I have a backward parabolic equation of the form:

\begin{equation} W_{\eta} + aW_{xx} - bW = 0 \end{equation}


\begin{equation} \lim_{\eta \rightarrow \infty}(x,\eta) = g(x) \end{equation}

were $x \in \mathbb{R}$, $\eta \geqslant 0$, and $a,b$ are positive constants.

Applying the following transformations:

\begin{align} W(x,\eta) &= U(x,t)e^{b\eta} \\ t &= a\eta \end{align}

we would get the backward heat equation below

\begin{equation} U_{t} = - U_{xx} \end{equation}

However, the transversality condition becomes a problem, since as $\eta \rightarrow \infty$, $e^{b\eta} \rightarrow \infty$.

Usually, if the terminal condition is of the form

\begin{equation} W(x,H) = g(x) \end{equation}

with $H$ finite, we could "reverse" it, that is, we could apply the following transformation:

\begin{equation} \nu = H - \eta \end{equation}

to obtain

\begin{equation} -W_{\nu} + aW_{xx} - bW = 0 \end{equation}


\begin{equation} W(x,0) = g(x) \end{equation}

which we can solve the traditional way (Fourier transform). However, as my terminal condition happens only at infinity I can't apply the reverse transformation above, thus I don't know how to overcome this problem. Any hint or reference?

  • $\begingroup$ Have you tried solving the equation with $U(x,T)=g(x)$ and let $T\to\infty$? $\endgroup$ – Dylan Apr 2 at 7:23
  • $\begingroup$ I have edited the question, now it reflects properly the problem I have. $\endgroup$ – Nicolas Pimentel de Souza Apr 2 at 14:48

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