Integration constants

Question

I have this equation with two integration constants:

$$u=\Bigl(\frac{n}{n+1}\Bigl)\cdot G^\frac{1}{n}\cdot (y+C_1)^\frac{1+n}{n}+C_2.$$

My boundary conditions are

$u=0$, $y=h$

$u=U$, $y=0$

and it yields

$$U=\frac{n\cdot G^\frac{1}{n}\cdot C_1^\frac{1+n}{n}}{(1+n)}+C_2$$

and

$$ 0=\frac{n\cdot G^\frac{1}{n}\cdot(h+C_1)^\frac{1+n}{n}}{(1+n)} + C_2.$$

How I can solve for $C_1$ and $C_2$ in term of other variables so I can find a generalised expression for $u$ in first equation?

Answer 1

From the first of the last two equations,

$$C_2=U-\frac{n\cdot G^\frac{1}{n}\cdot C_1^\frac{1+n}{n}}{(1+n)}$$

Substituting into the second equation:

$$\frac{n\cdot G^\frac{1}{n}\cdot(h+C_1)^\frac{1+n}{n}}{(1+n)}+U-\frac{n\cdot G^\frac{1}{n}\cdot C_1^\frac{1+n}{n}}{(1+n)}=0$$

Rearranging:

$$\frac{n}{1+n}G^{\frac{1}{n}}\left(C_1^\frac{1+n}{n}-(h+C_1)^\frac{1+n}{n}\right)=U$$

or

$$C_1^\frac{1+n}{n}-(h+C_1)^\frac{1+n}{n}=\frac{1+n}{n}\frac{U}{G^{\frac{1}{n}}}$$

which cannot be solved explicitly for $C_1$ (except for special values of $n$; e.g. $n=1$)

Answer 2

From $$\begin{cases}aC_1^\alpha+C_2&=U,\\a(C_1+h)^\alpha+C_2&=0\end{cases}$$

you draw

$$(C_1+h)^\alpha-C_1^\alpha=b.$$

Unfortunately, this is in general not solvable analytically. If $h$ is small compared to $C_1$ (but we don't know :( ), a first approximation is obtained with

$$(C_1+h)^\alpha-C_1^\alpha\approx\alpha C_1^{\alpha-1}h=b$$ and

$$C_1\approx\sqrt[\alpha-1]{\frac{b}{\alpha h}}.$$

Then improve with Newton's iterations.