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

We therefore set $$\begin{split} x &= -2\zeta + a\eta \\ y &= 3\zeta + b\eta \end{split}$$ with $a$ and $b$ chosen such that $$x + 2y = 4\zeta + (a + 2b)\eta = 4\zeta.$$ Any choice of $b$ other than $b = 0$ gives us a 1-1 transformation, but the choice $b = -\frac14$ gives the convenient $\eta = 3x + 2y$, and from the above we have $\zeta = \frac14(x+2y)$. The PDE is then reduced to $$u_\zeta + 5u = \tfrac12e^{4\zeta}$$ 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 $$u = g(\eta')e^{-5\zeta} + \tfrac{1}{18}e^{4\zeta}.$$

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

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