The general form of a second order scalar linear PDE in two independent variables is $\mathrm{\Delta }=a{u}_{x,x}+b{u}_{x,y}+c{u}_{y,y}+d{u}_y+e{u}_x+f=0$, where $u=u\left(x\&comma;y\right)$ is the unknown function, the coefficients $a,b,c,d,e,f$ are functions of the independent variables $x,y$ and $ac-\frac{b^2}{4}<0$. The method of Laplace (not to be confused with integral transform methods of the same name) is a method which, when successful, will yield the general closed-form solution to such equations, depending upon two arbitrary functions of a single variable.

The method works by transforming the original PDE $\mathrm{\Delta }=0$ into another second order scalar linear PDE $\mathrm{\Delta }\&apos;=0$ with the remarkable property that solutions of $\mathrm{\Delta }=0$ can be found from solutions of $\mathrm{\Delta }\&apos;=0$ by differentiations and simple linear algebraic manipulations. In favorable circumstances solutions to equation $\mathrm{\Delta }\&apos;=0$ can be found and this then leads to solutions of the original equation.  If solutions to $\mathrm{\Delta }\&apos;=0$ cannot be found, then one may iterate the process to generate a sequence of equations $\mathrm{\Delta }=0,\mathrm{\Delta }\&apos;=0,\mathrm{\Delta }\text{'}\text{'}=0,\mathrm{...},{\mathrm{\Delta }}^{\left(n\right)}=0$ with the property that solutions to $\mathrm{\Delta }=0$ can be constructed from solutions to ${\mathrm{\Delta }}^{\left(n\right)}=0$ by differentiations and simple linear algebraic manipulations. The third optional argument in the calling sequence to Laplace specifies the number of iterations the procedure will calculate in attempting to arrive at a PDE ${\mathrm{\Delta }}^{\left(n\right)}=0$ which can be integrated.  The default numberofiterations is 5.

The specific details of the method of Laplace are easiest to explain for equations of the type $u_{x,y}+{\mathrm{Au}}_x+{\mathrm{Bu}}_y+\mathrm{Cu}=0$ (although this special form is not required for the procedure). For an equation of this form, define the Laplace invariants $H=\frac{\mathrm{dA}}{\mathrm{dx}}+\mathrm{AB}-C$ and $K=\frac{\mathrm{dB}}{\mathrm{dy}}+\mathrm{AB}-C$. One can show that if either $H=0$ or $K=0$ then the PDE can be integrated directly by linear ODE methods. If both $H=0$ and $K=0$ then the PDE can be easily transformed to the wave equation $v_{x,y}=0$ and the solution thus found.

If $H\ne 0$ then one defines $v=u_y+\mathrm{Au}$ and finds that: [1] $v$ also satisfies an equation of the form $v_{x,y}+A\&apos;v_x+B\&apos;v_y+C\&apos;v=0$; and [2] the equation can be inverted to give $u=\frac{1}{H\left(v_x+B\&apos;v\right)}$. A similar transform can be defined if $K\ne 0$. See the examples for an explicit computation of these transforms.

$\mathrm{with}\left(\mathrm{PDEtools}\&comma;\mathrm{Laplace}\&comma;\mathrm{declare}\right)$

$\left[\mathrm{Laplace}\&comma;\mathrm{declare}\right]$

$\mathrm{declare}\left(u\left(x\&comma;y\right)\right)$

$u\left(x\&comma;y\right)\mathrm{will\; now\; be\; displayed\; as}u$

$k≔\left(n-2\right)\left(n-1\right)$

$k≔\left(n-2\right)\left(n-1\right)$

$\mathrm{PDE}\left[n\right]≔\mathrm{diff}\left(u\left(x\&comma;y\right)\&comma;x\&comma;y\right)-\frac{k}{{\left(x+y\right)}^2}u\left(x\&comma;y\right)$

${\mathrm{PDE}}_n≔u_{x,y}-\frac{\left(n-2\right)\left(n-1\right)u}{{\left(x+y\right)}^2}$

$\mathrm{PDE}\left[5\right]≔\mathrm{subs}\left(n=5\&comma;\mathrm{PDE}\left[n\right]\right)$

${\mathrm{PDE}}_5≔u_{x,y}-\frac{12u}{{\left(x+y\right)}^2}$

$\mathrm{Laplace}\left(\mathrm{PDE}\left[5\right]\&comma;u\left(x\&comma;y\right)\right)$

$u=\frac{{\left(x+y\right)}^3{\mathrm{f\_\_1}}_{x,x,x}+{\left(x+y\right)}^3{\mathrm{f\_\_2}}_{y,y,y}-12{\left(x+y\right)}^2{\mathrm{f\_\_1}}_{x,x}-12{\left(x+y\right)}^2{\mathrm{f\_\_2}}_{y,y}+60\left(x+y\right){\mathrm{f\_\_1}}_x+60\left(x+y\right){\mathrm{f\_\_2}}_y-120\mathrm{f\_\_1}\left(x\right)-120\mathrm{f\_\_2}\left(y\right)}{1440{\left(x+y\right)}^3}$

$\mathrm{PDE}\left[6\right]≔\mathrm{subs}\left(n=6\&comma;\mathrm{PDE}\left[n\right]\right)$

${\mathrm{PDE}}_6≔u_{x,y}-\frac{20u}{{\left(x+y\right)}^2}$

$\mathrm{Laplace}\left(\mathrm{PDE}\left[6\right]\&comma;u\left(x\&comma;y\right)\right)$

$\mathrm{Laplace}\left(\mathrm{PDE}\left[6\right]\&comma;u\left(x\&comma;y\right)\&comma;\mathrm{numberofiterations}=6\right)$

$u=\frac{{\left(x+y\right)}^4{\mathrm{f\_\_1}}_{x,x,x,x}+{\left(x+y\right)}^4{\mathrm{f\_\_2}}_{y,y,y,y}-20{\left(x+y\right)}^3{\mathrm{f\_\_1}}_{x,x,x}-20{\left(x+y\right)}^3{\mathrm{f\_\_2}}_{y,y,y}+180{\left(x+y\right)}^2{\mathrm{f\_\_1}}_{x,x}+180{\left(x+y\right)}^2{\mathrm{f\_\_2}}_{y,y}+\left(-840x-840y\right){\mathrm{f\_\_1}}_x+\left(-840x-840y\right){\mathrm{f\_\_2}}_y+1680\mathrm{f\_\_1}\left(x\right)+1680\mathrm{f\_\_2}\left(y\right)}{100800{\left(x+y\right)}^4}$

$\mathrm{PDE}\left[A\right]≔\mathrm{subs}\left(n=3\&comma;\mathrm{PDE}\left[n\right]\right)$

${\mathrm{PDE}}_A≔u_{x,y}-\frac{2u}{{\left(x+y\right)}^2}$

$\mathrm{PDE}\left[B\right]≔\mathrm{diff}\left(v\left(x\&comma;y\right)\&comma;x\&comma;y\right)+\frac{2}{x+y}\mathrm{diff}\left(v\left(x\&comma;y\right)\&comma;x\right)-\frac{2}{{\left(x+y\right)}^2}v\left(x\&comma;y\right)$

${\mathrm{PDE}}_B≔v_{x,y}+\frac{2v_x}{x+y}-\frac{2v\left(x\&comma;y\right)}{{\left(x+y\right)}^2}$

$\mathrm{PDE}\left[C\right]≔\mathrm{diff}\left(w\left(x\&comma;y\right)\&comma;x\&comma;y\right)+\frac{4}{x+y}\mathrm{diff}\left(w\left(x\&comma;y\right)\&comma;x\right)$

${\mathrm{PDE}}_C≔w_{x,y}+\frac{4w_x}{x+y}$

$L\left[\mathrm{AB}\right]≔u\&rarr;\frac{\partial }{\partial y}u$

$L_{\mathrm{AB}}≔u\&rarr;\frac{\partial }{\partial y}u$

$L\left[\mathrm{BA}\right]≔u\&rarr;\frac{1}{2}{\left(x\&plus;y\right)}^2\left(\frac{\partial }{\partial x}u\right)$

$L_{\mathrm{BA}}≔u\&rarr;\frac{1}{2}{\left(x\&plus;y\right)}^2\left(\frac{\partial }{\partial x}u\right)$

$L\left[\mathrm{BC}\right]≔u\&rarr;\frac{\partial }{\partial y}u\&plus;\frac{2u}{x\&plus;y}$

$L_{\mathrm{BC}}≔u\&rarr;\frac{\partial }{\partial y}u\&plus;\frac{2u}{x\&plus;y}$

$\mathrm{sol}\left[A\right]≔\mathrm{Laplace}\left(\mathrm{PDE}\left[A\right]\&comma;u\left(x\&comma;y\right)\right)$

${\mathrm{sol}}_A≔u=\frac{\left(x+y\right){\mathrm{f\_\_1}}_x+\left(x+y\right){\mathrm{f\_\_2}}_y-2\mathrm{f\_\_1}\left(x\right)-2\mathrm{f\_\_2}\left(y\right)}{2x+2y}$

$\mathrm{sol}\left[B\right]≔v\left(x\&comma;y\right)=L\left[\mathrm{AB}\right]\left(\mathrm{rhs}\left(\mathrm{sol}\left[A\right]\right)\right)$

${\mathrm{sol}}_B≔v\left(x\&comma;y\right)=\frac{{\mathrm{f\_\_1}}_x-{\mathrm{f\_\_2}}_y+\left(x+y\right){\mathrm{f\_\_2}}_{y,y}}{2x+2y}-\frac{2\left(\left(x+y\right){\mathrm{f\_\_1}}_x+\left(x+y\right){\mathrm{f\_\_2}}_y-2\mathrm{f\_\_1}\left(x\right)-2\mathrm{f\_\_2}\left(y\right)\right)}{{\left(2x+2y\right)}^2}$

$\mathrm{pdetest}\left(\&comma;\mathrm{PDE}\left[B\right]\right)$

$0$

$u\left(x\&comma;y\right)=L\left[\mathrm{BA}\right]\left(\mathrm{rhs}\left(\mathrm{sol}\left[B\right]\right)\right)$

$u=\frac{{\left(x+y\right)}^2\left(\frac{{\mathrm{f\_\_1}}_{x,x}+{\mathrm{f\_\_2}}_{y,y}}{2x+2y}-\frac{2\left({\mathrm{f\_\_1}}_x-{\mathrm{f\_\_2}}_y+\left(x+y\right){\mathrm{f\_\_2}}_{y,y}\right)}{{\left(2x+2y\right)}^2}-\frac{2\left(-{\mathrm{f\_\_1}}_x+\left(x+y\right){\mathrm{f\_\_1}}_{x,x}+{\mathrm{f\_\_2}}_y\right)}{{\left(2x+2y\right)}^2}+\frac{8\left(\left(x+y\right){\mathrm{f\_\_1}}_x+\left(x+y\right){\mathrm{f\_\_2}}_y-2\mathrm{f\_\_1}\left(x\right)-2\mathrm{f\_\_2}\left(y\right)\right)}{{\left(2x+2y\right)}^3}\right)}{2}$

$\mathrm{pdetest}\left(\&comma;\mathrm{PDE}\left[A\right]\right)$

$0$

$\mathrm{sol}\left[C\right]≔w\left(x\&comma;y\right)=L\left[\mathrm{BC}\right]\left(\mathrm{rhs}\left(\mathrm{sol}\left[B\right]\right)\right)$

${\mathrm{sol}}_C≔w\left(x\&comma;y\right)=\frac{\left(x+y\right){\mathrm{f\_\_2}}_{y,y,y}}{2x+2y}-\frac{4\left({\mathrm{f\_\_1}}_x-{\mathrm{f\_\_2}}_y+\left(x+y\right){\mathrm{f\_\_2}}_{y,y}\right)}{{\left(2x+2y\right)}^2}+\frac{8\left(\left(x+y\right){\mathrm{f\_\_1}}_x+\left(x+y\right){\mathrm{f\_\_2}}_y-2\mathrm{f\_\_1}\left(x\right)-2\mathrm{f\_\_2}\left(y\right)\right)}{{\left(2x+2y\right)}^3}+\frac{2\left(\frac{{\mathrm{f\_\_1}}_x-{\mathrm{f\_\_2}}_y+\left(x+y\right){\mathrm{f\_\_2}}_{y,y}}{2x+2y}-\frac{2\left(\left(x+y\right){\mathrm{f\_\_1}}_x+\left(x+y\right){\mathrm{f\_\_2}}_y-2\mathrm{f\_\_1}\left(x\right)-2\mathrm{f\_\_2}\left(y\right)\right)}{{\left(2x+2y\right)}^2}\right)}{x+y}$

$\mathrm{pdetest}\left(\&comma;\mathrm{PDE}\left[C\right]\right)$

$0$

$L\left[\mathrm{BC}\right]\left(v\left(x\&comma;y\right)\right)$

$v_y+\frac{2v\left(x\&comma;y\right)}{x+y}$

$\mathrm{pdsolve}\left(\right)$

$v\left(x\&comma;y\right)=\frac{\mathrm{f\_\_1}\left(x\right)}{{\left(x+y\right)}^2}$

$u\left(x\&comma;y\right)=L\left[\mathrm{BA}\right]\left(\mathrm{rhs}\left(\right)\right)$

$u=\frac{{\left(x+y\right)}^2\left(\frac{{\mathrm{f\_\_1}}_x}{{\left(x+y\right)}^2}-\frac{2\mathrm{f\_\_1}\left(x\right)}{{\left(x+y\right)}^3}\right)}{2}$

$\mathrm{pdetest}\left(\&comma;\mathrm{PDE}\left[A\right]\right)$

$0$