# Solution of PDE

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) : 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

• 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 $f(x,t)=a(x)+b(t)$ Apr 29, 2020 at 9:11
• The second equation is for $t \rightarrow \infty$ ?
– EDX
Apr 29, 2020 at 9:15

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 !).

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