# How to solve a 2nd order PDE with asymptotic boundary condition?

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: $$$$u(x,t) = x - x\ \text{erfc}(x/\sqrt{4kt}).$$$$ I cannot get to this solution!!!

My first attempt was to employ Laplace transform, $$$$\bar{u}(x,s) = \mathcal{L}_t u(x,t),$$$$ with the following transformed boundary conditions: $$$$\bar{u}(0,s)=0, \ \ \text{and} \ \ \bar{u}(x,s) = \frac{x}{s}\ \text{as} \ x \rightarrow \infty.$$$$ The solution was found to be $$$$\bar{u}(x,s) = A e^{-\sqrt{\frac{s}{k}}x} + Be^{+\sqrt{\frac{s}{k}}x} + \frac{x}{s}$$$$ However, both $$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 $$$$u(x,t) \equiv x + v(x,t), \ \ \ \text{or} \ \ \ u(x,t) \equiv x + x v(x,t),$$$$ where now $$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$$, $$$$u(x,t) = x + C\ \sqrt{\tfrac{2}{\pi}}\int^{\infty}_{0} e^{-ktq^2}\sin qx\ dq$$$$ where $$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$$. Also, $$$$\text{erf} (x) = \tfrac{2}{\pi} \int^{\infty}_{0} e^{-4q^2} \frac{\sin xq}{q}dq = \tfrac{2}{\pi} \int^{\infty}_{0} e^{-4q^2} \frac{d}{dx}(-\cos xq)dq$$$$ 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.

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