1 Introduction

The solution of the problem

$$\begin{aligned} {\Delta }{\varphi }=0\ \hbox {in} ~{{\mathcal O}}, {\varphi }(R_1,{\theta })=0, {\varphi }(R_2,{\theta })=V, \end{aligned}$$

where \({{\mathcal O}}=\{({\rho },{\theta }); 0<R_1<{\rho }<R_2, 0\le {\theta }<2\pi \}\), is given by

$$\begin{aligned} {\varphi }({\rho })=V\frac{ln({\rho }/R_1)}{ln(R_2/R_1)}. \end{aligned}$$

Moreover,

$$\begin{aligned} I=\int _{\{{\rho }=R_2\}}\frac{d{\varphi }}{dn}ds=\frac{2\pi V}{ln(R_2/R_1)}. \end{aligned}$$
(1.1)

If \(V\) represents the difference of potential applied to a metallic specimen \({{\mathcal O}}\), Eq. (1.1) tells us that the total current crossing \(\{{\rho }=R_2\}\) depends, in addition to \(V\), only on the ratio \(R_2/R_1\). In this paper we prove that this property of invariance remains true for much more complex domains and systems of PDE.

Let us consider the plane bounded doubly-connected domain \(\Omega \) with boundary formed by the two simple closed curves \({\Gamma }_1\) and \({\Gamma }_2\). We assume that physical state of \(\Omega \) is determined by \(n\) parameters \(u_i(x,y)\), \(i=1\ldots n\) via the \(n\) fluxes densities

$$\begin{aligned} {\mathbf{q}}_i=-\sum _{j=1}^na_{ij}(u_1\ldots u_n){\nabla }u_j. \end{aligned}$$
(1.2)

The functions \(a_{ij}(u_1\ldots u_n)\) are given and regular. The conservation laws

$$\begin{aligned} {\nabla }\cdot {\mathbf{q}}_i=0,\ i=1\ldots n \end{aligned}$$
(1.3)

are assumed to hold. Supposing Dirichlet’s boundary conditions for \(u_i\) on \({\Gamma }_1\) and \({\Gamma }_2\) we arrive at the following nonlinear boundary value problem

$$\begin{aligned} {\nabla }\cdot \left( \sum _{j=1}^na_{ij}(u_1\ldots u_n){\nabla }u_j \right) =0\quad {\ \hbox {in}\ \Omega }\end{aligned}$$
(1.4)
$$\begin{aligned} u_i=u_i^{(1)} {\ \hbox {on}\ {\Gamma }_1},u_i=u_i^{(2)} {\ \hbox {on}\ {\Gamma }_2},\quad i=1\ldots n, \end{aligned}$$
(1.5)

where \(u_i^{(1)}\) and \(u_i^{(2)}\) are \(2n\) functions given respectively on \({\Gamma }_1\) and \({\Gamma }_2\). In many cases what is of interest in the solutions of (1.4) and (1.5) are the \(n\) global quantities

$$\begin{aligned} I_i=\int _{{\Gamma }_1}{\mathbf{q}}_i\cdot {\mathbf{n}}\ ds, \end{aligned}$$
(1.6)

where \({\mathbf{n}}\) denotes the unit vector normal to the boundary of \(\Omega \) pointing outward with respect to \(\Omega \). In view of (1.3) we have

$$\begin{aligned} \int _{{\Gamma }_1}{\mathbf{q}}_i\cdot {\mathbf{n}}\ ds=-\int _{{\Gamma }_2}{\mathbf{q}}_i\cdot {\mathbf{n}}\ ds. \end{aligned}$$

Example 1

Consider a very long cylinder whose cross-section is the doubly-connected domain \(\Omega \). The cylinder is made of a material capable of conducting heat and electricity. Between the two lateral surfaces a difference of potential \(V\) is applied and they are kept at a constant temperature. The thermal conductivity \({\kappa }\) and the electric conductivity \({\sigma }\) are given functions of the temperature \(u\). In this problem are relevant the current density \({\mathbf{J}}\) given by the Ohm’s law

$$\begin{aligned} {\mathbf{J}}= {\mathbf{q}}_1=-{\sigma }(u){\nabla }{\varphi }\end{aligned}$$

and the density of the flow of energy [12]

$$\begin{aligned} {\mathbf{q}}_2=-{\kappa }(u){\nabla }u-{\varphi }{\sigma }(u){\nabla }{\varphi }. \end{aligned}$$

By the conservation of charge and energy we obtain the so-called thermistor problem, which has been thoroughly studied by many authors, [3, 4, 10] and recently by [6, 7] among others. In particular in [9] the invariance by conformal mappings of the problem is noted. The thermistor problem reads

$$\begin{aligned}&{\nabla }\cdot ({\sigma }(u){\nabla }{\varphi })=0\quad {\ \hbox {in}\ \Omega }\\&{\nabla }\cdot ({\kappa }(u){\nabla }u+{\varphi }{\sigma }(u){\nabla }{\varphi })=0\quad {\ \hbox {in}\ \Omega }\\&{\varphi }=0 {\ \hbox {on}\ {\Gamma }_1}, {\varphi }=V {\ \hbox {on}\ {\Gamma }_2},u=0 {\ \hbox {on}\ {\Gamma }_1}, u=0\quad {\ \hbox {on}\ {\Gamma }_2}. \end{aligned}$$

In this case

$$\begin{aligned} I_1=\int _{{\Gamma }_1}{\mathbf{q}}_1\cdot {\mathbf{n}}\ ds \end{aligned}$$

gives the total electric current crossing \({\Gamma }_1\) in the unit time.

Example 2

We consider, as a second example, the heat and mass transfer occurring in a Darcy’s flow in a porous medium occupying the cylinder of Example 1. The first flow is the velocity given by the Darcy’s law [1]

$$\begin{aligned} {\mathbf{v}}=-K{\nabla }p, \end{aligned}$$
(1.7)

where \(p(x,y)\) is the pressure. We take into account the Soret and Dufour’s effects [1], which in certain cases are not negligible [5]. Thus we have for the densities of the mass and heat flow

$$\begin{aligned} {\mathbf{q}}_m=-{\beta }{\nabla }c-S{\nabla }u+c{\mathbf{v}}, {\mathbf{q}}_h=-{\kappa }{\nabla }u-D{\nabla }c+u{\mathbf{v}}, \end{aligned}$$

where \(c(x,y)\) is the concentration, \(u(x,y)\) the temperature, \({\beta }\) the Fick’s coefficient, \(S\) and \(D\) the Soret and Dufour’s coefficients. They all are supposed, as \(K\), to be given positive functions of \(p,\ c\) and \(u\). If we take into account (1.7) we have

$$\begin{aligned} {\mathbf{q}}_m=-{\beta }{\nabla }c-S{\nabla }u-Kc{\nabla }p,\ {\mathbf{q}}_h=-{\kappa }{\nabla }u-D{\nabla }c-Ku{\nabla }p. \end{aligned}$$
(1.8)

By (1.3) we obtain from (1.7) and (1.8) the boundary value problem

$$\begin{aligned}&{\nabla }\cdot ( K(c,u,p){\nabla }p)=0 \quad {\ \hbox {in}\ \Omega }\\&{\nabla }\cdot ({\beta }(c,u,p){\nabla }c+S(c,u,p){\nabla }u+K(c,u,p)c{\nabla }p)=0\quad {\ \hbox {in}\ \Omega }\\&{\nabla }\cdot ({\kappa }(c,u,p){\nabla }u+D(c,u,p){\nabla }u+K(c,u,p)u{\nabla }p)=0\quad {\ \hbox {in}\ \Omega }\\&p=p^{(1)}, c=c^{(1)}, u=u^{(1)} {\ \hbox {on}\ {\Gamma }_1}, p=p^{(2)},\ c=c^{(2)}, u=u^{(2)}\ \quad {\ \hbox {on}\ {\Gamma }_2}. \end{aligned}$$

In this problem we are interested in the global quantities

$$\begin{aligned} Q_m=\int _{{\Gamma }_1}{\mathbf{q}}_m\cdot {\mathbf{n}}\ ds, Q_h=\int _{{\Gamma }_1}{\mathbf{q}}_h\cdot {\mathbf{n}}ds. \end{aligned}$$

In this paper we prove a property of invariance of the quantities \(I_i\) defined in (1.6) which, together with the notion of modulus of a doubly-connected domain (see[2, 8, 14]), permits to compute \(I_i\) reformulating the problem (1.4) and (1.5) in a simpler doubly-connected domain of the same modulus. In Sect. 2 this method is applied to the problem of Example 1 and in Sect. 3 to Example 2.

2 Invariance properties

Let \(w=f(z)\) be the conformal mapping \(f(z)={\Phi }(x,y)+\Psi (x,y), z=x+iy, w=X+iY\) such that \(|f'(z)|>0\) in \(\bar{\Omega }\). Let \({\omega }=f(\Omega )\), \({\gamma }_1=f({\Gamma }_1)\), \({\gamma }_2=f({\Gamma }_2)\). Assume \(u_i(x,y), i=1\ldots n\) to be a solution of problem (1.4), (1.5). We set \({{\mathcal U}}_i(X,Y)=u_i(x,y)\) and \(X={\Phi }(x,y)\), \(Y=\Psi (x,y)\). Using the Cauchy–Riemann equations and their consequences we find

$$\begin{aligned}&{\Delta }u_i=|f'(z)|^2{\Delta }{{\mathcal U}}_i\end{aligned}$$
(2.1)
$$\begin{aligned}&{\nabla }u_i\cdot {\nabla }u_j=|f'(z)|^2{\nabla }{{\mathcal U}}_i\cdot {\nabla }{{\mathcal U}}_j. \end{aligned}$$
(2.2)

Thus we have, using (2.1) and (2.2),

$$\begin{aligned} {\nabla }\cdot \left( \sum _{j=1}^na_{ij}(u_1\ldots u_n){\nabla }u_j\right)&= \sum _{j=1}^na_{ij}(u_1\ldots u_n){\Delta }u_j+\sum _{j, k=1}^n\frac{{\partial }a_{ij}}{{\partial }u_k}{\nabla }u_k\cdot {\nabla }u_j\nonumber \\&= |f'(z)|^2\left( \sum _{j=1}^na_{ij}({{\mathcal U}}_1\ldots {{\mathcal U}}_n){\Delta }{{\mathcal U}}_j{+}\!\!\!\sum _{j,k=1}^n\frac{{\partial }a_{ij}}{{\partial }u_k}{\nabla }{{\mathcal U}}_k\cdot {\nabla }{{\mathcal U}}_j\right) \nonumber \\&= |f'(z)|^2{\nabla }\cdot \left( \sum _{j=1}^na_{ij}({{\mathcal U}}_1\ldots {{\mathcal U}}_n){\nabla }{{\mathcal U}}_j\right) . \end{aligned}$$
(2.3)

Therefore the Eq. (1.4) is invariant under conformal mappings. Let \(z=\tilde{z}(s)\) be the parametric representation of the curve \({\Gamma }_1\) in term of the arc length \(s\) and \(w=f(\tilde{z}(s))\) the corresponding parametric representation of the curve \({\gamma }_1\) on the \(w\) plane. If \(S\) denotes the arc length on \({\gamma }_1\) we have

$$\begin{aligned} \frac{dS}{ds}=|f'(\tilde{z}(s))|. \end{aligned}$$

After a simple calculation, we have, if \({\mathbf{N}}\) denotes the unit vector normal to \({\omega }\),

$$\begin{aligned} \frac{du_i}{dn}=|f'(\tilde{z}(s))|\frac{d{{\mathcal U}}_i}{dN}. \end{aligned}$$

This gives

$$\begin{aligned} \int _{{\gamma }_1}a_{ij}({{\mathcal U}}_1\ldots {{\mathcal U}}_n)\frac{d{{\mathcal U}}_j}{dN}\ dS=\int _{{\Gamma }_1}a_{ij}(u_1\ldots u_n)\frac{du_j}{dn}\ ds. \end{aligned}$$

Therefore we have

$$\begin{aligned} I_i=\int _{{\Gamma }_1}{\mathbf{q}}_i\cdot {\mathbf{n}}ds=\int _{{\Gamma }_1}\sum _{j=1}^na_{ij}(u_1\ldots u_n)\frac{du_j}{dn}\ ds=\int _{{\gamma }_1}\sum _{j=1}^na_{ij}({{\mathcal U}}_1\ldots {{\mathcal U}}_n)\frac{d{{\mathcal U}}_j}{dN}\ dS. \end{aligned}$$

Summing up we have the following

Theorem 2.1

The boundary value problem

$$\begin{aligned} {\nabla }\cdot \left( \sum _{j=1}^na_{ij}(u_1\ldots u_n){\nabla }u_j\right) =0\ {\ \hbox {in}\ \Omega },u_i=u_i^{(1)}\ {\ \hbox {on}\ {\Gamma }_1},u_i=u_i^{(2)}\ {\ \hbox {on}\ {\Gamma }_2},\ i=1\ldots n \end{aligned}$$

and the quantities

$$\begin{aligned} I_i=\int _{{\Gamma }_1}{\mathbf{q}}_i\cdot {\mathbf{n}}\ ds,\ i=1\ldots n, \end{aligned}$$

are invariant under conformal mappings.

To apply this result we recall the definition and properties of the modulus of a doubly-connected domain [8]. If \(\Omega \) is a doubly-connected domain bounded by two non-degenerate curves it is always possible to map \(\Omega \) conformally in a one-to-one manner on the annulus \(1<|w|<{\mu }\). The number \({\mu }\), the modulus of \(\Omega \), is a characteristic constant of \(\Omega \), i.e. to every \(\Omega \) there corresponds one and only one number \({\mu }>1\). This determines a partition into equivalence classes of all doubly-connected domains, in particular all the annuli of radii \(R_2>R_1>0\), such that \(\frac{R_2}{R_1}={\mu }\) belongs to the same class of modulus \({\mu }\). Hence, by Theorem 2.1 the global fluxes \(I_i\) related to problem (1.4), (1.5) can be computed solving the same problem in an annulus of radii \(1\) and \({\mu }\).

3 Total flows in the problem of electric heating of a conductor

In this section we consider the problem of Example 1 i.e.

$$\begin{aligned} {\nabla }\cdot ({\sigma }(u){\nabla }{\varphi })=0 \quad {\ \hbox {in}\ \Omega }\end{aligned}$$
(3.1)
$$\begin{aligned} {\nabla }\cdot ({\kappa }(u){\nabla }u+{\varphi }{\sigma }(u){\nabla }{\varphi })=0\ \quad {\ \hbox {in}\ \Omega }\end{aligned}$$
(3.2)
$$\begin{aligned} {\varphi }=0 {\ \hbox {on}\ {\Gamma }_1}, {\varphi }=V\quad {\ \hbox {on}\ {\Gamma }_2}\end{aligned}$$
(3.3)
$$\begin{aligned} u=0 {\ \hbox {on}\ {\Gamma }_1}, u=0\quad {\ \hbox {on}\ {\Gamma }_2}. \end{aligned}$$
(3.4)

If \({\sigma }(u)\in C^1({\mathbf{R}^1})\), \({\kappa }(u)\in C^1({\mathbf{R}^1})\), \({\sigma }(u)>0\), \({\kappa }(u)>0\) and

$$\begin{aligned} \int _0^\infty \frac{{\kappa }(t)}{{\sigma }(t)}dt=\infty \end{aligned}$$
(3.5)

problem (3.1–3.4) has one and only one solution [3]. We wish to compute the total current

$$\begin{aligned} I=\int _{{\Gamma }_1}{\mathbf{J}}\cdot {\mathbf{n}}\ ds \end{aligned}$$

crossing the device for a generic doubly-connected domain of modulus \({\mu }\). Define

$$\begin{aligned} F(u)=\int _0^u\frac{{\kappa }(t)}{{\sigma }(t)}dt, {\theta }=\frac{{\varphi }^2}{2}+F(u). \end{aligned}$$

By (3.5) \(F\) maps one-to-one \([0,\infty )\) onto \([0,\infty )\). In terms of \({\theta }\), \({\varphi }\) and \(u\) we can restate problem (3.1–3.4) as follows

$$\begin{aligned}&{\nabla }\cdot ({\sigma }(u){\nabla }{\varphi })=0\quad {\ \hbox {in}\ \Omega }\nonumber \\&{\nabla }\cdot ({\sigma }(u){\nabla }{\theta })=0\quad {\ \hbox {in}\ \Omega }\nonumber \\&{\varphi }=0 {\ \hbox {on}\ {\Gamma }_1}, {\varphi }=V\quad {\ \hbox {on}\ {\Gamma }_2}\nonumber \\&{\theta }=0 {\ \hbox {on}\ {\Gamma }_1}, {\theta }=\frac{V^2}{2}\quad {\ \hbox {on}\ {\Gamma }_2}. \end{aligned}$$
(3.6)

The simple functional relation

$$\begin{aligned} {\theta }=\frac{V}{2}{\varphi }\end{aligned}$$
(3.7)

exists between \({\theta }\) and \({\varphi }\) (see [3]). Hence, we have

$$\begin{aligned} u=F^{-1}\left( \frac{V}{2}{\varphi }-\frac{{\varphi }^2}{2}\right) \end{aligned}$$
(3.8)

thus (3.6) can be rewritten

$$\begin{aligned} {\nabla }\cdot \left( {\sigma }\left( F^{-1}\left( \frac{V}{2}{\varphi }-\frac{{\varphi }^2}{2}\right) \right) {\nabla }{\varphi }\right) =0\quad {\ \hbox {in}\ \Omega } \end{aligned}$$
(3.9)

with the boundary conditions

$$\begin{aligned} {\varphi }=0 {\ \hbox {on}\ {\Gamma }_1}, {\varphi }=V\quad {\ \hbox {on}\ {\Gamma }_2}. \end{aligned}$$
(3.10)

To this problem we can apply the Kirchhoff’s reduction. More precisely, define

$$\begin{aligned} \psi =L({\varphi }), \hbox {where}\ L({\varphi })=\int _0^{\varphi }{\sigma }\left( F^{-1}\left( \frac{V}{2}\xi -\frac{\xi ^2}{2}\right) \right) d\xi . \end{aligned}$$

We have

$$\begin{aligned} {\nabla }\psi ={\sigma }\left( F^{-1}\left( \frac{V}{2}{\varphi }-\frac{{\varphi }^2}{2}\right) \right) {\nabla }{\varphi }\end{aligned}$$

and, in view of (3.9) and (3.10),

$$\begin{aligned} {\Delta }\psi =0 {\ \hbox {in}\ \Omega }, \psi =0 {\ \hbox {on}\ {\Gamma }_1}, \psi =L(V) {\ \hbox {on}\ {\Gamma }_2}. \end{aligned}$$

Moreover, recalling (3.8) we obtain

$$\begin{aligned} {\mathbf{J}}=-{\sigma }(u){\nabla }{\varphi }=-{\nabla }\psi . \end{aligned}$$
(3.11)

If \(v\) is the solution of the problem

$$\begin{aligned} {\Delta }v=0 {\ \hbox {in}\ \Omega }, v=0 {\ \hbox {on}\ {\Gamma }_1}, v=1 {\ \hbox {on}\ {\Gamma }_2} \end{aligned}$$

we have

$$\begin{aligned} \psi (x,y)=L(V)v(x,y). \end{aligned}$$

Hence, by (3.11)

$$\begin{aligned} {\mathbf{J}}=-L(V){\nabla }v \end{aligned}$$

and

$$\begin{aligned} I=\int _{{\Gamma }_1}{\mathbf{J}}\cdot {\mathbf{n}}\ ds=-L(V)\int _{{\Gamma }_1}\frac{dv}{dn}ds. \end{aligned}$$
(3.12)

On the other hand, \(I\) is invariant in the class of the doubly-connected domains of modulus \({\mu }\). It is therefore enough to compute \(\int _{{\Gamma }_1}\frac{dv}{dn}ds\) in the annulus of radii \(1\) and \({\mu }\). We easily find

$$\begin{aligned} \int _{{\Gamma }_1}\frac{dv}{dn}ds=-\frac{2\pi }{\ln {{\mu }}}. \end{aligned}$$

Hence, by (3.12)

$$\begin{aligned} I=\frac{2\pi L(V)}{\ln {{\mu }}}. \end{aligned}$$

This gives the total current crossing any doubly-connected domain of modulus \({\mu }\) if all the others data in problem (3.1–3.4) remain unchanged.

Remark 3.1

In problem (3.1–3.4) it is interesting to consider also the density of the heat flow as given by the Fourier’s law

$$\begin{aligned} {\mathbf{q}}_h=-{\kappa }(u){\nabla }u \end{aligned}$$

and the corresponding global quantities, a priori not necessarily equal,

$$\begin{aligned} Q_{{\Gamma }_1}=\int _{{\Gamma }_1}{\mathbf{q}}_h\cdot {\mathbf{n}}\ ds, Q_{{\Gamma }_2}=\int _{{\Gamma }_2}{\mathbf{q}}_h\cdot {\mathbf{n}}\ ds. \end{aligned}$$

We have

$$\begin{aligned} {\mathbf{q}}_2=-k(u){\nabla }u-{\varphi }{\sigma }(u){\nabla }{\varphi }=-{\sigma }(u){\nabla }{\theta }\end{aligned}$$

and, by (3.7),

$$\begin{aligned} {\mathbf{q}}_2=-{\sigma }(u)\frac{V}{2}{\nabla }{\varphi }=\frac{V}{2}{\mathbf{J}}. \end{aligned}$$

On the other hand,

$$\begin{aligned} {\mathbf{q}}_h={\mathbf{q}}_2-{\varphi }{\mathbf{J}}. \end{aligned}$$

Hence, in view of the condition \({\varphi }=0\) on \({\Gamma }_1\) and of (3.12)

$$\begin{aligned} Q_{{\Gamma }_1}=\int _{{\Gamma }_1}{\mathbf{q}}_h\cdot {\mathbf{n}}\ ds=\frac{V}{2}\int _{{\Gamma }_1}{\mathbf{J}}\cdot {\mathbf{n}}\ ds=\frac{\pi VL(V)}{\ln {{\mu }}}. \end{aligned}$$

Moreover, since \({\varphi }=V\) on \({\Gamma }_2\) and \(\int _{{\Gamma }_1}{\mathbf{J}}\cdot {\mathbf{n}}\ ds=-\int _{{\Gamma }_2}{\mathbf{J}}\cdot {\mathbf{n}}\ ds\) we have

$$\begin{aligned} Q_{{\Gamma }_2}=\int _{{\Gamma }_2}{\mathbf{q}}_h\cdot {\mathbf{n}}\ ds=-\frac{V}{2}\int _{{\Gamma }_2}{\mathbf{J}}\cdot {\mathbf{n}}\ ds=\frac{\pi VL(V)}{\ln {{\mu }}}. \end{aligned}$$

4 Invariance properties in the Soret-Dufour’s problem

Theorem 2.1 can be applied to the problem of Example 2 i.e.

$$\begin{aligned}&{\nabla }\cdot ( K(c,u,p){\nabla }p)=0\quad {\ \hbox {in}\ \Omega }\end{aligned}$$
(4.1)
$$\begin{aligned}&{\nabla }\cdot ({\beta }(c,u,p){\nabla }c+S(c,u,p){\nabla }u+K(c,u,p)c{\nabla }p)=0\quad {\ \hbox {in}\ \Omega }\end{aligned}$$
(4.2)
$$\begin{aligned}&{\nabla }\cdot ({\kappa }(c,u,p){\nabla }u+D(c,u,p){\nabla }c+K(c,u,p){\nabla }p)=0\quad {\ \hbox {in}\ \Omega }\end{aligned}$$
(4.3)
$$\begin{aligned}&p=p^{(1)},\ c=c^{(1)}, \ u=u^{(1)}\ {\ \hbox {on}\ {\Gamma }_1},\ p=p^{(2)},\ c=c^{(2)}, \ u=u^{(2)}\quad {\ \hbox {on}\ {\Gamma }_2}. \end{aligned}$$
(4.4)

Therefore the total flows of mass and heat depend on \(\Omega \) only via its modulus. In this section we consider a special case of problem (4.1–4.4). We suppose \({\beta }\), \({\kappa }\), \(D\), \(S\) and \(K\) to be positive constants. We have

Theorem 4.1

Let

$$\begin{aligned} p^{(1)}\ \hbox {and} \ p^{(2)}\in H^2(\Omega ),\ c^{(1)}, \ c^{(2)},\ u^{(1)}\ \hbox {and} \ u^{(2)}\in H^1(\Omega ). \end{aligned}$$
(4.5)

Suppose

$$\begin{aligned} \left( \frac{S+D}{2}\right) ^2<{\beta }{\kappa }. \end{aligned}$$
(4.6)

Then the problem

$$\begin{aligned}&{\nabla }\cdot ( K{\nabla }p)=0\quad {\ \hbox {in}\ \Omega }\end{aligned}$$
(4.7)
$$\begin{aligned}&{\nabla }\cdot ({\beta }{\nabla }c+S{\nabla }u+Kc{\nabla }p)=0\quad {\ \hbox {in}\ \Omega } \end{aligned}$$
(4.8)
$$\begin{aligned}&{\nabla }\cdot ({\nabla }u+D{\nabla }c+K{\nabla }p)=0\quad {\ \hbox {in}\ \Omega }\end{aligned}$$
(4.9)
$$\begin{aligned}&p=p^{(1)},\ c=c^{(1)}, u=u^{(1)} {\ \hbox {on}\ {\Gamma }_1}, p=p^{(2)},\ c=c^{(2)}, u=u^{(2)}\quad {\ \hbox {on}\ {\Gamma }_2} \end{aligned}$$
(4.10)

has one and only one solution. Moreover, if all the data are of class \(C^\infty \) then also the corresponding solution is of class \(C^\infty \).

Proof

We compute \(p(x,y)\) from the problem

$$\begin{aligned} {\Delta }p=0\ {\ \hbox {in}\ \Omega },\ p=p^{(1)}\ {\ \hbox {on}\ {\Gamma }_1},\ p=p^{(2)}\ {\ \hbox {on}\ {\Gamma }_2}. \end{aligned}$$
(4.11)

By (4.5) we have \(p\in H^2(\Omega )\). Denote \(c_b\) and \(u_b\) the solutions of the problems

$$\begin{aligned} {\Delta }c_b=0 \quad {\ \hbox {in}\ \Omega },\ c_b=c^{(1)} \quad {\ \hbox {on}\ {\Gamma }_1},\ c_b=c^{(2)} \quad {\ \hbox {on}\ {\Gamma }_2}\end{aligned}$$
(4.12)
$$\begin{aligned} {\Delta }u_b=0 \quad {\ \hbox {in}\ \Omega },\ u_b=u^{(1)} \quad {\ \hbox {on}\ {\Gamma }_1},\ u_b=u^{(2)} \quad {\ \hbox {on}\ {\Gamma }_2}. \end{aligned}$$
(4.13)

Setting \(h=c-c_b\) and \(z=u-u_b\) we restate problem (4.7–4.10) with homogeneous boundary conditions

$$\begin{aligned}&h\in H^1_0(\Omega ), {\nabla }\cdot ({\beta }{\nabla }h+S{\nabla }z+Kh{\nabla }p)=-{\nabla }\cdot (Kc_b{\nabla }p)\end{aligned}$$
(4.14)
$$\begin{aligned}&z\in H^1_0(\Omega ), {\nabla }\cdot ({\kappa }{\nabla }z+D{\nabla }h+Kz{\nabla }p)=-{\nabla }\cdot (Ku_b{\nabla }p). \end{aligned}$$
(4.15)

Define the bilinear form

$$\begin{aligned} a((h,z),(v,w))&= \int _{\Omega }({\beta }{\nabla }h\cdot {\nabla }v+S{\nabla }z\cdot {\nabla }v+Kh{\nabla }p\cdot {\nabla }v+{\kappa }{\nabla }z\cdot {\nabla }w\\&+\, D{\nabla }h\cdot {\nabla }w+Kz{\nabla }p\cdot {\nabla }w)dxdy, \end{aligned}$$

where \((h,z)\in H^1_0(\Omega )\times H^1_0(\Omega )\) and \((v,w)\in H^1_0(\Omega )\times H^1_0(\Omega )\). \(a((h,z),(v,w))\) is coercive and bounded. In fact, we have by (4.11)

$$\begin{aligned} \int _{\Omega }h{\nabla }p\cdot {\nabla }h\ dxdy=-\int _{\Omega }{\nabla }\cdot (h{\nabla }p)h\ dxdy=-\int _{\Omega }h{\nabla }h\cdot {\nabla }p\ dxdy. \end{aligned}$$

Thus

$$\begin{aligned} \int _{\Omega }h{\nabla }p\cdot {\nabla }h\ dxdy=0. \end{aligned}$$

Similarly

$$\begin{aligned} \int _{\Omega }z{\nabla }p\cdot {\nabla }z\ dxdy=0. \end{aligned}$$

Hence

$$\begin{aligned} a((h,z),(h,z))=\int _{\Omega }\bigl ({\beta }|{\nabla }h|^2+(S+D){\nabla }h\cdot {\nabla }z+{\kappa }|{\nabla }z|^2\bigl )dxdy. \end{aligned}$$

On the other hand, the matrix

$$\begin{aligned} \left( \begin{array}{l@{\quad }l@{\quad }l@{\quad }l} {\beta }&{} 0 &{} \frac{S+D}{2} &{} 0 \\ 0 &{} {\beta }&{} 0 &{} \frac{S+D}{2} \\ \frac{S+D}{2} &{} 0 &{} {\kappa }&{} 0 \\ 0 &{} \frac{S+D}{2} &{} 0 &{} {\kappa }\end{array} \right) \end{aligned}$$

of the quadratic form

$$\begin{aligned} {\beta }h_x^2+{\beta }h^2_y+{\kappa }z_x^2+{\kappa }z_y^2+(S+D)h_xz_x+(S+D)h_yz_y \end{aligned}$$

is definite positive since the determinants of the principal minors i.e.

$$\begin{aligned} {\beta }^2, {\beta }\left[ {\beta }{\kappa }-\left( \frac{S+D}{2}\right) ^2\right] ,\left[ {\beta }{\kappa }-\left( \frac{S+D}{2}\right) ^2\right] ^2 \end{aligned}$$

are all positive by (4.6). Hence there exists a positive constant \(L\) such that

$$\begin{aligned} a((h,z),(h,z))\ge L\int _{\Omega }\Bigl (|{\nabla }h|^2+|{\nabla }z|^2\Bigl )dxdy. \end{aligned}$$
(4.16)

It is also easy to prove that \(a((h,z),(v,w))\) is bounded. Let us define the linear functional of \((H^1_0(\Omega )\times H^1_0(\Omega ))'\)

$$\begin{aligned} \langle f,(v,w) \rangle =-K\int _{\Omega }\Bigl (c_b{\nabla }p\cdot {\nabla }v+u_b{\nabla }p\cdot {\nabla }w\Bigl )dxdy. \end{aligned}$$

We can rewrite (4.14), (4.15) as follows

$$\begin{aligned} (h,z)\in H^1_0(\Omega )\times H^1_0(\Omega ),\ a((h,z),(v,w))=\langle f,(v,w)\rangle \forall (v,w)\in H^1_0(\Omega )\times H^1_0(\Omega ). \end{aligned}$$

The Lax-Milgram lemma [13] can be applied and we conclude that (4.7–4.10) has one and only one solution. Using standard regularity results for elliptic system [11] this weak solution can be regularized.

Remark 4.1

If the condition \(\Bigl (\frac{S+D}{2}\Bigl )^2<{\beta }{\kappa }\) is not satisfied, problem (4.7–4.10) may not have solutions at all. This fact is already apparent in the one-dimensional counterpart of problem (4.7–4.10) i.e.

$$\begin{aligned} ({\beta }c'+Su'+ac)'=0, c(0)=c^{(1)},c(1)=c^{(2)}\\ ({\kappa }u'+Dc'+au)'=0, u(0)=u^{(1)},u(1)=u^{(2)}, \end{aligned}$$

where \(a\) is a given constant. In fact, suppose \({\beta }={\kappa }=D=S=a=1\). We have by difference \((c-u)'=0\) which is not always compatible with the boundary conditions \( c(0)=c^{(1)},c(1)=c^{(2)}\), \( u(0)=u^{(1)},u(1)=u^{(2)}\).

Let (4.6) be satisfied and let us take in problem (4.7–4.10) \( u^{(1)}, u^{(2)},\ c^{(1)},c^{(2)} \) as constants and \(p^{(1)}=0, p^{(2)}=\bar{P}\). We compute the total flows of mass and heat in a generic doubly-connected domain of modulus \({\mu }\). For this goal we solve the problem in the annulus of radii \(1\) and \({\mu }\). Denote \(\rho \) and \(\theta \) the polar coordinates. Since the solution is unique by Theorem 4.1 a solution which depends only on \({\rho }\) is the only possible one. Therefore, we have

$$\begin{aligned} p(\rho )=\frac{\bar{P}}{\ln {\mu }}\ln {\rho }. \end{aligned}$$
(4.17)

In polar coordinates the Eqs. (4.8) and (4.9) become

$$\begin{aligned}&\frac{d}{d{\rho }}\left( {\beta }{\rho }\frac{dc}{d{\rho }}+S{\rho }\frac{du}{d{\rho }}+mc\right) =0\end{aligned}$$
(4.18)
$$\begin{aligned}&\frac{d}{d{\rho }}\left( {\kappa }{\rho }\frac{du}{d{\rho }}+D{\rho }\frac{dc}{d{\rho }}+mu\right) =0, \end{aligned}$$
(4.19)

where

$$\begin{aligned} m=\frac{\bar{P}}{\ln {\mu }}. \end{aligned}$$

From (4.6) and the inequality \(\sqrt{DS}<\frac{S+D}{2}\) we obtain \(DS-{\beta }{\kappa }\ne 0\). This permits to solve the system of ordinary differential Eqs. (4.18), (4.19). We obtain

where

$$\begin{aligned} N=({\beta }-{\kappa })^2+4DS, L=\frac{m \left( {\beta }+{\kappa }+\sqrt{N}\right) }{2 \left( DS-{\beta }{\kappa }\right) },\ M=\frac{m \left( {\beta }+{\kappa }-\sqrt{N}\right) }{2 \left( DS-{\beta }{\kappa }\right) }. \end{aligned}$$

The four constants of integration \(A_1\), \(A_2\), \(A_3\) and \(A_4\) are in a one-to-one correspondence with the boundary values \( u^{(1)},\ u^{(2)},\ c^{(1)},c^{(2)} \). They can easily be computed explicitly. Thus we obtain for the total flows of mass \(Q_{m}\) and heat \(Q_{h}\) in an arbitrary domain of modulus \({\mu }\)