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I know we can transform a second order PDE into three standard forms. But how to deal with the remaining first order terms?

Particularly, how to solve the following PDE:

$$ u_{xy}+au_x+bu_y+cu+dx+ey+f=0 $$

update:

$a,b,c,d,e,f$ are all constant.

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  • $\begingroup$ $a$ , $b$ , $c$ , $d$ , $e$ and $f$ are constants or functions of $x$ and $y$ ? $\endgroup$ Nov 2, 2012 at 14:58
  • $\begingroup$ @doraemonpaul constant $\endgroup$
    – hxhxhx88
    Nov 3, 2012 at 2:59
  • $\begingroup$ First find and solve the special cases that can treat as ODE to solve. Then try to introduce some kind of variable transformations e.g. let $u(x,y)=f(x)g(y)v(x,y)$ to research the transforming ability to a PDE that can treat as ODE to solve. $\endgroup$ Nov 3, 2012 at 22:18
  • $\begingroup$ Whithout boundary conditions, it is hard to say. A trivial solution would be $$u(x,y) = -\frac{d}{c}x -\frac{e}{c}y -\frac{f}{c} + \frac{ad +be}{c^2}.$$ $\endgroup$
    – Pragabhava
    Nov 7, 2012 at 2:41

1 Answer 1

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Case $1$: $a=b=c=0$

Then $u_{xy}+dx+ey+f=0$

$u_{xy}=-dx-ey-f$

$u_x=\int(-dx-ey-f)~dy$

$u_x=C(x)-dxy-\dfrac{ey^2}{2}-fy$

$u=\int\left(C(x)-dxy-\dfrac{ey^2}{2}-fy\right)dx$

$u=C_1(x)+C_2(y)-\dfrac{dx^2y}{2}-\dfrac{exy^2}{2}-fxy$

Case $2$: $a\neq0$ and $b=c=0$

Then $u_{xy}+au_x+dx+ey+f=0$

Let $u_x=v$ ,

Then $u_{xy}=v_y$

$\therefore v_y+av+dx+ey+f=0$

$v_y+av=-dx-ey-f$

$(\exp(ay)v)_y=-dx\exp(ay)-ey\exp(ay)-f\exp(ay)$

$\exp(ay)v=\int(-dx\exp(ay)-ey\exp(ay)-f\exp(ay))~dy$

$\exp(ay)u_x=C(x)-\dfrac{dx\exp(ay)}{a}-\dfrac{ey\exp(ay)}{a}+\dfrac{e\exp(ay)}{a^2}-\dfrac{f\exp(ay)}{a}$

$u_x=C(x)\exp(-ay)-\dfrac{dx}{a}-\dfrac{ey}{a}+\dfrac{e}{a^2}-\dfrac{f}{a}$

$u=\int\left(C(x)\exp(-ay)-\dfrac{dx}{a}-\dfrac{ey}{a}+\dfrac{e}{a^2}-\dfrac{f}{a}\right)dx$

$u=C_1(x)\exp(-ay)+C_2(y)-\dfrac{dx^2}{2a}-\dfrac{exy}{a}+\dfrac{ex}{a^2}-\dfrac{fx}{a}$

Case $3$: $b\neq0$ and $a=c=0$

Then $u_{xy}+bu_y+dx+ey+f=0$

Let $u_y=v$ ,

Then $u_{xy}=v_x$

$\therefore v_x+bv+dx+ey+f=0$

$v_x+bv=-dx-ey-f$

$(\exp(bx)v)_x=-dx\exp(bx)-ey\exp(bx)-f\exp(bx)$

$\exp(bx)v=\int(-dx\exp(bx)-ey\exp(bx)-f\exp(bx))~dx$

$\exp(bx)u_y=C(y)-\dfrac{dx\exp(bx)}{b}+\dfrac{d\exp(bx)}{b^2}-\dfrac{ey\exp(bx)}{b}-\dfrac{f\exp(bx)}{b}$

$u_y=C(y)\exp(-bx)-\dfrac{dx}{b}+\dfrac{d}{b^2}-\dfrac{ey}{b}-\dfrac{f}{b}$

$u=\int\left(C(y)\exp(-bx)-\dfrac{dx}{b}+\dfrac{d}{b^2}-\dfrac{ey}{b}-\dfrac{f}{b}\right)dy$

$u=C_1(x)+C_2(y)\exp(-bx)-\dfrac{dxy}{b}+\dfrac{dy}{b^2}-\dfrac{ey^2}{2b}-\dfrac{fy}{b}$

Case $4$: $a,b,c\neq0$

Then $u_{xy}+au_x+bu_y+cu+dx+ey+f=0$

Try let $u=p(x)q(y)v$ ,

Then $u_x=p(x)q(y)v_x+p_x(x)q(y)v$

$u_y=p(x)q(y)v_y+p(x)q_y(y)v$

$u_{xy}=p(x)q(y)v_{xy}+p(x)q_y(y)v_x+p_x(x)q(y)v_y+p_x(x)q_y(y)v$

$\therefore p(x)q(y)v_{xy}+p(x)q_y(y)v_x+p_x(x)q(y)v_y+p_x(x)q_y(y)v+a(p(x)q(y)v_x+p_x(x)q(y)v)+b(p(x)q(y)v_y+p(x)q_y(y)v)+cp(x)q(y)v+dx+ey+f=0$

$p(x)q(y)v_{xy}+p(x)(q_y(y)+aq(y))v_x+(p_x(x)+bp(x))q(y)v_y+(p_x(x)q_y(y)+ap_x(x)q(y)+bp(x)q_y(y)+cp(x)q(y))v=-dx-ey-f$

Take $q_y(y)+aq(y)=0\Rightarrow q(y)=\exp(-ay)$ and $p_x(x)+bp(x)=0\Rightarrow p(x)=\exp(-bx)$ , the PDE becomes

$\exp(-bx-ay)v_{xy}+(c-ab)\exp(-bx-ay)v=-dx-ey-f$

$v_{xy}+(c-ab)v=-dx\exp(bx+ay)-ey\exp(bx+ay)-f\exp(bx+ay)$

Case $4a$: $c=ab$

Then $v_{xy}=-dx\exp(bx+ay)-ey\exp(bx+ay)-f\exp(bx+ay)$

$v_x=\int(-dx\exp(bx+ay)-ey\exp(bx+ay)-f\exp(bx+ay))~dy$

$v_x=C(x)-\dfrac{dx\exp(bx+ay)}{a}-\dfrac{ey\exp(bx+ay)}{a}+\dfrac{e\exp(bx+ay)}{a^2}-\dfrac{f\exp(bx+ay)}{a}$

$v=\int\left(C(x)-\dfrac{dx\exp(bx+ay)}{a}-\dfrac{ey\exp(bx+ay)}{a}+\dfrac{e\exp(bx+ay)}{a^2}-\dfrac{f\exp(bx+ay)}{a}\right)dx$

$\exp(bx+ay)u=C_1(x)+C_2(y)-\dfrac{dx\exp(bx+ay)}{ab}+\dfrac{d\exp(bx+ay)}{ab^2}-\dfrac{ey\exp(bx+ay)}{ab}+\dfrac{e\exp(bx+ay)}{a^2b}-\dfrac{f\exp(bx+ay)}{ab}$

$\exp(bx+ay)u=C_1(x)+C_2(y)-\dfrac{(dx+ey+f)\exp(bx+ay)}{ab}+\dfrac{d\exp(bx+ay)}{ab^2}+\dfrac{e\exp(bx+ay)}{a^2b}$

$u=C_1(x)\exp(-bx-ay)+C_2(y)\exp(-bx-ay)-\dfrac{dx+ey+f}{ab}+\dfrac{d}{ab^2}+\dfrac{e}{a^2b}$

$u=C_1(x)\exp(-ay)+C_2(y)\exp(-bx)-\dfrac{dx+ey+f}{ab}+\dfrac{d}{ab^2}+\dfrac{e}{a^2b}$

Hence the really difficult case is that when $c\neq ab$ . By letting $u=\exp(-bx-ay)v$ the PDE will reduce to $v_{xy}+(c-ab)v=-dx\exp(bx+ay)-ey\exp(bx+ay)-f\exp(bx+ay)$ , which is as headache as https://math.stackexchange.com/questions/218425 for finding its most general solution.

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    $\begingroup$ Most text books just show how to solve the standard second order PDE. I thought it was easy to handle linear terms which turns out to be not. Anyway, thank you very much! $\endgroup$
    – hxhxhx88
    Nov 7, 2012 at 14:32

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