**2. Generalizations of the Black-Scholes equation in the multidimensional case: (initial) boundary value problems**

Consider in **R**<sup>1</sup> *<sup>t</sup>* <sup>×</sup> **<sup>R</sup>***<sup>n</sup> <sup>x</sup>* the following generalization of the Black-Scholes equation:

$$Lu = u\_t + \sum\_{i,j=1}^n a\_{ij} \mathbf{x}\_i \mathbf{x}\_j u\_{\mathbf{x}\_i \mathbf{x}\_j} + \sum\_{i=1}^n b\_i \mathbf{x}\_i u\_{\mathbf{x}\_i} + cu = f(t, \mathbf{x}),\tag{1}$$

where 0 ≤ *t* ≤ *T* and *xj* ≥ 0, 1 ≤ *j* ≤ *n*.

This is the Cauchy problem:

2 Will-be-set-by-IN-TECH

are provided - among others - in [22] for the different types of standard barrier options, in [16] for simultaneous 'down' and 'up' barriers with exponential dependence on time, in [10] for two boundaries via Laplace transform, in [12] and [7] for partial barrier and rainbow options, in [17] for multi-asset options with an outside barrier, in [5] in a most comprehensive setting employing the image solution method. Many analytical formulas for barrier options

For analytical tractability most literature assumes that the barrier hitting is monitored in continuous time. However there exist some works dealing with the discrete version, i.e. barrier crossing is allowed only at some specific dates -typically at daily closings. (See [1] and [15], for a survey). Furthermore, a recent literature relaxes the Brownian motion assumption and considers a more general Lévy framework. For example, [4] study barrier options of European type assuming that the returns of the underlying asset follows a Lévy process from a wide class. They employ the Wiener-Hopf factorization method and elements of pseudodifferential calculus to solve the related boundary problem. This book chapter adopts a classical Black-Scholes framework. The problem of pricing barrier options is reducible to boundary value problems for a PDE of Black-Scholes type and with pre-specified boundaries. The value at the terminal time *T* is assigned, specifying the terminal payoff which is paid provided that an 'in' option is knocked in or an 'out' option is not knocked out during its lifetime. The option holder may be entitled or not to a rebate. From a mathematical point of view, the boundary condition can be inhomogeneous or homogeneous. While there are several types of barrier options, in this work we will focus on 'up' barriers in view of the relationships between the prices of different types of vanilla options (see [25]). Moreover, the case of floating barriers of exponential form can be easily accommodated by substitution of the relevant parameters (see [25], Chapter 11), thus we confine ourselves to the case of constant barriers. On the other hand, we work within a general framework that allows for multi-asset options, a generic payoff and rebate. Furthermore, we tackle some regularity questions and the problem of existence of generalized solutions. In Section 2 the (initial) boundary value problem is studied in a multidimensional framework generalizing the Black-Scholes equation and analytical solutions are obtained, while a comparison principle is provided in Section 4. Section 3 presents some applications in Finance: our general setting incorporates several known pricing expressions and, at the same time, allows to generate new valuation formulas. Section 5 and the Appendix study the existence and regularity of generalized solutions to the boundary value problems for a class of PDEs incorporating the Black-Scholes type. We build on the approach of Oleinik and Radkeviˇc and adapt the method to the PDEs of interest in the

**2. Generalizations of the Black-Scholes equation in the multidimensional**

*aijxixjuxixj* +

*<sup>x</sup>* the following generalization of the Black-Scholes equation:

*n* ∑ *i*=1

*bixiuxi* + *cu* = *f*(*t*, *x*), (1)

are collected also in handbooks (see [11], for example).

financial applications.

Consider in **R**<sup>1</sup>

**case: (initial) boundary value problems**

*Lu* = *ut* +

*n* ∑ *i*,*j*=1

*<sup>t</sup>* <sup>×</sup> **<sup>R</sup>***<sup>n</sup>*

$$\begin{cases} L\mu = f(t, \mathbf{x}), \\ \mu|\_{t=T} = \mu\_0(\mathbf{x}) \end{cases} \tag{2}$$
 
$$\begin{cases} x\_j \ge 0, & 1 \le j \le n \\ 0 \le t \le T \end{cases}$$

and this is the boundary value problem:

$$\begin{cases} \operatorname{Lu} = f \\ \operatorname{u}|\_{t=T} = \operatorname{u}\_0(\mathbf{x}) \\ \operatorname{u}|\_{\mathbf{x}\_j = a\_j} = g\_j(t, \mathbf{x})|\_{\mathbf{x}\_j = a\_{j'}}, 1 \le j \le n, \end{cases} \tag{3}$$

$$\begin{cases} 0 \le t \le T \\ 0 \le x\_j \le a\_{j'}, a\_j > 0 \\ 1 \le j \le n \end{cases}$$

In (1) *aij* = *aji* = *const*, *bi* = *const*, *c* = *const* and

$$\sum\_{\substack{i,j=1\\i,j=1}}^n a\_{i\overline{j}} \mathfrak{F}\_i \mathfrak{F}\_j \ge c\_0 |\mathfrak{F}|^2,\\ c\_0 = const > 0. \tag{4}$$

Our first step is to make in the non-hypoelliptic PDE *L* the change of the space variables:

$$\{y\_j = \ln \mathbf{x}\_j, 1 \le j \le n, \tau = T - t \Rightarrow \frac{\partial u}{\partial t} = -\frac{\partial u}{\partial \tau'}, y\_j \in \mathbf{R}^1\tag{5}$$

*∂u <sup>∂</sup>xi* <sup>=</sup> *<sup>e</sup>*−*yi <sup>∂</sup><sup>u</sup> ∂yi* , *<sup>∂</sup>*2*<sup>u</sup> <sup>∂</sup>xi∂xj* <sup>=</sup> *<sup>e</sup>*−*yi*−*yj* [ *<sup>∂</sup>*2*<sup>u</sup> <sup>∂</sup>yi∂yj* <sup>−</sup> *<sup>δ</sup>ij ∂u ∂yi* ], *δij* being the Kronecker symbol.

Thus, (1) takes the form:

$$\tilde{L}u = -\frac{\partial u}{\partial \tau} + \sum\_{i,j=1}^{n} a\_{ij} \frac{\partial^2 u}{\partial y\_i \partial y\_j} + \sum\_{i=1}^{n} \frac{\partial u}{\partial y\_i} (b\_i - a\_{ii}) + cu = f,\tag{6}$$

i.e.

$$\frac{\partial u}{\partial \tau} = \sum\_{i,j=1}^n a\_{ij} \frac{\partial^2 u}{\partial y\_i \partial y\_j} + \sum\_{i=1}^n \tilde{b}\_i \frac{\partial u}{\partial y\_i} + cu - f;\\ \tilde{b}\_i = b\_i - a\_{ii}$$

In the case (2) we have

$$\begin{cases} \tilde{\mathcal{L}}u = f, & 0 \le \tau \le T \\ u|\_{\tau=0} = \tilde{u}\_0(y) = u\_0(e^{y\_1}, \dots, e^{y\_n}), \ y \in \mathbb{R}^n, \end{cases} \tag{7}$$

while in the case (3)

$$\begin{cases} \tilde{\mathcal{L}}u = f\_{\prime} & 0 \le \tau \le T \\ u|\_{\tau=0} = \tilde{u}\_{0}(y) \\ u|\_{y\_{j}=\mathfrak{a}\_{j}} = \mathcal{g}\_{j}|\_{y\_{j}=\mathfrak{a}\_{j}} \end{cases} \tag{8}$$

#### 4 Will-be-set-by-IN-TECH 34 Risk Management – Current Issues and Challenges Boundary-Value Problems for Second Order PDEs Arising in Risk Management and Cellular Neural Networks Approach <sup>5</sup>

Denote *<sup>D</sup>* <sup>=</sup> {<sup>0</sup> <sup>≤</sup> *<sup>τ</sup>* <sup>≤</sup> *<sup>T</sup>*, <sup>−</sup><sup>∞</sup> <sup>&</sup>lt; *yj* <sup>≤</sup> *lnaj* <sup>=</sup> *<sup>a</sup>*˜*j*, 1 <sup>≤</sup> *<sup>j</sup>* <sup>≤</sup> *<sup>n</sup>*}, *xj* <sup>=</sup> *<sup>e</sup>yj* , 1 <sup>≤</sup> *<sup>j</sup>* <sup>≤</sup> *<sup>n</sup>* <sup>⇒</sup> *<sup>f</sup>*(*t*, *<sup>x</sup>*) = *<sup>f</sup>*(*<sup>T</sup>* <sup>−</sup> *<sup>τ</sup>*,*ey*<sup>1</sup> ,...,*eyn* ).

In (6) we make the change of the unknown function *<sup>u</sup>* : *<sup>u</sup>* <sup>=</sup> *<sup>v</sup>*(*τ*, *<sup>y</sup>*)*e*<sup>∑</sup> *<sup>α</sup>iyi*+*βτ* in (*τ*, *<sup>y</sup>*) <sup>∈</sup> *<sup>D</sup>*. Thus, after standard computations we get:

$$
\sigma v\_{\tau} + \beta v = \sum\_{i,j} a\_{ij} v\_{y\_i y\_j} + \sum\_{i,j} a\_{ij} (a\_i v\_{y\_j} + a\_j v\_{y\_i}) + \tag{9}
$$

$$\left[ + \sum\_{i=1}^{n} \tilde{b}\_{i} v\_{\mathcal{Y}\_{i}} + \sum\_{i,j=1}^{n} a\_{ij} a\_{i} a\_{j} v + \sum\_{i=1}^{n} \tilde{b}\_{i} a\_{i} v + \mathcal{c}v - f e^{-\sum a\_{l} y\_{l} - \beta \tau} \right]$$

Let us take

$$\beta = \sum\_{i,j} a\_{ij} \alpha\_i \alpha\_j + \sum\_i \tilde{b}\_i \alpha\_i + c \tag{10}$$

*In* being the unit matrix. Put now *<sup>C</sup>* = (*B*−1)<sup>∗</sup> <sup>⇒</sup> *<sup>C</sup>*<sup>∗</sup> <sup>=</sup> *<sup>B</sup>*−1. Then *<sup>C</sup>*∗*AC* <sup>=</sup> *In* <sup>⇒</sup>

*<sup>v</sup>*|*τ*=<sup>0</sup> <sup>=</sup> *<sup>v</sup>*0((*C*−1)∗*z*) <sup>≡</sup> *<sup>v</sup>*˜0(*z*), *<sup>z</sup>* <sup>∈</sup> **<sup>R</sup>***n*.

� **R***<sup>n</sup>*

�*π*(*<sup>τ</sup>* <sup>−</sup> <sup>Θ</sup>)]*<sup>n</sup> <sup>e</sup>*

*<sup>i</sup>*=1(*zi* <sup>−</sup> *<sup>λ</sup>i*)<sup>2</sup> (see [6] or [21]). Going back to the old coordinates (*τ*, *x*) and the old function *u* = *ve*<sup>∑</sup> *<sup>α</sup>iyi*+*βτ*, we find *<sup>u</sup>*(*t*, *<sup>x</sup>*)-the solution of (2); *<sup>t</sup>* <sup>=</sup> *<sup>T</sup>* <sup>−</sup> *<sup>τ</sup>*, *yj* <sup>=</sup> *lnxj*, *<sup>z</sup>* <sup>=</sup> *<sup>B</sup>*−1*<sup>y</sup>* <sup>=</sup> *<sup>B</sup>*−1(*lnx*1,..., *lnxn*); *<sup>u</sup>* <sup>=</sup>

**Remark 1.** To simplify the things, consider the quadratic form (elliptic) *Q* = *a*11*ξ*<sup>2</sup> + 2*a*12*ξη* +

12

� <sup>√</sup>*a*<sup>11</sup> <sup>√</sup>*a*<sup>12</sup>

<sup>0</sup> <sup>√</sup>*<sup>b</sup>*

leads to *<sup>Q</sup>* <sup>=</sup> *<sup>x</sup>*<sup>2</sup> <sup>+</sup> *<sup>y</sup>*2. Moreover, the first quadrant *<sup>ξ</sup>* <sup>≥</sup> 0, *<sup>η</sup>* <sup>≥</sup> 0 is transformed under the linear

�

� � � � �

Let us now consider the boundary value problem (8). The above-proposed procedure yields:

*a*<sup>11</sup>

and *l*<sup>2</sup> :

*ξ <sup>η</sup>* ),( *ξ <sup>η</sup>* )).

*a*<sup>11</sup>

, *D*−<sup>1</sup> =

*x* = <sup>√</sup>*a*<sup>12</sup> *<sup>a</sup>*<sup>11</sup> *η*

*<sup>y</sup>* <sup>=</sup> <sup>√</sup>

*<sup>a</sup>*<sup>11</sup> > 0. The change

� � *ξ*

*η*

⎛ ⎝ √ 1 *<sup>a</sup>*<sup>11</sup> <sup>−</sup> *<sup>a</sup>*<sup>12</sup> *a*<sup>11</sup> √*b*

*bη* ≥ 0

0 √ 1 *b* ⎞

⎠ into angle between

with opening *ϕ*0. Evidently,

�

˜ *f*1(Θ, *λ*)

*f*1(*τ*, *z*), 0 ≤ *τ* ≤ *T*

<sup>−</sup> <sup>|</sup>*z*−*λ*<sup>|</sup> 2

<sup>−</sup> <sup>|</sup>*z*−*λ*<sup>|</sup> 2 <sup>4</sup>(*τ*−Θ)*dλd*Θ,

<sup>4</sup>*<sup>τ</sup> dλ*+ (15)

Boundary-Value Problems for Second Order PDEs Arising in Risk Management and Cellular Neural Networks Approach

*v*˜0(*λ*)*e*

(14)

35

(16)

This way the change *<sup>y</sup>* = (*C*−1)∗*<sup>z</sup>* <sup>⇒</sup> *<sup>z</sup>* <sup>=</sup> *<sup>B</sup>*−1*<sup>y</sup>* transforms the Cauchy problem (12) to:

*k*=1 *∂*2*u ∂z*<sup>2</sup> *k* .

The solution of the Cauchy problem (14) is given by the formula

*<sup>v</sup>*(*τ*, *<sup>z</sup>*) = <sup>1</sup> (2 <sup>√</sup>*πτ*)*<sup>n</sup>*

> � *τ* 0

[2

<sup>12</sup> − *a*11*a*<sup>22</sup> < 0, *Q* = (*A*(

⎧ ⎨ ⎩

*∂v ∂τ* <sup>=</sup> <sup>∑</sup>*<sup>n</sup> k*=1 *∂*2*v ∂z*<sup>2</sup> *k* + ˜

+ � **R***<sup>n</sup>*

<sup>2</sup> = ∑*<sup>n</sup>*

*<sup>a</sup>*<sup>11</sup> (*a*11*<sup>ξ</sup>* <sup>+</sup> *<sup>a</sup>*12*η*)<sup>2</sup> <sup>+</sup> *<sup>b</sup>η*2; *<sup>b</sup>* <sup>=</sup> *<sup>a</sup>*<sup>22</sup> <sup>−</sup> *<sup>a</sup>*<sup>2</sup>

� *x*

� � � � �

*x* ≥ 0 *y* = 0

Consequently, the transformation *D* is not orthogonal for *a*<sup>12</sup> �= 0.

*y*

� =

� <sup>√</sup>*a*<sup>11</sup> <sup>√</sup>*a*<sup>12</sup>

<sup>0</sup> <sup>√</sup>*<sup>b</sup>*

*<sup>B</sup>*−1*A*(*B*−1)<sup>∗</sup> <sup>=</sup> *In* <sup>⇒</sup> *Mu* <sup>=</sup> <sup>∑</sup>*<sup>n</sup>*

*<sup>z</sup>* <sup>∈</sup> **<sup>R</sup>***n*, *<sup>λ</sup>* <sup>∈</sup> **<sup>R</sup>***<sup>n</sup>* ⇒ |*<sup>z</sup>* <sup>−</sup> *<sup>λ</sup>*<sup>|</sup>

*<sup>n</sup> eβ*(*T*−*t*).

*a*22*η*2, *a*<sup>11</sup> > 0, *a*<sup>22</sup> > 0, *a*<sup>2</sup>

transformation with matrix *D* =

the rays (straight lines ) *l*<sup>1</sup> :

(*D*−1)∗*AD*−<sup>1</sup> = *I*2.

Then *Q* = <sup>1</sup>

We shall concentrate now on (3), *n* = 2.

*vxα*<sup>1</sup> <sup>1</sup> ... *<sup>x</sup>α<sup>n</sup>*

and put *<sup>f</sup>*<sup>1</sup> <sup>=</sup> <sup>−</sup> *f e*<sup>−</sup> <sup>∑</sup>*<sup>i</sup> <sup>α</sup>iyi*−*βτ*. Put *<sup>A</sup>* = (*aij*)*<sup>n</sup> <sup>i</sup>*,*j*=1, *A*<sup>∗</sup> = *A*, *α* = (*α*1,..., *αn*). Then the scalar product (*Aα*, ∇*yv*) = ∑*i*,*<sup>j</sup> aijα<sup>j</sup> ∂v <sup>∂</sup>yi* = <sup>∑</sup>*i*,*<sup>j</sup> ajiα<sup>i</sup> ∂v <sup>∂</sup>yj* = <sup>∑</sup>*i*,*<sup>j</sup> aijα<sup>i</sup> ∂v ∂yj* , i.e. we assume that

$$2(A\mathfrak{a}\_{\prime}\nabla\_{y}v) + (\tilde{b}\_{\prime}\nabla\_{y}v) = 0 \iff \tag{11}$$

$$2A\mathfrak{a} + \tilde{b} = 0,$$

where ˜ *b* = (˜ *b*1,..., ˜ *bn*) is given, *detA* �= 0.

In conclusion we solve the algebraic system (11): *<sup>α</sup>* <sup>=</sup> <sup>−</sup><sup>1</sup> <sup>2</sup> *<sup>A</sup>*−1(˜ *b*) and then we define *β* by (10). This way (9) takes the form:

$$w\_{\tau} = \sum\_{i,j=1}^{n} a\_{ij} v\_{y\_iy\_j} + f\_1(\tau, y) \tag{12}$$

The Cauchy problem (12) has initial condition

$$v\_0(y) = v|\_{\tau=0} = \mathfrak{i}\_0(y)e^{-\sum a\_i y\_i}; \mathfrak{i}\_0 \equiv u\_0(e^{y\_1}, \dots, e^{y\_n}), y \in \mathbb{R}^n.$$

To find a formula (Poisson type) for the solution of the Cauchy problem (12), *v*|*τ*=<sup>0</sup> = *v*0(*y*) we must use some auxiliary results from the linear algebra. So let *Mu* = ∑*<sup>n</sup> <sup>i</sup>*,*j*=<sup>1</sup> *aijvyiyj* . Then the change of the independent variables *<sup>y</sup>* <sup>=</sup> *Bz* ⇐⇒ *<sup>z</sup>* <sup>=</sup> *<sup>B</sup>*−1*y*, *<sup>B</sup>*−<sup>1</sup> = (*βli*)*<sup>n</sup> <sup>l</sup>*,*i*=<sup>1</sup> leads to *∂*2 *<sup>∂</sup>yi∂yj* <sup>=</sup> <sup>∑</sup>*<sup>n</sup> <sup>k</sup>*,*l*=<sup>1</sup> *βliβkj ∂*2 *∂zk∂zl* , i.e.

$$M\mathfrak{u} = \sum\_{k,l=1}^n \left( \sum\_{i} (\sum\_{j} a\_{ij} \beta\_{kj}) \beta\_{li} \right) \frac{\partial^2 \mathfrak{u}}{\partial z\_k \partial z\_l}.$$

One can easily guess that ∑*i*(∑*<sup>j</sup> aijβkj*)*βli* = *c*˜*kl* are the elements of the matrix *B*−1*A*(*B*−1)<sup>∗</sup> and of course (*B*−1)<sup>∗</sup> = (*B*∗)−1. On the other hand consider the elliptic quadratic form (*Ax*, *x*) = (*C*∗*ACy*, *y*) after the nondegenerate change *x* = *Cy*. As we know one can find such a matrix *C* that

$$\mathbf{C}^\* A \mathbf{C} = I\_{\mathbf{n} \nu} \tag{13}$$

#### 34 Risk Management – Current Issues and Challenges Boundary-Value Problems for Second Order PDEs Arising in Risk Management and Cellular Neural Networks Approach <sup>5</sup> 35 Boundary-Value Problems for Second Order PDEs Arising in Risk Management and Cellular Neural Networks Approach

*In* being the unit matrix. Put now *<sup>C</sup>* = (*B*−1)<sup>∗</sup> <sup>⇒</sup> *<sup>C</sup>*<sup>∗</sup> <sup>=</sup> *<sup>B</sup>*−1. Then *<sup>C</sup>*∗*AC* <sup>=</sup> *In* <sup>⇒</sup> *<sup>B</sup>*−1*A*(*B*−1)<sup>∗</sup> <sup>=</sup> *In* <sup>⇒</sup> *Mu* <sup>=</sup> <sup>∑</sup>*<sup>n</sup> k*=1 *∂*2*u ∂z*<sup>2</sup> *k* .

This way the change *<sup>y</sup>* = (*C*−1)∗*<sup>z</sup>* <sup>⇒</sup> *<sup>z</sup>* <sup>=</sup> *<sup>B</sup>*−1*<sup>y</sup>* transforms the Cauchy problem (12) to:

$$\begin{cases} \frac{\partial v}{\partial \tau} = \sum\_{k=1}^{n} \frac{\partial^2 v}{\partial z\_k^2} + \vec{f}\_1(\tau, z), & 0 \le \tau \le T \\\ v|\_{\tau=0} = v\_0((C^{-1})^\* z) \equiv \vec{v}\_0(z), \; z \in \mathbb{R}^n. \end{cases} \tag{14}$$

The solution of the Cauchy problem (14) is given by the formula

$$v(\tau, z) = \frac{1}{(2\sqrt{\pi \tau})^n} \int\_{\mathbf{R}^n} \vec{v}\_0(\lambda) e^{-\frac{|z-\lambda|^2}{4\tau}} d\lambda + \tag{15}$$

$$+ \int\_{\mathbf{R}^n} \int\_0^\tau \frac{\vec{f}\_1(\Theta, \lambda)}{[2\sqrt{\pi(\tau-\Theta)}]^n} e^{-\frac{|z-\lambda|^2}{4(\tau-\Theta)}} d\lambda d\Theta\_\tau$$

*<sup>z</sup>* <sup>∈</sup> **<sup>R</sup>***n*, *<sup>λ</sup>* <sup>∈</sup> **<sup>R</sup>***<sup>n</sup>* ⇒ |*<sup>z</sup>* <sup>−</sup> *<sup>λ</sup>*<sup>|</sup> <sup>2</sup> = ∑*<sup>n</sup> <sup>i</sup>*=1(*zi* <sup>−</sup> *<sup>λ</sup>i*)<sup>2</sup> (see [6] or [21]).

Going back to the old coordinates (*τ*, *x*) and the old function *u* = *ve*<sup>∑</sup> *<sup>α</sup>iyi*+*βτ*, we find *<sup>u</sup>*(*t*, *<sup>x</sup>*)-the solution of (2); *<sup>t</sup>* <sup>=</sup> *<sup>T</sup>* <sup>−</sup> *<sup>τ</sup>*, *yj* <sup>=</sup> *lnxj*, *<sup>z</sup>* <sup>=</sup> *<sup>B</sup>*−1*<sup>y</sup>* <sup>=</sup> *<sup>B</sup>*−1(*lnx*1,..., *lnxn*); *<sup>u</sup>* <sup>=</sup> *vxα*<sup>1</sup> <sup>1</sup> ... *<sup>x</sup>α<sup>n</sup> <sup>n</sup> eβ*(*T*−*t*).

We shall concentrate now on (3), *n* = 2.

4 Will-be-set-by-IN-TECH

Denote *<sup>D</sup>* <sup>=</sup> {<sup>0</sup> <sup>≤</sup> *<sup>τ</sup>* <sup>≤</sup> *<sup>T</sup>*, <sup>−</sup><sup>∞</sup> <sup>&</sup>lt; *yj* <sup>≤</sup> *lnaj* <sup>=</sup> *<sup>a</sup>*˜*j*, 1 <sup>≤</sup> *<sup>j</sup>* <sup>≤</sup> *<sup>n</sup>*}, *xj* <sup>=</sup> *<sup>e</sup>yj* , 1 <sup>≤</sup> *<sup>j</sup>* <sup>≤</sup> *<sup>n</sup>* <sup>⇒</sup> *<sup>f</sup>*(*t*, *<sup>x</sup>*) =

In (6) we make the change of the unknown function *<sup>u</sup>* : *<sup>u</sup>* <sup>=</sup> *<sup>v</sup>*(*τ*, *<sup>y</sup>*)*e*<sup>∑</sup> *<sup>α</sup>iyi*+*βτ* in (*τ*, *<sup>y</sup>*) <sup>∈</sup> *<sup>D</sup>*.

*i*,*j*

*n* ∑ *i*=1 ˜

*aijαiα<sup>j</sup>* + ∑

*∂v*

2*Aα* + ˜

To find a formula (Poisson type) for the solution of the Cauchy problem (12), *v*|*τ*=<sup>0</sup> = *v*0(*y*)

One can easily guess that ∑*i*(∑*<sup>j</sup> aijβkj*)*βli* = *c*˜*kl* are the elements of the matrix *B*−1*A*(*B*−1)<sup>∗</sup> and of course (*B*−1)<sup>∗</sup> = (*B*∗)−1. On the other hand consider the elliptic quadratic form (*Ax*, *x*) = (*C*∗*ACy*, *y*) after the nondegenerate change *x* = *Cy*. As we know one can find such a matrix

*i* ˜

*<sup>∂</sup>yj* = <sup>∑</sup>*i*,*<sup>j</sup> aijα<sup>i</sup>*

*b* = 0,

*aij*(*αivyj* + *αjvyi*

*biαiv* <sup>+</sup> *cv* <sup>−</sup> *f e*<sup>−</sup> <sup>∑</sup> *<sup>α</sup>iyi*−*βτ*.

*∂v ∂yj*

<sup>2</sup> *<sup>A</sup>*−1(˜

; *<sup>u</sup>*˜0 <sup>≡</sup> *<sup>u</sup>*0(*ey*<sup>1</sup> ,...,*<sup>e</sup>*

*aijβkj*)*βli*) *<sup>∂</sup>*2*<sup>u</sup>*

*∂zk∂zl* .

*C*∗*AC* = *In*, (13)

)+ (9)

*biα<sup>i</sup>* + *c* (10)

, i.e. we assume that

*b*) and then we define *β* by (10).

*<sup>i</sup>*,*j*=<sup>1</sup> *aijvyiyj*

. Then

*<sup>l</sup>*,*i*=<sup>1</sup> leads to

*<sup>i</sup>*,*j*=1, *A*<sup>∗</sup> = *A*, *α* = (*α*1,..., *αn*). Then the scalar

*b*, ∇*yv*) = 0 ⇐⇒ (11)

*aijvyiyj* + *f*1(*τ*, *y*) (12)

*yn* ), *<sup>y</sup>* <sup>∈</sup> **<sup>R</sup>***n*.

*aijvyiyj* + ∑

*aijαiαjv* +

*<sup>f</sup>*(*<sup>T</sup>* <sup>−</sup> *<sup>τ</sup>*,*ey*<sup>1</sup> ,...,*eyn* ).

Let us take

where ˜

*∂*2 *<sup>∂</sup>yi∂yj* <sup>=</sup> <sup>∑</sup>*<sup>n</sup>*

*C* that

*b* = (˜

Thus, after standard computations we get:

+ *n* ∑ *i*=1 ˜ *bivyi* +

and put *<sup>f</sup>*<sup>1</sup> <sup>=</sup> <sup>−</sup> *f e*<sup>−</sup> <sup>∑</sup>*<sup>i</sup> <sup>α</sup>iyi*−*βτ*. Put *<sup>A</sup>* = (*aij*)*<sup>n</sup>*

The Cauchy problem (12) has initial condition

*∂*2 *∂zk∂zl*

, i.e.

product (*Aα*, ∇*yv*) = ∑*i*,*<sup>j</sup> aijα<sup>j</sup>*

*b*1,..., ˜

This way (9) takes the form:

*<sup>k</sup>*,*l*=<sup>1</sup> *βliβkj*

*v<sup>τ</sup>* + *βv* = ∑

*n* ∑ *i*,*j*=1

*∂v*

*bn*) is given, *detA* �= 0. In conclusion we solve the algebraic system (11): *<sup>α</sup>* <sup>=</sup> <sup>−</sup><sup>1</sup>

*<sup>v</sup>*0(*y*) = *<sup>v</sup>*|*τ*=<sup>0</sup> <sup>=</sup> *<sup>u</sup>*˜0(*y*)*e*<sup>−</sup> <sup>∑</sup>*<sup>i</sup> <sup>α</sup>iyi*

*Mu* =

*i*,*j*

*β* = ∑ *i*,*j*

*<sup>∂</sup>yi* = <sup>∑</sup>*i*,*<sup>j</sup> ajiα<sup>i</sup>*

<sup>2</sup>(*Aα*, <sup>∇</sup>*yv*)+(˜

*v<sup>τ</sup>* =

we must use some auxiliary results from the linear algebra. So let *Mu* = ∑*<sup>n</sup>*

*n* ∑ *k*,*l*=1 (∑ *i* (∑ *j*

the change of the independent variables *<sup>y</sup>* <sup>=</sup> *Bz* ⇐⇒ *<sup>z</sup>* <sup>=</sup> *<sup>B</sup>*−1*y*, *<sup>B</sup>*−<sup>1</sup> = (*βli*)*<sup>n</sup>*

*n* ∑ *i*,*j*=1 **Remark 1.** To simplify the things, consider the quadratic form (elliptic) *Q* = *a*11*ξ*<sup>2</sup> + 2*a*12*ξη* + *a*22*η*2, *a*<sup>11</sup> > 0, *a*<sup>22</sup> > 0, *a*<sup>2</sup> <sup>12</sup> − *a*11*a*<sup>22</sup> < 0, *Q* = (*A*( *ξ <sup>η</sup>* ),( *ξ <sup>η</sup>* )).

Then *Q* = <sup>1</sup> *<sup>a</sup>*<sup>11</sup> (*a*11*<sup>ξ</sup>* <sup>+</sup> *<sup>a</sup>*12*η*)<sup>2</sup> <sup>+</sup> *<sup>b</sup>η*2; *<sup>b</sup>* <sup>=</sup> *<sup>a</sup>*<sup>22</sup> <sup>−</sup> *<sup>a</sup>*<sup>2</sup> 12 *<sup>a</sup>*<sup>11</sup> > 0. The change

$$
\begin{pmatrix} \chi \\ \chi \end{pmatrix} = \begin{pmatrix} \sqrt{a\_{11}} \ \frac{a\_{12}}{\sqrt{a\_{11}}} \\ 0 \ \sqrt{b} \end{pmatrix} \begin{pmatrix} \xi \\ \eta \end{pmatrix} \tag{16}
$$

leads to *<sup>Q</sup>* <sup>=</sup> *<sup>x</sup>*<sup>2</sup> <sup>+</sup> *<sup>y</sup>*2. Moreover, the first quadrant *<sup>ξ</sup>* <sup>≥</sup> 0, *<sup>η</sup>* <sup>≥</sup> 0 is transformed under the linear transformation with matrix *D* = � <sup>√</sup>*a*<sup>11</sup> <sup>√</sup>*a*<sup>12</sup> *a*<sup>11</sup> <sup>0</sup> <sup>√</sup> *b* � , *D*−<sup>1</sup> = ⎛ ⎝ √ 1 *<sup>a</sup>*<sup>11</sup> <sup>−</sup> *<sup>a</sup>*<sup>12</sup> *a*<sup>11</sup> √*b* 0 √ 1 *b* ⎞ ⎠ into angle between the rays (straight lines ) *l*<sup>1</sup> : � � � � � *x* ≥ 0 *y* = 0 and *l*<sup>2</sup> : � � � � � *x* = <sup>√</sup>*a*<sup>12</sup> *<sup>a</sup>*<sup>11</sup> *η <sup>y</sup>* <sup>=</sup> <sup>√</sup> *bη* ≥ 0 with opening *ϕ*0. Evidently, (*D*−1)∗*AD*−<sup>1</sup> = *I*2.

Consequently, the transformation *D* is not orthogonal for *a*<sup>12</sup> �= 0.

Let us now consider the boundary value problem (8). The above-proposed procedure yields:

#### 6 Will-be-set-by-IN-TECH 36 Risk Management – Current Issues and Challenges Boundary-Value Problems for Second Order PDEs Arising in Risk Management and Cellular Neural Networks Approach <sup>7</sup>

$$\begin{cases} v\_{\tau} = \sum\_{i,j=1}^{2} a\_{ij} v\_{y\_i y\_j} + f\_1(\tau, y) \\ v|\_{\tau=0} = v\_0(y) = u\_0(e^{y\_1}, e^{y\_2}) e^{-\sum a\_i y\_i} \\ v|\_{y\_1=0\_1} = g\_1(T - \tau, e^{y\_1}, e^{y\_2})|\_{y\_1=\vec{a}\_1} e^{-\beta\tau} a\_1^{-a\_1} e^{-a\_2 y\_2} \equiv \tilde{g}\_1(\tau, y\_2) \\ v|\_{y\_2=\vec{a}\_2} = g\_2(T - \tau, e^{y\_1}, e^{y\_2})|\_{y\_2=\vec{a}\_2} e^{-\beta\tau} a\_2^{-a\_2} e^{-a\_1 y\_1} \equiv \tilde{g}\_2(\tau, y\_1) \end{cases} \tag{17}$$
 
$$\begin{cases} -\infty < y\_j < \ln a\_j = \vec{a}\_j \\ 0 \le \tau \le T \end{cases}$$

initial-boundary value problem for (19) with unknown function *w*(*τ*,*r*, *ϕ*):

*<sup>∂</sup>r*<sup>2</sup> <sup>+</sup> <sup>1</sup> *r ∂w <sup>∂</sup><sup>r</sup>* <sup>+</sup> <sup>1</sup> *r*2 *∂*2*w*

*g*˜1(*τ*,*r*)

*g*˜2(*τ*,*r*)

*r* ≥ 0, 0 ≤ *ϕ* ≤ *ϕ*0, *l*<sup>1</sup> : {*ϕ* = 0,*r* ≥ 0}, *l*<sup>2</sup> : {*ϕ* = *ϕ*0,*r* ≥ 0}, *r* ↔ *ξ*, *ϕ* ↔ *η*, 0 ≤ Θ ≤ *τ*,

*w*|*τ*=<sup>0</sup> = *w*0(*r*, *ϕ*)

*∂ϕ*<sup>2</sup> + *f*(*τ*,*r*, *ϕ*)

*∂η <sup>G</sup>*(*r*, *<sup>ϕ</sup>*, *<sup>ξ</sup>*, *<sup>η</sup>*, *<sup>τ</sup>* <sup>−</sup> <sup>Θ</sup>)]*η*=0*dξd*Θ<sup>−</sup>

*∂η <sup>G</sup>*(*r*, *<sup>ϕ</sup>*, *<sup>ξ</sup>*, *<sup>η</sup>*, *<sup>τ</sup>* <sup>−</sup> <sup>Θ</sup>)]*η*=*ϕ*<sup>0</sup> *<sup>d</sup>ξd*Θ<sup>+</sup>

*<sup>ϕ</sup>*<sup>0</sup> *<sup>ϕ</sup>sin <sup>n</sup><sup>π</sup>*

*dz* <sup>−</sup> (*z*<sup>2</sup> <sup>+</sup> *<sup>ν</sup>*2)*<sup>w</sup>* <sup>=</sup> 0, *<sup>ν</sup>* <sup>≥</sup> 0, (22)

<sup>0</sup> *w*0(*ξ*, *η*)*G*(*r*, *ϕ*, *ξ*, *η*, *τ*)*ξdξdη* = *w*0(*r*, *ϕ*), i.e.

*w*0(*ξ*, *η*)*G*(*r*, *ϕ*, *ξ*, *η*, *τ*)*ξdξdη*,

*f*(Θ, *ξ*, *η*)*G*(*r*, *ϕ*, *ξ*, *η*, *τ* − Θ)*ξdξdηd*Θ+ (21)

Boundary-Value Problems for Second Order PDEs Arising in Risk Management and Cellular Neural Networks Approach

*<sup>ϕ</sup>*<sup>0</sup> *η* and the modified Bessel

(20)

37

(23)

*<sup>w</sup><sup>τ</sup>* = *<sup>∂</sup>*2*<sup>w</sup>*

*<sup>w</sup>*|*ϕ*=<sup>0</sup> <sup>=</sup> ˜

*<sup>w</sup>*|*ϕ*=*ϕ*<sup>0</sup> <sup>=</sup> ˜

⎧ ⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

0

� ∞ 0 ˜ *g*˜1(Θ, *ξ*)

� ∞ 0 ˜ *g*˜2(Θ, *ξ*)

> + � *ϕ*<sup>0</sup> 0

� *ϕ*<sup>0</sup> 0

� ∞ 0

> 1 *ξ* [ *∂*

1 *ξ* [ *∂*

<sup>4</sup>*<sup>τ</sup>* ∑<sup>∞</sup>

*dw*

<sup>+</sup> × [0, *ϕ*0]), **R**<sup>+</sup> = {*ξ* ≥ 0}. *G* is the corresponding Green function.

� *ϕ*<sup>0</sup> 0 � ∞

formally *limτ*→+0*ξG*(*r*, *ϕ*, *ξ*, *η*, *τ*) = *δ*(*r* − *ξ*, *ϕ* − *η*) in the sense of Schwartz distributions

Formula (21) is given in [21], pages 182 and 166 or in [6], pp.498. The proof of (21) is based on

**Remark 3.** In the special case when *a*<sup>12</sup> = 0 in (16) we obtain (18) and after the change *τ* = *τ*,

*v*˜0(*z*)

*g*˜1(*τ*, *z*2)

*g*˜2(*τ*, *z*1)

*<sup>n</sup>*=<sup>1</sup> *I <sup>n</sup><sup>π</sup> ϕ*0 ( *rξ* <sup>2</sup>*<sup>τ</sup>* )*sin <sup>n</sup><sup>π</sup>*

� ∞ 0

*<sup>ϕ</sup>*0*<sup>τ</sup> <sup>e</sup>*<sup>−</sup> (*r*2+*ξ*2)

*<sup>z</sup>*<sup>2</sup> *<sup>d</sup>*2*<sup>w</sup> dz*<sup>2</sup> <sup>+</sup> *<sup>z</sup>*

the properties of the Bessel functions and Hankel transform.

� � � � � � � � � � � � *∂* ˜ *v*˜ *∂τ* <sup>=</sup> *<sup>∂</sup>*<sup>2</sup> ˜ *v*˜ *∂z*<sup>2</sup> 1 + *<sup>∂</sup>*<sup>2</sup> ˜ *v*˜ *∂z*<sup>2</sup> 2 + ˜˜ *f*1(*τ*, *z*)

˜ *<sup>v</sup>*˜|*τ*=<sup>0</sup> <sup>=</sup> ˜

˜ *<sup>v</sup>*˜|*z*1=<sup>0</sup> <sup>=</sup> ˜

˜ *<sup>v</sup>*˜|*z*2=<sup>0</sup> <sup>=</sup> ˜

0 ≤ *ξ* ≤ ∞, 0 ≤ *η* ≤ *ϕ*0, 0 < *ϕ*<sup>0</sup> < *π*. Then

+ � *τ* 0

− � *τ* 0

function *w* = *Iν*(*z*) satisfies the equation:

( *z* <sup>2</sup> )<sup>2</sup>*m*+*<sup>ν</sup> <sup>m</sup>*!Γ(*m*+*ν*+1) (see [2]).

**Remark 2.** One can see that *limτ*→+<sup>0</sup>

*<sup>λ</sup><sup>j</sup>* <sup>=</sup> <sup>√</sup>*ajjzj*, 1 <sup>≤</sup> *<sup>j</sup>* <sup>≤</sup> 2 (18) takes the form:

where *G*(*r*, *ϕ*, *ξ*, *η*, *τ*) = <sup>1</sup>

*Iν*(*z*) = ∑<sup>∞</sup>

*D*� (**R**<sup>1</sup> *m*=0

*<sup>w</sup>*(*τ*,*r*, *<sup>ϕ</sup>*) = � *<sup>τ</sup>*

The change � *λ<sup>j</sup>* = *a*˜*<sup>j</sup>* − *yj* ≥ 0, *j* = 1, 2 *<sup>τ</sup>* <sup>=</sup> *<sup>τ</sup>* in (17) yields: ⎧ ⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎩ *v*˜*<sup>τ</sup>* = ∑<sup>2</sup> *<sup>i</sup>*,*j*=<sup>1</sup> *aijv*˜*λiλ<sup>j</sup>* <sup>+</sup> ˜ *f*1(*τ*, *λ*) *<sup>v</sup>*˜|*τ*=<sup>0</sup> <sup>=</sup> *<sup>v</sup>*˜0(*λ*) = *<sup>v</sup>*0(*a*˜1 <sup>−</sup> *<sup>λ</sup>*1, *<sup>a</sup>*˜2 <sup>−</sup> *<sup>λ</sup>*2)*e*<sup>−</sup> <sup>∑</sup><sup>2</sup> *<sup>i</sup>*=<sup>1</sup> *αi*(*a*˜*i*−*λi*) *v*˜|*λ*1=<sup>0</sup> = *g*˜1(*τ*, *a*˜2 − *λ*2) *v*˜|*λ*2=<sup>0</sup> = *g*˜2(*τ*, *a*˜1 − *λ*1), (18)

<sup>Ω</sup> <sup>=</sup> {<sup>0</sup> <sup>≤</sup> *<sup>τ</sup>* <sup>≤</sup> *<sup>T</sup>*, *<sup>λ</sup><sup>j</sup>* <sup>≥</sup> 0, *<sup>j</sup>* <sup>=</sup> 1, 2}, <sup>Ω</sup> is a wedge with opening *<sup>π</sup>* 2 .

Now we use the linear transformation described in Remark 1, that maps the first quadrant *λ*<sup>1</sup> ≥ 0, *λ*<sup>2</sup> ≥ 0 onto the angle between the rays *l*<sup>1</sup> and *l*<sup>2</sup> in the plane 0*z*1*z*<sup>2</sup> and we obtain:

$$\begin{cases} w\_{\tau} = w\_{z\_1 z\_1} + w\_{z\_2 z\_2} + f(\tau, z) \\ w|\_{\tau=0} = w\_0(z) \\ w|\_{z\_1=0} = \tilde{g}\_1(\tau, z\_1), (\tau, z) \in \tilde{\Omega} \\ w|\_{l\_2} = \tilde{g}\_2(\tau, z\_1, z\_2)|\_{(z\_1, z\_2)} \in l\_{2\prime} \end{cases} \tag{19}$$

*l*<sup>1</sup> : � *<sup>z</sup>*<sup>1</sup> <sup>=</sup> <sup>0</sup> *z*<sup>2</sup> = √ *λ*2 *b* , *l*<sup>2</sup> : � *<sup>z</sup>*<sup>1</sup> = <sup>√</sup> *λ*1 *a*<sup>11</sup> *z*<sup>2</sup> = √−*<sup>a</sup>*<sup>12</sup> *ba*<sup>11</sup> *λ*1 , <sup>Ω</sup>˜ is a wedge with opening *<sup>ϕ</sup>*0, i.e. <sup>Ω</sup>˜ = [0, *<sup>T</sup>*] <sup>×</sup> <sup>Γ</sup>, <sup>Γ</sup> being the interior of the angle between *l*1, *l*2.

In fact, *<sup>λ</sup>* <sup>=</sup> *Bz* ⇐⇒ *<sup>z</sup>* <sup>=</sup> *<sup>B</sup>*−1*<sup>λ</sup>* and *<sup>B</sup>*−1*A*(*B*−1)<sup>∗</sup> <sup>=</sup> *<sup>I</sup>*<sup>2</sup> implies that <sup>∑</sup><sup>2</sup> *<sup>i</sup>*,*j*=<sup>1</sup> *aij ∂*2 *∂λi∂λ<sup>j</sup>* is transformed in *<sup>∂</sup>*<sup>2</sup> *∂z*<sup>2</sup> 1 + *<sup>∂</sup>*<sup>2</sup> *∂z*<sup>2</sup> 2 . According to Remark 1: (*D*−1)∗*AD*−<sup>1</sup> = *I*2. Taking *B*−<sup>1</sup> = (*D*−1)∗, i.e. *B* = *D*<sup>∗</sup> we obtain that {*λ*<sup>1</sup> ≥ 0, *λ*<sup>2</sup> ≥ 0} is mapped onto the angle *ϕ*<sup>0</sup> between the rays *l*1, *l*2. Of course, there are three possibilities: *ϕ*<sup>0</sup> = *<sup>π</sup>* <sup>2</sup> , 0 <sup>&</sup>lt; *<sup>ϕ</sup>*<sup>0</sup> <sup>&</sup>lt; *<sup>π</sup>* <sup>2</sup> , *<sup>π</sup>* <sup>2</sup> < *ϕ*<sup>0</sup> < *π*.

From now on we shall make polar coordinates change in (19): � *<sup>z</sup>*<sup>1</sup> <sup>=</sup> *rcos<sup>ϕ</sup> z*<sup>2</sup> = *rsinϕ* and to fix the ideas let 0 < *ϕ*<sup>0</sup> < *<sup>π</sup>* 2 , � *<sup>r</sup>* <sup>≥</sup> <sup>0</sup> *<sup>π</sup>* <sup>2</sup> <sup>−</sup> *<sup>ϕ</sup>*<sup>0</sup> <sup>≤</sup> *<sup>ϕ</sup>* <sup>≤</sup> *<sup>π</sup>* 2 , *ϕ*<sup>0</sup> is the angle between *l*<sup>2</sup> and *l*1.

The new change <sup>Φ</sup> <sup>=</sup> *<sup>ϕ</sup>* <sup>−</sup> ( *<sup>π</sup>* <sup>2</sup> <sup>−</sup> *<sup>ϕ</sup>*0) <sup>⇒</sup> <sup>0</sup> <sup>≤</sup> <sup>Φ</sup> <sup>≤</sup> *<sup>ϕ</sup>*<sup>0</sup> and *<sup>∂</sup> <sup>∂</sup>*<sup>Φ</sup> <sup>=</sup> *<sup>∂</sup> ∂ϕ* . To simplify the notation we shall write again (*r*, *ϕ*) instead of (*r*, Φ), 0 ≤ Φ ≤ *ϕ*0. Thus we have a wedge type initial-boundary value problem for (19) with unknown function *w*(*τ*,*r*, *ϕ*):

$$\begin{cases} w\_{\tau} = \frac{\partial^2 w}{\partial r^2} + \frac{1}{r} \frac{\partial w}{\partial r} + \frac{1}{r^2} \frac{\partial^2 w}{\partial \phi^2} + f(\tau, r, \rho) \\\\ w|\_{\tau=0} = w\_0(r, \rho) \\\\ w|\_{\rho=0} = \tilde{\mathcal{g}}\_1(\tau, r) \\\\ w|\_{\rho=\rho\_0} = \tilde{\mathcal{g}}\_2(\tau, r) \end{cases} \tag{20}$$

*r* ≥ 0, 0 ≤ *ϕ* ≤ *ϕ*0, *l*<sup>1</sup> : {*ϕ* = 0,*r* ≥ 0}, *l*<sup>2</sup> : {*ϕ* = *ϕ*0,*r* ≥ 0}, *r* ↔ *ξ*, *ϕ* ↔ *η*, 0 ≤ Θ ≤ *τ*, 0 ≤ *ξ* ≤ ∞, 0 ≤ *η* ≤ *ϕ*0, 0 < *ϕ*<sup>0</sup> < *π*. Then

$$w(\tau, r, \boldsymbol{\varrho}) = \int\_0^\tau \int\_0^{\rho\_0} \int\_0^\infty f(\Theta, \boldsymbol{\xi}, \boldsymbol{\eta}) G(r, \boldsymbol{\varrho}, \boldsymbol{\xi}, \boldsymbol{\eta}, \tau - \Theta) \tilde{\boldsymbol{\xi}} d\boldsymbol{\xi} d\boldsymbol{\eta} d\Theta + \tag{21}$$

$$+ \int\_0^\tau \int\_0^\infty \tilde{\boldsymbol{\xi}}\_1(\Theta, \boldsymbol{\xi}) \frac{1}{\tilde{\boldsymbol{\xi}}} \frac{\partial}{\partial \boldsymbol{\eta}} G(r, \boldsymbol{\varrho}, \boldsymbol{\xi}, \boldsymbol{\eta}, \tau - \Theta) \Big|\_{\boldsymbol{\eta} = 0} d\boldsymbol{\xi} d\Theta -$$

$$- \int\_0^\tau \int\_0^\infty \tilde{\boldsymbol{\xi}}\_2(\Theta, \boldsymbol{\xi}) \frac{1}{\tilde{\boldsymbol{\xi}}} \frac{\partial}{\partial \boldsymbol{\eta}} G(r, \boldsymbol{\varrho}, \boldsymbol{\xi}, \boldsymbol{\eta}, \tau - \Theta) \Big|\_{\boldsymbol{\eta} = \boldsymbol{\varrho} \boldsymbol{\eta}} d\boldsymbol{\xi} d\Theta +$$

$$+ \int\_0^{\rho\_0} \int\_0^\infty w\_0(\boldsymbol{\xi}, \boldsymbol{\eta}) G(r, \boldsymbol{\varrho}, \boldsymbol{\xi}, \boldsymbol{\eta}, \tau) \tilde{\boldsymbol{\xi}} d\boldsymbol{\xi} d\boldsymbol{\eta},$$

where *G*(*r*, *ϕ*, *ξ*, *η*, *τ*) = <sup>1</sup> *<sup>ϕ</sup>*0*<sup>τ</sup> <sup>e</sup>*<sup>−</sup> (*r*2+*ξ*2) <sup>4</sup>*<sup>τ</sup>* ∑<sup>∞</sup> *<sup>n</sup>*=<sup>1</sup> *I <sup>n</sup><sup>π</sup> ϕ*0 ( *rξ* <sup>2</sup>*<sup>τ</sup>* )*sin <sup>n</sup><sup>π</sup> <sup>ϕ</sup>*<sup>0</sup> *<sup>ϕ</sup>sin <sup>n</sup><sup>π</sup> <sup>ϕ</sup>*<sup>0</sup> *η* and the modified Bessel function *w* = *Iν*(*z*) satisfies the equation:

$$z^2\frac{d^2w}{dz^2} + z\frac{dw}{dz} - (z^2 + \nu^2)w = 0, \nu \ge 0,\tag{22}$$

*Iν*(*z*) = ∑<sup>∞</sup> *m*=0 ( *z* <sup>2</sup> )<sup>2</sup>*m*+*<sup>ν</sup> <sup>m</sup>*!Γ(*m*+*ν*+1) (see [2]).

6 Will-be-set-by-IN-TECH

−∞ < *yj* < *lnaj* = *a*˜*<sup>j</sup>*

*f*1(*τ*, *λ*)

Now we use the linear transformation described in Remark 1, that maps the first quadrant *λ*<sup>1</sup> ≥ 0, *λ*<sup>2</sup> ≥ 0 onto the angle between the rays *l*<sup>1</sup> and *l*<sup>2</sup> in the plane 0*z*1*z*<sup>2</sup> and we obtain:

*w<sup>τ</sup>* = *wz*1*z*<sup>1</sup> + *wz*2*z*<sup>2</sup> + *f*(*τ*, *z*)

*<sup>w</sup>*|*z*1=<sup>0</sup> <sup>=</sup> *<sup>g</sup>*˜1(*τ*, *<sup>z</sup>*1),(*τ*, *<sup>z</sup>*) <sup>∈</sup> <sup>Ω</sup>˜ *w*|*l*<sup>2</sup> = *g*˜2(*τ*, *z*1, *z*2)|(*z*1,*z*2) ∈ *l*2,

i.e. *B* = *D*<sup>∗</sup> we obtain that {*λ*<sup>1</sup> ≥ 0, *λ*<sup>2</sup> ≥ 0} is mapped onto the angle *ϕ*<sup>0</sup> between the rays

*w*|*τ*=<sup>0</sup> = *w*0(*z*)

In fact, *<sup>λ</sup>* <sup>=</sup> *Bz* ⇐⇒ *<sup>z</sup>* <sup>=</sup> *<sup>B</sup>*−1*<sup>λ</sup>* and *<sup>B</sup>*−1*A*(*B*−1)<sup>∗</sup> <sup>=</sup> *<sup>I</sup>*<sup>2</sup> implies that <sup>∑</sup><sup>2</sup>

2

<sup>2</sup> <sup>−</sup> *<sup>ϕ</sup>*0) <sup>⇒</sup> <sup>0</sup> <sup>≤</sup> <sup>Φ</sup> <sup>≤</sup> *<sup>ϕ</sup>*<sup>0</sup> and *<sup>∂</sup>*

we shall write again (*r*, *ϕ*) instead of (*r*, Φ), 0 ≤ Φ ≤ *ϕ*0. Thus we have a wedge type

<sup>1</sup> *<sup>e</sup>*−*α*2*y*<sup>2</sup> <sup>≡</sup> *<sup>g</sup>*˜1(*τ*, *<sup>y</sup>*2)

(17)

(18)

(19)

*<sup>i</sup>*,*j*=<sup>1</sup> *aij*

*∂ϕ* . To simplify the notation

*∂*2 *∂λi∂λ<sup>j</sup>*

and to fix the

is

<sup>2</sup> *<sup>e</sup>*−*α*1*y*<sup>1</sup> <sup>≡</sup> *<sup>g</sup>*˜2(*τ*, *<sup>y</sup>*1)

*<sup>i</sup>*=<sup>1</sup> *αi*(*a*˜*i*−*λi*)

2 .

, <sup>Ω</sup>˜ is a wedge with opening *<sup>ϕ</sup>*0, i.e. <sup>Ω</sup>˜ = [0, *<sup>T</sup>*] <sup>×</sup> <sup>Γ</sup>, <sup>Γ</sup>

<sup>2</sup> , *<sup>π</sup>*

<sup>2</sup> < *ϕ*<sup>0</sup> < *π*.

� *z*<sup>1</sup> = *rcosϕ z*<sup>2</sup> = *rsinϕ*

. According to Remark 1: (*D*−1)∗*AD*−<sup>1</sup> = *I*2. Taking *B*−<sup>1</sup> = (*D*−1)∗,

<sup>2</sup> , 0 <sup>&</sup>lt; *<sup>ϕ</sup>*<sup>0</sup> <sup>&</sup>lt; *<sup>π</sup>*

, *ϕ*<sup>0</sup> is the angle between *l*<sup>2</sup> and *l*1.

*<sup>∂</sup>*<sup>Φ</sup> <sup>=</sup> *<sup>∂</sup>*

*<sup>i</sup>*,*j*=<sup>1</sup> *aijvyiyj* + *f*1(*τ*, *y*) *<sup>v</sup>*|*τ*=<sup>0</sup> <sup>=</sup> *<sup>v</sup>*0(*y*) = *<sup>u</sup>*0(*ey*<sup>1</sup> ,*ey*<sup>2</sup> )*e*<sup>−</sup> <sup>∑</sup> *<sup>α</sup>iyi*

�

*<sup>τ</sup>* <sup>=</sup> *<sup>τ</sup>* in (17) yields:

*v*˜|*λ*1=<sup>0</sup> = *g*˜1(*τ*, *a*˜2 − *λ*2) *v*˜|*λ*2=<sup>0</sup> = *g*˜2(*τ*, *a*˜1 − *λ*1),

<sup>Ω</sup> <sup>=</sup> {<sup>0</sup> <sup>≤</sup> *<sup>τ</sup>* <sup>≤</sup> *<sup>T</sup>*, *<sup>λ</sup><sup>j</sup>* <sup>≥</sup> 0, *<sup>j</sup>* <sup>=</sup> 1, 2}, <sup>Ω</sup> is a wedge with opening *<sup>π</sup>*

⎧ ⎪⎪⎪⎨

⎪⎪⎪⎩

*λ*1 *a*<sup>11</sup> *z*<sup>2</sup> = √−*<sup>a</sup>*<sup>12</sup> *ba*<sup>11</sup> *λ*1

From now on we shall make polar coordinates change in (19):

<sup>2</sup> <sup>−</sup> *<sup>ϕ</sup>*<sup>0</sup> <sup>≤</sup> *<sup>ϕ</sup>* <sup>≤</sup> *<sup>π</sup>*

� *<sup>z</sup>*<sup>1</sup> = <sup>√</sup>

*l*1, *l*2. Of course, there are three possibilities: *ϕ*<sup>0</sup> = *<sup>π</sup>*

� *<sup>r</sup>* <sup>≥</sup> <sup>0</sup> *<sup>π</sup>*

being the interior of the angle between *l*1, *l*2.

*∂z*<sup>2</sup> 1 + *<sup>∂</sup>*<sup>2</sup> *∂z*<sup>2</sup> 2

2 ,

*<sup>i</sup>*,*j*=<sup>1</sup> *aijv*˜*λiλ<sup>j</sup>* <sup>+</sup> ˜

*<sup>v</sup>*|*y*1=*a*˜1 <sup>=</sup> *<sup>g</sup>*1(*<sup>T</sup>* <sup>−</sup> *<sup>τ</sup>*,*ey*<sup>1</sup> ,*ey*<sup>2</sup> )|*y*1=*a*˜1 *<sup>e</sup>*−*βτ <sup>a</sup>*−*α*<sup>1</sup>

*<sup>v</sup>*|*y*2=*a*˜2 <sup>=</sup> *<sup>g</sup>*2(*<sup>T</sup>* <sup>−</sup> *<sup>τ</sup>*,*ey*<sup>1</sup> ,*ey*<sup>2</sup> )|*y*2=*a*˜2 *<sup>e</sup>*−*βτ <sup>a</sup>*−*α*<sup>2</sup>

0 ≤ *τ* ≤ *T*

*<sup>v</sup>*˜|*τ*=<sup>0</sup> <sup>=</sup> *<sup>v</sup>*˜0(*λ*) = *<sup>v</sup>*0(*a*˜1 <sup>−</sup> *<sup>λ</sup>*1, *<sup>a</sup>*˜2 <sup>−</sup> *<sup>λ</sup>*2)*e*<sup>−</sup> <sup>∑</sup><sup>2</sup>

⎧ ⎪⎪⎪⎪⎪⎪⎨

*v<sup>τ</sup>* = ∑<sup>2</sup>

*λ<sup>j</sup>* = *a*˜*<sup>j</sup>* − *yj* ≥ 0, *j* = 1, 2

*v*˜*<sup>τ</sup>* = ∑<sup>2</sup>

⎧ ⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

⎪⎪⎪⎪⎪⎪⎩

�

The change

*l*<sup>1</sup> :

� *<sup>z</sup>*<sup>1</sup> <sup>=</sup> <sup>0</sup> *z*<sup>2</sup> = √ *λ*2 *b* , *l*<sup>2</sup> :

transformed in *<sup>∂</sup>*<sup>2</sup>

ideas let 0 < *ϕ*<sup>0</sup> < *<sup>π</sup>*

The new change <sup>Φ</sup> <sup>=</sup> *<sup>ϕ</sup>* <sup>−</sup> ( *<sup>π</sup>*

**Remark 2.** One can see that *limτ*→+<sup>0</sup> � *ϕ*<sup>0</sup> 0 � ∞ <sup>0</sup> *w*0(*ξ*, *η*)*G*(*r*, *ϕ*, *ξ*, *η*, *τ*)*ξdξdη* = *w*0(*r*, *ϕ*), i.e. formally *limτ*→+0*ξG*(*r*, *ϕ*, *ξ*, *η*, *τ*) = *δ*(*r* − *ξ*, *ϕ* − *η*) in the sense of Schwartz distributions *D*� (**R**<sup>1</sup> <sup>+</sup> × [0, *ϕ*0]), **R**<sup>+</sup> = {*ξ* ≥ 0}. *G* is the corresponding Green function.

Formula (21) is given in [21], pages 182 and 166 or in [6], pp.498. The proof of (21) is based on the properties of the Bessel functions and Hankel transform.

**Remark 3.** In the special case when *a*<sup>12</sup> = 0 in (16) we obtain (18) and after the change *τ* = *τ*, *<sup>λ</sup><sup>j</sup>* <sup>=</sup> <sup>√</sup>*ajjzj*, 1 <sup>≤</sup> *<sup>j</sup>* <sup>≤</sup> 2 (18) takes the form:

$$\begin{cases} \frac{\partial \tilde{v}}{\partial \tau} = \frac{\partial^2 \tilde{v}}{\partial z\_1^2} + \frac{\partial^2 \tilde{v}}{\partial z\_2^2} + \tilde{f}\_1(\tau, z) \\\\ \tilde{v}|\_{\tau=0} = \tilde{v}\_0(z) \\\\ \tilde{v}|\_{z\_1=0} = \tilde{\tilde{g}}\_1(\tau, z\_2) \\\\ \tilde{\psi}|\_{z\_2=0} = \tilde{\tilde{g}}\_2(\tau, z\_1) \end{cases} \tag{23}$$

#### 8 Will-be-set-by-IN-TECH 38 Risk Management – Current Issues and Challenges Boundary-Value Problems for Second Order PDEs Arising in Risk Management and Cellular Neural Networks Approach <sup>9</sup>

<sup>0</sup> <sup>≤</sup> *<sup>τ</sup>* <sup>≤</sup> *<sup>T</sup>*, *zj* <sup>≥</sup> 0, 1 <sup>≤</sup> *<sup>j</sup>* <sup>≤</sup> 2. Certainly, *<sup>ϕ</sup>*<sup>0</sup> <sup>=</sup> *<sup>π</sup>* 2 .

According to [21]:

$$
\tilde{\sigma}(\tau, z) = \int\_0^\tau \int\_0^\infty \int\_0^\infty \tilde{f}\_1(\Theta, \tilde{\xi}, \eta) G(\tau - \Theta, z\_1, z\_2, \tilde{\xi}, \eta)) d\tilde{\xi} d\eta d\Theta + \tag{24}
$$

$$
+ \int\_0^\infty \int\_0^\infty \tilde{v}\_0(\tilde{\xi}, \eta) G(\tau, z\_1, z\_2, \tilde{\xi}, \eta) d\tilde{\xi} d\eta +
$$

$$
+ \int\_0^\tau \int\_0^\infty \tilde{\xi}\_1(\Theta, \eta) [\frac{\partial}{\partial \tilde{\xi}} G(\tau - \Theta, z\_1, z\_2, \tilde{\xi}, \eta)]\_{\tilde{\xi} = 0} d\eta d\Theta +
$$

$$
+ \int\_0^\tau \int\_0^\infty \tilde{\xi}\_2(\Theta, \tilde{\xi}) [\frac{\partial}{\partial \eta} G(\tau - \Theta, z\_1, z\_2, \tilde{\xi}, \eta)]\_{\eta = 0} d\xi d\Theta,
$$

$$
= \int\_0^\infty \int\_0^\infty \tilde{\xi}\_1(\Theta, \tilde{\xi}) [\frac{\partial}{\partial \eta} G(\tau - \Theta, z\_1, z\_2, \tilde{\xi}, \eta)]\_{\eta = 0} d\xi d\Theta,
$$

**Definition 1.** *The CNN is a*

*1). CNN cell dynamics;*

*3). Boundary conditions; 4). Initial conditions.*

neighborhood.

appropriate *A*-template:



*a). 2-, 3-, or n- dimensional array of*

*c). most interactions are local within a finite radius r, and*

*d). all state variables are continuous valued signals.*

characterized by its CNN cell dynamics :

*b). mainly identical dynamical systems, called cells, which satisfies two properties:*

**Definition 2.** *An M* × *M cellular neural network is defined mathematically by four specifications:*

*2). CNN synaptic law which represents the interactions (spatial coupling) within the neighbor cells;*

Now in terms of definition 2 we can present the dynamical systems describing CNNs. For a general CNN whose cells are made of time-invariant circuit elements, each cell *C*(*ij*) is

*x*˙*ij* = −*g*(*xij*, *uij*, *I*

the neighbor cell *C*(*i* + *k*, *j* + *l*) are specified by a CNN synaptic law:

+ *B*˜

*I s*

where *xij* <sup>∈</sup> **<sup>R</sup>***m*, *uij* is usually a scalar. In most cases, the interactions (spatial coupling) with

<sup>+</sup> *<sup>A</sup>*˜*ij*,*kl* <sup>∗</sup> *fkl*(*xij*, *xi*<sup>+</sup>*k*,*j*+*l*) +

*ij*,*kl* ∗ *ui*<sup>+</sup>*k*,*j*+*l*(*t*).

The first term *Aij*,*klxi*<sup>+</sup>*k*,*j*+*<sup>l</sup>* of (26) is simply a linear feedback of the states of the neighborhood nodes. The second term provides an arbitrary nonlinear coupling, and the third term accounts for the contributions from the external inputs of each neighbor cell that is located in the *Nr*

It is known [24] that some autonomous CNNs represent an excellent approximation to nonlinear partial differential equations (PDEs). The intrinsic space distributed topology makes the CNN able to produce real-time solutions of nonlinear PDEs. There are several ways to approximate the Laplacian operator in discrete space by a CNN synaptic law with an

*A*<sup>1</sup> = (1, −2, 1),

⎞ ⎠ .

⎛ ⎝

*A*<sup>2</sup> =

*s*

*ij* = *Aij*,*klxi*<sup>+</sup>*k*,*j*+*<sup>l</sup>* + (26)

*ij*), (25)

Boundary-Value Problems for Second Order PDEs Arising in Risk Management and Cellular Neural Networks Approach

39

where the Green function *G*(*τ*, *z*1, *z*2, *ξ*, *η*) = <sup>1</sup> *πτ* [*e*<sup>−</sup> (*z*1−*ξ*)<sup>2</sup> *<sup>τ</sup>* <sup>−</sup> *<sup>e</sup>*<sup>−</sup> (*z*1+*ξ*)<sup>2</sup> *<sup>τ</sup>* ] <sup>×</sup> [*e*<sup>−</sup> (*z*2−*η*)<sup>2</sup> *<sup>τ</sup>* <sup>−</sup> *<sup>e</sup>*<sup>−</sup> (*z*2+*η*)<sup>2</sup> <sup>4</sup>*<sup>τ</sup>* ].
