Variational Analysis and Integration of the (2 + 1) Fourth-Order Time-Dependent Biharmonic Equation via Energy and Momentum Conservation

Abstract

We consider the fourth-order PDE $u_{xxxx}+2u_{xxyy}+u_{yyyy}-u_{tt}=h\left(u\right)$. The Lie and Noether symmetry generators are constructed, and we reduce the PDE to simpler ODEs. Furthermore, we use some well-known methods to compute the conserved vectors associated with the PDE. An analysis of reduced ordinary differential equations (ODEs), invariant solutions, and their physical interpretations is presented.

1. Introduction

The fourth-order partial differential equation has various applications in beam and plate theories, fluid dynamics, image processing, phase field models, quantum mechanics, elasticity and shell theory, inter alia (see [1,2]) and the range of references that describe the methods adopted in dealing with the analysis of the underlying equations.

Menglian Li, Omid Nikan, Wenlin Qiu, and Da Xu studied the two-dimensional Burgers-type equation arising in fluid turbulent flows [3]. Bluman et al. briefly introduced time-independent Lie symmetries of the fourth-order linear equation, $u_{xxxx}+2u_{xxyy}+u_{yyyy}=0$ by [4], while Kara et al. [5] studied $u_{xxxx}+2u_{xxyy}+u_{yyyy}=f\left(u\right)$; a nonlinear and nonhomogeneous fourth-order equation using Lie symmetries, Noether symmetries, and conservation laws. Moreover, in [6], symmetries of the equation ${\displaystyle \frac{d^{4}y}{d{x}^4}}=f\left(y\right)$ are studied. We consider the fourth-order time-dependent PDE $u_{xxxx}+2u_{xxyy}+u_{yyyy}-u_{tt}=h\left(u\right)$ and determine Lie, Noether symmetries, and conservation laws.

Variational symmetries or invariance of an underlying variational functional are widely studied in these references [4,7,8,9,10,11,12]. The invariance approach is used to reduce the system of equations to an equivalent system of simpler form. Furthermore, double reductions of the differential equation are obtained by means of Noether symmetries [13,14]. Herein, conserved vectors are calculated by using the celebrated Noether’s theorem [13,14].

The biharmonic equation has been widely studied in elasticity theory, particularly for modeling thin plate deformation, with foundational work by Love [15] and Timoshenko [16]. Variational methods have been studied for nonlinear biharmonic problems where critical point theory, Sobolev embeddings, and energy minimization techniques have been used to prove existence and qualitative properties of solutions [17,18,19,20]. These approaches often focus on existence, multiplicity, and regularity rather than closed-form solutions.

In recent studies, symmetry analysis has gained attention for exploring higher-order dispersive systems [21,22] though many results remain numerical or rely on perturbation methods. The present work focuses on Lie symmetries, double reduction via the Noether approach, and traveling wave transformations to derive explicit analytical solutions of nonlinear biharmonic-type PDEs arising from variational principles. The exact solutions are obtained, ranging from quartic to modulated and cubic traveling waves. These solutions offer direct physical interpretation and complement the existing literature by providing tractable, interpretable models for wave propagation, elastic deformation, and related phenomena.

2. Determining Equations

Consider the following Lagrangian equation:

$$L=h\left(u\right)-{\displaystyle \frac{1}{2}}{u}_{xx}^2-{\displaystyle \frac{1}{2}}{u}_{yy}^2-u_{xy}^2-{\displaystyle \frac{1}{2}}{u}_t^2$$

(1)

giving rise to the PDE

$$h^{\prime }\left(u\right)-u_{xxxx}-u_{yyyy}-2u_{xxyy}+u_{tt}=0$$

(2)

or using ${\nabla }^2={\displaystyle \frac{{\partial }^2}{\partial x^2}}+{\displaystyle \frac{{\partial }^2}{\partial y^2}}$, this equation can be put in the following form

$${\nabla }^2\left({\nabla }^{2}u\right)=u_{tt}+g\left(u\right),\mathrm{where}{h}^{\prime }\left(u\right)=g\left(u\right).$$

(3)

We consider the equation in (2) with infinitesimal generators written as follows:

$$\mathcal{X}=\xi {\displaystyle \frac{\partial }{\partial x}}+\zeta {\displaystyle \frac{\partial }{\partial y}}+\tau {\displaystyle \frac{\partial }{\partial t}}+\eta {\displaystyle \frac{\partial }{\partial u}}.$$

(4)

The higher order prolongations of the vector field $\mathcal{X}$ in (4) are given as follows:

$$\begin{array}{cc}\hfill {\mathcal{X}}^{\left[4\right]}=& \mathcal{X}+{\eta }_x^{\left(1\right)}{\displaystyle \frac{\partial }{\partial u_x}}+{\eta }_y^{\left(1\right)}{\displaystyle \frac{\partial }{\partial u_y}}+{\eta }_{xx}^{\left(2\right)}{\displaystyle \frac{\partial }{\partial u_{xx}}}+\dots \dots \dots \hfill \\ \hfill & +{\eta }_{xxyy}^{\left(4\right)}{\displaystyle \frac{\partial }{\partial u_{xxyy}}}+{\eta }_{yyyy}^{\left(4\right)}{\displaystyle \frac{\partial }{\partial u_{yyyy}}}.\hfill \end{array}$$

(5)

from which the invariance condition is

$${\mathcal{X}}^{\left[4\right]}{\left(u_{xxxx}+2u_{xxyy}+u_{yyyy}-u_{tt}-g\left(u\right)\right)}_{u_{xxxx}+2u_{xxyy}+u_{yyyy}-u_{tt}=g\left(u\right)}=0.$$

(6)

or

$${\left({\eta }_{xxxx}^{\left(4\right)}+2{\eta }_{xxxx}^{\left(4\right)}+{\eta }_{yyyy}^{\left(4\right)}-{\eta }_{tt}^{\left(2\right)}-\eta g^{\prime }\left(u\right)\right)}_{u_{xxxx}+2u_{xxyy}+u_{yyyy}-u_{tt}=g\left(u\right)}=0.$$

(7)

2.1. Symmetry Classification

Following detailed calculations, we study important cases for the ‘external force’, g.

Firstly, if g depends on u arbitrarily, then from Equation (7), we obtain PDEs, written as follows:

$$\begin{array}{c}{\xi }_x=0, {\xi }_t=0, {\xi }_u=0, {\xi }_y=-{\zeta }_x, {\zeta }_y=0, {\zeta }_t=0, {\zeta }_u=0\hfill \end{array}$$

(8)

$$\begin{array}{c}{\tau }_x=0, {\tau }_y=0, {\tau }_t=0, {\tau }_u=0, \eta =0.\hfill \end{array}$$

(9)

From the above PDEs, we have the following:

$$\begin{array}{c}\hfill \tau =c_0, \xi =-c_{1}y+c_2, \zeta =c_{1}x+c_3, \eta =0.\end{array}$$

(10)

Therefore, in the case of arbitrary $h\left(u\right)$, we obtain the following symmetry generators:

$${\mathcal{X}}_1={\displaystyle \frac{\partial }{\partial x}}, {\mathcal{X}}_2={\displaystyle \frac{\partial }{\partial y}}, {\mathcal{X}}_3={\displaystyle \frac{\partial }{\partial t}}, {\mathcal{X}}_4=-y{\displaystyle \frac{\partial }{\partial x}}+x{\displaystyle \frac{\partial }{\partial y}}.$$

(11)

The following commutation table gives the closed form of the algebra associated with the generators:

$\left[{\mathcal{X}}_i,{\mathcal{X}}_j\right]$	${\mathcal{X}}_1$	${\mathcal{X}}_2$	${\mathcal{X}}_3$	${\mathcal{X}}_4$
${\mathcal{X}}_1$	0	0	0	${\mathcal{X}}_2$
${\mathcal{X}}_2$	0	0	0	$-{\mathcal{X}}_1$
${\mathcal{X}}_3$	0	0	0	0
${\mathcal{X}}_4$	$-{\mathcal{X}}_2$	${\mathcal{X}}_1$	0	0

Case I. If $g=0$, the determining equations are as follows:

$$\begin{array}{c}\hfill {\eta }_{uu}=0, {\eta }_{xu}=0, {\eta }_{yu}=0, {\eta }_{tu}=0, {\eta }_{xxxx}^{\left(4\right)}={\eta }_{tt}^{\left(2\right)}-2{\eta }_{xxyy}^{\left(4\right)}-{\eta }_{yyyy}^{\left(4\right)},\end{array}$$

(12)

$$\begin{array}{c}\hfill {\xi }_u=0, {\xi }_t=0, {\xi }_x={\displaystyle \frac{1}{2}}{\tau }_t, {\zeta }_u=0,{\zeta }_t=0, {\zeta }_y={\displaystyle \frac{1}{2}}{\tau }_t, {\zeta }_{xx}=0,\end{array}$$

(13)

$$\begin{array}{c}\hfill {\tau }_u=0, {\tau }_x=0, {\tau }_y=0, {\tau }_{tt}=0.\end{array}$$

(14)

From the above equations, we obtain the following infinitesimals:

$$\begin{array}{cc}\hfill & \tau =c_{0}t+c_1, \xi ={\displaystyle \frac{1}{2}}{c}_{0}x+c_{2}y+c_3, \zeta ={\displaystyle \frac{1}{2}}{c}_{0}y-c_{2}x+c_4, \eta =c_{5}u.\hfill \end{array}$$

(15)

The principal algebra is given by the following:

$$\begin{array}{cc}\hfill & {\mathcal{X}}_1={\displaystyle \frac{\partial }{\partial x}}, {\mathcal{X}}_2={\displaystyle \frac{\partial }{\partial y}}, {\mathcal{X}}_3={\displaystyle \frac{\partial }{\partial t}}, {\mathcal{X}}_4=y{\displaystyle \frac{\partial }{\partial x}}-x{\displaystyle \frac{\partial }{\partial y}},\hfill \\ \hfill & {\mathcal{X}}_5={\displaystyle \frac{1}{2}}x{\displaystyle \frac{\partial }{\partial x}}+{\displaystyle \frac{1}{2}}y{\displaystyle \frac{\partial }{\partial y}}+t{\displaystyle \frac{\partial }{\partial t}}, {\mathcal{X}}_6=u{\displaystyle \frac{\partial }{\partial u}}.\hfill \end{array}$$

(16)

The commutation table in this case is written as follows:

$\left[{\mathcal{X}}_i,{\mathcal{X}}_j\right]$	${\mathcal{X}}_1$	${\mathcal{X}}_2$	${\mathcal{X}}_3$	${\mathcal{X}}_4$	${\mathcal{X}}_5$	${\mathcal{X}}_6$
${\mathcal{X}}_1$	0	0	0	$-{\mathcal{X}}_2$	${\displaystyle \frac{1}{2}}{\mathcal{X}}_1$	0
${\mathcal{X}}_2$	0	0	0	${\mathcal{X}}_1$	${\displaystyle \frac{1}{2}}{\mathcal{X}}_2$	0
${\mathcal{X}}_3$	0	0	0	0	${\mathcal{X}}_3$	0
${\mathcal{X}}_4$	${\mathcal{X}}_2$	$-{\mathcal{X}}_1$	0	0	0	0
${\mathcal{X}}_5$	$-{\displaystyle \frac{1}{2}}{\mathcal{X}}_1$	$-{\displaystyle \frac{1}{2}}{\mathcal{X}}_2$	$-{\mathcal{X}}_3$	0	0	0
${\mathcal{X}}_6$	0	0	0	0	0	0

Case II. If $g=u$, we have the following conservationitesimals:

$$\begin{array}{c}\hfill \tau =c_0,\xi =-c_{1}y+c_2,\zeta =c_{1}x+c_3,\eta =c_{4}u.\end{array}$$

(17)

The corresponding symmetry generators are as follows:

$$\begin{array}{cc}\hfill & {\mathcal{X}}_1={\displaystyle \frac{\partial }{\partial x}},{\mathcal{X}}_2={\displaystyle \frac{\partial }{\partial y}},{\mathcal{X}}_3={\displaystyle \frac{\partial }{\partial t}},{\mathcal{X}}_4=-y{\displaystyle \frac{\partial }{\partial x}}+x{\displaystyle \frac{\partial }{\partial y}},{\mathcal{X}}_5=u{\displaystyle \frac{\partial }{\partial u}}.\hfill \end{array}$$

(18)

The Lie algebra in case II is five-dimensional and the corresponding commutator table of symmetry generators is given below:

$\left[{\mathcal{X}}_i,{\mathcal{X}}_j\right]$	${\mathcal{X}}_1$	${\mathcal{X}}_2$	${\mathcal{X}}_3$	${\mathcal{X}}_4$	${\mathcal{X}}_5$
${\mathcal{X}}_1$	0	0	0	${\mathcal{X}}_2$	0
${\mathcal{X}}_2$	0	0	0	$-{\mathcal{X}}_1$	0
${\mathcal{X}}_3$	0	0	0	0	0
${\mathcal{X}}_4$	$-{\mathcal{X}}_2$	${\mathcal{X}}_1$	0	0	0
${\mathcal{X}}_5$	0	0	0	0	0

Case III. If $g=e^u$, we obtain the following symmetry generators:

$$\begin{array}{cc}\hfill & {\mathcal{X}}_1={\displaystyle \frac{\partial }{\partial x}},{\mathcal{X}}_2={\displaystyle \frac{\partial }{\partial y}},{\mathcal{X}}_3={\displaystyle \frac{\partial }{\partial t}},{\mathcal{X}}_4=-y{\displaystyle \frac{\partial }{\partial x}}+x{\displaystyle \frac{\partial }{\partial y}},{\mathcal{X}}_5={\displaystyle \frac{1}{2}}x{\displaystyle \frac{\partial }{\partial x}}+{\displaystyle \frac{1}{2}}y{\displaystyle \frac{\partial }{\partial y}}+t{\displaystyle \frac{\partial }{\partial t}}-2{\displaystyle \frac{\partial }{\partial u}}.\hfill \end{array}$$

Here in case III, the Lie algebra is five-dimensional. The commutation relations corresponding to these symmetry generators are given below:

$\left[{\mathcal{X}}_i,{\mathcal{X}}_j\right]$	${\mathcal{X}}_1$	${\mathcal{X}}_2$	${\mathcal{X}}_3$	${\mathcal{X}}_4$	${\mathcal{X}}_5$
${\mathcal{X}}_1$	0	0	0	${\mathcal{X}}_2$	${\displaystyle \frac{1}{2}}{\mathcal{X}}_1$
${\mathcal{X}}_2$	0	0	0	$-{\mathcal{X}}_1$	${\displaystyle \frac{1}{2}}{\mathcal{X}}_2$
${\mathcal{X}}_3$	0	0	0	0	${\mathcal{X}}_3$
${\mathcal{X}}_4$	$-{\mathcal{X}}_2$	${\mathcal{X}}_1$	0	0	${\displaystyle \frac{1}{2}}{\mathcal{X}}_4$
${\mathcal{X}}_5$	$-{\displaystyle \frac{1}{2}}{\mathcal{X}}_1$	$-{\displaystyle \frac{1}{2}}{\mathcal{X}}_2$	$-{\mathcal{X}}_3$	0	$-{\displaystyle \frac{1}{2}}{\mathcal{X}}_4$

2.2. Noether Symmetries

Consider the following Lagrangian equation:

$$L=h\left(u\right)-{\displaystyle \frac{1}{2}}{u}_t^2-{\displaystyle \frac{1}{2}}{u}_{xx}^2-{\displaystyle \frac{1}{2}}{u}_{yy}^2-u_{xy}^2.$$

(19)

The symmetry generator $\mathcal{X}$ given in (4) is often known as a variational symmetry (Noether symmetry) if the functional $\int \int \int Ldxdydt$ is invariant under the vector field. The vector field $\mathcal{X}$ with zero gauge satisfies the following constraint:

$${\mathcal{X}}^{\left[2\right]}L+\left(D_t\tau +D_x\xi +D_y\zeta \right)L=0.$$

(20)

The Noether symmetries for the case in which the gauge is taken as zero are referred to as ‘variational symmetries’. The vector field $\mathcal{X}$, prolonged to the second-order, is written as follows

$$\begin{array}{cc}\hfill & {\mathcal{X}}^{\left[2\right]}=\mathcal{X}+{\eta }_x^{\left(1\right)}{\displaystyle \frac{\partial }{\partial u_x}}+{\eta }_y^{\left(1\right)}{\displaystyle \frac{\partial }{\partial u_y}}+{\eta }_t^{\left(1\right)}{\displaystyle \frac{\partial }{\partial u_t}}+{\eta }_{xx}^{\left(2\right)}{\displaystyle \frac{\partial }{\partial u_{xx}}}+{\eta }_{xy}^{\left(2\right)}{\displaystyle \frac{\partial }{\partial u_{xy}}}+{\eta }_{yy}^{\left(2\right)}{\displaystyle \frac{\partial }{\partial u_{yy}}}.\hfill \end{array}$$

(21)

For arbitrary $h\left(u\right)$ we obtain the following Noether symmetries:

$$\begin{array}{cc}\hfill & V_1={\displaystyle \frac{\partial }{\partial x}},V_2={\displaystyle \frac{\partial }{\partial y}},V_3={\displaystyle \frac{\partial }{\partial t}},V_4=y{\displaystyle \frac{\partial }{\partial x}}-x{\displaystyle \frac{\partial }{\partial y}}.\hfill \end{array}$$

(22)

The only nontrivial case that yields an additional Noether symmetry is $h\left(u\right)=e^u$, in which case Noether symmetry is written as follows

$$\begin{array}{c}\hfill V_5={\displaystyle \frac{1}{2}}x{\displaystyle \frac{\partial }{\partial x}}+{\displaystyle \frac{1}{2}}y{\displaystyle \frac{\partial }{\partial y}}+t{\displaystyle \frac{\partial }{\partial t}}-2{\displaystyle \frac{\partial }{\partial u}}.\end{array}$$

(23)

We, thus, conclude that the principal Lie algebra of Lie symmetry generators contains an algebra of Noether symmetries. The additional Lie symmetry generator for the case $g=u$, viz., ${\mathcal{X}}_5$ in case II, is not a Noether symmetry. However, for the $g=e^u$ case, it admits an additional conservation law. These aspects, will be discussed in the following section.

3. Conservation Laws

The conservation laws are expressed in the notion of conserved vectors, written as follows:

$$D_t{T}^t+D_x{T}^x+D_y{T}^y=0$$

(24)

subject to $\mathcal{E}(t,x,y,u,u_t,u_x,u_y,u_{xx},\dots )=0$.

Next, we present the Noether’s theorem and use it to calculate the conserved vectors. The operator $\mathcal{X}$ in (4) is a variational symmetry, in particular with zero gauge, if it optimizes the following function:

$$\int \int \int Ldtdxdy.$$

Moreover, the vector $T^x,T^y$ and $T^t$ are written as follows:

$$\begin{array}{cc}\hfill & T^x=L\xi +w\left({\displaystyle \frac{\partial L}{\partial u_x}}-D_x{\displaystyle \frac{\partial L}{\partial u_{xx}}}-D_y{\displaystyle \frac{\partial L}{\partial u_{xy}}}\right)+D_{x}w{\displaystyle \frac{\partial L}{\partial u_{xx}}}+D_{y}w{\displaystyle \frac{\partial L}{\partial u_{xy}}},\hfill \\ \hfill & T^y=L\zeta +w\left({\displaystyle \frac{\partial L}{\partial u_y}}-D_x{\displaystyle \frac{\partial L}{\partial u_{xy}}}-D_y{\displaystyle \frac{\partial L}{\partial u_{yy}}}\right)+D_{x}w{\displaystyle \frac{\partial L}{\partial u_{xy}}}+D_{y}w{\displaystyle \frac{\partial L}{\partial u_{yy}}},\hfill \\ \hfill & T^t=L\tau +w{\displaystyle \frac{\partial L}{\partial u_t}},\hfill \end{array}$$

(25)

where $w=\eta -u_x\xi -u_y\zeta -u_t\tau$ is the characteristic of $\mathcal{X}$.

Arbitrary $h\left(u\right)$

In general, we obtain the following:

i. $w=-u_x$ linear momentum in x:

$$\begin{array}{cc}\hfill & T^x=L-u_x\left(u_{xxx+2u_{xyy}}\right)+u_{xx}^2+2u_{xy}^2\hfill \\ \hfill & T^y=-u_x\left(2u_{xxy+u_{yyy}}\right)+2u_{xx}{u}_{xy}+u_{xy{u}_{yy}}\hfill \\ \hfill & T^t=u_x{u}_t\hfill \end{array}$$

(26)

ii. $w=-u_y$ linear momentum in y:

$$\begin{array}{cc}\hfill & T^x=-u_y\left(u_{xxx+2u_{xyy}}\right)+u_{xx}{u}_{xy}+2u_{xy}{u}_{yy}\hfill \\ \hfill & T^y=L-u_y\left(u_{yyy+2u_{xxy}}\right)+2u_{xy}^2+u_{yy}^2\hfill \\ \hfill & T^t=u_y{u}_t\hfill \end{array}$$

(27)

iii. $w=-u_t$ energy:

$$\begin{array}{cc}\hfill & T^x=-u_t(u_{xxx}+2u_{xyy})+u_{xt}{u}_{xx}+2u_{yt}{u}_{xy}\hfill \\ \hfill & T^y=-u_t(2u_{xxy}+u_{yyy})+2u_{xt}{u}_{xy}+u_{yt}{u}_{yy}\hfill \\ \hfill & T^t=L+u_t^2\hfill \\ \hfill & =h\left(u\right)-{\displaystyle \frac{1}{2}}{u}_{xx}^2-{\displaystyle \frac{1}{2}}{u}_{yy}^2-u_{xy}^2+{\displaystyle \frac{1}{2}}{u}_t^2\hfill \end{array}$$

(28)

iv. $w=y{u}_x-x{u}_y$—angular momentum:

$$\begin{array}{cc}\hfill & T^x=yL+w(u_{xxx}+2u_{xyy})-u_{xx}(y{u}_{xx}-x{u}_{xy}-u_y)\hfill \\ \hfill & -2u_{xy}(u_y+y{u}_{xy}-x{u}_{yy})\hfill \\ \hfill & T^y=-xL+w(u_{yyy}+2u_{xxy})+u_{yy}(y{u}_{xy}-x{u}_{yy}+u_x)\hfill \\ \hfill & -2u_{xy}(u_y-y{u}_{xx}+x{u}_{xy})\hfill \\ \hfill & T^t=-w{u}_t\hfill \end{array}$$

(29)

For $h=\gamma e^u$, we obtain, additionally, $w=-2-{\displaystyle \frac{1}{2}}x{u}_x-{\displaystyle \frac{1}{2}}y{u}_y-t{u}_t$, $L=\gamma e^u-{\displaystyle \frac{1}{2}}{u}_t^2-{\displaystyle \frac{1}{2}}{u}_{xx}^2-{\displaystyle \frac{1}{2}}{u}_{yy}^2-u_{xy}^2$ and $(T^x,T^y,T^t)$ is determined similarly from the forms given above.

$$\begin{array}{cc}\hfill & T^x={\displaystyle \frac{1}{2}}xL+w(u_{xxx}+2u_{xyy})+{\displaystyle \frac{1}{2}}{u}_{xx}(x{u}_{xx}+y{u}_{xy}+2t{u}_{xt}+u_x)\hfill \\ \hfill & +u_{xy}(u_y+x{u}_{xy}+y{u}_{yy}+2t{u}_{yt})\hfill \\ \hfill & T^y={\displaystyle \frac{1}{2}}yL+w(u_{yyy}+2u_{xxy})+{\displaystyle \frac{1}{2}}{u}_{yy}(x{u}_{xy}+y{u}_{yy}+2t{u}_{yt}+u_y)\hfill \\ \hfill & +u_{xy}(u_x+x{u}_{xx}+y{u}_{xy}+2t{u}_{xt})\hfill \\ \hfill & T^t=tL-w{u}_t\hfill \end{array}$$

(30)

4. Reductions

We suppose the traveling wave transformation $\zeta =\mu x-\nu y-t$ and transform $(T^x,T^y,T^t)$ to $({\mathsf{\Phi }}^{\rho },{\mathsf{\Phi }}^{\chi },{\mathsf{\Phi }}^{\zeta })$ with two translations via symmetries, written as follows:

$$\begin{array}{c}\hfill Y_1=({\mu }^2+{\nu }^2){\partial }_t+\mu {\partial }_x+\nu {\partial }_y\mathrm{and}{Y}_2=\nu {\partial }_x-\mu {\partial }_y.\end{array}$$

(31)

From $Y_1\rho =1$ and $Y_2\chi =1$, otherwise it is zero, we obtain the following:

$$\begin{array}{c}\hfill \rho =c^{2}t,\chi ={\displaystyle \frac{1}{\nu }}x,\end{array}$$

(32)

where $c^2={\displaystyle \frac{1}{{\mu }^2+{\nu }^2}}$. Thus, the following can be shown:

$$\begin{array}{cc}\hfill & {\mathsf{\Phi }}^{\rho }=c^2{T}^t,\hfill \\ \hfill & {\mathsf{\Phi }}^{\chi }={\displaystyle \frac{1}{\nu }}{T}^x,\hfill \\ \hfill & {\mathsf{\Phi }}^{\zeta }=-T^t+\mu T^x-\nu T^y\hfill \end{array}$$

(33)

Since each of the Noether symmetries ${\mathcal{X}}_1$–${\mathcal{X}}_4$ is ‘associated’ with each T in (26)–(28), one can proceed to ‘double reduction’ of (2) (see [13,23] for details).

As a first, illustrative case, we consider the ‘energy conserved’ vector in (28) to perform the reduction, written as follows:

$$\begin{array}{cc}\hfill & {\mathsf{\Phi }}^{\rho }=c^2[h\left(u\right)-{\displaystyle \frac{1}{2}}{u}_{xx}^2-{\displaystyle \frac{1}{2}}{u}_{yy}^2-u_{xy}^2+{\displaystyle \frac{1}{2}}{u}_t^2],\hfill \\ \hfill & {\mathsf{\Phi }}^{\chi }={\displaystyle \frac{1}{\nu }}[-u_t(u_{xxx}+2u_{xyy})+u_{xt}{u}_{xx}+2u_{yt}{u}_{xy}],\hfill \\ \hfill & {\mathsf{\Phi }}^{\zeta }=-[h\left(u\right)-{\displaystyle \frac{1}{2}}{u}_{xx}^2-{\displaystyle \frac{1}{2}}{u}_{yy}^2-u_{xy}^2+{\displaystyle \frac{1}{2}}{u}_t^2]\hfill \\ \hfill & +\mu [-u_t(u_{xxx}+2u_{xyy})+u_{xt}{u}_{xx}+2u_{yt}{u}_{xy}]\hfill \\ \hfill & -\nu [-u_t(2u_{xxy}+u_{yyy})+2u_{xt}{u}_{xy}+u_{yt}{u}_{yy}]\hfill \end{array}$$

(34)

in which $(x,y,t,u)$ should be transformed to $(\rho ,\chi ,\zeta ,U)$, $U=u$. Finally, ${\mathsf{\Phi }}^{\zeta }=k$ is (2) doubly reduced, viz., written as follows:

$$\begin{array}{cc}\hfill & -[h\left(u\right)-{\displaystyle \frac{1}{2}}{u}_{xx}^2-{\displaystyle \frac{1}{2}}{u}_{yy}^2-u_{xy}^2+{\displaystyle \frac{1}{2}}{u}_t^2]\hfill \\ \hfill & +\mu [-u_t(u_{xxx}+2u_{xyy})+u_{xt}{u}_{xx}+2u_{yt}{u}_{xy}]\hfill \\ \hfill & -\nu [-u_t(2u_{xxy}+u_{yyy})+2u_{xt}{u}_{xy}+u_{yt}{u}_{yy}]=k\hfill \end{array}$$

(35)

so that

$$\begin{array}{cc}\hfill & -[h\left(U\right)-{\displaystyle \frac{1}{2}}{\mu }^4{U}^{\prime \prime 2}-{\displaystyle \frac{1}{2}}{\nu }^4{U}^{\prime \prime 2}+{\mu }^2{\nu }^2{U}^{\prime \prime 2}+{\displaystyle \frac{1}{2}}{U}^{\prime 2}]\hfill \\ \hfill & +\mu [U^{\prime }({\mu }^3{U}^{\prime \prime \prime }+2\mu {\nu }^2{U}^{\prime \prime \prime })-{\mu }^3{U}^{\prime \prime 2}-2\mu {\nu }^2{U}^{\prime \prime 2}]\hfill \\ \hfill & -\nu [U^{\prime }(-2\nu {\mu }^2{U}^{\prime \prime \prime }-{\nu }^3{U}^{\prime \prime \prime })+2{\mu }^2\nu U^{\prime \prime 2}+{\nu }^3{U}^{\prime \prime 2}]=k\hfill \end{array}$$

(36)

or

$$\begin{array}{cc}\hfill & -h\left(U\right)-{\displaystyle \frac{1}{2}}({\mu }^4+6{\mu }^2{\nu }^2+{\nu }^4)U^{\prime \prime 2}+({\mu }^4+4{\mu }^2{\nu }^2+{\nu }^4)U^{\prime }{U}^{\prime \prime \prime }=k.\hfill \end{array}$$

(37)

1. For a strict energy solution to (2), we would consider the ‘energy conserved’ vector in (28) with ${\mathcal{X}}_3={\partial }_t$.

2. For double reduction by rotation (angular momentum), we would perform an equivalent calculation using symmetry ${\mathcal{X}}_4=x{\partial }_y-y{\partial }_x$ and its associate conserved vector in (29). Here, we obtain the following:

$$\begin{array}{c}\hfill \zeta =x^2+y^2,\rho =t,\chi =arcsin{\displaystyle \frac{y}{\sqrt{x^2+y^2}}}.\end{array}$$

(38)

3. For $h=e^u$, ${\mathcal{X}}_5$ in Case III is associated with (30) from which a solution of (2) would be an exact scaling in $(x,y,t)$ and translation in u combined solution.

5. Analysis of Reduced ODE

A classification of exact solutions and the forms of solutions obtained under different assumptions for the nonlinear function $h\left(U\right)$ in the double reduction framework is presented in

Table 1. Summary of symmetry reductions and corresponding compact solution forms.

Case	$\mathit{h}\left(\mathit{U}\right)$	Symmetry/Ansatz	Solution Form	Type
I	$U\left(\zeta \right)$	$\zeta =\mu x-\nu y-t$	$u\left(x,y,t\right)={\displaystyle \frac{{\left(\mu x-\nu y-t\right)}^4}{-24{\mu }^4+48{\mu }^2{\nu }^2-24{\nu }^4}}$	Quartic traveling wave
II	$U^{\prime }\left(\zeta \right)$	Symmetry reduction	$u_1\left(x,y,t\right)=$ (58)	Modulated wave
III	$U^{\prime }\left(\zeta \right)$	$p\left(\zeta \right)=k{\zeta }^n$	$u_2\left(x,y,t\right)={\displaystyle \frac{1}{12}}{\displaystyle \frac{{(\mu x-\nu y-t)}^3}{{\mu }^2{\nu }^2}}+c_1.$	Cubic traveling wave

. Case I assumes a linear potential $h\left(U\right)=U\left(\zeta \right)$ and yields a quartic traveling wave solution under the transformation $\zeta =\mu x-\nu y-t$ representing a localized bending wave commonly seen in elastic plate dynamics. Case II uses $h\left(U\right)=U^{\prime }\left(\zeta \right)$ and employs Lie symmetry reduction to produce a modulated nonlinear wave $u_1(x,y,t)$ whose explicit form is given in Equation (58). Case III also uses $h\left(U\right)=U^{\prime }\left(\zeta \right)$ but with a power-law ansatz $p\left(\zeta \right)=k{\zeta }^n$, leading to a cubic traveling wave $u_2(x,y,t)$ with an asymmetric profile.

5.1. Analysis of Reduced ODE for Case I, When $h\left(U\right)=U\left(\zeta \right)$

The ODE (37) can be re-written as follows:

$$U\left(\zeta \right)-{\displaystyle \frac{1}{2}}B{\left({\displaystyle \frac{{\mathrm{d}}^2}{\mathrm{d}{\zeta }^2}}U\left(\zeta \right)\right)}^2+A\left({\displaystyle \frac{\mathrm{d}}{\mathrm{d}\zeta }}U\left(\zeta \right)\right){\displaystyle \frac{{\mathrm{d}}^3}{\mathrm{d}{\zeta }^3}}U\left(\zeta \right)=0$$

(39)

by substituting $g\left(U\right)=U\left(\zeta \right)$, ${\mu }^4+4{\mu }^2{\nu }^2+{\nu }^4=A$ and ${\mu }^4+6{\mu }^2{\nu }^2+{\nu }^4=B$. ODE (39) is has the following two symmetries given by the following:

$$\begin{array}{c}\hfill {\mathcal{X}}_1={\displaystyle \frac{\partial }{\partial x}},{\mathcal{X}}_2=u{\displaystyle \frac{\partial }{\partial u}}+{\displaystyle \frac{1}{4}}x{\displaystyle \frac{\partial }{\partial x}}.\end{array}$$

(40)

ODE (39) under symmetry ${\mathcal{X}}_1={\displaystyle \frac{\partial }{\partial x}}$ further reduced to the following nonlinear ODE:

$$\begin{array}{c}\hfill A\left({\displaystyle \frac{{\mathrm{d}}^2}{\mathrm{d}{\alpha }^2}}w\left(\alpha \right)\right){\left(w\left(\alpha \right)\right)}^3+{\left(w\left(\alpha \right)\right)}^2\left(A-{\displaystyle \frac{B}{2}}\right){\left({\displaystyle \frac{\mathrm{d}}{\mathrm{d}\alpha }}w\left(\alpha \right)\right)}^2+\alpha =0.\end{array}$$

(41)

under the invariant variables $u\left(x\right)=w\left(\alpha \right)x+c_1$ and $\alpha =u\left(x\right)$. ODE (41) has Lie symmetry, written as follows:

$$\begin{array}{c}\hfill {\mathcal{X}}_1={\displaystyle \frac{3}{4}}u{\displaystyle \frac{\partial }{\partial u}}+x{\displaystyle \frac{\partial }{\partial x}}.\end{array}$$

(42)

and has a reduced form which is also nonlinear ODE and requires a complete analysis to solve.

Now, we look for some particular form of solutions for the ODE (39) by considering the value of $U\left(\zeta \right)=a{\zeta }^k$, where a and k are constants. By substituting the value of $U\left(\zeta \right)=a{\zeta }^k$ and its derivatives in ODE (39), we obtain the following:

$$\begin{array}{c}\hfill a\left(a\left(\left(A-B/2\right)k-2A+B/2\right)k^2\left(k-1\right){\zeta }^k+{\zeta }^4\right){\zeta }^{k-4}=0.\end{array}$$

(43)

For the Equation (43) to hold for all $\zeta$, we obtain $k=4$ by matching exponents of $\zeta$. Equation (43) is further reduced to the following:

$$\begin{array}{c}\hfill a\left(48a\left(2A-3/2B\right){\zeta }^4+{\zeta }^4\right)=0.\end{array}$$

(44)

That provides us the value of $a=-{\left(96A-72B\right)}^{-1}$ and finally we obtain the solution of the ODE (39) given by the following:

$$\begin{array}{c}\hfill U\left(\zeta \right)={\displaystyle \frac{{\zeta }^4}{72B-96A}}\end{array}$$

(45)

Now for the invariant solution of the Euler–Lagrange Equation (2), we substitute the traveling wave transformation $\zeta =\mu x-\nu y-t$ and $U=u$ back, from which we have the following:

$$\begin{array}{c}\hfill u\left(x,y,t\right)={\displaystyle \frac{{(\mu x-\nu y-t)}^4}{72B-96A}},\end{array}$$

(46)

where

$$\begin{array}{c}\hfill {\mu }^4+4{\mu }^2{\nu }^2+{\nu }^4=A,\end{array}$$

(47)

$$\begin{array}{c}\hfill {\mu }^4+6{\mu }^2{\nu }^2+{\nu }^4=B.\end{array}$$

(48)

The 3D plot of the solution is calculated as follows:

$$\begin{array}{c}\hfill u\left(x,y,t\right)={\displaystyle \frac{{\left(\mu x-\nu y-t\right)}^4}{-24{\mu }^4+48{\mu }^2{\nu }^2-24{\nu }^4}}.\end{array}$$

(49)

and is shown in the Figure 1.

Solution $u\left(x,y,t\right)$ is presented via invariance analysis, and its physical interpretations depict a traveling disturbance moving along a direction determined by $\mu$ and $\nu$. This form naturally arises in the context of elasticity theory, specifically in the bending dynamics of thin plates governed by the biharmonic operator. Here u can be interpreted as the transverse displacement of an elastic plate, with the solution corresponding to a localized bending wave. In wave propagation models this solution captures dispersive behavior seen in surface waves where higher-order spatial derivatives play a role. From a quantum mechanical perspective, the governing equation resembles those in higher-order effective field theories where the polynomial traveling wave solution models excitations in structured media.

5.2. Analysis of Reduced ODE for Case II, When $h\left(U\right)={\displaystyle \frac{\mathrm{d}}{\mathrm{d}\zeta }}U\left(\zeta \right)$

The ODE (37) can be further modified to the following:

$${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\zeta }}U\left(\zeta \right)-{\displaystyle \frac{1}{2}}B{\left({\displaystyle \frac{{\mathrm{d}}^2}{\mathrm{d}{\zeta }^2}}U\left(\zeta \right)\right)}^2+A\left({\displaystyle \frac{\mathrm{d}}{\mathrm{d}\zeta }}U\left(\zeta \right)\right){\displaystyle \frac{{\mathrm{d}}^3}{\mathrm{d}{\zeta }^3}}U\left(\zeta \right)=0$$

(50)

by substituting $h\left(U\right)={\displaystyle \frac{\mathrm{d}}{\mathrm{d}\zeta }}U\left(\zeta \right)$, ${\mu }^4+4{\mu }^2{\nu }^2+{\nu }^4=A$ and ${\mu }^4+6{\mu }^2{\nu }^2+{\nu }^4=B$. ODE (50) can be reduce by one order after substituting the following:

$${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\zeta }}U\left(\zeta \right)=v\left(\zeta \right).$$

The ODE (50) takes the form given by the following:

$$\begin{array}{c}\hfill -v\left(\zeta \right)-{\displaystyle \frac{1}{2}}B{\left({\displaystyle \frac{\mathrm{d}}{\mathrm{d}\zeta }}v\left(\zeta \right)\right)}^2+Av\left(\zeta \right){\displaystyle \frac{{\mathrm{d}}^2}{\mathrm{d}{\zeta }^2}}v\left(\zeta \right)=0\end{array}$$

(51)

The ODE (51) is nonlinear and has the following symmetries:

$$\begin{array}{c}\hfill {\mathcal{X}}_1={\displaystyle \frac{\partial }{\partial \zeta }},{\mathcal{X}}_2=v{\displaystyle \frac{\partial }{\partial v}}+{\displaystyle \frac{1}{2}}\zeta {\displaystyle \frac{\partial }{\partial \zeta }}.\end{array}$$

(52)

5.2.1. Reduction Under ${\mathcal{X}}_1$

Reduction in ODE (51) under symmetry ${\mathcal{X}}_1={\displaystyle \frac{\partial }{\partial \zeta }}$ provides us with the following reduced ODE:

$$\begin{array}{c}\hfill -\alpha -{\displaystyle \frac{1}{2}}B{\left(w\left(\alpha \right)\right)}^2+A\alpha \left({\displaystyle \frac{\mathrm{d}}{\mathrm{d}\alpha }}w\left(\alpha \right)\right)w\left(\alpha \right)=0,\end{array}$$

(53)

that has similarity variables $v\left(\zeta \right)=w\left(\alpha \right)\zeta +c_2$ and $\alpha =v\left(\zeta \right)$. ODE (53) can easily be solved and has the solution $w\left(\alpha \right)$ given by the following:

$$\begin{array}{c}\hfill w\left(\alpha \right)=\pm \left({\displaystyle \frac{\sqrt{\left(A-B\right)\left({\alpha }^{\frac{B}{A}}{c}_{1}A-{\alpha }^{\frac{B}{A}}{c}_{1}B+2\alpha \right)}}{A-B}}\right),\end{array}$$

(54)

and $v\left(\zeta \right)$ then has the form

$$\begin{array}{c}\hfill v\left(\zeta \right)={\displaystyle \frac{\zeta }{A-B}}\sqrt{\left(A-B\right)\left({\alpha }^{{\displaystyle \frac{B}{A}}}{c}_{1}A-{\alpha }^{{\displaystyle \frac{B}{A}}}{c}_{1}B+2\alpha \right)}+c_2.\end{array}$$

(55)

Finally, the solution $U\left(\zeta \right)=\int v\left(\zeta \right)\mathrm{d}\zeta$ of ODE (50) can be presented by the following:

$$\begin{array}{c}\hfill U\left(\zeta \right)={\displaystyle \frac{{\zeta }^2}{2(A-B)}}\sqrt{\left(A-B\right)\left({\alpha }^{{\displaystyle \frac{B}{A}}}{c}_{1}A-{\alpha }^{{\displaystyle \frac{B}{A}}}{c}_{1}B+2\alpha \right)}+c_2\left(\zeta \right).\end{array}$$

(56)

Thus, the invariant solution of the Euler–Lagrange Equation (2) can be obtained by traveling wave transformation $\zeta =\mu x-\nu y-t$ and $U=u$, from which we obtain the following:

$$\begin{array}{c}\hfill u\left(x,y,t\right)={\displaystyle \frac{1}{2}}{\displaystyle \frac{{\left(\mu x-\nu y-t\right)}^2}{A-B}}\sqrt{\left(A-B\right)\left({\alpha }^{{\displaystyle \frac{B}{A}}}{c}_{1}A-{\alpha }^{{\displaystyle \frac{B}{A}}}{c}_{1}B+2\alpha \right)}+c_2\left(\mu x-\nu y-t\right),\end{array}$$

(57)

The final form of the solution is calculated as follows:

$$\begin{array}{c}\hfill u_1\left(x,y,t\right)={\displaystyle \frac{-\mu x+\nu y+t}{4{\mu }^2{\nu }^2}}\left(2\left(-\mu x+\nu y+t\right)\sqrt{{\mu }^2{\nu }^2\left({\mu }^2{\nu }^2{c}_1{\alpha }^{{\displaystyle \frac{{\mu }^4+6{\mu }^2{\nu }^2+{\nu }^4}{{\mu }^4+4{\mu }^2{\nu }^2+{\nu }^4}}}-\alpha \right)}+4c_2{\mu }^2{\nu }^2\right)\end{array}$$

(58)

can be presented in Figure 2 and the behavior of the solution (58) over time can be observed.

The solution $u_1\left(x,y,t\right)$ is presented via invariance analysis and its graphical representations illustrate a nonlinear traveling wave with time-varying amplitude and profile. The structure $-\mu x+\nu y+t$ indicates a propagating wave-front while the nonlinear amplitude modulation controlled by parameters $\alpha ,c_1,c_2$ reflects the effects of medium nonlinearity. This behavior is consistent with real-world models of dispersive waves in nonlinear media such as optical pulses in fibers or internal waves in fluids. In elasticity theory the solution can be interpreted as a dynamic bending mODE of a thin elastic plate undergoing nonlinear deformation. The time evolution visualized in the plots illustrates how such solutions can capture physically relevant evolving responses in systems with both spatial stiffness and inertial effects. These properties also align with features seen in effective field theories where higher-order spatial dynamics govern quantum or thermal excitations.

5.2.2. Reduction Under ${\mathcal{X}}_2$

Reduction of ODE (51) under symmetry ${\mathcal{X}}_1={\displaystyle \frac{\partial }{\partial \zeta }}$ gives us the following reduced ODE:

$$\begin{array}{c}\hfill 2A\alpha \left({\displaystyle \frac{\mathrm{d}}{\mathrm{d}\alpha }}w\left(\alpha \right)\right)w\left(\alpha \right)-2A\alpha w\left(\alpha \right)-B{\left(w\left(\alpha \right)\right)}^2-2\alpha x^2=0,\end{array}$$

(59)

that has similarity variables $v\left(\zeta \right)={\displaystyle \frac{1}{2}}w\left(\alpha \right){\zeta }^2+c_1$ and ${\zeta }^2\alpha =v\left(\zeta \right)$. ODE (59) has solution $w\left(\alpha \right)$ in the integral form given by the following:

$$\begin{array}{cc}& -{\displaystyle \frac{1}{\sqrt{\alpha }w\left(\alpha \right)}}(-{\int }^{-2{\displaystyle \frac{\sqrt{\alpha }}{w\left(\alpha \right)}}}{(-{\vartheta }^2+2A-2B)}^{-{\displaystyle \frac{A}{2A-2B}}}{\vartheta }^{{\displaystyle \frac{A}{A-B}}}d\vartheta \sqrt{\alpha }w\left(\alpha \right)\\ & +{(-2{\displaystyle \frac{\sqrt{\alpha }}{w\left(\alpha \right)}})}^{{\displaystyle \frac{A}{A-B}}}(w\left(\alpha \right)2^{-{\displaystyle \frac{A}{2A-2B}}}-2^{{\displaystyle \frac{A-2B}{2A-2B}}}\alpha ){\left({\displaystyle \frac{(A-B){\left(w\left(\alpha \right)\right)}^2-2\alpha }{{\left(w\left(\alpha \right)\right)}^2}}\right)}^{-{\displaystyle \frac{A}{2A-2B}}}\\ & -c_1\sqrt{\alpha }w\left(\alpha \right))=0.\end{array}$$

(60)

ODE (59) is nonlinear and requires a comprehensive analysis for its solution. ODE (59) can also be studied using numerical techniques.

5.3. Solution of the Form $p\left(\zeta \right)=k{\zeta }^n$ for ODE (50)

ODE (50) can be reduced by one order with the substitution ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}\zeta }}U\left(\zeta \right)=p\left(\zeta \right)$, from which we obtain the following:

$$\begin{array}{c}\hfill -p\left(\zeta \right)-{\displaystyle \frac{1}{2}}B{\left({\displaystyle \frac{\mathrm{d}}{\mathrm{d}\zeta }}p\left(\zeta \right)\right)}^2+Ap\left(\zeta \right){\displaystyle \frac{{\mathrm{d}}^2}{\mathrm{d}{\zeta }^2}}p\left(\zeta \right)=0.\end{array}$$

(61)

Considering $p\left(\zeta \right)=k{\zeta }^n$ and substituting it in the ODE (61), we obtain the following:

$$\begin{array}{c}\hfill \left(kn\left(\left(A-B/2\right)n-A\right){\zeta }^n-{\zeta }^2\right){\zeta }^{n-2}k=0,\end{array}$$

(62)

For Equation (62) to hold for all $\zeta$, we obtain $n=2$ by matching the exponents of $\zeta$. That provides us with the value of $k={\left(2A-2B\right)}^{-1}$ and finally we obtain the solution of the ODE (50) given by the following:

$$\begin{array}{c}\hfill U\left(\zeta \right)={\displaystyle \frac{1}{6}}{\displaystyle \frac{{\zeta }^3}{A-B}}+c_1,\end{array}$$

(63)

or equivalently can be written as

$$\begin{array}{c}\hfill U\left(\zeta \right)={\displaystyle \frac{1}{12}}{\displaystyle \frac{{\zeta }^3}{{\mu }^2{\nu }^2}}+c_1.\end{array}$$

(64)

Finally, the solution of the Euler–Lagrange Equation (2) can be obtained by traveling wave transformation $\zeta =\mu x-\nu y-t$ and $U=u$, so we have the following:

$$\begin{array}{c}\hfill u_2\left(x,y,t\right)={\displaystyle \frac{1}{12}}{\displaystyle \frac{{(\mu x-\nu y-t)}^3}{{\mu }^2{\nu }^2}}+c_1.\end{array}$$

(65)

can be presented in Figure 3 and the behavior of the solution can be observed over time. In Figure 3 2D plot of the solution $u_2\left(x,y,t\right)$ is also presented, which shows the behavior of the solution over time.

The solution $u_2(x,y,t)$ obtained represents a cubic-profile traveling wave. This solution propagates in space with an asymmetric wavefront as seen in both the 2D and 3D plots in Figure 4. Physically, this form can model nonlinear wave propagation in dispersive media where wave profiles steepen and travel without dispersion. It may also describe deformation fronts in thin elastic structures such as plates or membranes, especially in cases involving non-uniform stiffness or nonlinear stress–strain relationships. The evolving behavior over time illustrates how such solutions can capture the dynamics of localized unidirectional wave-like disturbances that arise in both mechanical and wave-based systems.

6. Conclusions

In this work, the fourth-order time-dependent partial differential equation is studied from the standpoint of symmetries and conservation laws. These results generalize Noether’s theorem. In addition, we reduce the partial differential equation and analyze the solutions of reduced equations.