solve the ODE above and then take the inverse Fourier transform $G(x,y,t) = \frac{1}{2\pi}\int_{-\infty}^\infty \hat{G}(k,y,t)e^{-ikx}{\rm d}k$ using the hint. To get it on a convenient form try completing the square in the exponential $Ak^2 + Bk + C = A\left(k+\frac{B}{2A}\right)^2 - \frac{B^2}{4A}$ and shift the integration variable $u = k + \frac{B}{2A}$.

Taking $\tau = i t$ in the PDE gives us the Heat equation $u_\tau = u_{xx}$ so the solution you should find for $G(x,y,t)$ should be the Heat kernel with the replacement $t\to i t$.


Following the comment by @winther, the Green (or Green's) function $G(x,x';t)$ satisfies

$$G_{xx}(x,x';t)+iG_t(x,x';t)=0 \tag 1$$

with $G(x,x';0)=\delta(x-x')$.

Taking the Fourier transform, $g(k,x';t)=\int_{-\infty}^{\infty}G(x,x';t)e^{ikx}\,dx$, of $(1)$ reveals

$g_t(k,x';t)+ik^2g(k,x',t)=0$ and $g(k,x';0)=e^{ikx'}$.

Hence, we see that $g(k,x';t)=e^{i(kx'-k^2t)}$ and

$$\begin{align} G(x,x';t)&=\frac{1}{2\pi}\int_{-\infty}^\infty e^{i(kx'-k^2t)}e^{-ikx}\,dk\\\\ &=\frac1{2\pi}\int_{-\infty}^\infty e^{-i(x-x'+k^2t)}\,dk\\\\ &=e^{i(x-x')^2/(4t)}\frac{1}{2\pi}\int_{-\infty}^\infty e^{-it(k+(x-x')/(2t))^2}\,dk\\\\ &=e^{i(x-x')^2/(4t)}\frac{1}{2\pi\sqrt t}\int_{-\infty}^\infty e^{-ik^2}\,dk\\\\ &=e^{i(x-x')^2/(4t)}\frac{1}{2\pi\sqrt t}\,\sqrt \pi e^{-i\pi/4}\\\\ &=\frac{e^{i(x-x')^2/(4t)}}{\sqrt{4\pi it}}\tag 2 \end{align}$$

Using $(2)$, we find the solution to the problem of interest is given by

$$u(x,t)=\int_{-\infty}^\infty f(x') \frac{e^{i(x-x')^2/(4t)}}{\sqrt{4\pi it}}\,dx'$$


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