Prove mathematically (using variables, not numbers) that kx>kz for hydraulic conductivity - topic: Groundwater Hydrology

hexacordoK 2021-09-19 Answered
Prove mathematically (using variables, not numbers) that kx>kz for hydraulic conductivity - topic: Groundwater Hydrology
\(\displaystyle{K}_{{z}}={\frac{{{d}}}{{{\frac{{{d}_{{1}}}}{{{K}_{{1}}}}}+{\frac{{{d}_{{2}}}}{{{K}_{{2}}}}}+\dot{{s}}+{\frac{{{d}_{{n}}}}{{{K}_{{n}}}}}}}}\)
\(\displaystyle{K}_{{x}}={\frac{{{K}_{{1}}{d}_{{1}}+{K}_{{2}}{d}_{{2}}+\dot{{s}}+{K}_{{n}}{d}_{{n}}}}{{{d}}}}\)

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Expert Answer

comentezq
Answered 2021-09-20 Author has 10200 answers
Step 1 Let \(\displaystyle{d}={d}_{{1}}+{d}_{{2}}+\dot{{s}}{d}_{{n}}\)
From Cauchy Schwarz inequality , we have
\(\displaystyle{\left(\sqrt{{{\frac{{{d}_{{1}}}}{{{k}_{{1}}}}}}}\right)}^{{2}}+{\left(\sqrt{{{\frac{{{d}_{{2}}}}{{{k}_{{2}}}}}}}\right)}^{{2}}+\dot{{s}}+{\left(\sqrt{{{\frac{{{d}_{{n}}}}{{{k}_{{n}}}}}}}\right)}^{{2}}\)
\(\displaystyle{\left({\left(\sqrt{{{k}_{{1}}{d}_{{1}}}}\right)}^{{2}}{\left(\sqrt{{{k}_{{2}}{d}_{{2}}}}\right)}^{{2}}+\dot{{s}}{\left(\sqrt{{{k}_{{n}}{d}_{{n}}}}\right)}^{{2}}\right)}\)
\(\displaystyle\geq{\left(\sqrt{{{\frac{{{d}_{{1}}}}{{{k}_{{1}}}}}{k}_{{1}}{d}_{{1}}}}+\sqrt{{{\frac{{{d}_{{2}}}}{{{k}_{{2}}}}}{k}_{{2}}{d}_{{2}}}}+\dot{{s}}+\sqrt{{{\frac{{{d}_{{n}}}}{{{k}_{{n}}}}}{d}_{{n}}{k}_{{n}}}}\right)}^{{2}}\)
\(\displaystyle\Rightarrow{\left({\frac{{{d}_{{1}}}}{{{k}_{{1}}}}}+{\frac{{{d}_{{2}}}}{{{k}_{{2}}}}}+\dot{{s}}+{\frac{{{d}_{{n}}}}{{{k}_{{n}}}}}\right)}{\left({k}_{{1}}{d}_{{1}}+{k}_{{2}}{d}_{{2}}+\dot{{s}}+{k}_{{n}}{d}_{{n}}\right)}\geq{\left({d}_{{1}}+{d}_{{2}}+\dot{{s}}+{d}_{{n}}\right)}^{{2}}={d}^{{2}}\)
Step 2 \(\displaystyle\Rightarrow{\frac{{{k}_{{1}}{d}_{{1}}+{k}_{{2}}{d}_{{2}}+\dot{{s}}+{k}_{{n}}{d}_{{n}}}}{{{d}}}}\geq{\frac{{{d}}}{{{\frac{{{d}_{{1}}}}{{{k}_{{1}}}}}+{\frac{{{d}_{{2}}}}{{{k}_{{2}}}}}+\dot{{s}}+{\frac{{{d}_{{n}}}}{{{k}_{{n}}}}}}}}\)
\(\displaystyle\Rightarrow{k}_{{x}}\geq{k}_{{z}}\)
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