{"id":1385,"date":"2026-05-12T21:47:42","date_gmt":"2026-05-12T21:47:42","guid":{"rendered":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/chapter\/box-1-derivation-of-the-radial-flow-equation\/"},"modified":"2026-09-02T18:55:46","modified_gmt":"2026-09-02T18:55:46","slug":"box-1-derivation-of-the-radial-flow-equation","status":"publish","type":"chapter","link":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/chapter\/box-1-derivation-of-the-radial-flow-equation\/","title":{"raw":"Box 1 - Derivation of the Radial Flow Equation","rendered":"Box 1 – Derivation of the Radial Flow Equation"},"content":{"raw":"
\r\n
\r\n\r\nIn Section 8, we stated the two\u2011dimensional groundwater flow equation with respect to Cartesian coordinate system [latex]x,\\ y[\/latex] is as shown in Equation\u00a0Box\u00a0(1-1).\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial^{2}h}{{\\partial x}^{2}} + \\frac{\\partial^{2}h}{{\\partial y}^{2}} = \\frac{S}{T}\\frac{\\partial h}{\\partial t}[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u20111)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nwhere (parameter dimensions are dark green font with mass as M, length as L, time as T):\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n\r\n\r\n\r\n\r\n
[latex]h[\/latex]<\/th>\r\n=<\/th>\r\nhydraulic head (L)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
[latex]S[\/latex]<\/td>\r\n=<\/td>\r\nstorativity (:)<\/td>\r\n<\/tr>\r\n
[latex]T[\/latex]<\/td>\r\n=<\/td>\r\ntransmissivity (L2<\/sup>T\u20111<\/sup>)<\/td>\r\n<\/tr>\r\n
[latex]t[\/latex]<\/td>\r\n=<\/td>\r\ntime (T)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
\r\n\r\nFlow to a well is radial; in other words, flow lines are converging on the well from all sides (Figure\u00a0Box\u00a01\u20111).\r\n\r\n<\/div>\r\n
\r\n\r\nBecause the flow lines converge, the equipotential lines that define the cone of depression around the well get closer together near the well. In this case, we cannot use the Cartesian form of the equation, and therefore need to express the equation in terms of a polar coordinate system.\r\n\r\n<\/div>\r\n
\r\n\r\n\"\"\r\n\r\n<\/div>\r\n
\r\n
Figure\u00a0Box\u00a01\u20111\u00a0\u2011\u00a0<\/strong>Radial flow to a well.<\/blockquote>\r\n<\/div>\r\n
\r\n\r\nIn a polar coordinate system, a point is defined by its location with respect to a fixed central pole (Figure\u00a0Box\u00a01\u20112). The location of a point is defined by its radial distance from the pole ([latex]r[\/latex]) and the angle (\ud835\udf03) between the radial line and some fixed polar axis (e.g., north).\r\n\r\n<\/div>\r\n
\r\n\r\n\"\"\r\n
\r\n
\r\n\r\nFigure\u00a0Box\u00a01\u20112\u00a0\u2011\u00a0<\/strong>Polar coordinate system.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n\r\nThe derivation of the transformation of the equation in Cartesian coordinates to polar coordinates follows below.\r\n\r\n<\/div>\r\n
\r\n\r\nThe Pythagorean theorem as shown by Equation\u00a0Box\u00a0(1\u20112).\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]r = \\left( x^{2} + y^{2} \\right)^{\\frac{1}{2}}[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u20112)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nwhere (parameter dimensions are dark green font with mass as M, length as L, time as T):\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n\r\n\r\n
[latex]r[\/latex]<\/th>\r\n=<\/th>\r\nradial distance from the central pole (L)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
[latex]x,\\ y[\/latex]<\/td>\r\n=<\/td>\r\nCartesian coordinates (L)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
\r\n\r\n\"\"\r\n
\r\n
\r\n\r\nFigure\u00a0Box\u00a01\u20113\u00a0\u2011\u00a0<\/strong>Definition of terms in Equation\u00a0Box\u00a01\u20112.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n\r\nIf we assume that [latex]h(x,y)\\ = \\ h(r)[\/latex], then\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial h}{\\partial x} = \\frac{\\partial h}{\\partial r}\\frac{\\partial r}{\\partial x}[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u20113)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nand\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial h}{\\partial y} = \\frac{\\partial h}{\\partial r}\\frac{\\partial r}{\\partial y}[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u20114)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nThe second partial of [latex]h[\/latex] with respect to [latex]x[\/latex] is\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial}{\\partial x}\\left( \\frac{\\partial h}{\\partial x} \\right) = \\frac{\\partial}{\\partial x}\\left\\lbrack \\frac{\\partial h}{\\partial r}\\frac{\\partial r}{\\partial x} \\right\\rbrack[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u20115)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nUsing the product rule, we can remove the [latex]\\partial h\/\\partial r[\/latex] and take the second derivative of [latex]r[\/latex] with respect to [latex]x[\/latex] to get [latex]\\left( \\frac{\\partial h}{\\partial r} \\right)\\left( \\frac{\\partial^{2}r}{{dx}^{2}} \\right)[\/latex] . This product rule also gives us [latex]\\left( \\frac{\\partial r}{\\partial x} \\right)\\ \\left( \\frac{\\partial}{\\partial x} \\right) [\/latex]of [latex]\\left( \\frac{\\partial h}{\\partial r} \\right)[\/latex], or\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial}{\\partial x}\\left( \\frac{\\partial h}{\\partial x} \\right) = \\frac{\\partial h}{\\partial r}\\left\\lbrack \\frac{\\partial^{2}r}{{\\partial x}^{2}} \\right\\rbrack + \\frac{\\partial r}{\\partial x}\\left\\lbrack \\frac{\\partial}{\\partial x}\\frac{\\partial h}{\\partial r} \\right\\rbrack[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u20116)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nNow we use the chain rule on the terms in the brackets at the end of the equation. Recognizing that [latex]h[\/latex] is a function of [latex]r[\/latex], which itself is a function of [latex]x[\/latex] and [latex]y[\/latex]\u2014that is, [latex]h\\lbrack r(x,y)\\rbrack[\/latex]\u2014we get [latex]\\partial^{2}r\/{dx}^{2}[\/latex]times an additional [latex]\\partial r\/\\partial x[\/latex] out of those brackets, or\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial}{\\partial x}\\left( \\frac{\\partial h}{\\partial x} \\right) = \\frac{\\partial h}{\\partial r}\\left\\lbrack \\frac{\\partial^{2}r}{{\\partial x}^{2}} \\right\\rbrack + \\frac{\\partial r}{\\partial x}\\left\\lbrack \\frac{\\partial^{2}h}{\\partial r^{2}}\\frac{\\partial r}{\\partial x} \\right\\rbrack[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u20117)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nTherefore, the second partial derivative of [latex]h[\/latex] with respect to [latex]x[\/latex] is:\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial\\left( \\frac{\\partial h}{\\partial x} \\right)}{\\partial x} = \\frac{\\partial^{2}h}{{\\partial x}^{2}} = \\frac{\\partial h}{\\partial r}\\frac{\\partial^{2}r}{{\\partial x}^{2}} + \\frac{\\partial r}{\\partial x}\\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{\\partial r}{\\partial x} = \\frac{\\partial h}{\\partial r}\\frac{\\partial^{2}r}{{\\partial x}^{2}} + \\frac{\\partial^{2}h}{{dr}^{2}}\\left( \\frac{\\partial r}{\\partial x} \\right)^{2}[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u20118)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nand the second partial derivative of [latex]h[\/latex] with respect to [latex]y[\/latex] is:\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial\\left( \\frac{\\partial h}{\\partial y} \\right)}{\\partial y} = \\frac{\\partial^{2}h}{{\\partial y}^{2}} = \\frac{\\partial h}{\\partial r}\\frac{\\partial^{2}r}{{\\partial y}^{2}} + \\frac{\\partial r}{\\partial y}\\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{\\partial r}{\\partial y} = \\frac{\\partial h}{\\partial r}\\frac{\\partial^{2}r}{{\\partial y}^{2}} + \\frac{\\partial^{2}h}{{dr}^{2}}\\left( \\frac{\\partial r}{\\partial y} \\right)^{2}[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u20119)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nNow, the partial derivative of [latex]r[\/latex] with respect to [latex]x[\/latex] is:\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial r}{\\partial x} = \\frac{1}{2}\\left( x^{2} + y^{2} \\right)^{\\frac{- 1}{\\ \\ \\ 2}}2x = \\frac{1}{2}\\frac{1}{r}2x = \\frac{x}{r}[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u201110)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nNext, we use the quotient rule to get the second derivative of [latex]r[\/latex] with respect to [latex]x[\/latex]. The quotient rule says that:\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\left( \\frac{f}{g} \\right) = \\frac{f'g - fg'}{g}[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u201111)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nso, if [latex]\\partial r\/\\partial x\\ = \\ x\/r,[\/latex] then [latex]\\frac{\\partial^{2}r}{{\\partial x}^{2}}[\/latex] is as below (because [latex]f\u2019\\ = \\ \\partial x\/\\partial x\\ = \\ 1[\/latex]).\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial^{2}r}{{\\partial x}^{2}} = \\frac{r - x\\left( \\frac{\\partial r}{\\partial x} \\right)}{r^{2}} = \\frac{r(r) - r\\left( x\\left( \\frac{x}{r} \\right) \\right)}{r\\left( r^{2} \\right)} = \\frac{r^{2} - x^{2}}{r^{3}}[\/latex]<\/th>\r\nEquation\u00a0Box\u00a0(1\u201112)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nbut since\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]r = \\left( x^{2} + y^{2} \\right)^{\\frac{1}{2}}\\ and\\ r^{2} = x^{2} + y^{2}[\/latex]<\/th>\r\n(Equation\u00a0Box\u00a01\u201113)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nand\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]y^{2} = r^{2} - x^{2}[\/latex]<\/th>\r\n(Equation\u00a0Box\u00a01\u201114)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n<\/div>\r\n
\r\n\r\nThe second partial with respect to x is:\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial^{2}r}{{\\partial x}^{2}} = \\frac{y^{2}}{r^{3}}[\/latex]<\/th>\r\n(Equation\u00a0Box\u00a01\u201115)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nAnd, similarly, the partials of [latex]r[\/latex] with respect to [latex]y[\/latex] are:\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial r}{\\partial y} = \\frac{y}{r}[\/latex]<\/th>\r\n(Equation\u00a0Box\u00a01\u201116)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nand\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial^{2}r}{{\\partial y}^{2}} = \\frac{x^{2}}{r^{3}}[\/latex]<\/th>\r\n(Equation\u00a0Box\u00a01\u201117)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nWe can substitute into the expression for the second partial of h<\/em> with respect to x<\/em> and y<\/em> to obtain:\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial^{2}h}{{\\partial x}^{2}} = \\frac{\\partial h}{\\partial r}\\frac{y^{2}}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{x^{2}}{r^{2}}[\/latex]<\/th>\r\n(Equation\u00a0Box\u00a01\u201118)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial^{2}h}{{\\partial y}^{2}} = \\frac{\\partial h}{\\partial r}\\frac{x^{2}}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{y^{2}}{r^{2}}[\/latex]<\/th>\r\n(Equation\u00a0Box\u00a01\u201119)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nTherefore, we have:\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial^{2}h}{{\\partial x}^{2}} + \\frac{\\partial^{2}h}{{\\partial y}^{2}}\\ \\ \\ \\ \\ \\ = \\frac{\\partial h}{\\partial r}\\frac{y^{2}}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{x^{2}}{r^{2}} + \\frac{\\partial h}{\\partial r}\\frac{x^{2}}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{y^{2}}{r^{2}} = \\frac{\\partial h}{\\partial r}\\frac{\\left( x^{2} + y^{2} \\right)}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{\\left( x^{2} + y^{2} \\right)}{r^{2}} = \\frac{\\partial h}{\\partial r}\\frac{r^{2}}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{r^{2}}{r^{2}}[\/latex]\r\n\r\n[latex]= \\frac{1}{r}\\frac{\\partial h}{\\partial r} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}[\/latex]<\/th>\r\n(Equation\u00a0Box\u00a01\u201120)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nand substituting for the left side of Equation Box (1-1):\r\n\r\n<\/div>\r\n<\/colgroup>\r\n\r\n\r\n
<\/th>\r\n[latex]\\frac{\\partial^{2}h}{{\\partial r}^{2}} + \\frac{1}{r}\\frac{\\partial h}{\\partial r} = \\frac{S}{T}\\frac{\\partial h}{\\partial t}[\/latex]<\/th>\r\n(Equation\u00a0Box\u00a01\u201121)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<\/table>\r\n
\r\n\r\nReturn to where text linked to Box\u00a01<\/strong><\/span><\/a>\u2ba5\r\n\r\n<\/div>\r\n<\/div>","rendered":"
\n
\n

In Section 8, we stated the two\u2011dimensional groundwater flow equation with respect to Cartesian coordinate system [latex]x,\\ y[\/latex] is as shown in Equation\u00a0Box\u00a0(1-1).<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial^{2}h}{{\\partial x}^{2}} + \\frac{\\partial^{2}h}{{\\partial y}^{2}} = \\frac{S}{T}\\frac{\\partial h}{\\partial t}[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u20111)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

where (parameter dimensions are dark green font with mass as M, length as L, time as T):<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n\n\n\n\n
[latex]h[\/latex]<\/th>\n=<\/th>\nhydraulic head (L)<\/th>\n<\/tr>\n<\/thead>\n
[latex]S[\/latex]<\/td>\n=<\/td>\nstorativity (:)<\/td>\n<\/tr>\n
[latex]T[\/latex]<\/td>\n=<\/td>\ntransmissivity (L2<\/sup>T\u20111<\/sup>)<\/td>\n<\/tr>\n
[latex]t[\/latex]<\/td>\n=<\/td>\ntime (T)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\n

Flow to a well is radial; in other words, flow lines are converging on the well from all sides (Figure\u00a0Box\u00a01\u20111).<\/p>\n<\/div>\n

\n

Because the flow lines converge, the equipotential lines that define the cone of depression around the well get closer together near the well. In this case, we cannot use the Cartesian form of the equation, and therefore need to express the equation in terms of a polar coordinate system.<\/p>\n<\/div>\n

\n

\"\"<\/p>\n<\/div>\n

\n

Figure\u00a0Box\u00a01\u20111\u00a0\u2011\u00a0<\/strong>Radial flow to a well.<\/p><\/blockquote>\n<\/div>\n

\n

In a polar coordinate system, a point is defined by its location with respect to a fixed central pole (Figure\u00a0Box\u00a01\u20112). The location of a point is defined by its radial distance from the pole ([latex]r[\/latex]) and the angle (\ud835\udf03) between the radial line and some fixed polar axis (e.g., north).<\/p>\n<\/div>\n

\n

\"\"<\/p>\n

\n
\n

Figure\u00a0Box\u00a01\u20112\u00a0\u2011\u00a0<\/strong>Polar coordinate system.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

\n

The derivation of the transformation of the equation in Cartesian coordinates to polar coordinates follows below.<\/p>\n<\/div>\n

\n

The Pythagorean theorem as shown by Equation\u00a0Box\u00a0(1\u20112).<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]r = \\left( x^{2} + y^{2} \\right)^{\\frac{1}{2}}[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u20112)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

where (parameter dimensions are dark green font with mass as M, length as L, time as T):<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n\n\n
[latex]r[\/latex]<\/th>\n=<\/th>\nradial distance from the central pole (L)<\/th>\n<\/tr>\n<\/thead>\n
[latex]x,\\ y[\/latex]<\/td>\n=<\/td>\nCartesian coordinates (L)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\n

\"\"<\/p>\n

\n
\n

Figure\u00a0Box\u00a01\u20113\u00a0\u2011\u00a0<\/strong>Definition of terms in Equation\u00a0Box\u00a01\u20112.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

\n

If we assume that [latex]h(x,y)\\ = \\ h(r)[\/latex], then<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial h}{\\partial x} = \\frac{\\partial h}{\\partial r}\\frac{\\partial r}{\\partial x}[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u20113)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

and<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial h}{\\partial y} = \\frac{\\partial h}{\\partial r}\\frac{\\partial r}{\\partial y}[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u20114)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

The second partial of [latex]h[\/latex] with respect to [latex]x[\/latex] is<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial}{\\partial x}\\left( \\frac{\\partial h}{\\partial x} \\right) = \\frac{\\partial}{\\partial x}\\left\\lbrack \\frac{\\partial h}{\\partial r}\\frac{\\partial r}{\\partial x} \\right\\rbrack[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u20115)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

Using the product rule, we can remove the [latex]\\partial h\/\\partial r[\/latex] and take the second derivative of [latex]r[\/latex] with respect to [latex]x[\/latex] to get [latex]\\left( \\frac{\\partial h}{\\partial r} \\right)\\left( \\frac{\\partial^{2}r}{{dx}^{2}} \\right)[\/latex] . This product rule also gives us [latex]\\left( \\frac{\\partial r}{\\partial x} \\right)\\ \\left( \\frac{\\partial}{\\partial x} \\right)[\/latex]of [latex]\\left( \\frac{\\partial h}{\\partial r} \\right)[\/latex], or<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial}{\\partial x}\\left( \\frac{\\partial h}{\\partial x} \\right) = \\frac{\\partial h}{\\partial r}\\left\\lbrack \\frac{\\partial^{2}r}{{\\partial x}^{2}} \\right\\rbrack + \\frac{\\partial r}{\\partial x}\\left\\lbrack \\frac{\\partial}{\\partial x}\\frac{\\partial h}{\\partial r} \\right\\rbrack[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u20116)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

Now we use the chain rule on the terms in the brackets at the end of the equation. Recognizing that [latex]h[\/latex] is a function of [latex]r[\/latex], which itself is a function of [latex]x[\/latex] and [latex]y[\/latex]\u2014that is, [latex]h\\lbrack r(x,y)\\rbrack[\/latex]\u2014we get [latex]\\partial^{2}r\/{dx}^{2}[\/latex]times an additional [latex]\\partial r\/\\partial x[\/latex] out of those brackets, or<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial}{\\partial x}\\left( \\frac{\\partial h}{\\partial x} \\right) = \\frac{\\partial h}{\\partial r}\\left\\lbrack \\frac{\\partial^{2}r}{{\\partial x}^{2}} \\right\\rbrack + \\frac{\\partial r}{\\partial x}\\left\\lbrack \\frac{\\partial^{2}h}{\\partial r^{2}}\\frac{\\partial r}{\\partial x} \\right\\rbrack[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u20117)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

Therefore, the second partial derivative of [latex]h[\/latex] with respect to [latex]x[\/latex] is:<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial\\left( \\frac{\\partial h}{\\partial x} \\right)}{\\partial x} = \\frac{\\partial^{2}h}{{\\partial x}^{2}} = \\frac{\\partial h}{\\partial r}\\frac{\\partial^{2}r}{{\\partial x}^{2}} + \\frac{\\partial r}{\\partial x}\\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{\\partial r}{\\partial x} = \\frac{\\partial h}{\\partial r}\\frac{\\partial^{2}r}{{\\partial x}^{2}} + \\frac{\\partial^{2}h}{{dr}^{2}}\\left( \\frac{\\partial r}{\\partial x} \\right)^{2}[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u20118)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

and the second partial derivative of [latex]h[\/latex] with respect to [latex]y[\/latex] is:<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial\\left( \\frac{\\partial h}{\\partial y} \\right)}{\\partial y} = \\frac{\\partial^{2}h}{{\\partial y}^{2}} = \\frac{\\partial h}{\\partial r}\\frac{\\partial^{2}r}{{\\partial y}^{2}} + \\frac{\\partial r}{\\partial y}\\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{\\partial r}{\\partial y} = \\frac{\\partial h}{\\partial r}\\frac{\\partial^{2}r}{{\\partial y}^{2}} + \\frac{\\partial^{2}h}{{dr}^{2}}\\left( \\frac{\\partial r}{\\partial y} \\right)^{2}[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u20119)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

Now, the partial derivative of [latex]r[\/latex] with respect to [latex]x[\/latex] is:<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial r}{\\partial x} = \\frac{1}{2}\\left( x^{2} + y^{2} \\right)^{\\frac{- 1}{\\ \\ \\ 2}}2x = \\frac{1}{2}\\frac{1}{r}2x = \\frac{x}{r}[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u201110)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

Next, we use the quotient rule to get the second derivative of [latex]r[\/latex] with respect to [latex]x[\/latex]. The quotient rule says that:<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\left( \\frac{f}{g} \\right) = \\frac{f'g - fg'}{g}[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u201111)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

so, if [latex]\\partial r\/\\partial x\\ = \\ x\/r,[\/latex] then [latex]\\frac{\\partial^{2}r}{{\\partial x}^{2}}[\/latex] is as below (because [latex]f\u2019\\ = \\ \\partial x\/\\partial x\\ = \\ 1[\/latex]).<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial^{2}r}{{\\partial x}^{2}} = \\frac{r - x\\left( \\frac{\\partial r}{\\partial x} \\right)}{r^{2}} = \\frac{r(r) - r\\left( x\\left( \\frac{x}{r} \\right) \\right)}{r\\left( r^{2} \\right)} = \\frac{r^{2} - x^{2}}{r^{3}}[\/latex]<\/th>\nEquation\u00a0Box\u00a0(1\u201112)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

but since<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]r = \\left( x^{2} + y^{2} \\right)^{\\frac{1}{2}}\\ and\\ r^{2} = x^{2} + y^{2}[\/latex]<\/th>\n(Equation\u00a0Box\u00a01\u201113)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

and<\/p>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]y^{2} = r^{2} - x^{2}[\/latex]<\/th>\n(Equation\u00a0Box\u00a01\u201114)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n<\/div>\n
\n

The second partial with respect to x is:<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial^{2}r}{{\\partial x}^{2}} = \\frac{y^{2}}{r^{3}}[\/latex]<\/th>\n(Equation\u00a0Box\u00a01\u201115)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

And, similarly, the partials of [latex]r[\/latex] with respect to [latex]y[\/latex] are:<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial r}{\\partial y} = \\frac{y}{r}[\/latex]<\/th>\n(Equation\u00a0Box\u00a01\u201116)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

and<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial^{2}r}{{\\partial y}^{2}} = \\frac{x^{2}}{r^{3}}[\/latex]<\/th>\n(Equation\u00a0Box\u00a01\u201117)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

We can substitute into the expression for the second partial of h<\/em> with respect to x<\/em> and y<\/em> to obtain:<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial^{2}h}{{\\partial x}^{2}} = \\frac{\\partial h}{\\partial r}\\frac{y^{2}}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{x^{2}}{r^{2}}[\/latex]<\/th>\n(Equation\u00a0Box\u00a01\u201118)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial^{2}h}{{\\partial y}^{2}} = \\frac{\\partial h}{\\partial r}\\frac{x^{2}}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{y^{2}}{r^{2}}[\/latex]<\/th>\n(Equation\u00a0Box\u00a01\u201119)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

Therefore, we have:<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial^{2}h}{{\\partial x}^{2}} + \\frac{\\partial^{2}h}{{\\partial y}^{2}}\\ \\ \\ \\ \\ \\ = \\frac{\\partial h}{\\partial r}\\frac{y^{2}}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{x^{2}}{r^{2}} + \\frac{\\partial h}{\\partial r}\\frac{x^{2}}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{y^{2}}{r^{2}} = \\frac{\\partial h}{\\partial r}\\frac{\\left( x^{2} + y^{2} \\right)}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{\\left( x^{2} + y^{2} \\right)}{r^{2}} = \\frac{\\partial h}{\\partial r}\\frac{r^{2}}{r^{3}} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}\\frac{r^{2}}{r^{2}}[\/latex]<\/p>\n

[latex]= \\frac{1}{r}\\frac{\\partial h}{\\partial r} + \\frac{\\partial^{2}h}{{\\partial r}^{2}}[\/latex]<\/th>\n

(Equation\u00a0Box\u00a01\u201120)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

and substituting for the left side of Equation Box (1-1):<\/p>\n<\/div>\n\n\n\n\n<\/colgroup>\n\n\n
<\/th>\n[latex]\\frac{\\partial^{2}h}{{\\partial r}^{2}} + \\frac{1}{r}\\frac{\\partial h}{\\partial r} = \\frac{S}{T}\\frac{\\partial h}{\\partial t}[\/latex]<\/th>\n(Equation\u00a0Box\u00a01\u201121)<\/th>\n<\/tr>\n<\/thead>\n<\/table>\n
\n

Return to where text linked to Box\u00a01<\/strong><\/span><\/a>\u2ba5<\/p>\n<\/div>\n<\/div>\n","protected":false},"author":8,"menu_order":111,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1385","chapter","type-chapter","status-publish","hentry"],"part":1528,"_links":{"self":[{"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/pressbooks\/v2\/chapters\/1385","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/wp\/v2\/users\/8"}],"version-history":[{"count":5,"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/pressbooks\/v2\/chapters\/1385\/revisions"}],"predecessor-version":[{"id":2429,"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/pressbooks\/v2\/chapters\/1385\/revisions\/2429"}],"part":[{"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/pressbooks\/v2\/parts\/1528"}],"metadata":[{"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/pressbooks\/v2\/chapters\/1385\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/wp\/v2\/media?parent=1385"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/pressbooks\/v2\/chapter-type?post=1385"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/wp\/v2\/contributor?post=1385"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/books.gw-project.org\/basic-hydrogeology-an-introduction-to-the-fundamentals-of-groundwater-science\/wp-json\/wp\/v2\/license?post=1385"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}