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I came across the following diffusion problem in ''Myint-U, Lokenath Debnath, Linear Partial Differential Equations for Scientists and Engineers (2007, Birkhäuser)'' on page 526 (Problem 31):

Solve (by means of an apt integral transform): \begin{align} &u_t = k u_{xx}, \ \ 0 < x < +\infty, \ \ \ t>0, \\ &u(x,0)=x \ \ \ \ x>0,\\ &u(0,t)\ =0, \ \ u(x,t) \rightarrow x\ \text{uniformly in $t$ as}\ x \rightarrow \infty, \ \ t>0. \end{align}

The solution is given on page 743: \begin{equation} u(x,t) = x - x\ \text{erfc}(x/\sqrt{4kt}). \end{equation} I cannot get to this solution!!!

My first attempt was to employ Laplace transform, \begin{equation} \bar{u}(x,s) = \mathcal{L}_t u(x,t), \end{equation} with the following transformed boundary conditions: \begin{equation} \bar{u}(0,s)=0, \ \ \text{and} \ \ \bar{u}(x,s) = \frac{x}{s}\ \text{as} \ x \rightarrow \infty. \end{equation} The solution was found to be \begin{equation} \bar{u}(x,s) = A e^{-\sqrt{\frac{s}{k}}x} + Be^{+\sqrt{\frac{s}{k}}x} + \frac{x}{s} \end{equation} However, both $A$ and $B$ should vanish in order to satisfy the boundary conditions, leaving only the solution $u(x,t)=x$, which seems trivial.

I sense that I should not replace the asymptotic behavior of the solution with a boundary condition, yet I do not know how to proceed further.

I also tried to perform assorted change of functions, as \begin{equation} u(x,t) \equiv x + v(x,t), \ \ \ \text{or} \ \ \ u(x,t) \equiv x + x v(x,t), \end{equation} where now $v(x,t)$ should vanish as $x\rightarrow\infty$ for all $t$, but the followed up ODE's would not result in the desired answer using the Laplace transform. I could get the following solution, though, with use of infinite sine Fourier transform when $u(x,t) \equiv x + v(x,t)$ and $v(0,t)=0,\ v(\infty,t) = 0$, \begin{equation} u(x,t) = x + C\ \sqrt{\tfrac{2}{\pi}}\int^{\infty}_{0} e^{-ktq^2}\sin qx\ dq \end{equation} where $C$ could be any constant (I didn't manage to find its true value, and it seems there's a degree of freedom for the asymptotic behavior, and I think it's because I have used the obviously false statement that v(x,t) should asymptotically tend to $0$ as $x$ blows up). This solution captures a tinge of the essence of the problem, because $\sin qx$ oscillates rapidly in large $x$, and $u(x,t)$ behaves as $x$ for all $t$. Also, \begin{equation} \text{erf} (x) = \tfrac{2}{\pi} \int^{\infty}_{0} e^{-4q^2} \frac{\sin xq}{q}dq = \tfrac{2}{\pi} \int^{\infty}_{0} e^{-4q^2} \frac{d}{dx}(-\cos xq)dq \end{equation} which somehow implies that I should have used cosine Fourier Transform, which was not feasible.

I would be grateful if somebody could provide me with a solution to the problem and also an explanation on what I may have done wrong, and what details I do not take into account.

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  • $\begingroup$ Please read the description of the asymptotics tag. It is not applicable here I believe. $\endgroup$
    – user79161
    Jan 17, 2021 at 14:34
  • $\begingroup$ Corrected, thanks. @user79161. $\endgroup$
    – Bjaam
    Jan 17, 2021 at 14:59

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