Solve $2A{\frac{1-\sqrt{w}}{\log{w}}}=1$ in terms of Lambert W function.

Question

I have tried it in this way: $$2A(1-\sqrt{w})=\log{w}$$ $$w\exp(2A\sqrt{w})=\exp{2A}$$ $$A^2w\exp(2A\sqrt{w})=(A\exp{A})^2$$ $$A^2w=W^2(A\cdot \exp{A})$$ $$w=A^-2W^2(A\cdot \exp{A})$$ Is this solution right?

Answer 1

To get the full extent of the solutions, follow Claude Leibovici's advice.

Note that if you set $y=w^{1/2}$, then $y^2=w$, but because of the squaring you are going to miss solutions, so you need to set up two equations, one for the positive and one for the negative root of the substitution:

$$\begin{align} \frac{2A(1-y)}{\ln(y^2)}&=1\,\,\,\,(1)\\ \frac{2A(1+y)}{\ln(y^2)}&=1\,\,\,\,(2) \end{align}$$

After simplifying you get:

$$\begin{align} y&=\exp(A(1-y))\,\,\,(3)\\ y&=\exp(A(1+y))\,\,\,(4) \end{align}$$

which now have the visible solutions via the standard Lambert algebraic proccess:

$$ \begin{align} (3)\implies y&=\frac{W(A\cdot\exp(A))}{A}=1\,\,\,\,\text{(for $A>0$, W is a partial inverse of }z\cdot\exp(z))\,\,\,(5)\\ (4)\implies y&=\frac{W(-A\cdot\exp(A))}{-A}\,\,\,(6) \end{align}$$

from which upon reversing the substitution $y=w^{1/2}$ you get:

$$\begin{align} (5)\implies w_1&=1\\ (6)\implies w_2&=\left(\frac{W(-A\cdot\exp(A))}{-A}\right)^2 \end{align}$$

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$\mathbf{Addendum}$-1

If you are wondering about whether $w_2$ simplifies further, yes, it does and it equals 1, because the numerator of the fraction in the parenthesis equals $-A$ again. For details please refer to this answer of mine from an older question by Gottfried Helms.

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$\mathbf{Addendum}$-2 (second root clarification)

I was wrong on my first Addendum concerning the second root. It is true that:

$$W(|A|\exp(|A|)=|A|$$

but this does not apply in this case because the signs are alternating, so:

$$\frac{W(-A\cdot\exp(A))}{-A}$$

$\mathbf{cannot}$ be simplified further.

To conclude, the solutions are $w_1=1$ (for $A>0$), and $w_2=\left(\frac{W(-A\cdot\exp(A))}{-A}\right)^2$ (for $A<0$), as above in (6).

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$\mathbf{Addendum}$-3 (addressing additional solutions)

It is possible that $w_2$ may be complex for all the branches of $W$. Here's an example with Maple code:

W := proc (k, w) options operator, arrow; LambertW(k, w) end proc
sol2 := proc (k, A) options operator, arrow; W(k, -A*exp(A))^2/A^2 end proc
for k from -3 to 3 do evalf(sol2(k, 2)) end do #for A=2

outputs:

-49.98644048 - 0.3118949792 I

-15.63286398 - 2.455783917 I

-0.5299074982 - 1.853237059 I

-0.5299074982 + 1.853237059 I

-15.63286398 + 2.455783917 I

-49.98644048 + 0.3118949792 I

-104.0602403 - 3.292630660 I

So, in the case of $A=2$ all solutions $w_2$ are complex. You can check that these are indeed solutions, by substituting back to the original equation.

If you have more questions, comment and I will add more.

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$\mathbf{Addendum}$-4 (addressing your last comment and correcting a typo)

The curves for the ranges of $W(k,z)$ are exactly:

$$ C_k(y) = \begin{cases} -y\cot(y)+yi, &y\in(2k\pi,(2k+1)\pi) \text{, if $k\ge 0$} \\ -y\cot(y)+yi, &y\in((2k+1)\pi,(2k+2)\pi)\text{, if $k<0$} \end{cases} $$

In Maple they can be plotted as:

restart;
with(plots):
f:=y->-y*cot(y)+y*I;
p0 := complexplot(f(y), y = 0 .. Pi-.2, scaling = constrained);
p01 := complexplot(f(y), y = -Pi+.2 .. 0, scaling = constrained);
p1 := complexplot(f(y), y = 2*Pi+.5 .. 3*Pi-.5, scaling = constrained);
p12 := complexplot(f(y), y = -3*Pi+.5 .. -2*Pi-.5, scaling = constrained);
p2 := complexplot(f(y), y = 4*Pi+.8 .. 5*Pi-.8, scaling = constrained);
p23 := complexplot(f(y), y = -5*Pi+.8 .. -4*Pi-.8, scaling = constrained);
display(p0, p01, p1, p12, p2, p23);

I am just breaking the whole thing into segments and subtracting a bit from the ranges to adjust it for full screen.

Sorry, I don't know MatLab, so I have no idea how you could port this. In Maple they are just plots on the Complex plane.

(Corrected a typo, also. Note that the function is $f(y)=-y\cot(y)+yi$, $\mathbf{not}$ $f(y)=\cot(y)+yi$ as I had on my code).