solves a second order linear PDE in 2 independent variables using the method of Laplace
Laplace(PDE, F, numberofiterations = ...)
a linear partial differential equation in two independent variables
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
The general form of a second order scalar linear PDE in two independent variables is Δ=a⁢ux,x+b⁢ux,y+c⁢uy,y+d⁢uy+e⁢ux+f⁢=0, where u=u⁡x,y is the unknown function, the coefficients a,b,c,d,e,f are functions of the independent variables x,y and a⁢c−b24<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 Δ=0 into another second order scalar linear PDE Δ'⁢=0 with the remarkable property that solutions of Δ=0 can be found from solutions of Δ'⁢=0 by differentiations and simple linear algebraic manipulations. In favorable circumstances solutions to equation Δ'⁢=0 can be found and this then leads to solutions of the original equation. If solutions to Δ'⁢=0 cannot be found, then one may iterate the process to generate a sequence of equations Δ=0,Δ'=0,Δ''=0,...,Δn=0 with the property that solutions to Δ=0 can be constructed from solutions to Δn=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 Δn=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 ux,y+Aux+Buy+Cu=0 (although this special form is not required for the procedure). For an equation of this form, define the Laplace invariants H=dAdx+AB−C and K=dBdy+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 vx,y=0 and the solution thus found.
If H≠0 then one defines v=uy+Au and finds that:  v also satisfies an equation of the form vx,y+A'⁢vx+B'⁢vy+C'⁢v=0; and  the equation can be inverted to give u=1H⁡vx+B'⁢v. A similar transform can be defined if K≠0. See the examples for an explicit computation of these transforms.
The PDE ux,y−k⁢ux+y2=0 is known to be integrable in n steps if k=n−2⁡n−1.
u⁡x,y⁢will now be displayed as⁢u
k ≔ n−2⁢n−1
PDEn ≔ ∂2∂y⁢∂x⁢u⁡x,y−k⁢u⁡x,yx+y2
PDE5 ≔ subs⁡n=5,PDEn
For n=6, Laplace returns NULL since the default number of iterations is 5.
PDE6 ≔ subs⁡n=6,PDEn
To obtain the solution in this example use the optional argument numberofiterations.
We analyze here the case n=3 to show some of the details of the method. We define a sequence of three PDEs, PDEA, PDEB and PDEC. We wish to solve PDEA. The PDEs PDEB and PDEC are generated by the method of Laplace. We also define three maps which we denote by LAB, LBA and LBC. These are also prescribed by the method of Laplace.
PDEA ≔ subs⁡n=3,PDEn
PDEB ≔ ∂2∂y⁢∂x⁢v⁡x,y+2⁢∂∂x⁢v⁡x,yx+y−2⁢v⁡x,yx+y2
PDEC ≔ ∂2∂y⁢∂x⁢w⁡x,y+4⁢∂∂x⁢w⁡x,yx+y
LAB ≔ u→∂∂y⁢u
LBA ≔ u→x+y2⁢∂∂x⁢u2
LBC ≔ u→∂∂y⁢u+2⁢ux+y
Let's show that if u is a solution to PDEA, then v⁡x,y=LAB⁡u is a solution to PDEB.
solA ≔ Laplace⁡PDEA,u⁡x,y
solB ≔ v⁡x,y=LAB⁡rhs⁡solA
Also, if v⁡x,y is a solution to PDEB, then u=LBA⁡v⁡x,y is a solution to PDEA.
Finally, if v⁡x,y is a solution to PDEB, then w⁡x,y=LBC⁡v⁡x,y is a solution to PDEC.
solC ≔ w⁡x,y=LBC⁡rhs⁡solB
Now, remarkably, we start with the zero solution to PDEC, integrate the equation w⁡x,y=LBC⁡v⁡x,y to find v⁡x,y and apply LBA to find u:
So this is the solution to PDEA
A similar sequence of PDEs and transformations can be constructed to find a solution depending on an arbitrary function of y.
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