Find the general solution of the given PDE

[chwala]

Homework Statement

This is a question set by myself;

Find the general solution of

##-4u_x+6u_y+10u=e^{x+2y}##

Relevant Equations

Method of characteristic.

My take;

##ξ=-4x+6y## and ##η=6x+4y##

it follows that,

##52u_ξ +10u=e^{x+2y}##

for the homogenous part; we shall have the general solution;

$$u_h=e^{\frac{-5}{26} ξ} f{η }$$

now we note that

$$e^{x+2y}=e^{\frac{8ξ+η}{26}}$$

that is from solving the simultaneous equation;

##ξ=-4x+6y## and ##η=6x+4y## where ##x=\dfrac{12ξ-18η}{156}## and ##y=\dfrac{3ξ+2η}{26}##

for the inhomogenous part, we shall have;

$$u_p = e^{\frac{-5}{26}ξ}\int \frac{e^{\frac{8ξ+η}{26}}⋅e^{\frac{5}{26}ξ}}{52} dξ$$$$u_p = \frac{e^{\frac{13ξ+η}{26}}⋅e^{\frac{-5}{26}ξ}}{26} $$

$$u(ξ,η)=u_h+u_p=e^{\frac{-5}{26} ξ} \left[ f(η)+\frac{e^{\frac{13ξ+η}{26}}}{26}\right]$$

$$u(x,y)=
e^{\frac{-5}{26} (-4x+6y)} \left[ f(6x+4y)+\frac{1}{26} {e^{\frac{-46x+82y}{26}}}\right]$$your insight is welcome or alternative approach to this problem.

 

[pasmith]

First divide by 2: [tex]
-2 u_x + 3u_y + 5u = \tfrac12 e^{x+ 2y}.[/tex] We want to change to variables [itex](\zeta, \eta)[/itex] with [itex]x_\zeta = -2[/itex] and [itex]y_\zeta = 3[/itex] so that [tex]-2u_x + 3u_y = u_xx_\zeta + u_yy_\zeta = u_\zeta.[/tex] Note that you did the opposite, which was to set [tex]
-4 u_x + 6u_y = u_x\xi_x + u_y \xi_y[/tex] where the right hand side is not the multivariate chain rule for [itex]u_\xi[/itex].

We therefore set [tex]\begin{split}
x &= -2\zeta + a\eta \\ y &= 3\zeta + b\eta \end{split} [/tex] with [itex]a[/itex] and [itex]b[/itex] chosen such that [tex]
x + 2y = 4\zeta + (a + 2b)\eta = 4\zeta.[/tex] Any choice of [itex]b[/itex] other than [itex]b = 0[/itex] gives us a 1-1 transformation, but the choice [itex]b = -\frac14[/itex] gives the convenient [itex]\eta = 3x + 2y[/itex], and from the above we have [itex]\zeta = \frac14(x+2y)[/itex]. The PDE is then reduced to [tex]
u_\zeta + 5u = \tfrac12e^{4\zeta}[/tex] which can be solved by an integrating factor.

 

[chwala]

@pasmith I'll need to go through your approach...cheers man! Just curious are we going to get same solutions? Your integrating factor seems different just by looking at it...

 

[pasmith]

I get [tex]
u = g(\eta')e^{-5\zeta} + \tfrac{1}{18}e^{4\zeta}.[/tex]

I think you have made an error: If [itex]\xi = -4x + 6y[/itex] and [itex]\eta = 6x + 4y[/itex] then [tex]x = (3\eta - 2\xi)/26,\qquad y = (3\xi + 2\eta)/26.[/tex] Your result of [tex]
x = \frac{12\xi - 18\eta}{156} = \frac{2\xi - 3\eta}{26}[/tex] is out by a factor of -1, so your calculation of [itex]x + 2y[/itex] is incorrect; you should have [tex]
x + 2y = \frac{7\eta + 4\xi}{26}.[/tex] Note that in terms of my [itex]\zeta, \eta'[/itex] you have [tex]\begin{split}
\eta &= 2\eta' \\
\xi &= 26\zeta - \frac{7}{2}\eta'.\end{split}[/tex] Now solving [tex]
u_\xi + \frac{5}{26}u = \frac{1}{52}e^{(7\eta +4\xi)/26}[/tex] gives [tex]\begin{split}
u &= f(\eta)e^{-5\xi/26} + \frac{1}{52}\frac{26}{9}e^{(7\eta+4\xi)/26} \\
&= f(\eta)e^{-5\xi/26} + \frac{1}{18}e^{(7\eta + 4\xi)/26}.\end{split}[/tex] We agree on the second term, since [tex]4\zeta = x + 2y = \frac{7\eta + 4\xi}{26}.[/tex] For the first term, we have [tex]
f(\eta)e^{-5\xi/26} = \overbrace{f(2\eta')e^{35\eta'/52}}^{\equiv g(\eta')}e^{-5\zeta}[/tex] as required.

I suspect you will agree that the numbers in my solution are simpler than in yours.