Linear partial differential equation solved by shift operator-like

Question

I try to find a solution to this partial differential equation (in a thermodynamic problem again):

$$ \left[ \partial_x^2 + a(t) \partial_x - \partial_t \right] w(x,t) = 0 $$

Where $a$ and $w$ are real valued functions, $x > 0$ and $t > 0$ position and time. Do we have general solution for arbitrary $a(t)$ ?

HINT/TRACK

Related to this topic, when we remind that for this simple equation:

$$ \left( \partial_x + \frac{1}{c} \partial_t \right) f(x,t) = 0$$

The general solution is (using the shift operator and an arbitrary function $h(t)$):

$$ f(x,t) = e^{- \frac{x}{c} \partial_t } h(t) = h(t-x/c) $$

So my idea was to use also shift operator $e^{\partial_t}$. So I pose:

$$ w = e^{x \hat{s}(t)} D(t) $$

where $\hat{s}(t)$ is an operator on $t$, $D(t)$ function of $t$. Then I replace in the equation and obtained:

$$ ( \hat{s}^2 + a \hat{s} - \partial_t ) D(t) = 0 $$

I solve the quadratic operator equation on $\hat{s}(t)$ and obtained:

$$ \hat{s}_{1,2} = - \frac{a}{2} \pm \frac{1}{2} \sqrt{a^2 + 4 \partial_t} = - \frac{a}{2} \pm \hat{k} $$

with $\hat{k} = 1/2 \sqrt{a^2 + 4 \partial_t} $. Where I suppose I can evaluate the square root of the operator by applying an expansion as for the shift operator.

Now here comes the problem: $a(t)$ and $\hat{k}(t)$ does not always commute (only if $a$ is a constant). So the 2 independent solutions are somehow difficult to express, intuitively we should think to (for example):

$$ w(x,t) = e^{- x \frac{a}{2}} e^{x \hat{k} } D_1(t) + e^{- x \frac{a}{2}} e^{- x \hat{k} } D_2(t) = w_1(x,t) + w_2(x,t) $$

So from here things begin to get tricky since we are dealing with operators and Poisson's Bracket, because for example for $w_1$ we obtain:

$$ \left[ \partial_x^2 + a(t) \partial_x + \partial_t \right] w_1(x,t) = \left[\frac{a(t)}{2},\hat{k} \right] w_1(x,t) \neq 0 $$

Do you have an idea for this track (maybe Lie algebra or stuff like that?)?

Thanks for reading

Answer 1

Your approach is the right idea, you want to reduce this convection-diffusion equation to a heat equation by "factoring out" the transport dynamics induced by the $a(t)\partial_x w$ term. If define the transformed solution $g(y,t) := w(y-\mu(t),t)$, where $\mu$ is unknown, by the chain rule we have

$$\partial_x w = \partial_y g, \ \partial^2_x w = \partial^2_y g$$

and more importantly

$$\partial_t g(y,t) = (\partial_t w)(y-\mu(t),t) - \dot{\mu}(t)\partial_y g(y,t)$$

Hence,

$$\begin{align}(\partial_x^2+a(t)\partial_x)w(y-\mu(t),t) &= (\partial_tw)(y-\mu(t),t) \\ \implies (\partial_y^2+a(t)\partial_y)g(y,t) &= \partial_t g(y,t)+\dot{\mu}(t)\partial_y g(y,t)\end{align}$$

Or equivalently,

$$(\partial_y^2-\partial_t)g(y,t) = (a(t)-\dot{\mu}(t))\partial_yg(y,t)$$

Which is a heat equation if $\dot{\mu} = a$, so one can pick $\mu(t) = \int_{-\infty}^t a(s)ds$ (where $a$ is extended to be zero outside of $[0,\infty)$).