Symmetries of the Black-Scholes equation

We study the algebro–geometrical structure of the Black–Scholes equation.

The computation of the symmetry group of a partial differential equation has proven itself to be a very useful tool in Classical Mathematical Physics (see e.g. [Harrison-Estabrook 1971]), as well as in Euclidean Quantum Mechanics (see e.g. [Lescot-Zambrini 2004] and [Lescot-Zambrini 2008]). It was therefore natural to try and use the same method in Financial Mathematics.

After setting the general framework (§2), and performing some preliminary reductions (§3), we determine (§4) the isovectors for the Black-Scholes equation in a way broadly similar to the one used for the backward heat equation with potential term in the second of the two aforementioned joint papers with J.-C. Zambrini. Our computation turns out to suggest Black and Scholes’ original solution method ([Black-Scholes 1973]) of their equation; in particular, the quantities $r-\frac{{\sigma}^{2}}{2}$ and $r+\frac{{\sigma}^{2}}{2}$ appear naturally in this context. As corollaries, we determine (§5) the structure of the Lie algebra of the symmetry group of the equation, then (§6) we obtain some interesting transformations on the solutions.

We shall be concerned with the classical Black-Scholes equation :

$$\displaystyle\frac{\partial C}{\partial t}+\displaystyle\frac{1}{2}\sigma^{2}S^{2}\displaystyle\frac{\partial^{2}C}{\partial S^{2}}+rS\displaystyle\frac{\partial C}{\partial S}-rC=0\,\,(\mathcal{E})$$

for the price $C(t,S)$ of a call option with maturity $T$ and strike price $K$ on an underlying asset satisfying $S_{t}=S$ (see [Black-Scholes 1973], where $\sigma$ is denoted by $v$, $C$ by $w$, and $S$ by $x$). As is well–known ([Black-Scholes 1973], p.646), the same equation is satisfied by the price of a put option.

We assume $\sigma>0$, and define

$$\tilde{r}:=r-\displaystyle\frac{\sigma^{2}}{2}\,\,$$

and

$$\tilde{s}:=r+\displaystyle\frac{\sigma^{2}}{2}\,\,.$$

It is useful to remark that

$$\displaystyle\frac{\tilde{r}^{2}}{2\sigma^{2}}+r=\displaystyle\frac{\tilde{s}^{2}}{2\sigma^{2}}\,\,.$$

We intend to determine the isovectors for $(\mathcal{E})$, using the method applied, in [Harrison-Estabrook 1971], pp. 657–658 (see also [Lescot-Zambrini 2004], pp.189–192) to the heat equation, and in [Lescot-Zambrini 2008], §3, to the (backward) heat equation with a potential term.

Let us set $x=\ln(S)$ and

$$\varphi(t,x):=C(t,e^{x})=C(t,S)\,\,;$$

then $\varphi$ is defined on $\mathbf{R}_{+}\times\mathbf{R}$. One has

$$\displaystyle\frac{\partial C}{\partial S}=\displaystyle\frac{1}{S}\displaystyle\frac{\partial\varphi}{\partial x}\,\,,$$

$$\displaystyle\frac{\partial^{2}C}{\partial S^{2}}=-\displaystyle\frac{1}{S^{2}}\displaystyle\frac{\partial\varphi}{\partial x}+\displaystyle\frac{1}{S^{2}}\displaystyle\frac{\partial^{2}\varphi}{\partial x^{2}}$$

and

$$\displaystyle\frac{\partial C}{\partial t}=\displaystyle\frac{\partial\varphi}{\partial t}\,\,.$$

Equation $(\mathcal{E})$ is therefore equivalent to the following equation in $\varphi$ :

$$\displaystyle\frac{\partial\varphi}{\partial t}+\displaystyle\frac{\sigma^{2}}{2}(\displaystyle\frac{\partial^{2}\varphi}{\partial x^{2}}-\displaystyle\frac{\partial\varphi}{\partial x})+r\displaystyle\frac{\partial\varphi}{\partial x}-r\varphi=0\,\,(\mathcal{E}_{1})\,\,,$$

that is :

$$\displaystyle\frac{\partial\varphi}{\partial t}+\displaystyle\frac{\sigma^{2}}{2}\displaystyle\frac{\partial^{2}\varphi}{\partial x^{2}}+\tilde{r}\displaystyle\frac{\partial\varphi}{\partial x}-r\varphi=0\,\,(\mathcal{E}_{2})\,\,.$$

Let us set $A=\displaystyle\frac{\partial\varphi}{\partial x}$ and $B=\displaystyle\frac{\partial\varphi}{\partial t}$, and consider thenceforth $t$, $x$, $\varphi$, $A$ and $B$ as independent variables . Then $(\mathcal{E}_{2})$ is equivalent to the vanishing, on the five–dimensional manifold $M=\mathbf{R}_{+}\times\mathbf{R}^{4}$ of $(t,x,\varphi,A,B)$, of the following system of differential forms :

(3.1)

$\displaystyle\alpha=d\varphi-Adx-Bdt\,\,,$

(3.2)

$\displaystyle d\alpha=-dAdx-dBdt\,\,,$

and

(3.3)

$\displaystyle\beta=(B+\tilde{r}A-r\varphi)dxdt+\displaystyle\frac{1}{2}\sigma^{2}dAdt\,\,.$

Let $I$ denote the ideal of $\Lambda T^{*}(M)$ generated by $\alpha$, $d\alpha$ and $\beta$ ; as

(3.4)		$\displaystyle d\beta$	$\displaystyle=$	$\displaystyle(dB+\tilde{r}dA-rd\varphi)dxdt$
			$\displaystyle=$	$\displaystyle d\alpha(dx-\tilde{r}dt)+\alpha(-rdxdt)\in I\,\,,$

$I$ is a differential ideal of $\Lambda T^{*}(M)$. By definition (see [Harrison-Estabrook 1971]), an isovector for $(\mathcal{E}_{2})$ is a vector field

(3.5)

$\displaystyle N=N^{t}\displaystyle\frac{\partial}{\partial t}+N^{x}\displaystyle\frac{\partial}{\partial x}+N^{\varphi}\displaystyle\frac{\partial}{\partial\varphi}+N^{A}\displaystyle\frac{\partial}{\partial A}+N^{B}\displaystyle\frac{\partial}{\partial B}$

such that

(3.6)

$\displaystyle\mathcal{L}_{N}(I)\subseteq I\,\,.$

Using the formal properties of the Lie derivative ([Harrison-Estabrook 1971], p.654), one easily proves that the set $\mathcal{G}$ of these isovectors constitutes a Lie algebra (for the usual bracket of vector fields).

In order to determine $\mathcal{G}$, we may use a trick first explained in

[Harrison-Estabrook 1971], p.657, that applies in all situations in which there is only one $1$–form among the given generators of the ideal $I$ (see also [Lescot-Zambrini 2008], p.211).

Let $N\in\mathcal{G}$ ; as $\mathcal{L}_{N}(I)\subseteq I$, one has $\mathcal{L}_{N}(\alpha)\in I=<\alpha,d\alpha,\beta>$, whence there is a $0$–form (i.e. a function) $\lambda$ such that $\mathcal{L}_{N}(\alpha)=\lambda\alpha$. Let us define

(3.7)

$\displaystyle F:=N\rfloor\alpha=N^{\varphi}-AN^{x}-BN^{t}\,\,.$

This can be rewritten as

(3.8)

$\displaystyle\lambda\alpha=\mathcal{L}_{N}(\alpha)=N\rfloor d\alpha+d(N\rfloor\alpha)=N\rfloor d\alpha+dF\,\,,$

whence

(3.9)

$\displaystyle N\rfloor d\alpha=\lambda\alpha-dF\,\,,$

that is

(3.10)

$\displaystyle N\rfloor(-dAdx-dBdt)=\lambda\alpha-dF$

i.e.

	$\displaystyle-N^{A}dx+N^{x}dA-N^{B}dt+N^{t}dB$
		$\displaystyle=$	$\displaystyle\lambda\alpha-dF$
		$\displaystyle=$	$\displaystyle\lambda(d\varphi-Adx-Bdt)-dF\,\,.$

Whence (letters as lower indices indicating differentiation, as usual)

$$(*)\left\{\begin{array}[]{lr}-N^{B}=-\lambda B-F_{t}\\ -N^{A}=-\lambda A-F_{x}\\ 0=\lambda-F_{\varphi}\\ N^{x}=-F_{A}\\ N^{t}=-F_{B}\,\,.\end{array}\right.$$

Using the third equation, we can eliminate $\lambda$ and obtain

$$(**)\left\{\begin{array}[]{lr}N^{t}=-F_{B}\\ N^{x}=-F_{A}\\ N^{\varphi}=F-AF_{A}-BF_{B}\\ N^{A}=F_{x}+AF_{\varphi}\\ N^{B}=F_{t}+BF_{\varphi}\,\,.\end{array}\right.$$

Conversely, the existence of a function $F(t,x,\varphi,A,B)$ such that the above equations hold clearly implies that $\mathcal{L}_{N}(\alpha)=F_{\varphi}\alpha\in I$ ; but then

$$\mathcal{L}_{N}(d\alpha)=d(\mathcal{L}_{N}(\alpha))\in d(I)\subseteq I\,\,,$$

and there only remains to be satisfied the condition

$$\mathcal{L}_{N}(\beta)\in I\,\,.$$

The last condition in the previous paragraph can be stated as

(4.1)

$\displaystyle\mathcal{L}_{N}(\beta)=\rho\alpha+\xi d\alpha+\omega\beta\,\,,$

for $\rho$ a $1$–form, $\xi$ a $0$–form and $\omega$ a $0$–form. Let $D$ denote the coefficient of $d\varphi$ in $\rho$ ; replacing $\rho$ by $\rho-D\alpha$ (which doesn’t affect the validity of $(4.1)$ as $\alpha^{2}=0$), we may assume that $D=0$. Setting then

(4.2)

$\displaystyle\rho=R_{1}dt+R_{2}dx+R_{3}dA+R_{4}dB\,\,,$

(4.3)

$\displaystyle\xi=R_{5}\,\,,$

and

(4.4)

$\displaystyle\omega=R_{6}\,\,,$

we shall obtain a system of ten equations in $F$ ; we shall then eliminate $R_{1}$,…,$R_{6}$.

Identifying the coefficients of, in that order, $dtdx$, $dtd\varphi$, $dtdA$, $dtdB$, $dxd\varphi$, $dxdA$, $dxdB$, $d\varphi dA$, $d\varphi dB$ and $dAdB$ yields the following system :

(4.5)				$\displaystyle rN^{\varphi}-\tilde{r}N^{A}-N^{B}+(r\varphi-\tilde{r}A-B)N_{x}^{x}+(r\varphi-\tilde{r}A-B)N_{t}^{t}-\displaystyle\frac{1}{2}\sigma^{2}N_{x}^{A}$
			$\displaystyle=$	$\displaystyle-AR_{1}+BR_{2}-(B+\tilde{r}A-r\varphi)R_{6}$

(4.6)

$\displaystyle(r\varphi-\tilde{r}A-B)N_{\varphi}^{x}-\displaystyle\frac{1}{2}\sigma^{2}N_{\varphi}^{A}=R_{1}$

(4.7)

$\displaystyle(r\varphi-\tilde{r}A-B)N_{A}^{x}-\displaystyle\frac{1}{2}\sigma^{2}N_{A}^{A}-\displaystyle\frac{1}{2}\sigma^{2}N_{t}^{t}=BR_{3}-\displaystyle\frac{1}{2}\sigma^{2}R_{6}$

(4.8)

$\displaystyle(r\varphi-\tilde{r}A-B)N_{B}^{x}-\displaystyle\frac{1}{2}\sigma^{2}N_{B}^{A}=BR_{4}+R_{5}$

(4.9)

$\displaystyle(B+\tilde{r}A-r\varphi)N_{\varphi}^{t}=R_{2}$

(4.10)

$\displaystyle(B+\tilde{r}A-r\varphi)N_{A}^{t}-\displaystyle\frac{1}{2}\sigma^{2}N_{x}^{t}=AR_{3}+R_{5}$

(4.11)

$\displaystyle(B+\tilde{r}A-r\varphi)N_{B}^{t}=AR_{4}$

(4.12)

$\displaystyle-\displaystyle\frac{1}{2}\sigma^{2}N_{\varphi}^{t}=-R_{3}$

(4.13)

$\displaystyle 0=-R_{4}$

(4.14)

$\displaystyle\displaystyle\frac{1}{2}\sigma^{2}N_{B}^{t}=0\,\,.$

Equation $(4.13)$ gives $R_{4}=0$ ; then $(4.8)$ defines $R_{5}$. Equation $(4.14)$ is equivalent to $N_{B}^{t}=0$ ; if that be the case, then $(4.11)$ holds automatically. Now $(4.9)$ defines $R_{2}$, $(4.12)$ defines $R_{3}$, $(4.6)$ defines $R_{1}$ and $(4.7)$ defines $R_{6}$. We are left with equations $(4.5)$ and $(4.10)$ and the condition $N_{B}^{t}=0$.

Let us begin with the last mentioned ; it is equivalent to $F_{BB}=0$ : $F$ is affine in $B$, i.e. $F=c+Bd$, where $c$ and $d$ depend only on $(t,x,\varphi,A)$. Now $(**)$ can be rewritten as

$$(***)\left\{\begin{array}[]{lr}N^{t}=-d\\ N^{x}=-c_{A}-Bd_{A}\\ N^{\varphi}=c-Ac_{A}-ABd_{A}\\ N^{A}=c_{x}+Bd_{x}+Ac_{\varphi}+ABd_{\varphi}\\ N^{B}=c_{t}+Bd_{t}+Bc_{\varphi}+B^{2}d_{\varphi}\,\,.\end{array}\right.$$

From $(4.12)$ follows

(4.15)

$\displaystyle R_{3}=-\displaystyle\frac{1}{2}\sigma^{2}d_{\varphi}\,\,,$

and $(4.8)$ yields

(4.16)

$\displaystyle R_{5}=(B+\tilde{r}A-r\varphi)d_{A}-\displaystyle\frac{1}{2}\sigma^{2}(d_{x}+Ad_{\varphi})\,\,.$

Now we can rewrite $(4.10)$ as

			$\displaystyle(B+\tilde{r}A-r\varphi)(-d_{A})-\displaystyle\frac{1}{2}\sigma^{2}(-d_{x})$
		$\displaystyle=$	$\displaystyle(B+\tilde{r}A-r\varphi)d_{A}-\displaystyle\frac{1}{2}\sigma^{2}(d_{x}+Ad_{\varphi})\,\,.$

Comparing the coefficients of $B$ on both sides gives $-d_{A}=d_{A}$, that is $d_{A}=0$, i.e. $d$ depends only upon $(t,x,\varphi)$. Now $(4.17)$ becomes

(4.18)

$\displaystyle\displaystyle\frac{1}{2}\sigma^{2}d_{x}=-\displaystyle\frac{1}{2}\sigma^{2}(d_{x}+Ad_{\varphi})$

that is

(4.19)

$\displaystyle 2d_{x}=-Ad_{\varphi}\,\,.$

Differentiating with respect to $A$ leads to

(4.20)

$\displaystyle 0=2d_{Ax}=-d_{\varphi}\,\,,$

whence $d_{\varphi}=0$ and $d_{x}=-\displaystyle\frac{1}{2}Ad_{\varphi}=0$ : $d$ depends only upon $t$. Then $(***)$ becomes

$$(****)\left\{\begin{array}[]{lr}N^{t}=-d\\ N^{x}=-c_{A}\\ N^{\varphi}=c-Ac_{A}\\ N^{A}=c_{x}+Ac_{\varphi}\\ N^{B}=c_{t}+Bd_{t}+Bc_{\varphi}\,\,,\\ \end{array}\right.$$

the new unknowns being a function $c(t,x,\varphi,A)$ and a function $d(t)$, and we still have to satisfy equation $(4.5)$. Now $(4.9)$ implies $R_{2}=0$, and $(4.6)$ gives

(4.21)

$\displaystyle R_{1}=-(r\varphi-\tilde{r}A-B)c_{A\varphi}-\displaystyle\frac{1}{2}\sigma^{2}(c_{x\varphi}+Ac_{\varphi\varphi})\,\,.$

Equation $(4.7)$ now becomes

			$\displaystyle(r\varphi-\tilde{r}A-B)(-c_{AA})-\displaystyle\frac{1}{2}\sigma^{2}(c_{xA}+c_{\varphi}+Ac_{A\varphi})-\displaystyle\frac{1}{2}\sigma^{2}(-d_{t})$
		$\displaystyle=$	$\displaystyle-\displaystyle\frac{1}{2}\sigma^{2}R_{6}\,\,,$

that is :

(4.23)

$\displaystyle R_{6}=-d_{t}+c_{xA}+c_{\varphi}+Ac_{A\varphi}+\displaystyle\frac{2}{\sigma^{2}}(B+\tilde{r}A--r\varphi)(-c_{AA})\,\,.$

Eliminating $R_{1}$, $R_{2}$ and $R_{6}$ turns $(4.5)$ into

			$\displaystyle r(c-Ac_{A})-\tilde{r}(c_{x}+Ac_{\varphi})-(c_{t}+Bd_{t}+Bc_{\varphi})+(r\varphi-\tilde{r}A-B)(-c_{Ax})$
		$\displaystyle+$	$\displaystyle(r\varphi-\tilde{r}A-B)(-d_{t})-\displaystyle\frac{1}{2}\sigma^{2}(c_{xx}+Ac_{\varphi x})$
		$\displaystyle=$	$\displaystyle-A(-(r\varphi-\tilde{r}A-B)c_{A,\varphi}-\displaystyle\frac{1}{2}\sigma^{2}(c_{x\varphi}+Ac_{\varphi\varphi}))$
			$\displaystyle-(B+\tilde{r}A-r\varphi)(-d_{t}+c_{xA}+c_{\varphi}+Ac_{A\varphi}+\displaystyle\frac{2}{\sigma^{2}}(B+\tilde{r}A-r\varphi)(-c_{AA}))\,\,.$

Both members of $(4.24)$ are second–order polynomials in $B$ ; equating the coefficients of $B^{2}$ gives $c_{AA}=0$, whence $c$ is affine in $A$ : $c=e+Af$ with $e$ and $f$ functions of $(t,x,\varphi)$.

Equating the coefficients of $B$ yields

(4.25)

$\displaystyle-d_{t}-c_{\varphi}+c_{Ax}+d_{t}=-Ac_{A\varphi}-(-d_{t}+c_{Ax}+c_{\varphi}+Ac_{A\varphi})$

that is :

(4.26)

$\displaystyle d_{t}=2c_{Ax}+2Ac_{A\varphi}=2f_{x}+2Af_{\varphi}\,\,.$

Differentiating the last equality with respect to $A$ gives $f_{\varphi}=0$ (that is, $f$ is a function of $(t,x)$), and then we get that $d_{t}=2f_{x}$, i.e.

(4.27)

$\displaystyle f=\displaystyle\frac{1}{2}d^{{}^{\prime}}(t)x+\mu(t)$

for some function $\mu$ of $t$ alone.

Equating the constant terms in $B$ now gives us

			$\displaystyle re-\tilde{r}(e_{x}+Af_{x}+Ae_{\varphi})-(e_{t}+Af_{t})+(r\varphi-\tilde{r}A)(-f_{x})$
			$\displaystyle+(r\varphi-\tilde{r}A)(-d_{t})-\displaystyle\frac{1}{2}\sigma^{2}(e_{xx}+Af_{xx}+Ae_{\varphi x})$
		$\displaystyle=$	$\displaystyle A\displaystyle\frac{1}{2}\sigma^{2}(e_{x\varphi}+Ae_{\varphi\varphi})-(\tilde{r}A-r\varphi)(-d_{t}+f_{x}+e_{\varphi})\,\,.$

Both sides of equation $(4.28)$ are polynomials in $A$ with coefficients depending only upon $(t,x,\varphi)$. Identifying the terms in $A^{2}$ leads us to $e_{\varphi\varphi}=0$, whence $e=g+h\varphi$ with $g$ and $h$ depending only upon $(t,x)$. The unknowns are now $d$ and $\mu$ (functions of $t$ alone) and $g$ and $h$ (functions of $(t,x)$).

Identifying the coefficients of $A$ on both sides of $(4.29)$ yields

			$\displaystyle-\tilde{r}(f_{x}+e_{\varphi})-f_{t}+\tilde{r}f_{x}+\tilde{r}d_{t}-\displaystyle\frac{1}{2}\sigma^{2}(f_{xx}+e_{\varphi x})$
		$\displaystyle=$	$\displaystyle\displaystyle\frac{1}{2}\sigma^{2}e_{x\varphi}-\tilde{r}(-d_{t}+f_{x}+e_{\varphi})\,\,,$

that is

(4.30)

$\displaystyle-f_{t}-\displaystyle\frac{1}{2}\sigma^{2}f_{xx}=\sigma^{2}h_{x}-\tilde{r}f_{x}$

or

(4.31)

$\displaystyle\sigma^{2}h_{x}=\tilde{r}\displaystyle\frac{1}{2}d^{{}^{\prime}}(t)-\displaystyle\frac{1}{2}d^{{}^{\prime\prime}}(t)x-\mu^{{}^{\prime}}(t)\,\,,$

that is

(4.32)

$\displaystyle h(t,x)=\displaystyle\frac{\tilde{r}}{2\sigma^{2}}d^{{}^{\prime}}(t)x-\displaystyle\frac{d^{{}^{\prime\prime}}(t)}{4\sigma^{2}}x^{2}-\displaystyle\frac{\mu^{{}^{\prime}}(t)}{\sigma^{2}}x+k(t)\,\,,$

for some function $k$ of $t$ alone.

The constant term gives

(4.33)

$\displaystyle re-\tilde{r}e_{x}-e_{t}-r\varphi f_{x}-r\varphi d_{t}-\displaystyle\frac{1}{2}\sigma^{2}e_{xx}=r\varphi(-d_{t}+f_{x}+e_{\varphi})\,\,.$

According to the relation $d_{t}=2f_{x}$, this becomes

(4.34)

$\displaystyle rg-\tilde{r}g_{x}-\tilde{r}\varphi h_{x}-g_{t}-\varphi h_{t}-r\varphi d_{t}-\displaystyle\frac{1}{2}\sigma^{2}g_{xx}-\displaystyle\frac{1}{2}\sigma^{2}\varphi h_{xx}=0\,\,.$

Equating the constant term (in $\varphi$) of $(4.34)$ to zero gives that $g$ is a solution of $(\mathcal{E}_{2})$.

The value of the term in $\varphi$ means that

(4.35)

$\displaystyle-\tilde{r}h_{x}-h_{t}-rd_{t}-\displaystyle\frac{1}{2}\sigma^{2}h_{xx}=0\,\,,$

that is

	$\displaystyle-\tilde{r}(\displaystyle\frac{\tilde{r}}{2\sigma^{2}}d^{{}^{\prime}}(t)-\displaystyle\frac{d^{{}^{\prime\prime}}(t)}{2\sigma^{2}}x-\displaystyle\frac{\mu^{{}^{\prime}}(t)}{\sigma^{2}})$
	$\displaystyle-(\displaystyle\frac{\tilde{r}}{2\sigma^{2}}d^{{}^{\prime\prime}}(t)x-\displaystyle\frac{d^{{}^{\prime\prime\prime}}(t)}{4\sigma^{2}}x^{2}-\displaystyle\frac{\mu^{{}^{\prime\prime}}(t)}{\sigma^{2}}x+k^{{}^{\prime}}(t))-rd^{{}^{\prime}}(t)-\displaystyle\frac{1}{2}\sigma^{2}(-\displaystyle\frac{d^{{}^{\prime\prime}}(t)}{2\sigma^{2}})=0.$

This is a polynomial equation in $x$ with functions of $t$ as coefficients. Considering the coefficient of $x^{2}$ gives $d^{{}^{\prime\prime\prime}}(t)=0$, that is $d(t)=C_{1}t^{2}+C_{2}t+C_{3}$ for constants $C_{1}$, $C_{2}$ and $C_{3}$.

Equating the terms in $x$ leads to

(4.37)

$\displaystyle\displaystyle\frac{\tilde{r}}{2\sigma^{2}}d^{{}^{\prime\prime}}(t)-\displaystyle\frac{\tilde{r}}{2\sigma^{2}}d^{{}^{\prime\prime}}(t)+\displaystyle\frac{\mu^{{}^{\prime\prime}}(t)}{\sigma^{2}}=0\,\,,$

that is

(4.38)

$\displaystyle\mu^{{}^{\prime\prime}}(t)=0$

or

(4.39)

$\displaystyle\mu(t)=C_{4}t+C_{5}$

($C_{4}$, $C_{5}$ real constants).

We are left with the constant term in $x$ :

(4.40)

$\displaystyle-\displaystyle\frac{\tilde{r}^{2}}{2\sigma^{2}}d^{{}^{\prime}}(t)+\tilde{r}\displaystyle\frac{\mu^{{}^{\prime}}(t)}{\sigma^{2}}-k^{{}^{\prime}}(t)-rd^{{}^{\prime}}(t)+\displaystyle\frac{d^{{}^{\prime\prime}}(t)}{4}=0\,\,,$

or

	$\displaystyle k(t)$	$\displaystyle=$	$\displaystyle-(\displaystyle\frac{\tilde{r}^{2}}{2\sigma^{2}}+r)d(t)+\tilde{r}\displaystyle\frac{\mu(t)}{\sigma^{2}}+\displaystyle\frac{d^{{}^{\prime}}(t)}{4}+C_{6}$
		$\displaystyle=$	$\displaystyle-\displaystyle\frac{\tilde{s}^{2}}{2\sigma^{2}}d(t)+\tilde{r}\displaystyle\frac{\mu(t)}{\sigma^{2}}+\displaystyle\frac{d^{{}^{\prime}}(t)}{4}+C_{6}$

for some constant $C_{6}$. Therefore

	$\displaystyle f(t,x)$	$\displaystyle=$	$\displaystyle\displaystyle\frac{1}{2}d^{{}^{\prime}}(t)x+\mu(t)$
		$\displaystyle=$	$\displaystyle\displaystyle\frac{1}{2}x(2C_{1}t+C_{2})+C_{4}t+C_{5}$

and

	$\displaystyle h(t,x)$	$\displaystyle=$	$\displaystyle\displaystyle\frac{\tilde{r}}{2\sigma^{2}}x(2C_{1}t+C_{2})-\displaystyle\frac{C_{1}}{2\sigma^{2}}x^{2}$
		$\displaystyle-$	$\displaystyle\displaystyle\frac{C_{4}x}{\sigma^{2}}-\displaystyle\frac{\tilde{s}^{2}}{2\sigma^{2}}(C_{1}t^{2}+C_{2}t+C_{3})$
		$\displaystyle+$	$\displaystyle\tilde{r}\displaystyle\frac{(C_{4}t+C_{5})}{\sigma^{2}}+\displaystyle\frac{2C_{1}t+C_{2}}{4}+C_{6}\,\,.$

We have established

For $u$ a solution of $(\mathcal{E}_{2})$, let $N_{u}$ denote the isovector defined by $g=u$ and $C_{1}=...=C_{6}=0$, and, for $1\leq i\leq 6$, let $N_{i}$ denote the isovector defoned by by $g=0$, $C_{i}=1$ and $C_{j}=0$ for $j\neq i$.

The function $g$ and $h$ are determined by the isovector $N$ :

$$g=N^{\varphi}-\varphi\displaystyle\frac{\partial N^{\varphi}}{\partial\varphi}$$

and

$$h=\displaystyle\frac{\partial N^{\varphi}}{\partial\varphi}\,\,;$$

we shall denote them repectively by $g_{N}$ and $h_{N}$. As seen in §4, $h_{N}$ is a function of $(t,x)$.