# I Second order PDE with variable coefficients

1. Jul 18, 2017

### grquanti

Hello,
I have an equation of the form:

$\partial_t f(x,t)+a\partial_x^2 f(x,t)+g(x)\partial_xf(x,t)=0$

(In my particular case $g(x)=kx$ with $k>0$ and $a=2k=2g'(x)$)

I'd like to know if there is some general technique that i can use to solve my problem (for example: in the first order case I know the method of characteristics is widely used).

P.s. I have searched for the answer in the forum, but I can't find it. If a similar question was already answered,please, forgive me.​

2. Jul 18, 2017

### Orodruin

Staff Emeritus
I suggest you use the method of characteristics to rewrite the first order part as a single partial derivative. You can then deal with the second order derivative later on. You should get a differential equation reminiscent of the heat equation.

3. Jul 19, 2017

### grquanti

Forgive me, I'm not an expert (actually I don't know nothing about the method of characteristics), can you be a little more explicit?

If it's what you mean, my starting point is

$\partial_t f(x,t) = k\partial_x^2 xf(x,t)$

with $k>0$ (there is an error in the first post: when putted in the homogeneous form, the $+$ singns should be $-$.
For instance, it's a Fokker-Planck equation with no drift.

4. Jul 19, 2017

### Orodruin

Staff Emeritus
Perfect opportunity to learn then. If given a partial differential equation on the form $\partial_t f + \vec v(\vec x,t) \cdot \nabla f = \ldots$, where $\vec v$ can be regarded as a velocity field you can introduce new variables $s$ and $\vec \xi$ such that the left-hand side is equal to $\partial_s f$ by finding the characteristics of the velocity field $\vec v$, i.e., its flow lines. You do this by letting $s = t$ and solving the resulting ordinary differential equation $\partial{\vec x}/\partial s = \vec v$ (note that the resulting $\vec x$ will generally depend on both $\vec \xi$ and $s$!). In your case, your space is one-dimensional and so you can do away with all the vector arrows. I suggest trying to introduce these new variables and writing down the differential equation in terms of them.

5. Jul 21, 2017

### grquanti

If I understand, you mean:

from

$\partial_t f(x,t) -2a\partial_x f(x,t) = ax\partial_x^2 f(x,t)$

I take $s$ such that

$\partial_s f(x,t) = \partial_t f(x,t) -2a\partial_x f(x,t)$

This implies

$\frac{d}{ds}t=1$
$\frac{d}{ds}x=-2a$

So that

$t(s)=s+c_1$
$x(s)=-2as+c_2$

and I'm left with

$\partial_s f(x(s),t(s))=ax(s)\partial_x^2f(x(s),t(s))$

But I can't understand how to use the parameter $ξ$ and, more in general, I have seen the method is applied when the right hand side of the equation is independent from the solution: if the right hand side is a function $g(x,t)$ , in the characteristic equations I should also consider

$\frac{d}{ds}z=g(x(s),t(x))$

but in my case the function $g$ is $g=ax\partial_x^2 f$ which depends on $f$

Where are my mistakes?

6. Jul 21, 2017

### Orodruin

Staff Emeritus
This is not necessarily true. Note that
$$\partial_x = \frac{\partial s}{\partial x} \partial_s + \frac{\partial \xi}{\partial x}\partial_\xi.$$
This is the reason to call both $s$ and $\xi$ something different from $t$ and $x$ - to keep track of which variables are which.

A good choice would be to let $\xi = x$ for $s = 0$. Also, you can choose to let your integration constant $c_1$ be equal to 0 as this just represents a shift in $s$. What does the differential equation $\partial x/\partial s = -2ax$ really tell you? (Don't forget the $x$! As you defined the problem, it is here and not in the second order term.).