# Evolution of a Geothermal System with Fluid Re-Injection
## 3 Steady-State Solutions
### Case 1. _Recherage Due to Liquid Groundwater Recharge and a Single Vapor Extraction/Injection Well_
In this case \(\mathfrak{R}_{\ell}^{(i)}=0\), \(D_{\ell}\) is given by \(1+\Gamma-\sigma\) (since \(T_{1}=\sigma\)), and \(\beta_{v}\), given by (30), has the form  
\[\frac{P}{T}\left(\frac{\partial P}{\partial T}-\eta\frac{P}{T}\right)=-\hat{ \beta}_{\mathrm{v}}\theta\left(T-T_{\mathrm{o}}\right)+C_{1}\, \tag{47}\]6  
Therefore, Equation (29), 1 becomes  
\[\frac{P}{T}\left(\frac{\partial P}{\partial T}-\eta\frac{P}{T}\right)=-\hat{ \beta}_{\mathrm{v}}\theta\left(T-T_{\mathrm{o}}\right)+C_{1}\, \tag{47}\]7  
Since \(B_{v}\mathfrak{R}_{v}=\hat{\beta}_{v}\), after one integration, we obtain  
\[\frac{P}{T}\left(\frac{\partial P}{\partial T}-\eta\frac{P}{T}\right)=-\hat{ \beta}_{\mathrm{v}}\theta\left(T-T_{\mathrm{o}}\right)+C_{1}\, \tag{47}\]8  
where \(C_{1}\) is the integration constant. Setting again \(z=P^{2}/2\), we get  
\[\frac{P}{T}\left(\frac{\partial P}{\partial T}-\eta\frac{P}{T}\right)=-\hat{ \beta}_{\mathrm{v}}\theta\left(T-T_{\mathrm{o}}\right)+C_{1}\, \tag{47}\]9  
whose general solution is  
\[z^{\prime}-2\eta\frac{z}{T}=C_{1}T-\hat{\beta}_{\mathrm{v}}T\theta\left(T-T_{ \mathrm{o}}\right),\]0 \[z^{\prime}-2\eta\frac{z}{T}=C_{1}T-\hat{\beta}_{\mathrm{v}}T\theta\left(T-T_{ \mathrm{o}}\right),\]1  
where \(C_{2}\) is the second integration constant. Proceeding as in Section 4.1 (i.e., imposing the boundary conditions in Equations (29), 2, and (29), 3) we obtain  
\[z^{\prime}-2\eta\frac{z}{T}=C_{1}T-\hat{\beta}_{\mathrm{v}}T\theta\left(T-T_{ \mathrm{o}}\right),\]2 \[z^{\prime}-2\eta\frac{z}{T}=C_{1}T-\hat{\beta}_{\mathrm{v}}T\theta\left(T-T_{ \mathrm{o}}\right),\]3 \[z^{\prime}-2\eta\frac{z}{T}=C_{1}T-\hat{\beta}_{\mathrm{v}}T\theta\left(T-T_{ \mathrm{o}}\right),\]4where \(A\) (\(\sigma\)) is given by Equation (40), which reduces to Equation (48) when \(B_{v}=0\) and to Equation (39) when \(\hat{\beta}_{v}=0\). Next,  
\[z^{\prime}-2\eta\frac{z}{T}=C_{1}T-\hat{\beta}_{\mathrm{v}}T\theta\left(T-T_{ \mathrm{o}}\right),\]5 \[z^{\prime}-2\eta\frac{z}{T}=C_{1}T-\hat{\beta}_{\mathrm{v}}T\theta\left(T-T_{ \mathrm{o}}\right),\]6 \[z^{\prime}-2\eta\frac{z}{T}=C_{1}T-\hat{\beta}_{\mathrm{v}}T\theta\left(T-T_{ \mathrm{o}}\right),\]7  
which gives (49) when \(B_{v}=0\). Thus, recalling Equations (34), (42) and (50), we have  
\[z^{\prime}-2\eta\frac{z}{T}=C_{1}T-\hat{\beta}_{\mathrm{v}}T\theta\left(T-T_{ \mathrm{o}}\right),\]8  
where  
\[z^{\prime}-2\eta\frac{z}{T}=C_{1}T-\hat{\beta}_{\mathrm{v}}T\theta\left(T-T_{ \mathrm{o}}\right),\]9  
Function \(H\) monotonically decreases and is negative. Therefore, \(0>\mathcal{H}(\sigma)>\mathcal{H}(1+\Gamma)\approx-0.015\). **Remark 4**. : _Recalling what we emphasized in Remark 3, in order to obtain physically meaningful solutions to Equation (38), we need \(\Omega\) to remain negative when we change \(\hat{\beta}_{v}\) and \(B_{v}\). For fixed \(T_{o}\in(1,\mathfrak{T}_{o})\) and \(\hat{\beta}_{v}>\beta_{cr}(\mathfrak{T}_{o})\); this implies_  
\[z\left(T\right)=C_{2}T^{2\eta}-\frac{C_{1}T^{2}}{2\left(\eta-1\right)}-\frac{ \hat{\beta}_{\mathrm{v}}}{2\left(\eta-1\right)}\left[T^{2\eta}\int_{1}^{T} \tau^{2-2\eta}\delta\left(\tau-T_{\mathrm{o}}\right)d\tau-T^{2}\theta\left(T-T _{\mathrm{o}}\right)\right],\]0  
_Figure 3 shows the level lines (the white boxes indicate some values) of \(B_{v,cr}\) for all \(\sigma\in(1,1+\Gamma)\) and \(\hat{\beta}_{v}\) varying over 10 orders of magnitude from negative to positive values, when \(\mathfrak{T}_{o}=1.0005\). To handle a specific case, one fixes \(\hat{\beta}_{v}>\beta_{cr}(\mathfrak{T}_{o})\) and uses \(\sup\limits_{\sigma>T_{o}}B_{cr}(\sigma,\hat{\beta}_{v},\mathfrak{T}_{o})\) as a critical value. We notice that only non-negative values of \(B_{v}\), which is the meteoric recharge, are of physical interest; when \(\hat{\beta}_{v}\geq 0\), \(\sup\limits_{\sigma>T_{o}}B_{v,cr}\) is negative (\(\approx\)\(-1.9\times 10^{4}\)) and Equation (54) does not need to be invoked. For \(\hat{\beta}_{v}\) negative (extraction), \(\sup\limits_{\sigma>T_{o}}B_{v,cr}\) continues to be negative as long as \(\hat{\beta}_{v}>-1.13\times 10^{5}\).