Energy Integral.

We multiply equation (4.77) and (4.78) respectively by 2u and (U²-u²), so we have

$$% latex2html id marker 1024 \mbox{(\ref{eq:ch234b})} \rightar... ...frac{\partial u}{\partial x}-2vu\frac{\partial u}{\partial y},$$ (94)

$$% latex2html id marker 1039 \mbox{(\ref{eq:ch234b})} \rightar... ...ial u}{\partial x}+(U^{2}-u^{2})\frac{\partial v}{\partial y},$$ (95)

and by adding them we obtain

$$2\nu\left(\frac{\partial u}{\partial y}\right)^{2}-2\nu\frac{... ...al x}(U^{2}u-u^{3})+\frac{\partial}{\partial y}(vU^{2}-vu^{2})$$ (96)

If we integrate the equation above with respect to y from 0 to $\infty$, and if we take into account equations (4.79) to (4.83), we obtain

$$\frac{2D}{\rho} = \frac{\partial}{\partial t}(U^{2}\delta_{2}... ...ial t}\delta_{1}+\frac{\partial}{\partial x}(U^{3}\delta_{3}),$$ (97)

since v(U²-u²) and $\partial u/\partial y$ both tend to zero as y tends to infinite and the term (vU²)_y=0 is zero because it is assumed that there is no suction on the plate. The energy integral may also be regarded as an equation for the ``kinetic energy defect'' $\frac{1}{2}\rho(U^{2}-u^{2})$ per unit volume, namely

$$\frac{\partial}{\partial t}\int_{0}^{\infty}\frac{1}{2}\rho(U... ... = D+\rho\frac{\partial U}{\partial t}\int_{0}^{\infty}(U-u)dy$$ (98)