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Let $u(x,t)$ be the solution of the equation $$ \frac{\partial{^2u}}{\partial{x^2}}=\frac{\partial{u}}{\partial{t}}$$ which tends to zero as $t\rightarrow\infty$ and has the value $\cos(x)$ when $t=0$.

Then which of the following is true?

$1.\;u=\sum^\infty_{n=1}a_n\sin(nx+b_n)e^{-nt}$, where $a_n,b_n$ are arbitrary constants.

$2.\;u=\sum^\infty_{n=1}a_n\sin(nx+b_n)e^{-n^2t}$, where $a_n,b_n$ are arbitrary constants.

$3.\;u=\sum^\infty_{n=1}a_n\cos(nx+b_n)e^{-nt}$, where $a_n$ are not all zeros and $b_n=0$

$4.\;u=\sum^\infty_{n=1}a_n\cos(nx+b_n)e^{-n^2t}$, where $a_1\neq0,\;a_n=0 \text{ for } n>1,b_n=0 \text{ for } n\geq1$.

Michael Hardy
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1 Answers1

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HINT:

First, show that neither $(1)$ nor $(3)$ satisfies the PDE $u_{xx}=u_t$.

Then, let $t=0$ in Cases $(2)$ and $(4)$ and see which of these proposed solutions satisfies $u(x,0) =\cos x$.

SPOILER ALERT: Scroll over the highlighted area to reveal the solution.

Note that $u$ satisfies the PDE $u_{xx}=u_t$. For Proposed Solution $(1)$, we have $$u_{xx}=-\sum_{n=1}^{\infty}n^2a_n\sin(nx+b_n)e^{-nt}$$while $$u_{t}=-\sum_{n=1}^{\infty}na_n\sin(nx+b_n)e^{-nt}$$Clearly, $u_{xx}\ne u_t$ for $(1)$. Similarly, Proposed Solution $(3)$ fails to satisfy the PDE. Next, we note that for Proposed Solution $(2)$ we have $$u(x,0)=\sum_{n=1}^{\infty}a_n\sin(nx+b_n)$$But, this series cannot equal $\cos x$ for arbitrary $a_n$ and $b_n$. However, Proposed Solution $(4)$ satisfies the PDE and the initial condition. Therefore, it is the solution to the problem!

Mark Viola
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