I'm trying to solve the following initial value problem.

$\begin{cases} u_t + u_x - u_{xx} &= -u, \quad \text{on} \quad \mathbb R \times \mathbb R_+\\ u(x,0) &= \frac{1}{4\pi} e^\frac{-x^2}{4}, \quad x \in \mathbb R \end{cases}$

I know how to solve the heat equation $u_t = u_{xx}$, but I have no idea how to get started with this. So far -inspired by the initial value- I have tried substituting and checking various forms of $\frac{e^\frac{-x^2}{4t}}{\sqrt{4{\pi}t}}$ to no avail. Is there a general technique that can be used in this case?

  • $\begingroup$ Hi, I have tried to let u(x,t) = e^(ax)*v(x,t) and variations of this format to solve the equation in terms of v(x,t) but to no avail. Could you give me some hints to find that function? $\endgroup$ – Bread Apr 22 '16 at 16:50
  • $\begingroup$ I think the method of characteristics should work. :) $\endgroup$ – Andrew Apr 22 '16 at 16:55
  • $\begingroup$ Oops my bad, it should be u_xx instead! forgotten to check the formatting. $\endgroup$ – Bread Apr 22 '16 at 16:57
  • $\begingroup$ Oh. Well at least this way the boundary conditions make intuitive sense.. Have you tried looking for the solution in the form of $u(x,t) = X(x)T(t)$? $\endgroup$ – Andrew Apr 22 '16 at 17:03
  • $\begingroup$ To me, that method would not work as I do not have the Boundary conditions :( Hence, I would have to use the usual diffusion method. $\endgroup$ – Bread Apr 22 '16 at 17:05

You should be able to transform the above equation into the heat equation by the following substitution: $$ u(x,t) = w(x,t) e^{\frac{1}{2}x-(1+\frac{1}{4})t}$$ In fact (if $\alpha\neq 0$), the general \begin{align} u_t - \alpha u_{xx} + cu_x = -\lambda u \tag 1 \label{eq1} \end{align} equation can be seen to be equivalent to the (one dimensional) heat equation $$w_t = \alpha w_{xx}$$ using the substitution $$u(x,t) = w(x,t) \operatorname{exp}\left (\frac{c}{2\alpha}x - \left (\lambda + \frac{c^2}{4\alpha}\right )t\right)$$ I'll include a quick proof here for the sake of completeness: $$A(x,t) := \operatorname{exp}\left (\frac{c}{2\alpha}x - \left (\lambda + \frac{c^2}{4\alpha}\right )t\right)\quad\text{so}\quad u = wA$$ Then \begin{align} \color {red}{u_t} &= w_tA - \left(\lambda + \frac{c^2}{4\alpha} \right)wA \\ \color {green}{cu_x} &= c\left(w_x A + \frac{c}{2\alpha}wA \right)\\ \color {blue}{\alpha u_{xx}} &= \alpha \left( w_{xx}A + \frac{c}{\alpha}w_xA + \frac{c^2}{4\alpha^2}wA \right) \end{align} Substituting back into $(\ref{eq1})$ and dividing the equation by the nonzero $A$, we get \begin{align} -\lambda w &= \color {red}{w_t - \lambda w - \frac{c^2}{4\alpha}w} \color {blue} {- \alpha w_{xx} - cw_x - \frac{c^2}{4\alpha}w} \color{green}{+ cw_x + \frac{c^2}{2\alpha}w}\\ 0 &= \color {red}{w_t} - \color {blue}{\alpha w_{xx}} \end{align}

  • $\begingroup$ Sure thing, happy to help. :) $\endgroup$ – Andrew Apr 22 '16 at 19:06

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