Symmetries of the Black-Scholes-Merton equation for European options
L.N. Bakirova1, M.A. Shurygina2, V.V. Shurygin, jr.3

1 ICL Group, Kazan, Russia<EMAIL_ADDRESS>
2 Kazan National Research Technical University named after A.N. Tupolev–KAI,
Kazan, Russia<EMAIL_ADDRESS>
3 N.I. Lobachevskii Institute of Mathematics and Mechanics, Kazan Federal
University, Kazan, Russia<EMAIL_ADDRESS>corresponding author

###### Abstract

The aim of the present paper is the clarification of the result of A.
Paliathanasis, K. Krishnakumar, K.M. Tamizhmani and P.G.L. Leach on the
symmetry Lie algebra of the Black-Scholes-Merton equation for European
options.

Keywords: symmetries of PDEs, Black-Scholes-Merton equation.

2020 MSC: 35B06.

##  1 Main result

The Black-Scholes-Merton (BSM) [1, 2, 5] model is one of the most important
concepts in modern financial theory. It is used for the valuation of stock
options, taking into account the impact of time and other risk factors.

The classical BSM model is described by a second order PDE

$u_{t}+\frac{1}{2}\sigma^{2}x^{2}u_{xx}+rxu_{x}-ru=0.$

In recent years the number of papers is devoted to the determining the Lie
algebra of symmetries of some PDEs that generalize this model. First of all we
mention the papers of R.K. Gazizov, and N.H. Ibragimov [4] and O. Sinkala,
P.G.L. Leach, J. OHara [7]. The paper of Y. Bozhkov and S. Dimas [3] solves
the problem of group classification of the generalized BSM equation

$u_{t}+\frac{1}{2}\sigma^{2}x^{2}u_{xx}+rxu_{x}+f(u)=0,$

where $f(u)$ is an arbitrary smooth function.

A. Paliathanasis, K. Krishnakumar, K.M. Tamizhmani, P.G.L. Leach [6]
considered the BSM equation for European options with stochastic volatility
for which the premium term depends only upon the return-to-risk ratio. It has
the form

$\dfrac{1}{2}f^{2}(y)x^{2}u_{xx}+\rho\beta
xf(y)u_{xy}+\dfrac{1}{2}\beta^{2}u_{yy}+rxu_{x}+\left(\alpha(m-y)-\beta\rho\dfrac{\mu-r}{f(y)}\right)u_{y}-ru+u_{t}=0,$
(1)

where $t$, $x$, $y$ are independent variables, $u=u(t,x,y)$ is the value of
the option, $f(y)$ is an arbitrary smooth function, and $r$, $\rho$, $m$,
$\mu$, $\alpha$, $\beta$ are real parameters with $|\rho|<1$.

The authors calculate the Lie algebra of symmetries of this equation in the
case when $f(y)={\rm const}$, and claim that in the case of a non-constant
function $f(y)$ this Lie algebra is the direct sum of a three-dimensional
commutative Lie algebra and the infinite dimensional commutative Lie algebra
which corresponds to the solutions on the equation. It should be mentioned
that there are some misprints in their paper.

In the present paper we add some corrections to this result. We show that this
PDE admits additional symmetries also in the case when

$f(y)=\frac{k}{y-m},~{}~{}{k}={\rm const}.$

For this case we denote

$g=2\left(\alpha+\dfrac{\rho\beta(\mu-r)}{k}\right).$

The computations were performed using the Maple packages DifferentialGeometry
and JetCalculus by I.M. Anderson.

Theorem. For arbitrary function $f(y)$ the equation (1) admits the Lie
symmetries

$X_{1}=\dfrac{\partial}{\partial t},~{}~{}X_{2}=x\dfrac{\partial}{\partial
x},~{}~{}X_{3}=u\dfrac{\partial}{\partial
u},~{}~{}X_{b}=b(x,\,y,\,t)\dfrac{\partial}{\partial u},$

where $b(x,y,t)$ is the solution of (1).

Moreover, in the case when $f=f_{0}={\rm const}$, it admits three additional
symmetries

$\displaystyle X_{4}=e^{-\alpha t}\dfrac{\partial}{\partial y},$
$\displaystyle X_{5}=f_{0}^{2}\left(\rho^{2}+\alpha
t\right)x\dfrac{\partial}{\partial x}+f_{0}\rho\beta\dfrac{\partial}{\partial
y}+\dfrac{1}{2}\alpha\left(t(f_{0}^{2}-2r)+2\ln
x\right)u\dfrac{\partial}{\partial u},$ $\displaystyle X_{6}=2\beta
f_{0}^{2}\rho e^{\alpha t}x\dfrac{\partial}{\partial
x}+\beta^{2}f_{0}e^{\alpha t}\dfrac{\partial}{\partial y}-2e^{\alpha
t}\left(\alpha f_{0}(m-y)+\beta\rho(r-\mu)\right)u\dfrac{\partial}{\partial
u}.$

In the case $f(y)=\dfrac{k}{y-m}$ and $g\neq 0$ the equation (1) takes the
form

$\dfrac{1}{2}f^{2}(y)x^{2}u_{xx}+\rho\beta
xf(y)u_{xy}+\dfrac{1}{2}\beta^{2}u_{yy}+rxu_{x}+\frac{1}{2}g(m-y)u_{y}-ru+u_{t}=0$

and admits two additional symmetries

$X_{4}=e^{-gt}\left(rx\dfrac{\partial}{\partial
x}-\dfrac{1}{2}g(y-m)\dfrac{\partial}{\partial y}+\dfrac{\partial}{\partial
t}+ur\dfrac{\partial}{\partial u}\right),$
$X_{5}=e^{gt}\left(\dfrac{x}{\beta}\left(\rho gk+\beta
r\right)\dfrac{\partial}{\partial
x}+\dfrac{1}{2}g(y-m)\dfrac{\partial}{\partial y}+\dfrac{\partial}{\partial
t}+\dfrac{u}{2\beta^{2}}\left(g^{2}(y-m)^{2}+\beta^{2}(2r-g)\right)\dfrac{\partial}{\partial
u}\right).$

In the case $f(y)=\dfrac{k}{y-m}$ and $g=0$ the equation (1) takes the form

$\dfrac{1}{2}f^{2}(y)x^{2}u_{xx}+\rho\beta
xf(y)u_{xy}+\dfrac{1}{2}\beta^{2}u_{yy}+rxu_{x}-ru+u_{t}=0$

and admits two additional symmetries

$X_{4}=\dfrac{1}{2\beta}\left(\rho k+2\beta
rt\right)x\dfrac{\partial}{\partial x}+\dfrac{y-m}{2}\dfrac{\partial}{\partial
y}+t\dfrac{\partial}{\partial t}+rtu\dfrac{\partial}{\partial u},$
$X_{5}=\dfrac{xt}{\beta}\left(\rho k+\beta rt\right)\dfrac{\partial}{\partial
x}+t\left(y-m\right)\dfrac{\partial}{\partial
y}+t^{2}\dfrac{\partial}{\partial
t}+\dfrac{u}{2\beta^{2}}\left((y-m)^{2}+\beta^{2}(2rt^{2}-t)\right)\dfrac{\partial}{\partial
u}.$

##  2 Invariant solutions

In this section, we apply the Lie symmetries in order to reduce the equation
(1) and to construct the invariant solutions. To do this, we need to use
2-dimensional Lie subalgebras. As it is mentioned in [6], solutions in which
$u$ does not depend upon some of the independent variables, are not
interesting. Therefore, the reductions are performed with modified symmetry
vectors like $X_{1}+k_{1}X_{3}$, $X_{2}+k_{2}X_{3}$ and some other.

###  2.1 The case $f=f_{0}={\rm const}$

In this subsection we repeat the results of A. Paliathanasis et al. with minor
corrections. The Lie Brackets of the Lie algebra are given in the table.

$[X_{i},X_{j}]$ | $X_{1}$ | $X_{2}$ | $X_{3}$ | $X_{4}$ | $X_{5}$ | $X_{6}$
---|---|---|---|---|---|---
$X_{1}$ | 0 | 0 | 0 | $-\alpha X_{4}$ | $f_{0}^{2}\alpha X_{2}+\alpha(\frac{1}{2}f_{0}^{2}-r)X_{3}$ | $\alpha X_{6}$
$X_{2}$ | 0 | 0 | 0 | 0 | $\alpha X_{3}$ | 0
$X_{3}$ | 0 | 0 | 0 | 0 | 0 | 0
$X_{4}$ | $\alpha X_{4}$ | 0 | 0 | 0 | 0 | $2\alpha f_{0}X_{3}$
$X_{5}$ | $-f_{0}^{2}\alpha X_{2}-\alpha(\frac{1}{2}f_{0}^{2}-r)X_{3}$ | $-\alpha X_{3}$ | 0 | 0 | 0 | 0
$X_{6}$ | $-\alpha X_{6}$ | 0 | 0 | $-2\alpha f_{0}X_{3}$ | 0 | 0

Throughout this subsection we denote $Y_{1}=X_{1}+k_{1}X_{3}$,
$Y_{2}=X_{2}+k_{2}X_{3}$.

1) Algebra $\\{Y_{1},Y_{2}\\}$.

The invariant solution has the form $u(x,y,t)=e^{k_{1}t}x^{k_{2}}w(y)$, where
$w(y)$ satisfies the equation

$\beta^{2}f_{0}w^{\prime\prime}+2(\rho\beta
f_{0}^{2}k_{2}+\alpha(m-y)f_{0}+\beta\rho(r-\mu))w^{\prime}+((k_{2}-1)(f_{0}^{2}k_{2}+2r)+2k_{1})w=0.$

Its solution is expressed via Kummer functions $M$, $U$, so finally we get:

$u(x,y,t)=e^{k_{1}t}x^{k_{2}}(C_{1}M(\gamma,\tfrac{1}{2},z)+C_{2}U(\gamma,\tfrac{1}{2},z)),$

where

$z=\frac{(k_{2}\rho\beta f_{0}^{2}+\beta\rho(r-\mu)-\alpha
f_{0}(y-m))^{2}}{\alpha\beta^{2}f_{0}^{2}},~{}~{}\gamma=\frac{k_{2}(1-k_{2})f_{0}^{2}+2r(1-k_{2})-2k_{1}}{4\alpha}.$

2) Algebra $\\{Y_{2},X_{4}+kX_{3}\\}$.

The invariant solution has the form $u(x,y,t)=\exp(ke^{\alpha
t}y)x^{k_{2}}w(t)$, where $w(t)$ satisfies the equation

$2f_{0}w^{\prime}+\left(\beta^{2}k^{2}f_{0}e^{2\alpha
t}+2k\beta\rho\left(k_{2}f_{0}^{2}+\frac{\alpha
mf_{0}}{\rho\beta}-(\mu-r)\right)e^{\alpha
t}+(k_{2}-1)f_{0}(k_{2}f_{0}^{2}+2r)\right)w=0.$

Its solution is

$w(t)=C\exp\left(-\frac{\beta^{2}k^{2}}{4\alpha}e^{2\alpha
t}-\frac{k\beta\rho}{\alpha f_{0}}e^{\alpha
t}\left(f_{0}^{2}k_{2}+\frac{\alpha
f_{0}m}{\beta\rho}-\mu+r\right)-\frac{1}{2}t(k_{2}-1)(f_{0}^{2}k_{2}+2r)\right).$

3) Algebra $\\{Y_{2},X_{6}+kX_{3}\\}$.

The invariant solution has the form

$u(x,y,t)=\exp\left(\dfrac{\alpha}{\beta^{2}}y^{2}-\left(\dfrac{2\alpha
f_{0}m+ke^{-\alpha
t}}{\beta^{2}f_{0}}+\dfrac{2\rho(r-\mu+f_{0}^{2}k_{2})}{\beta
f_{0}}\right)y\right)x^{k_{2}}w(t),$

where $w(t)$ satisfies the equation

$2\beta^{2}f_{0}^{2}w^{\prime}+\left(f_{0}^{2}\beta^{2}(2\alpha+(k_{2}-1)(f_{0}^{2}k_{2}+2r))+2k(k_{2}\beta
f_{0}^{2}\rho+\alpha f_{0}m+\beta\rho(r-\mu))e^{-\alpha t}+k^{2}e^{-2\alpha
t}\right)w=0.$

Its solution is

$w(t)=C\exp\Bigl{(}\frac{1}{4\alpha\beta^{2}f_{0}^{2}}\bigl{(}k^{2}e^{-2\alpha
t}+4k(k_{2}\beta f_{0}^{2}\rho+\alpha f_{0}m+\beta\rho(r-\mu))e^{-\alpha
t}-\\\
-2t\alpha\beta^{2}f_{0}^{2}(2\alpha+(k_{2}-1)(f_{0}^{2}k_{2}+2r))\bigr{)}\Bigr{)}.$

4) Algebra $\\{X_{5},X_{4}+kX_{3}\\}$.

The invariant solution has the form

$u(x,y,t)=\exp\left(kye^{\alpha
t}+\frac{\alpha\ln^{2}x}{2f_{0}^{2}(\rho^{2}+\alpha
t)}\right)x^{\psi(t)}w(t),$

where

$\psi(t)=-\frac{2k\beta\rho f_{0}e^{\alpha
t}+(2r-f_{0}^{2})t\alpha}{2f_{0}^{2}(\rho^{2}+\alpha t)}.$

The function $w(t)$ is the solution of the linear first order ODE

$8f_{0}^{2}(\rho^{2}+\alpha t)^{2}w^{\prime}+\Bigl{(}4e^{2\alpha
t}k^{2}f_{0}^{2}\beta^{2}(\alpha^{2}t^{2}+\rho^{2}-\rho^{4})\,+\\\ +4e^{\alpha
t}\bigl{(}kf_{0}\alpha^{2}(\beta f_{0}^{2}\rho+2f_{0}\alpha
m-2\rho\beta\mu)t^{2}+k\alpha\rho^{2}f_{0}(\beta f_{0}^{2}\rho+4\alpha
f_{0}m+2\beta\rho(r-2\mu))t\,+\\\ +(2kf_{0}^{2}\alpha
m\rho^{4}-2k\beta\rho^{3}f_{0}(\rho^{2}(\mu-r)+r)-1)\bigr{)}-\\\
-\bigl{(}\alpha^{2}(2r+f_{0}^{2})^{2}t^{2}+2\alpha(f_{0}^{4}\rho^{2}+2f_{0}^{2}(2r\rho^{2}-\alpha)+4r^{2}\rho^{2})t+4f_{0}^{2}\rho^{2}(2r\rho^{2}-\alpha)\bigr{)}\Bigr{)}w=0.$

The final form of $w$ is too cumbersome to give it here.

5) Algebra $\\{X_{5},X_{6}\\}$.

The invariant solution has the form

$u(x,y,t)=x^{\psi(t)}\exp(\varphi(x,y,t))w(t),$

where we denoted

$\psi(t)=\frac{\bigl{(}t(r-\frac{1}{2}f_{0}^{2})\beta-2\rho
f_{0}(m-y)\bigr{)}\alpha+2\beta\rho^{2}(\mu-r)}{f_{0}^{2}(\rho^{2}-\alpha
t)\beta},$
$\varphi(x,y,t)=\alpha\beta^{2}\ln^{2}x-2yf_{0}\bigl{(}(\rho^{2}+\alpha
t)(2m-y)\alpha
f_{0}+\alpha\beta\rho(f_{0}^{2}-2\mu)t+2\rho^{3}\beta(r-\mu)\bigr{)}.$

The function $w(t)$ has the form

$w(t)=\frac{C\exp\left(\dfrac{\xi(t)}{\alpha\beta^{2}f_{0}^{2}(\alpha
t-\rho^{2})}\right)}{\sqrt{\alpha t-\rho^{2}}},$

where

$\xi(t)=\frac{1}{8}\Bigl{(}\alpha\beta^{2}(\alpha
t-\rho^{2})\bigl{(}(2r+f_{0}^{2})^{2}-8f_{0}^{2}\alpha\bigr{)}+\rho^{2}\bigl{(}\beta\rho(f_{0}^{2}-4\mu+2r)+4\alpha
f_{0}m\bigr{)}^{2}\Bigr{)}.$

###  2.2 The case $f=\dfrac{k}{y-m}$, $g\neq 0$

For convenience we will use
$g=2\left(\alpha+\dfrac{\rho\beta(\mu-r)}{k}\right)$ instead of $\alpha$. The
Lie Brackets of the Lie algebra are given in the table.

$[X_{i},X_{j}]$ | $X_{1}$ | $X_{2}$ | $X_{3}$ | $X_{4}$ | $X_{5}$
---|---|---|---|---|---
$X_{1}$ | 0 | 0 | 0 | $-gX_{4}$ | $gX_{5}$
$X_{2}$ | 0 | 0 | 0 | 0 | 0
$X_{3}$ | 0 | 0 | 0 | 0 | 0
$X_{4}$ | $gX_{4}$ | 0 | 0 | 0 | $X$
$X_{5}$ | $-gX_{5}$ | 0 | 0 | $-X$ | 0

where we denoted

$X=2gX_{1}+\Bigl{(}\frac{g^{2}\rho
k}{\beta}+2r\Bigr{)}X_{2}-\frac{1}{2}g(g-4r)X_{3}.$

1) Algebra $\\{X_{1}+aX_{3},X_{2}+bX_{3}\\}$.

The invariant solution has the form

$u(x,y,t)=e^{at}x^{b}w(y),$

where $w(y)$ satisfies the second order ODE

$\beta^{2}(y-m)^{2}w^{\prime\prime}+(2\beta\rho
bk-g(y-m)^{2})(y-m)w^{\prime}+\bigl{(}2(a+r(b-1))(y-m)^{2}+k^{2}b(b-1)\bigr{)}w=0.$

Its solution can be expressed via Whittaker functions $M_{\kappa\nu}$,
$W_{\kappa\nu}$:

$w(y)=\exp\left(\frac{g(y-m)^{2}}{4\beta^{2}}\right)(y-m)^{\gamma}\left(M_{\kappa\nu}\left(\frac{g(y-m)^{2}}{2\beta^{2}}\right)+W_{\kappa\nu}\left(\frac{g(y-m)^{2}}{2\beta^{2}}\right)\right),~{}~{}\gamma=-\frac{1}{2}+\frac{kb\rho}{\beta},$

and

$\kappa=\frac{(\beta+2kb\rho)g+4\beta(a+r(b-1))}{4g\beta},~{}~{}\nu=\frac{1}{4\beta}\sqrt{\beta^{2}+4k(k-\rho\beta)b-4k^{2}b^{2}(1-\rho^{2})}.$

2) Algebra $\\{X_{1}+aX_{3},X_{4}\\}$

The invariant solution has the form

$u(x,y,t)=e^{at}x^{1-a/r}w(h(t,y)),~{}~{}h(t,y)=x^{g}(y-m)^{2r}.$

The function $w(h)$ satisfies the second order Euler equation

$r^{2}(4\beta^{2}r^{2}+g^{2}k^{2}+4rkg\rho\beta)h^{2}w^{\prime\prime}-\\\
-r(k^{2}g(2a-g(r+1))+4r\rho\beta(a-g(r+1))k-2r^{2}\beta^{2}(2r-1))hw^{\prime}+ak^{2}(a-r)w=0.$

3) Algebra $\\{X_{2}+bX_{3},X_{4}\\}$

The invariant solution has the form

$u(x,y,t)=x^{b}(y-m)^{\gamma}w(h(t,y)),~{}~{}h(t,y)=e^{gt}(y-m)^{2},~{}~{}\gamma=\frac{2r(b-1)}{g}.$

The function $w(h)$ satisfies the second order Euler equation

$4\beta^{2}g^{2}h^{2}w^{\prime\prime}+2g\beta(g(2bk\rho+\beta)+4r\beta(b-1))hw^{\prime}+\\\
+(b(k^{2}g^{2}+4r^{2}\beta^{2}+4kgr\rho\beta)-2r\beta^{2}(2r+g))w=0.$

4) Algebra $\\{X_{2}+bX_{3},X_{5}+pX_{3}\\}$.

The invariant solution has the form

$u(x,y,t)=x^{b}(y-m)^{\gamma}\exp(\varphi(t,y))w(h(t,y)),$

where we denoted

$h(t,y)=(y-m)^{2}e^{-gt},~{}~{}\varphi(t,y)=-\frac{g^{2}y(2m-y)+2p\beta^{2}e^{-gt}}{2g\beta^{2}};~{}~{}\gamma=-\frac{(g-2r+2rb)\beta+2b\rho
kg}{g\beta}.$

The function $w(h)$ satisfies the second order ODE

$4g^{2}\beta^{2}h^{2}w^{\prime\prime}-2\beta g(g\beta+2b\rho kg+4\beta
r(b-1))hw^{\prime}+(2pg^{2}h+\\\
+6g\beta^{2}rb-8\beta^{2}r^{2}b+4\beta^{2}r^{2}b^{2}-6g\beta^{2}r+k^{2}b^{2}g^{2}-k^{2}bg^{2}+4b^{2}\rho\beta
kgr+\\\ +2g^{2}\beta^{2}+4\beta^{2}r^{2}+4g^{2}\beta b\rho k-4b\rho\beta
rkg)w=0.$

Its solution is

$w(h)=h^{\gamma}\left(C_{1}J_{n}\left(\frac{\sqrt{2ph}}{\beta}\right)+C_{2}Y_{n}\left(\frac{\sqrt{2ph}}{\beta}\right)\right),$

where

$\gamma=\frac{(4r(b-1)+3g)\beta+2bkg\rho}{4g\beta},~{}~{}n=\frac{1}{2\beta}\sqrt{(4(\rho^{2}-1)k^{2}b^{2}-4k(4\beta\rho-k)b+\beta^{2}}$

and $J_{n}$, $Y_{n}$ are the Bessel functions.

5) Algebra $\\{X_{2}+aX_{1},X_{4}\\}$.

The invariant solution has the form

$u(x,y,t)=(y-m)^{-2r/g}w(h(x,y,t)),~{}~{}h(x,y,t)=e^{gt}x^{-ag}(y-m)^{2(1-ar)}.$

The function $w(h)$ satisfies the second order Euler equation

$g^{2}((4\beta^{2}r^{2}+4kgr\rho\beta+k^{2}g^{2})a^{2}-4\beta(gk\rho+2\beta
r)a+4\beta^{2})h^{2}w^{\prime\prime}+\\\
+g(g(4\beta^{2}r^{2}+4kgr\rho\beta+k^{2}g^{2})a^{2}+(k^{2}g^{2}+8\beta^{2}r^{2}+4kgr\rho\beta-6gr\beta^{2}-4kg^{2}\rho\beta)a+2g\beta^{2})hw^{\prime}+\\\
+2r\beta^{2}(2r+g)w=0.$

6) Algebra $\\{X_{2}+aX_{1},X_{5}\\}$.

The invariant solution has the form

$u(x,y,t)=(y-m)^{-1+2r/g}\exp\left(\frac{gy(y-2m)}{2\beta^{2}}\right)w(h(x,y,t)),$

where

$h(x,y,t)=\frac{(y-m)^{\gamma}e^{\beta gt}}{x^{a\beta
g}},~{}~{}\gamma=2(agk\rho+ar\beta-\beta).$

The function $w(h)$ satisfies the second order Euler equation

$g^{4}\beta^{2}(a^{2}(4r^{2}\beta^{2}+4rgk\beta\rho+k^{2}g^{2})-4a\beta(gk\rho+2\beta
r)+4\beta^{2})h^{2}w^{\prime\prime}+\\\
+g^{3}\beta(4g(ar-1)^{2}\beta^{3}+2(ar-1)(2ag^{2}k\rho+4r-3g)\beta^{2}+agk(akg^{2}+4\rho(r-g))\beta+k^{2}g^{2}a)hw^{\prime}+\\\
+(g-r)(g-2r)w=0.$

###  2.3 The case $f=\dfrac{k}{y-m}$, $g=0$

The Lie Brackets of the Lie algebra are given in the table.

$[X_{i},X_{j}]$ | $X_{1}$ | $X_{2}$ | $X_{3}$ | $X_{4}$ | $X_{5}$
---|---|---|---|---|---
$X_{1}$ | 0 | 0 | 0 | $X_{1}+rX_{2}+rX_{3}$ | $-\frac{1}{2}X_{3}+2X_{4}$
$X_{2}$ | 0 | 0 | 0 | 0 | 0
$X_{3}$ | 0 | 0 | 0 | 0 | 0
$X_{4}$ | $-X_{1}-rX_{2}-rX_{3}$ | 0 | 0 | 0 | $X_{5}$
$X_{5}$ | $\frac{1}{2}X_{3}-2X_{4}$ | 0 | 0 | $-X_{5}$ | 0

1) Algebra $\\{X_{1}+aX_{3},X_{2}+bX_{3}\\}$.

The invariant solution has the form

$u(x,y,t)=e^{at}x^{b}w(y),$

where $w(y)$ satisfies the second order ODE

$\beta^{2}(y-m)^{2}w^{\prime\prime}+2\beta\rho
bk(y-m)w^{\prime}+\bigl{(}2(a+r(b-1))(y-m)^{2}+k^{2}b(b-1)\bigr{)}w=0.$

Its solution can be expressed via Whittaker functions:

$w(y)=(y-m)^{-bk\rho/\beta}(C_{1}M_{0,\nu}(z)+C_{2}W_{0,\nu}(z)),$

where

$z=\frac{2\sqrt{(2-2b)r-2a}(y-m)}{\beta},~{}~{}\nu=\frac{\sqrt{4b(1-b+b\rho^{2})k^{2}+\beta(\beta-4bk\rho)}}{2\beta}.$

2) Algebra $\\{X_{2}+bX_{3},X_{4}\\}$.

The invariant solution has the form

$u(x,y,t)=x^{b}(y-m)^{\gamma}\exp(\varphi(t,y))w(h(t,y)),$

where we denoted

$h(t,y)=\frac{t}{(y-m)^{2}},~{}~{}\varphi(t,y)=\frac{r(b-1)(2m-y)ty}{(y-m)^{2}},~{}~{}\gamma=-\frac{b\rho
k}{\beta}.$

The function $w(h)$ satisfies the second order ODE

$4\beta^{2}h^{2}w^{\prime\prime}+2(4rm^{2}\beta^{2}(b-1)h^{2}+3\beta^{2}h+1)w^{\prime}+\\\
+\bigl{(}4m^{4}r^{2}\beta^{2}(b-1)^{2}h^{2}+6rm^{2}\beta^{2}(b-1)h+(k^{2}b+2m^{2}r)(b-1)+bk\rho(\beta-
bk\rho)\bigr{)}w=0.$

Its solution can be expressed via Kummer functions:

$w(h)=e^{-m^{2}hr(b-1)}h^{-\gamma}\left(C_{1}M\left(\gamma,2\gamma+\frac{1}{2},\frac{1}{2\beta^{2}h}\right)+C_{1}U\left(\gamma,2\gamma+\frac{1}{2},\frac{1}{2\beta^{2}h}\right)\right),$

where

$\gamma=\frac{\beta+\sqrt{4(\rho^{2}-1)k^{2}b^{2}+4k(k-4\rho\beta)b+\beta^{2}}}{4\beta}.$

3) Algebra $\\{X_{2}+bX_{3},X_{5}\\}$.

The invariant solution has the form

$u(x,y,t)=x^{b}(y-m)^{\gamma}\exp(\varphi(t,y))w(h(t,y)),$

where we denoted

$h(t,y)=\frac{t}{y-m},~{}~{}\varphi(t,y)=\frac{y(y-m)}{2\beta
t}-\frac{r(b-1)ty}{y-m},~{}~{}\gamma=-\frac{1}{2}-\frac{b\rho k}{\beta}.$

The function $w(h)$ satisfies the second order ODE

$4\beta^{4}h^{4}w^{\prime\prime}-4\beta^{2}h^{2}(2\beta^{2}rm(b-1)h^{2}-3\beta^{2}h+m)w^{\prime}+\bigl{(}4\beta^{4}r^{2}m^{2}(b-1)^{2}h^{4}-12\beta^{4}rm(b-1)h^{3}+\\\
+\beta^{2}(4kb(k(b-1)-kb\rho^{2}-\rho\beta)+4m^{2}r(b-1)+3\beta^{2})h^{2}-2\beta^{2}mh+m^{2}\bigr{)}w=0.$

Its solution is

$w(h)=(C_{1}h^{\lambda_{1}}+C_{2}h^{\lambda_{2}})\exp\left(\frac{m(2r\beta^{2}(b-1)h^{2}-1)}{\beta^{2}h}\right),$

where $\lambda_{1}$, $\lambda_{2}$ are the roots of the quadratic equation

$\lambda^{2}+2\lambda+\frac{1}{4\beta^{2}}\bigl{(}4(1-\rho^{2})b^{2}k^{2}+3\beta^{2}-4k^{2}b+4bk\beta\rho\bigr{)}=0.$

4) Algebra $\\{X_{4},X_{5}\\}$.

The invariant solution has the form

$u=x\sqrt{rt}\exp\left(\frac{(y-m)^{2}}{2\beta^{2}t}\right)(y-m)^{-1-k\rho/\beta}w(h),$

where $h(t,x,y)=k\rho\ln(y-m)+\beta(rt-\ln x)$, and the function $w(h)$
satisfies the second order ODE with the constant coefficients

$k^{2}\beta^{2}(1-\rho^{2})w^{\prime\prime}+k\beta(k+\beta\rho-2k\rho^{2})w^{\prime}+(\beta+k\rho)(2\beta-k\rho)w=0.$

Author Contributions: Vadim V. Shurygin, jr. determined the problem and the
methods for the solution, checked the results and wrote the paper. Landysh N.
Bakirova calculated the Lie algebra of symmetries. Marina A. Shurygina
computed the invariant solutions.

Conflicts of Interest: The authors declare no conflict of interest.

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