# How to solve $\frac{\partial u(x,t)}{\partial t}=-axu(x,t)-bx\frac{\partial u(x,t)}{\partial x}+cx^2 \frac{\partial^2 u(x,t)}{\partial x^2}$?

Can someone give me a hint how to solve this pde.

$$\frac{\partial u(x,t)}{\partial t}=-a\,x\,u(x,t)-b\,x\,\frac{\partial u(x,t)}{\partial x}+c\,x^2\, \frac{\partial^2 u(t,x)}{\partial x^2}$$

Where ${a,b,c}$ are constants.

I was looking at the method of characteristics but I am not sure if it can be applied.

• Maple hint: $u(x,t)=\frac{a x^{\frac{b}{2 c}} \left(\sqrt{-\frac{a}{c}} c \sqrt{x} J_{1-\frac{b}{c}}\left(2 \sqrt{-\frac{a}{c}} \sqrt{x}\right) c_1+b J_{-\frac{b}{c}}\left(2 \sqrt{-\frac{a}{c}} \sqrt{x}\right) c_1+\left(\sqrt{-\frac{a}{c}} c \sqrt{x} Y_{1-\frac{b}{c}}\left(2 \sqrt{-\frac{a}{c}} \sqrt{x}\right)+b Y_{-\frac{b}{c}}\left(2 \sqrt{-\frac{a}{c}} \sqrt{x}\right)\right) c_2\right)}{\left(-\frac{a}{c}\right)^{3/2} c^2}$ where $J_x$ and $Y_x$ Bessel function of the first kind and second kind. – Mariusz Iwaniuk Mar 27 '18 at 22:35
• @MariuszIwaniuk: Can anyone derive that? – Hans Mar 28 '18 at 2:32
• I'm not strong in solving Pdes that way I use CAS system to do that.Why waste the time to solve how you can answer in a second with MAPLE or Mathematica? – Mariusz Iwaniuk Mar 28 '18 at 7:46

## 1 Answer

$$\frac{\partial u(x,t)}{\partial t}=-a\,x\,u(x,t)-b\,x\,\frac{\partial u(x,t)}{\partial x}+c\,x^2\, \frac{\partial^2 u(t,x)}{\partial x^2} \tag 1$$ The method of separation of variables seems easier in this case.

We look first for particular solutions on the form $u(x,t)=X(x)T(t)$. $$XT'=-axXT-bxX'T+cx^2X''T$$ $$\frac{T'}{T}=-ax-bx\frac{X'}{X}+cx^2\frac{X''}{X}=\lambda$$ $$\frac{T'}{T}=\lambda \quad\text{leads to}\quad T(t)=c_1 e^{\lambda t}$$ $cx^2X''-bxX'-(ax+\lambda)X=0\quad$ is an ODE of Bessel kind, leading to : $$X(x)=c_2 x^{-(b+c)/(2c)}I_{\pm\nu}\left(2\sqrt{\frac{ax}{c}} \right)$$ $I_\nu$ is the modified Bessel function of first kind and order $\nu=\sqrt{\left(\frac{b+c}{c}\right)^2+4\frac{\lambda}{c}}$ .

A particular solution of EDP $(1)$ is : $$u=e^{\lambda t}x^{-(b+c)/(2c)}I_{\pm\nu}\left(2\sqrt{\frac{ax}{c}} \right) \tag 2$$ This particular solution depends on the parameter $\lambda$. The general solution is any linear combination of particular solutions with different $\lambda$. $$u(x,t)=\int f(\lambda)e^{\lambda t}x^{-(b+c)/(2c)}I_{\nu_{(\lambda)}} \left(2\sqrt{\frac{ax}{c}} \right)d\lambda +\int g(\lambda)e^{\lambda t}x^{-(b+c)/(2c)}I_{-\nu_{(\lambda)}}\left(2\sqrt{\frac{ax}{c}} \right)d\lambda$$ $f(\lambda)$ and $g(\lambda)$ are two arbitray functions.$\quad \nu_{(\lambda)}=\sqrt{\left(\frac{b+c}{c}\right)^2+4\frac{\lambda}{c}}$.

Note : The result from MAPLE reported by Mariusz Iwaniuk in comment is not the general solution : The variable $t$ doesn't appears in it. It corresponds to the particular case $\lambda=0$ of above Eq.$(2)$ , where the variable $2\sqrt{\frac{ax}{c}}$ becomes $2\sqrt{\frac{-ax}{c}}$ and thus the modified Bessel function becomes the Bessel function $J_\nu$.

Note : The arbitrary functions $f(\lambda)$ and $g(\lambda)$ have to be determined according to the boundary conditions (which are not specified in the wording of the question). This is in fact the most difficult part of the job. If some boundary conditions where specified, the method of characteristics might becomes easier than the above method of separation of variables, which in many cases leads afterwards to hard calculus in order to determine the arbitrary functions. All depends of the kind of boundary conditions.

• Could you pick some simple boundary conditions i.e.: $u(-5,t)=0, u(5,t)=0$, and $u(x,0)=1$ so we can compare the answer with a numerical solution? Also, this was sort of a bad question from my part. The original equation should be a sub-form of the Faynman-Kac formula (en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula). – Edv Beq Apr 5 '18 at 2:43