Todai Entrance Exam: Math 2019 – Problem 2

(1)

Let u = \sqrt{a}(x + id), so du = \sqrt{a}dx:

    \[ \int_{-\infty}^{\infty} \exp(-a(x + id)^{2})dx = \frac{1}{\sqrt{a}}\int_{-\infty}^{\infty} \exp(-u^{2})du = \sqrt{\frac{\pi}{a}} \]

(2)

(i)

From Eq. (2.1), we have:

    \[ \frac{\partial^2 U}{\partial t^2} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{\partial^2 u}{\partial t^2}\exp(-ikx)dx = \frac{c^{2}}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{\partial^2 u}{\partial x^2}\exp(-ikx)dx \]

Remember that:

    \[ uv = \int \frac{\mathrm{d} u}{\mathrm{d} x}vdx + \int \frac{\mathrm{d} v}{\mathrm{d} x}udx  \]

So:

    \[ \frac{\partial^2 U}{\partial t^2} = \frac{c^{2}}{\sqrt{2\pi}}\left ( \frac{\partial u}{\partial x}\exp(-ikx) + ik\int_{-\infty}^{\infty} \frac{\partial u}{\partial x}\exp(-ikx)dx \right ) \]

    \[ \frac{\partial^2 U}{\partial t^2} = \frac{c^{2}}{\sqrt{2\pi}}\left ( \frac{\partial u}{\partial x}\exp(-ikx) + ik \left ( u\exp(-ikx) + ik\int_{-\infty}^{\infty}u\exp(-ikx)dx \right ) \right ) \]

    \[ \frac{\partial^2 U}{\partial t^2} = \frac{c^{2}}{\sqrt{2\pi}}\left ( \frac{\partial u}{\partial x}\exp(-ikx) + ik \left ( u\exp(-ikx) + ik\sqrt{2\pi}U \right ) \right ) \]

    \[ \frac{\partial^2 U}{\partial t^2}  = \frac{c^{2}}{\sqrt{2\pi}}\frac{\partial u}{\partial x}\exp(-ikx) + \frac{c^{2}ik}{\sqrt{2\pi}}u\exp(-ikx) - (ck)^{2}U \]

Notice that:

    \[ \frac{c^{2}}{\sqrt{2\pi}}\frac{\partial u}{\partial x}\exp(-ikx) - \frac{c^{2}ik}{\sqrt{2\pi}}u\exp(-ikx) = \frac{c^{2}}{\sqrt{2\pi}}\frac{\partial u\exp(-ikx)}{\partial x} \]

So:

    \[ \frac{\partial^2 U}{\partial t^2}  = \frac{c^{2}}{\sqrt{2\pi}}\frac{\partial u\exp(-ikx)}{\partial x} + 2\frac{c^{2}ik}{\sqrt{2\pi}}u\exp(-ikx) - (ck)^{2}U \]

    \[ \int_{-\infty}^{\infty}\frac{\partial^2 U}{\partial t^2}dx  = \int_{-\infty}^{\infty}\frac{c^{2}}{\sqrt{2\pi}}\frac{\partial u\exp(-ikx)}{\partial x}dx + \int_{-\infty}^{\infty}2\frac{c^{2}ik}{\sqrt{2\pi}}u\exp(-ikx)dx  - \int_{-\infty}^{\infty}(ck)^{2}Udx \]

Working…

(ii)

I speculate the general form of the solution is: F(k)(C_{1}\cos(kct) + C_{2}\sin(kct)). Eq. (2.3) implies that:

    \[ \frac{\partial U}{\partial t}(x, 0) = 0 \]

So C_{2} = 0.

(iii)

    \[ U(k, 0) = F(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} u(x, 0)\exp(-ikx)dx = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp(-ax^{2})\exp(-ikx)dx = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp(-ax^{2} - ikx)dx \]

    \[ U(k, 0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp\left (-a\left (x + \frac{ik}{a} \right )^{2} - \frac{k^{2}}{4a} \right )dx = \frac{\exp(-\frac{k^{2}}{4a})}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp\left (-a\left (x + \frac{ik}{a} \right )^{2} \right )dx = \frac{1}{\sqrt{2a}}\exp \left (-\frac{k^{2}}{4a} \right ) \]

So:

    \[ F(k) = \frac{1}{\sqrt{2a}}\exp \left (-\frac{k^{2}}{4a} \right ) \]

(3)

    \[ u(x, t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2a}}\exp \left (-\frac{k^{2}}{4a} \right )\cos(kct)\exp(ikx)dk = \frac{1}{2\sqrt{a\pi}}\int_{-\infty}^{\infty}\frac{e^{ikct} + e^{-ikct}}{2}\exp\left ( -\frac{k^{2}}{4a} + ikx \right )dk \]

    \[ u(x, t) = \frac{1}{4\sqrt{a\pi}}\left ( \int_{-\infty}^{\infty}\exp\left ( -\frac{k^{2}}{4a} + (ix + ict)k \right )dk  + \int_{-\infty}^{\infty}\exp\left ( -\frac{k^{2}}{4a} + (ix - ict)k  \right )dk \right ) \]

    \[ u(x, t) = \frac{\exp(-a(x + ct)^{2})}{4\sqrt{a\pi}}\int_{-\infty}^{\infty}\exp\left ( -\frac{1}{4a}(k - 2ai(x + ct))^{2} \right )dk  + \frac{\exp(-a(x - ct)^{2})}{4\sqrt{a\pi}}\int_{-\infty}^{\infty}\exp\left ( -\frac{1}{4a}(k - 2ai(x - ct))^{2}  \right )dk \]

    \[ u(x, t) = \frac{\exp(-a(x + ct)^{2}) + \exp(-a(x - ct)^{2})}{2} \]


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