I would like to find the closed-form solution (for $x > 0$) to
$$\frac{\partial u}{\partial t} = a \frac{\partial^2 u}{\partial x^2} +b \frac{\partial u}{\partial x} + (cx+d) u,$$ $$u(0,\,x) = 1,\;\; \frac{\partial u(t,\,0)}{\partial x} = 0,$$
where $a,\,b,\,c,\,d \in \mathbb{R}\backslash\{0\}$ are arbitrary constants. Though I looked in several books, I was not able to find a suitable approach with the given boundary/initial conditions. Any help would be appreciated!
On
Let \begin{align} u(x,t) = e^{- \lambda t} f(x) \end{align} to obtain the differential equation, in $x$, \begin{align} a f^{''} + b f^{'} + (c x + d + \lambda) f = 0. \end{align} Now let \begin{align} f(x) = e^{-(b x)/(2a)} g(x) \end{align} to obtain the differential equation \begin{align} g^{''} + \left( \frac{c}{a} \ x + \frac{d + \lambda}{a} - \frac{b^{2}}{4 a^{2}} \right) g = 0. \end{align} This is Airy's differential equation in a slightly different form and has the solution \begin{align} g(x) = c_{1} Ai\left(- \frac{\beta + \alpha x}{(-\alpha)^{2/3}} \right) + c_{2} Bi\left(- \frac{\beta + \alpha x}{(-\alpha)^{2/3}} \right), \end{align} where $\alpha = c/a$ and $\beta = (d+\lambda)/a - b^{2}/(4 a^{2})$.
The general solution is then \begin{align} u(x,t) = e^{- \lambda t} \ e^{- bx/2a} \left[ c_{1} Ai\left(- \frac{\beta + \alpha x}{(-\alpha)^{2/3}} \right) + c_{2} Bi\left(- \frac{\beta + \alpha x}{(-\alpha)^{2/3}} \right) \right] \end{align} where $\alpha = c/a$ and $\beta = (d+\lambda)/a - b^{2}/(4 a^{2})$.
The remainder of the problem is using the initial conditions to find the constants $c_{1}$, $c_{2}$ and the separation constant $\lambda$.
On
Sorry for the late response. This is what the Polyanin book suggests:
Consider $$u(t,\,x) = w(t,\,z) \exp\left( ctx-\frac{b}{2a}x+\frac{1}{3}ac^2t^3 + \left(d-\frac{b^2}{4a}\right)t \right),\;\;\;z=x+act^2$$.
Then $$\partial_t w = a \partial_{zz} w$$.
However, the book makes no mention of boundary conditions. I would appreciate any tips!
Thanks, Bryon
The Laplace transform with respect to $t$ leads to the differential equation $$sY \left( x \right) -1=a\;{\frac {{\rm d}^{2}}{ {\rm d}{x}^{2}}}Y \left( x \right) +b\;{\frac {\rm d}{{\rm d}x}}Y \left( x \right) + ( c x+d)\; Y \left( x \right) $$ which has solutions involving Airy functions and integrals of Airy functions times exponentials (not surprising since the Airy differential equation is $w'' = x w$). I would be surprised if the inverse Laplace transform could be expressed in closed form.