I'd like to find the solution of this PDE :

$$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}+D\frac{\partial^2 f}{\partial x^2} $$

with $D$ a constant such that at infinity $t\rightarrow +\infty$ the solution is the solution of the equation is the solution of the ODE : $$\frac{\partial f}{\partial x}+D\frac{\partial^2 f}{\partial x^2}=0 $$

whose solution looks like this when you calculate it (it's an analytical formula) :

enter image description here

I'm wondering if I could find an analytical solution to my equation. $f(x,t)=F(x)+G(t)$ is indeed an obivous solution. But I'm wondering if to some extent one could impose some initial conditions : $f(x,t=0)=f_0(x)$.

EDIT: I found that what I'm looking for is the solution of the so called Mason-Weaver equation. It can be found here

  • 1
    $\begingroup$ I did not try to do the computations, but did you try to write the solution in the form $f(x,t)=a(x)b(t)$ and derive a possible expression for $a$ and $b$ in such a way that $f$ solves the PDE? Sometimes this variable splitting is a good approach...otherwise you can test even $f(x,t)=a(x)+b(t)$ $\endgroup$
    – Dadeslam
    Apr 29, 2020 at 9:11
  • 2
    $\begingroup$ The second equation is for $t \rightarrow \infty $ ? $\endgroup$
    – EDX
    Apr 29, 2020 at 9:15

2 Answers 2


Observe we have that \begin{align} \partial_t f = L f, \ \ f(0, x) = f_0(x) \end{align} then it follows \begin{align} f(t,x) = e^{tL}f_0. \end{align} Here, we see that \begin{align} L= \frac{\partial}{\partial x}+ D\frac{\partial^2}{\partial x^2}=: L_1+L_2. \end{align} Note that $[L_1, L_2] =0$, i.e. they commute. Then it follows \begin{align} e^{tL}f_0 = e^{tL_1}e^{tL_2}f_0. \end{align} Note that $g:=e^{tL_2}f_0$ solves the heat equation \begin{align} \frac{\partial g}{\partial t} = D\frac{\partial^2 g}{\partial x^2} \end{align} with initial condition $g(0, x) = f_0(x)$. Hence you can write out explicitly $g(t,x)$. Finally, we see that \begin{align} f(t,x) =e^{tL_1} g(x, t) = g(x+t, t) \end{align} since $e^{tL_1}g(x) = g(x+t)$.


First we wil suppose we can separate :

$$ f(x,t)=g(x)h(t)$$

The first equation becomes :

$$ \dfrac{dh}{dt}g=h(\dfrac{dg}{dx}+D\dfrac{d^2g}{dx^2}) \ (1)$$

So at the infinity the equation becomes:

$$ h(\infty)(\dfrac{dg}{dx}+D\dfrac{d^2g}{dx^2})= 0 \ (2)\\\ \\\dfrac{dh}{dt}_{t=\infty}=0 \ (3) $$

Solving for $g$ assuming $h(\infty)$ isn't null (depend on your physical modeling for coherence).

$$ \exists (A,B) \in \mathbb{R}^2, \ g:x\rightarrow A+Be^{-\frac{x}{D}}$$

Then from $(1)$ :

$$ \dfrac{dh}{dt}g=0 $$

We have $h$ constant (coherent with $(3)$) or $g$ null (evinced otherwise the problem has no utility !).

$$ f : (x,t) \rightarrow A'+B'e^{-\frac{x}{D}} $$


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