We assumed we had a holonomic system and the $q$ ‘s were all independent, so this equation holds for arbitrary virtual displacements $\delta q_{j}$, and therefore
$$
\frac{d}{d t} \frac{\partial T}{\partial \dot{q}{j}}-\frac{\partial T}{\partial q{j}}-Q_{j}=0
$$
Now let us restrict ourselves to forces given by a potential, with $\vec{F}{i}=$ $-\vec{\nabla}{i} U({\vec{r}}, t)$, or
$$
Q_{j}=-\sum_{i} \frac{\partial \vec{r}{i}}{\partial q{j}} \cdot \vec{\nabla}{i} U=-\left.\frac{\partial \tilde{U}({q}, t)}{\partial q{j}}\right|_{t}
$$
Notice that $Q_{j}$ depends only on the value of $U$ on the constrained surface.
Also, $U$ is independent of the $\dot{q}{i}$ ‘s, so $$ \frac{d}{d t} \frac{\partial T}{\partial \dot{q}{j}}-\frac{\partial T}{\partial q_{j}}+\frac{\partial U}{\partial q_{j}}=0=\frac{d}{d t} \frac{\partial(T-U)}{\partial \dot{q}{j}}-\frac{\partial(T-U)}{\partial q{j}}
$$
or
$$
\frac{d}{d t} \frac{\partial L}{\partial \dot{q}{j}}-\frac{\partial L}{\partial q{j}}=0
$$
This is Lagrange’s equation, which we have now derived in the more general context of constrained systems.
MATH3977 COURSE NOTES ：
$$
\tilde{u}=\dot{i}=0 \quad \text { as } z \rightarrow \infty
$$
Hence
$$
\tilde{u}=C_{3} e^{-(1+i) z \delta_{E}}+C_{4} e^{-(1-i) z / \delta_{L}},
$$
while from (4.3.13)
$$
\tilde{v}=-i C_{3} e^{-(1+i) z^{\prime} \delta E}+i C_{4} e^{-(1-i) z / \delta_{E}} .
$$
On $z=0$,
$$
\begin{aligned}
&\tilde{v}=0 \
&\tilde{u}=-U
\end{aligned}
$$
so that
$$
C_{3}=C_{4}=-\frac{U}{2}
$$