Laplace - Maple Help

PDEtools

 Laplace
 solves a second order linear PDE in 2 independent variables using the method of Laplace

 Calling Sequence Laplace(PDE, F, numberofiterations = ...)

Parameters

 PDE - a linear partial differential equation in two independent variables F - the unknown of the PDE numberofiterations = ... - optional - the right hand side is a positive integer limiting the number of iterations used to tackle the PDE

Description

 • 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 $u=u\left(x,y\right)$ 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 $\mathrm{\Delta }'=0$ with the remarkable property that solutions of $\mathrm{\Delta }=0$ can be found from solutions of $\mathrm{\Delta }'=0$ by differentiations and simple linear algebraic manipulations. In favorable circumstances solutions to equation $\mathrm{\Delta }'=0$ can be found and this then leads to solutions of the original equation.  If solutions to $\mathrm{\Delta }'=0$ cannot be found, then one may iterate the process to generate a sequence of equations $\mathrm{\Delta }=0,\mathrm{\Delta }'=0,\mathrm{\Delta }\text{'}\text{'}=0,\mathrm{...},{\mathrm{\Delta }}^{\left(n\right)}=0$ 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 ${v}_{x,y}+A'{v}_{x}+B'{v}_{y}+C'v=0$; and [2] the equation can be inverted to give $u=\frac{1}{H\left({v}_{x}+B'v\right)}$. A similar transform can be defined if $K\ne 0$. See the examples for an explicit computation of these transforms.

Examples

The PDE ${u}_{x,y}-\frac{ku}{{\left(x+y\right)}^{2}}=0$ is known to be integrable in $n$ steps if $k=\left(n-2\right)\left(n-1\right)$.

 > $\mathrm{with}\left(\mathrm{PDEtools},\mathrm{Laplace},\mathrm{declare}\right)$
 $\left[{\mathrm{Laplace}}{,}{\mathrm{declare}}\right]$ (1)
 > $\mathrm{declare}\left(u\left(x,y\right)\right)$
 ${u}{}\left({x}{,}{y}\right){}{\mathrm{will now be displayed as}}{}{u}$ (2)
 > $k≔\left(n-2\right)\left(n-1\right)$
 ${k}{≔}\left({n}{-}{2}\right){}\left({n}{-}{1}\right)$ (3)
 > $\mathrm{PDE}\left[n\right]≔\mathrm{diff}\left(u\left(x,y\right),x,y\right)-\frac{k}{{\left(x+y\right)}^{2}}u\left(x,y\right)$
 ${{\mathrm{PDE}}}_{{n}}{≔}{{u}}_{{x}{,}{y}}{-}\frac{\left({n}{-}{2}\right){}\left({n}{-}{1}\right){}{u}}{{\left({x}{+}{y}\right)}^{{2}}}$ (4)
 > $\mathrm{PDE}\left[5\right]≔\mathrm{subs}\left(n=5,\mathrm{PDE}\left[n\right]\right)$
 ${{\mathrm{PDE}}}_{{5}}{≔}{{u}}_{{x}{,}{y}}{-}\frac{{12}{}{u}}{{\left({x}{+}{y}\right)}^{{2}}}$ (5)
 > $\mathrm{Laplace}\left(\mathrm{PDE}\left[5\right],u\left(x,y\right)\right)$
 ${u}{=}\frac{{\left({x}{+}{y}\right)}^{{3}}{}{{\mathrm{_F1}}}_{{x}{,}{x}{,}{x}}{+}{\left({x}{+}{y}\right)}^{{3}}{}{{\mathrm{_F2}}}_{{y}{,}{y}{,}{y}}{-}{12}{}{\left({x}{+}{y}\right)}^{{2}}{}{{\mathrm{_F1}}}_{{x}{,}{x}}{-}{12}{}{\left({x}{+}{y}\right)}^{{2}}{}{{\mathrm{_F2}}}_{{y}{,}{y}}{+}\left({60}{}{x}{+}{60}{}{y}\right){}{{\mathrm{_F1}}}_{{x}}{+}\left({60}{}{x}{+}{60}{}{y}\right){}{{\mathrm{_F2}}}_{{y}}{-}{120}{}{\mathrm{_F1}}{}\left({x}\right){-}{120}{}{\mathrm{_F2}}{}\left({y}\right)}{{1440}{}{\left({x}{+}{y}\right)}^{{3}}}$ (6)

For $n=6$, Laplace returns NULL since the default number of iterations is 5.

 > $\mathrm{PDE}\left[6\right]≔\mathrm{subs}\left(n=6,\mathrm{PDE}\left[n\right]\right)$
 ${{\mathrm{PDE}}}_{{6}}{≔}{{u}}_{{x}{,}{y}}{-}\frac{{20}{}{u}}{{\left({x}{+}{y}\right)}^{{2}}}$ (7)
 > $\mathrm{Laplace}\left(\mathrm{PDE}\left[6\right],u\left(x,y\right)\right)$

To obtain the solution in this example use the optional argument numberofiterations.

 > $\mathrm{Laplace}\left(\mathrm{PDE}\left[6\right],u\left(x,y\right),\mathrm{numberofiterations}=6\right)$
 ${u}{=}\frac{{\left({x}{+}{y}\right)}^{{4}}{}{{\mathrm{_F1}}}_{{x}{,}{x}{,}{x}{,}{x}}{+}{\left({x}{+}{y}\right)}^{{4}}{}{{\mathrm{_F2}}}_{{y}{,}{y}{,}{y}{,}{y}}{-}{20}{}{\left({x}{+}{y}\right)}^{{3}}{}{{\mathrm{_F1}}}_{{x}{,}{x}{,}{x}}{-}{20}{}{\left({x}{+}{y}\right)}^{{3}}{}{{\mathrm{_F2}}}_{{y}{,}{y}{,}{y}}{+}{180}{}{\left({x}{+}{y}\right)}^{{2}}{}{{\mathrm{_F1}}}_{{x}{,}{x}}{+}{180}{}{\left({x}{+}{y}\right)}^{{2}}{}{{\mathrm{_F2}}}_{{y}{,}{y}}{+}\left({-}{840}{}{x}{-}{840}{}{y}\right){}{{\mathrm{_F1}}}_{{x}}{+}\left({-}{840}{}{x}{-}{840}{}{y}\right){}{{\mathrm{_F2}}}_{{y}}{+}{1680}{}{\mathrm{_F1}}{}\left({x}\right){+}{1680}{}{\mathrm{_F2}}{}\left({y}\right)}{{100800}{}{\left({x}{+}{y}\right)}^{{4}}}$ (8)

We analyze here the case $n=3$ to show some of the details of the method. We define a sequence of three PDEs, ${\mathrm{PDE}}_{A}$, ${\mathrm{PDE}}_{B}$ and ${\mathrm{PDE}}_{C}$. We wish to solve ${\mathrm{PDE}}_{A}$. The PDEs ${\mathrm{PDE}}_{B}$ and ${\mathrm{PDE}}_{C}$ are generated by the method of Laplace. We also define three maps which we denote by ${L}_{\mathrm{AB}}$, ${L}_{\mathrm{BA}}$ and ${L}_{\mathrm{BC}}$. These are also prescribed by the method of Laplace.

 > $\mathrm{PDE}\left[A\right]≔\mathrm{subs}\left(n=3,\mathrm{PDE}\left[n\right]\right)$
 ${{\mathrm{PDE}}}_{{A}}{≔}{{u}}_{{x}{,}{y}}{-}\frac{{2}{}{u}}{{\left({x}{+}{y}\right)}^{{2}}}$ (9)
 > $\mathrm{PDE}\left[B\right]≔\mathrm{diff}\left(v\left(x,y\right),x,y\right)+\frac{2}{x+y}\mathrm{diff}\left(v\left(x,y\right),x\right)-\frac{2}{{\left(x+y\right)}^{2}}v\left(x,y\right)$
 ${{\mathrm{PDE}}}_{{B}}{≔}{{v}}_{{x}{,}{y}}{+}\frac{{2}{}{{v}}_{{x}}}{{x}{+}{y}}{-}\frac{{2}{}{v}{}\left({x}{,}{y}\right)}{{\left({x}{+}{y}\right)}^{{2}}}$ (10)
 > $\mathrm{PDE}\left[C\right]≔\mathrm{diff}\left(w\left(x,y\right),x,y\right)+\frac{4}{x+y}\mathrm{diff}\left(w\left(x,y\right),x\right)$
 ${{\mathrm{PDE}}}_{{C}}{≔}{{w}}_{{x}{,}{y}}{+}\frac{{4}{}{{w}}_{{x}}}{{x}{+}{y}}$ (11)
 > $L\left[\mathrm{AB}\right]≔u→\frac{\partial }{\partial y}u$
 ${{L}}_{{\mathrm{AB}}}{≔}{u}{→}\frac{{\partial }}{{\partial }{y}}{}{u}$ (12)
 > $L\left[\mathrm{BA}\right]≔u→\frac{1}{2}{\left(x+y\right)}^{2}\left(\frac{\partial }{\partial x}u\right)$
 ${{L}}_{{\mathrm{BA}}}{≔}{u}{→}\frac{{1}}{{2}}{}{\left({x}{+}{y}\right)}^{{2}}{}\left(\frac{{\partial }}{{\partial }{x}}{}{u}\right)$ (13)
 > $L\left[\mathrm{BC}\right]≔u→\frac{\partial }{\partial y}u+\frac{2u}{x+y}$
 ${{L}}_{{\mathrm{BC}}}{≔}{u}{→}\frac{{\partial }}{{\partial }{y}}{}{u}{+}\frac{{2}{}{u}}{{x}{+}{y}}$ (14)

Let's show that if $u\left(x,y\right)$ is a solution to ${\mathrm{PDE}}_{A}$, then $v\left(x,y\right)={L}_{\mathrm{AB}}\left(u\left(x,y\right)\right)$ is a solution to ${\mathrm{PDE}}_{B}$.

 > $\mathrm{sol}\left[A\right]≔\mathrm{Laplace}\left(\mathrm{PDE}\left[A\right],u\left(x,y\right)\right)$
 ${{\mathrm{sol}}}_{{A}}{≔}{u}{=}\frac{\left({x}{+}{y}\right){}{{\mathrm{_F1}}}_{{x}}{+}\left({x}{+}{y}\right){}{{\mathrm{_F2}}}_{{y}}{-}{2}{}{\mathrm{_F1}}{}\left({x}\right){-}{2}{}{\mathrm{_F2}}{}\left({y}\right)}{{2}{}{x}{+}{2}{}{y}}$ (15)
 > $\mathrm{sol}\left[B\right]≔v\left(x,y\right)=L\left[\mathrm{AB}\right]\left(\mathrm{rhs}\left(\mathrm{sol}\left[A\right]\right)\right)$
 ${{\mathrm{sol}}}_{{B}}{≔}{v}{}\left({x}{,}{y}\right){=}\frac{{{\mathrm{_F1}}}_{{x}}{-}{{\mathrm{_F2}}}_{{y}}{+}\left({x}{+}{y}\right){}{{\mathrm{_F2}}}_{{y}{,}{y}}}{{2}{}{x}{+}{2}{}{y}}{-}\frac{{2}{}\left(\left({x}{+}{y}\right){}{{\mathrm{_F1}}}_{{x}}{+}\left({x}{+}{y}\right){}{{\mathrm{_F2}}}_{{y}}{-}{2}{}{\mathrm{_F1}}{}\left({x}\right){-}{2}{}{\mathrm{_F2}}{}\left({y}\right)\right)}{{\left({2}{}{x}{+}{2}{}{y}\right)}^{{2}}}$ (16)
 > $\mathrm{pdetest}\left(,\mathrm{PDE}\left[B\right]\right)$
 ${0}$ (17)

Also, if $v\left(x,y\right)$ is a solution to ${\mathrm{PDE}}_{B}$, then $u\left(x,y\right)={L}_{\mathrm{BA}}\left(v\left(x,y\right)\right)$ is a solution to ${\mathrm{PDE}}_{A}$.

 > $u\left(x,y\right)=L\left[\mathrm{BA}\right]\left(\mathrm{rhs}\left(\mathrm{sol}\left[B\right]\right)\right)$
 ${u}{=}\frac{{\left({x}{+}{y}\right)}^{{2}}{}\left(\frac{{{\mathrm{_F1}}}_{{x}{,}{x}}{+}{{\mathrm{_F2}}}_{{y}{,}{y}}}{{2}{}{x}{+}{2}{}{y}}{-}\frac{{2}{}\left({{\mathrm{_F1}}}_{{x}}{-}{{\mathrm{_F2}}}_{{y}}{+}\left({x}{+}{y}\right){}{{\mathrm{_F2}}}_{{y}{,}{y}}\right)}{{\left({2}{}{x}{+}{2}{}{y}\right)}^{{2}}}{-}\frac{{2}{}\left({-}{{\mathrm{_F1}}}_{{x}}{+}\left({x}{+}{y}\right){}{{\mathrm{_F1}}}_{{x}{,}{x}}{+}{{\mathrm{_F2}}}_{{y}}\right)}{{\left({2}{}{x}{+}{2}{}{y}\right)}^{{2}}}{+}\frac{{8}{}\left(\left({x}{+}{y}\right){}{{\mathrm{_F1}}}_{{x}}{+}\left({x}{+}{y}\right){}{{\mathrm{_F2}}}_{{y}}{-}{2}{}{\mathrm{_F1}}{}\left({x}\right){-}{2}{}{\mathrm{_F2}}{}\left({y}\right)\right)}{{\left({2}{}{x}{+}{2}{}{y}\right)}^{{3}}}\right)}{{2}}$ (18)
 > $\mathrm{pdetest}\left(,\mathrm{PDE}\left[A\right]\right)$
 ${0}$ (19)

Finally, if $v\left(x,y\right)$ is a solution to ${\mathrm{PDE}}_{B}$, then $w\left(x,y\right)={L}_{\mathrm{BC}}\left(v\left(x,y\right)\right)$ is a solution to ${\mathrm{PDE}}_{C}$.

 > $\mathrm{sol}\left[C\right]≔w\left(x,y\right)=L\left[\mathrm{BC}\right]\left(\mathrm{rhs}\left(\mathrm{sol}\left[B\right]\right)\right)$
 ${{\mathrm{sol}}}_{{C}}{≔}{w}{}\left({x}{,}{y}\right){=}\frac{\left({x}{+}{y}\right){}{{\mathrm{_F2}}}_{{y}{,}{y}{,}{y}}}{{2}{}{x}{+}{2}{}{y}}{-}\frac{{4}{}\left({{\mathrm{_F1}}}_{{x}}{-}{{\mathrm{_F2}}}_{{y}}{+}\left({x}{+}{y}\right){}{{\mathrm{_F2}}}_{{y}{,}{y}}\right)}{{\left({2}{}{x}{+}{2}{}{y}\right)}^{{2}}}{+}\frac{{8}{}\left(\left({x}{+}{y}\right){}{{\mathrm{_F1}}}_{{x}}{+}\left({x}{+}{y}\right){}{{\mathrm{_F2}}}_{{y}}{-}{2}{}{\mathrm{_F1}}{}\left({x}\right){-}{2}{}{\mathrm{_F2}}{}\left({y}\right)\right)}{{\left({2}{}{x}{+}{2}{}{y}\right)}^{{3}}}{+}\frac{{2}{}\left(\frac{{{\mathrm{_F1}}}_{{x}}{-}{{\mathrm{_F2}}}_{{y}}{+}\left({x}{+}{y}\right){}{{\mathrm{_F2}}}_{{y}{,}{y}}}{{2}{}{x}{+}{2}{}{y}}{-}\frac{{2}{}\left(\left({x}{+}{y}\right){}{{\mathrm{_F1}}}_{{x}}{+}\left({x}{+}{y}\right){}{{\mathrm{_F2}}}_{{y}}{-}{2}{}{\mathrm{_F1}}{}\left({x}\right){-}{2}{}{\mathrm{_F2}}{}\left({y}\right)\right)}{{\left({2}{}{x}{+}{2}{}{y}\right)}^{{2}}}\right)}{{x}{+}{y}}$ (20)
 > $\mathrm{pdetest}\left(,\mathrm{PDE}\left[C\right]\right)$
 ${0}$ (21)

Now, remarkably, we start with the zero solution to ${\mathrm{PDE}}_{C}$, integrate the equation $w\left(x,y\right)={L}_{\mathrm{BC}}\left(v\left(x,y\right)\right)$ to find $v\left(x,y\right)$ and apply ${L}_{\mathrm{BA}}$ to find $u\left(x,y\right)$:

 > $L\left[\mathrm{BC}\right]\left(v\left(x,y\right)\right)$
 ${{v}}_{{y}}{+}\frac{{2}{}{v}{}\left({x}{,}{y}\right)}{{x}{+}{y}}$ (22)
 > $\mathrm{pdsolve}\left(\right)$
 ${v}{}\left({x}{,}{y}\right){=}\frac{{\mathrm{_F1}}{}\left({x}\right)}{{\left({x}{+}{y}\right)}^{{2}}}$ (23)

So this is the solution to ${\mathrm{PDE}}_{A}$

 > $u\left(x,y\right)=L\left[\mathrm{BA}\right]\left(\mathrm{rhs}\left(\right)\right)$
 ${u}{=}\frac{{\left({x}{+}{y}\right)}^{{2}}{}\left(\frac{{{\mathrm{_F1}}}_{{x}}}{{\left({x}{+}{y}\right)}^{{2}}}{-}\frac{{2}{}{\mathrm{_F1}}{}\left({x}\right)}{{\left({x}{+}{y}\right)}^{{3}}}\right)}{{2}}$ (24)
 > $\mathrm{pdetest}\left(,\mathrm{PDE}\left[A\right]\right)$
 ${0}$ (25)

A similar sequence of PDEs and transformations can be constructed to find a solution depending on an arbitrary function of y.