Todai Entrance Exam: Math 2017 – Problem 2

(1)

    \[ \int_{0}^{1}\sin(n\pi x)\sin(m\pi x)dx = \frac{1}{2}\left (\int_{0}^{1}\cos(n\pi x - m\pi x)dx - \int_{0}^{1}\cos(n\pi x + m\pi x)dx \right ) \]

    \[ \int_{0}^{1}\sin(n\pi x)\sin(m\pi x)dx = \frac{1}{2}\left ( \frac{1}{n\pi + m\pi}\sin(n\pi x + m\pi x) \big |^{1}_{0} - \frac{1}{n\pi - m\pi}\sin(n\pi x - m\pi x) \big |^{1}_{0} \right ) \]

    \[ \int_{0}^{1}\sin(n\pi x)\sin(m\pi x)dx = \frac{1}{2}\left (\frac{1}{(n + m)\pi}\sin((m+n)\pi) -  \frac{1}{(n - m)\pi}\sin((n-m)\pi)  \right ) \]

(2)

    \[ \frac{\partial u}{\partial t} = \frac{\partial^{2} u}{\partial x^{2}} \Leftrightarrow \varepsilon (x)\frac{\mathrm{d} \tau }{\mathrm{d} t} = \tau (t)\frac{\mathrm{d}^{2} \varepsilon }{\mathrm{d} x^{2}} \Leftrightarrow \frac{1}{\tau (t)}\frac{\mathrm{d} \tau }{\mathrm{d} t} = \frac{1}{\varepsilon (x)}\frac{\mathrm{d}^{2} \varepsilon }{\mathrm{d} x^{2}} \]

Let f(x) = \frac{1}{\varepsilon (x)}\frac{\mathrm{d}^{2} \varepsilon }{\mathrm{d} x^{2}} and g(t) = \frac{1}{\tau (t)}\frac{\mathrm{d} \tau }{\mathrm{d} t}. So, thanks to the given fact:

    \[ f(x) = g(t) = C \Leftrightarrow \frac{1}{\tau (t)}\frac{\mathrm{d} \tau }{\mathrm{d} t} = \frac{1}{\varepsilon (x)}\frac{\mathrm{d}^{2} \varepsilon }{\mathrm{d} x^{2}} = C \]

(3)

    \[ \int \frac{1}{\tau (t)}\frac{\mathrm{d} \tau }{\mathrm{d} t}dt = \int Cdt \Leftrightarrow \ln {\tau (t)} = Ct \Leftrightarrow \tau (t) = e^{Ct} \]

    \[ \frac{\mathrm{d}^{2} \varepsilon }{\mathrm{d} x^{2}} - C\varepsilon (x) = 0 \]

We solve the characteristic equation of the above homogenous second order linear differential equations:

    \[ r^{2} - C = 0 \Leftrightarrow \begin{Bmatrix} r = \pm \sqrt{C} & \text{ if } & C \geq 0\\  r = \pm i\sqrt{-C} & \text{ if } & C < 0 \end{matrix} \]

So:

    \[ \begin{matrix} \varepsilon (x) = c_{1}e^{\sqrt{C}x} + c_{2}e^{-\sqrt{C}x} & \text{ if }  & C \geq 0 \\  \varepsilon (x) = c_{1}\cos(\sqrt{-C}x) + c_{2}\sin(\sqrt{-C}x) & \text{ if }  & C < 0 \end{matrix} \]

If C > 0:

    \[ u(0, t) = u(1, t) = 0 \Leftrightarrow e^{Ct}(c_{1} + c_{2}) = e^{Ct}(c_{1}e^{\sqrt{C}} + c_{2}e^{-\sqrt{C}}) = 0 \Leftrightarrow \begin{Bmatrix} \frac{c_{2}}{c_{1}} = - 1\\  \frac{c_{2}}{c_{1}} = - e^{2\sqrt{C}} \end{matrix} \Leftrightarrow \begin{Bmatrix} c_{1} + c_{2} = 0\\  C = 0 \end{matrix} \Leftrightarrow u(x, t) = 0 \]

which is the excluded solution.

If C < 0:

    \[ u(0, t) = u(1, t) = 0 \Leftrightarrow e^{Ct}c_{1} = e^{Ct}(c_{1}\cos(\sqrt{-C}) + c_{2}\sin(\sqrt{-C})) = 0 \Leftrightarrow \begin{Bmatrix} c_{1} = 0\\  \sqrt{-C} = n\pi  \end{matrix} \Leftrightarrow u(x, t) = c_{2}e^{-n^{2}\pi ^{2}t}\sin(n\pi x) \]

Note that if c_{2} = 0 in this case then u(x, t) = 0, which is the excluded solution.

If c_{2} = 0 then u_{n}(x, t) = u(x, t), so \alpha = -n^{2}\pi ^{2}, \beta = n\pi.

(4)

How to find c_{n} so that:

    \[ \sum_{n=1}^{\infty}c_{n}\sin(n\pi x) = x(1 - x) \]

?

Also notice that:

    \[ \sin(n\pi x) + \sin \left (\left (1 + k\frac{1}{x} \right )\pi x \right ) = 0 \]


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