Dtyjlui

Question:

How to solve the differential equation

$$v - eta frac{dv}{deta} = vfrac{d^2v}{deta^2} + bigg(frac{dv}{deta}bigg)^2$$

Attempt:

This equation looks quite disgusting to me. I tried writing it in the form

$$-eta^2 frac{d}{deta}bigg(frac{v}{eta}bigg) = frac{d}{deta}bigg( vfrac{dv}{deta}bigg)$$

but that doesn't seem to help.

Honestly, I am at a loss at how to tackle this equation (not even sure if there is an easy solution for this).

Note:

This differential equation actually arose as part of my attempt to find a similarity solution for the PDE

$$frac{partial u}{partial t} = frac{partial}{partial x}bigg(ufrac{partial u}{partial x}bigg)$$

Letting $u(x,t) = t^a v(eta)$ where $eta = x/t^b$ and $a,b$ are constants to be specified later on, we get

begin{align}
& frac{partial u}{partial t} = frac{partial}{partial x}bigg(ufrac{partial u}{partial x}bigg)\
implies & frac{partial u}{partial t} = ufrac{partial ^2u}{partial x^2} + bigg(frac{partial u}{partial x}bigg)^2 \
implies & at^{a - 1}v - bxt^{a-b-1}v' = t^{2a-2b}vv'' + t^{2a-2b} v'^2 \
implies & at^{a - 1}v - beta t^{a-1}v' = t^{2a-2b}vv'' + t^{2a-2b} v'^2
end{align}

So we need $a=b=1$ in order for this to be a similarity solution. The equation thus becomes

$$v - eta frac{dv}{deta} = vfrac{d^2v}{deta^2} + bigg(frac{dv}{deta}bigg)^2$$

which I am unable to solve.

What makes you think that the equation $v - eta frac{dv}{deta} = vfrac{d^2v}{deta^2} + bigg(frac{dv}{deta}bigg)^2$ can be solved in terms of a limited number of standard functions ?

In fact this arduous equation results from a guess about the existence of solutions of the PDE $$frac{partial u}{partial t} = frac{partial}{partial x}bigg(ufrac{partial u}{partial x}bigg)$$
on the particular form $u(x,t) = t^a vleft(x/t^bright)$.

More simply, why not looking for solutions of the form :
$$u(x,t)=f(x)g(t)$$
$$f(x)g'(t)=f(x)f''(x)(g(t))^2+(f'(x))^2(g(t))^2$$
$$frac{g'(t)}{(g(t))^2}=f''(x)+frac{f'(x))^2}{f(x)}=lambda=text{constant}$$
because a function of $t$ can be equal to a function of $x$ only if those fonctions are constant.

Solving $frac{g'(t)}{(g(t))^2}=lambda$ leads to
$$g(t)=-frac{1}{lambda t+c_1}$$

$f''(x)+frac{f'(x))^2}{f(x)}=lambda$ is an autonomous ODE. The usual method to solve this kind of ODE leads to :
$$f(x)=left(lambda e^{-c_2(x-x_0)/2}+frac{1}{c_2^2}e^{c_2(x-x_0)/2} right)^2$$
This method of separation of variables leads to a family of solutions with four parameters $lambda, c_0,c_1,c_2$ :
$$u(x,t)=-frac{1}{lambda t+c_1}left(lambda e^{-c_2(x-x_0)/2}+frac{1}{c_2^2}e^{c_2(x-x_0)/2} right)^2$$
Of course, other solutions of the PDE cannot derive from linear combination of those particular solutions since the PDE is not linear.