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What is the analytical solution for the following diffusion partial differential equation (initial value problem)? $$\frac{\partial f}{\partial t} = (ax^2+b)\frac{\partial f}{\partial x}+\frac{\partial^2 f}{\partial x^2},$$ 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 $x$: $$k\tilde f= (ax^2+b)\frac{d\tilde f}{d x}+\frac{d^2 \tilde f}{d x^2}.$$ 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?

Hans
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  • Have you tried separation of variable techniques? – Mhenni Benghorbal Dec 10 '13 at 05:04
  • By the way, you should take the Laplace transform w.r.to $t$ not Fourier transform. – Mhenni Benghorbal Dec 10 '13 at 05:06
  • @MhenniBenghorbal: Thank you for the comments, but have you seen the phrase "separate the variables" in the question? Also the form of the ODE equation is pretty clear the separation of variable method or equivalently (which the question has already stated as much explicitly) the Fourier (Laplace) transform has been applied. Regarding the Laplace transform, it is not too much different from a Fourier transform either in form or principle. They are taking different contours in the complex plane. – Hans Dec 10 '13 at 05:23
  • Offcourse, you need to take Laplce transform w.r.to $t$ not Fourier one. – Mhenni Benghorbal Dec 10 '13 at 05:39
  • @MhenniBenghorbal: Could you please state the reason for, instead of repeating, "of course, you need to take Laplace"? Now, the operator on $\tilde f(x)$ being Self-adjoint, given zero boundary condition for $x$ does make the eigenvalue $k$ real, and the transform on $t$ Laplace. But there is no need to restrict the transform to either Fourier or Laplace a-priori, as there is no essential difference between Fourier and Laplace transforms, aside from the particular region in the complex the integration contour resides, and the accompanying concern for convergence. – Hans Dec 10 '13 at 07:01
  • @MhenniBenghorbal: Besides, the objective is to solve for this particular Sturm-Liouville eigenvalue equation of $\tilde f$. – Hans Dec 10 '13 at 07:04
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    The solutions of ODE :

    kf=(ax+b)f'+f''

    are linear combinations of Hermite polynomial and Kummer function (confluent hypergeometric)

    The solutions of the ODE considered here :

    kf=(ax²+b)f'+f''

    certainly involve special functions of higher level. I don't think that they are actually standard and/or referenced special functions

     – JJacquelin Dec 10 '13 at 10:19
  • @JJacquelin: But see Sergio's solution below referring to Heun's Triconfluent equation. – Hans Dec 10 '13 at 23:14

1 Answers1

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Using Maple we get the solution is: $$f(x,t)=F1(x)\cdot F2(t)$$ where $F1$ and $F2$ are functions such that $$F1_{xx}=c_1\cdot F1-(a\cdot x^2-b)F1_x \quad\text{ and } $$$$F2_t=c_1\cdot F2, $$ where $c_1$ is an arbitrary constant. The ODE for $F1$ has an "explicit" solution in terms of the Heun Triconfluent function (very ugly! Maple is not making some simplifications because it is assuming a and b are complex numbers) and the second ODE is just $F2(t)= c_2 \exp(c1\cdot t)$.

$F1(x) = c_3\cdot HeunT\left(-3^{2/3}\cdot a^2\cdot c_1/(a^2)^{4/3}, -3\cdot \sqrt{a^2}/a, a\cdot b\cdot 3^{1/3}/(a^2)^{2/3}, (1/3)\cdot 3^{2/3}\cdot (a^2)^{1/6}\cdot x\right)\cdot \exp\left(-(1/6)\cdot x\cdot (a\cdot x^2+3\cdot b)\cdot ((a^2)^{1/6}\cdot a+(a^2)^{2/3})/(a^2)^{2/3}\right)+c_4\cdot HeunT\left(-3^{2/3}\cdot a^2\cdot c_1/(a^2)^{4/3}, 3\cdot \sqrt{a^2}/a, a\cdot b\cdot 3^{1/3}/(a^2)^{2/3}, -(1/3)\cdot 3^{2/3}\cdot (a^2)^{1/6}\cdot x\right)\cdot \exp\left((1/6)\cdot x\cdot (a\cdot x^2+3\cdot b)\cdot ((a^2)^{1/6}\cdot a-(a^2)^{2/3})/(a^2)^{2/3}\right)$

Sergio Parreiras
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  • Thank you, Sergio. But I am wondering where the exponential function in your F1(x) expression come from. I thought the equation for $\tilde f(x)$ was already in the Heun Triconfluent function form. – Hans Dec 10 '13 at 23:14
  • I'm not sure. Maple's definition of HeunT: $HeunT(\alpha, \beta, \gamma, z) = DESol({diff(Y(z), z, z)-(3z^2+\gamma)(diff(Y(z), z))-((-\beta+3)z-\alpha)Y(z)}, {Y(z)}, {Y(0) = 1, (D(Y))(0) = 0})$ my guess is the $\exp$ appears because we did not specify the initial conditions that HeunT requires... – Sergio Parreiras Dec 10 '13 at 23:59