Eufisky - The lost book

## 一个偏微分方程求解

$$\frac{\partial J}{\partial t}=\frac{1}{4}\left( \frac{\partial J}{\partial x} \right) ^2-x^2-\frac{1}{2}x^4 \tag 1$$
$$\frac{\partial^2 J}{\partial t\partial x}=\frac{1}{2} \frac{\partial J}{\partial x}\frac{\partial^2 J}{\partial x^2} -2x-2x^3$$
Change of function :
$$\frac{\partial J}{\partial x}=u(x,t)\quad\to\quad \frac{\partial u}{\partial t} -\frac{1}{2} u\frac{\partial u}{\partial x}= -2x-2x^3 \tag 2$$
Characteristic system of equations :
$$\frac{dt}{1}=\frac{dx}{-\frac{1}{2} u}=\frac{du}{-2x-2x^3}$$
First family of characteristic curves, from $\quad -2\frac{dx}{ u}=\frac{du}{-2x-2x^3} :$
$$2udu-(8x+8x^3)dx=0 \quad\to\quad u^2-4x^2-2x^4=c_1$$
Second family of characteristic curves, from $\frac{dt}{1}=\frac{dx}{-\frac{1}{2} u} :$
$$dx+\frac{u}{2}dt=0=dx+\frac{\sqrt{c_1+4x^2+2x^4}}{2}dt$$
$$dt+\frac{2dx}{\sqrt{c_1+4x^2+2x^4}}=0 \quad\to\quad t+\int\frac{2dx}{\sqrt{c_1+4x^2+2x^4}}=c_2$$
The integral can be expressed on closed form. The formula involves a special function, namely the Elliptic Integral of the first kind :
http://www.wolframalpha.com/input/?i=integrate+2%2Fsqrt(C%2B4+x%5E2%2B2+x%5E4)&x=0&y=0

[![enter image description here][1]][1]

In interest of space and in order to make easier the writing, this big formula will be symbolized as :
$$\int\frac{2dx}{\sqrt{c_1+4x^2+2x^4}}=\Psi\left(c_1,x\right)$$
Where $\Psi$ is the above known function. Thus the second family of characteristic curves is :
$$t+\Psi\left(c_1,x\right)=c_2$$

The general solution of the PDE $(2)$ is expressed on the form of the implicit equation :
$$F\left(\left(u^2-4x^2-2x^4\right) \:,\: \left(t+\Psi\left(u^2-4x^2-2x^4\:,\:x\right)\right) \right)=0$$
where $F$ is any differentiable function of two variables.

The function $F$ might be determined according to a boundary condition which has to be derived from a given boundary condition of Eq.$(1)$. Nevertheless it appears doubtful to find a closed form for $F$ considering the complicated function $\Psi$.

Supposing that the function $F$ be determined, which is optimistic, a more difficult step comes after, to go from $u(x,t)$ to $J(x,y)$, which suppose possible to find a closed form for $\int u(x,t)dx$.

[1]: https://i.stack.imgur.com/DLyqp.jpg