Another diffusion partial differential equation, or Sturm-Liouville eigenvalue ODE

Question

What is the solution for the following diffusion partial differential equation (initial value problem)? $$\frac{\partial f}{\partial t} = \pm\frac{\partial f}{\partial x}+(ax+b)^2\frac{\partial^2 f}{\partial x^2},$$ where $a$ and $b$ are real number constants.

We can separate the variables or take the Fourier transform $\tilde f(x)$ of $f$ in the time domain $t$, and turn the above into an ordinary differential equation eigenvalue problem in $x$: $$k\tilde f= \pm\frac{d\tilde f}{d x}+(ax+b)^2\frac{d^2 \tilde f}{d x^2}.$$ where $k$ can be views as an eigenvalue for the differential operator on the left hand side. Now we can further transform this into the Sturm-Liouville form.

However, I can not immediate recognize a transformation that can turn the above into a known form that admits an analytic solution. Can someone help?

Answer 1

Maple finds solutions of the ODE (with +) in terms of Kummer U and M functions: $$f \left( x \right) = \left( ax+b \right) ^{-{\frac {\sqrt {{a}^{2 }+4\,k}-a}{2a}}} \left( C_{{1}}\;{\text{KummerU}\left({\frac {\sqrt {{a}^{2 }+4\,k}-a}{2a}},\,{\frac {a+\sqrt {{a}^{2}+4\,k}}{a}},\,{\frac {1}{ \left( ax+b \right) a}}\right)}+C_{{2}}\;{\text{KummerM}\left({\frac { \sqrt {{a}^{2}+4\,k}-a}{2a}},\,{\frac {a+\sqrt {{a}^{2}+4\,k}}{a}},\,{ \frac {1}{ \left( ax+b \right) a}}\right)} \right) $$

EDIT:

If you take your ODE $(ax+b)^2 f''(x) + f'(x) - k f(x) = 0$ and do the change of dependent and independent variables $f(x) = z^c u(z)$, $x = \dfrac{1}{a^2 z} - \dfrac{b}{a}$ with constant $c$ to be determined, you get

$$ z\; u''(z) + (2c+2 - z) \; u'(z) + \left(\dfrac{a^2 c^2 + a^2 c - k}{a^2 z} - c \right) u = 0$$

Set $c$ such that $a^2 c^2 + a^2 c - k = 0$, the above equation turns into the Kummer equation

$$ z \; u'' + (\nu - z) u' - \mu u = 0$$ with $\nu = 2c+2$ and $\mu = c$.

Answer 2

Ok, you want to find any nontrivial solution of the ODE.

Note first that if $\alpha = 0$, then the answer is trivial. Because it will be simple linear homogeneous equation with constant coefficients.

Assume that $\alpha \neq 0$. As far as I understand, you can take almost any $k$ (with except $k =0$), since it doesn't depend on initial conditions. Let us try to find a solution in the form $$ \tilde{f} = x^2 + \beta x + \gamma. $$ Hence, $$ \frac{d \tilde{f}}{dx} = 2 x + \beta, \quad \frac{d^2 \tilde{f}}{dx^2} = 2. $$ Substituting these equalities into your expression, we get $$ k x^2 + \beta k x + \gamma k = \pm (2 x + \beta) + 2 (a^2 x^2 + 2 a b x + b^2). $$ Therefore, $$ (2a^2 -k) x^2 + (4 a b \pm 2 - \beta k)x + (2 b^2 \pm \beta - \gamma k) = 0. $$ Now we want expression in every bracket to be zero. For the first bracket we take $k = 2 a^2$. After that we can find $\beta$ from the second bracket, and finally $\gamma$ from the third. Hence, function $\tilde{f} = x^2 + \beta x + \gamma$ satisfies your ODE, that is, $\tilde{f}$ is a solution.