Explicit solution for ODE

Question

I’ve been attempting to find a solution to the following differential equation but to no avail $$2q^{2}h\frac{d^{2} C}{d q^{2}}-h(1+2(v+1)q)\frac{d C}{d q}=\ln (C) - \ln{(h\exp({2h(v+1)}))}$$

Subject to the boundary condition $$C(0)=\frac{\exp(2h(1+v))-1}{2(v+1)}$$

Is it possible to find a closed form solution for this equation?

Does anyone know a reference to a paper discussing ODEs of the form $f’’(x)+f’(x)+\ln(f(x))=0$ ?

Thanks, J

Answer 1

This is not an answer, but I hope this might be useful.

I consider the asymptotic case $q \gg 1$, which greatly simplifies the equation. In fact, I have obtained a canonical form of the general Abel equation, which is a well known nonlinear ODE.

I don't use the (single?) boundary condition so far, but it may hint at another way to solve the problem.

$\mathbf{1)}$ We write the equation in a simplified form by introducing a function:

$$f(C)=\frac{1}{2h} \ln \left( \frac{C}{h} \right)-(v+1)$$

Dividing the equation by $2h$ we obtain:

$$q^2 C''-\left( \frac{1}{2}+(v+1)q \right)C'-f(C)=0 \tag{1}$$

Now, as I said, we consider the case $q \gg 1$ which allows us to get rid of $\frac{1}{2}$ inside the bracket. The simplified equation reads:

$$q^2 C''-(v+1)q ~C'-f(C)=0 \tag{2}$$

$\mathbf{2)}$ Now we will consider the inverse function $q(C)$ instead, such that:

$$C'=\frac{1}{q'}, \qquad C''=-\frac{q''}{q'^3}$$

After substitution the equation reads:

$$q^2 q''+(v+1) q q'^2+f(C) q'^3=0$$

$\mathbf{3)}$ Now we will introduce a new function:

$$q(C)=e^{s(C)}, \qquad q'=s'e^{s}, \qquad q''=(s''+s'^2)e^{s}$$

After substitution and simplifications the equation becomes:

$$s''+(v+2) s'^2+f(C) s'^3=0$$

Introducing another function:

$$s'=p$$

We can write:

$$p'+(v+2) p^2+f(C) p^3=0 \tag{3}$$

$\mathbf{4)}$ Introducing another function:

$$p=\frac{1}{w}, \qquad p'=-\frac{w'}{w^2}$$

We obtain the equation:

$$w w'-(v+2) w-f(C)=0 $$

And after changing the variable:

$$x=(v+2) C, \qquad g(x)=\frac{f(C)}{v+2}$$

We finally have the canonical form of Abel equation linked at the top of the post:

$$w \frac{dw}{dx}-w-g(x)=0 \tag{4}$$

Where:

$$g(x)=\frac{1}{2h(v+2)} \ln \left( \frac{x}{h(v+2)} \right)-\frac{v+1}{v+2}$$

Or we can rewrite it as:

$$g(x)= a \ln x-b$$

Unfortunately, I haven't found the solution to this equation neither in the linked file, nor in Mathematica.

Equation (3) is also called Abel equation, and it is provided in that form at the Wikipedia page. The page also links a paper claiming to solve the general Abel equation, so you could look that up.

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Quite another idea (still not sure what it leads to):

• Differentiate the original equation w.r.t. $q$:

$$4h q C''+2h q^2 C'''-2h(v+1) C'-h(1+2(v+1) q) C''-\frac{C'}{C}=0$$

Simplifying, we obtain:

$$q^2 C'''+\left((1-v)q-\frac{1}{2} \right) C''-(1+v) C'-\frac{C'}{2hC}=0$$

If I have any more ideas, I will update this post.