How to reduce this differential equation into first-order differential equation?

Question

How can we reduce this differential equation into first-order differential equation?

$\left( {\frac {\rm d}{{\rm d}z}}U \left( z \right) \right) ^{3}{ \frac {{\rm d}^{2}}{{\rm d}{z}^{2}}}U \left( z \right) + \left( { \frac {\rm d}{{\rm d}z}}U \left( z \right) \right) {\frac {{\rm d}^{4 }}{{\rm d}{z}^{4}}}U \left( z \right) - \left( {\frac {{\rm d}^{2}}{ {\rm d}{z}^{2}}}U \left( z \right) \right) {\frac {{\rm d}^{3}}{ {\rm d}{z}^{3}}}U \left( z \right) =0 $

Answer 1

Set $V=U'$, then $$ V^3V'+VV'''-V'V''=0. $$ The last two terms look like the numerator of a quotient derivative $$ VV'+\left(\frac{V''}{V}\right)'=0 $$ Set $W=\frac12V^2+\frac{V''}{V}$, then $W'=0$ and $$ V''+\frac12V^3=VW $$ So if the components of the state vector are $X=[U,V,V',W]^T$, then $$ X'=[V,V', VW-\frac12V^3,0]^T, $$ that is $$ X'=\pmatrix{X_1'\\X_2'\\X_3'\\X_4'} =F(x)=\pmatrix{X_2\\X_3\\X_2X_4-X_2^3/2\\0} $$ Compared to the simpler first order system using the derivatives vector to $U$, this requires a non-trivial transformation of the initial value for $W(z_0)$. The advantage here is that the right side has no divisions by variable expressions. One could also eliminate $W$ as component and just use it as constant (depending on the initial value), thus reducing the dimension of the system.

Answer 2

As shown by Lutz Lehmann, \begin{align} \frac{(U')^2}{2}+\frac{U'''}{U'}=C_1, \end{align} Multiplying by $U'U''$ and integrating \begin{align} \frac{(U')^4}{8}+\frac{(U'')^2}{2}=C_1\frac{(U')^2}{2}+C_2^2,\\\\ 2U''=\sqrt{(U')^4+2C_1(U')^2+C_2^2}, \end{align} which is separable \begin{align} \int \frac{2 \mathrm dU'}{\sqrt{-(U')^4+2C_1(U')^2+C_2^2}}+C_3\equiv \mathcal I(U')=z, \end{align} which is the first order differential equation you requested. $\mathcal I$ involves the elliptic integral of the first kind, here's a link to WA's solution.

The solution to equations of the form $z=\mathcal I(U')$ can be found via the method of integration by differentiation. Denote $U'=\xi$, where $\xi$ will be our parameter in the parametric solution. Taking a derivative of our first order differential equation, substituting in our relation for $\xi$, and integrating we find the solution to be \begin{align} U(\xi)=\int \xi \mathcal I'(\xi)\mathrm d\xi+C_4,\quad z(\xi)=\mathcal I(\xi). \end{align}

The integral for $U(\xi)$ comes out as

\begin{align} U(\xi)-C_4=-\mathrm{arctan}\left(\frac{C_1-\xi^2}{\sqrt{-\xi^4+2C_1\xi^2+C_2^2}}\right)\longrightarrow \xi=\pm\sqrt{C_1+ C_2\sin(U+C_4)}, \end{align} So then our parametric solution can be reduced to an implicit one if we desire (which is not usually the case with this method) \begin{align} z=\mathcal I\left(\pm\sqrt{C_1+C_2\sin(U+C_4)}\right), \quad\text{where}\quad \mathcal I(s)=\int\frac{2 \mathrm ds}{\sqrt{-s^4+2C_1 s^2+C_2^2}}. \end{align}