Problem: Let $u(x,y)$ be the solution of the equation $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$, which tends to zero as $y\to \infty$ and has the value $\sin x$ when $y=0$. Then

  1. $u=\sum\limits_{n=1}^\infty a_n \sin(nx+b_n) e^{-ny}$, where $a_n$ are arbitrary and $b_n$ are non-zero constant

  2. $u=\sum\limits_{n=1}^\infty a_n \sin(nx+b_n) e^{-n^2y}$, where $a_1=1$ and $a_n(n>1), b_n$ are non zero constant.

  3. $u=\sum\limits_{n=1}^\infty a_n \sin(nx+b_n) e^{-ny}$, where $a_1=1, a_n=0$ for $n>1$ and $b_n=0$ for $n\geq 1$

  4. $u=\sum\limits_{n=1}^\infty a_n \sin(nx+b_n) e^{-n^2y}$, where $b_n=0$ for $n\geq 0$ and $a_n$ are all nonzero.

I seem the option 3. is the correct by taking $u(x,t)=X(x)T(t)$. This implies $\frac{X''}{X}=-\frac{T''}{T}=-\alpha^2 (\alpha>0)$, where $X''=\frac{d^2X}{dx^2}$. But I am not sure whether my answer is correct or not. Please help to solve the problem.

Thanks in advance.

  • 1
    $\begingroup$ That is correct. To convince yourself show that 2. and 4. does not satisfy the PDE (it has to satisfy it term by term - just insert into PDE and check) and 1. does not need to satisfy the initial condition (take e.g. $a_n=0$ for $n>1$ and pick $b_1$ such that when $y=0$ you don't get $\sin(x)$). $\endgroup$ – Winther Dec 21 '15 at 17:49
  • $\begingroup$ @Winther, thank you very much. +1 for u $\endgroup$ – Warrior Dec 21 '15 at 17:50

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