4
$\begingroup$

Let $u(x,t)$ be the bounded solution of $\frac{\partial u}{\partial t} -\frac{\partial ^2 u}{\partial x^2}=0$ with $u(x, 0)=\dfrac{e^ {2x} -1}{e^{2x} +1}.$ Then $\lim _{t \rightarrow \infty}u(1, t)=$

$$(A)-\frac{1}{2}~~~~~~(B)\frac{1}{2}~~~~~~(C)-1~~~~~~(D)1.$$

By separation of variable the solution of the above PDE is: $$u(x, t)=\begin{cases} e^{- \lambda ^2 t}(c_1 \cos(\lambda x)+c_2 \sin(\lambda x)) & ~~~\text{if}~k=-{\lambda}^2<0 ,\\ e^{\lambda ^2 t}(c_1 e^{\lambda x}+c_2 e^{\lambda x}) & ~~~\text{if}~k={\lambda}^2>0 ,\\ c_1x+c_2 & ~~~\text{if}~k=0 .\\ \end{cases}$$ where $k$ is separation constant. But when $t=0$, I cannot compare the solution with given $u(x,0)$, that's why I cannot find $\lim _{t \rightarrow \infty}u(1, t).$ Please help.

$\endgroup$
1
  • $\begingroup$ “Can’t manipulate the given data, please help”—can you show us what you tried and where and why you think it went wrong? It seems you know the procedure to solve the heat equation via separation of variables and Fourier series, so my hunch is you’re stuck on some integration for computing the coefficients, is this so? $\endgroup$ Commented Jan 12, 2018 at 15:33

1 Answer 1

1
$\begingroup$

Hint:

You have formulated the answer to the PDE and it's almost true except that because of constants in it and your equation being linear if there are two answers for this PDE so is any of their linear combination, that's why you need to integrate your answer to attain the most general case: $$u(x,t)=\int_{-\infty}^{\infty}(A(\lambda)\cos(\lambda x)+B(\lambda)\sin(\lambda x))e^{-\lambda^2t}d\lambda$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .