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$ax^2\frac{\partial^2 v}{\partial x^2}+bx\frac{\partial v}{\partial x}+c\frac{\partial^2 v}{\partial y^2}=10x^2+9x+6$ where $a,b,c$ are constants,

initial conditions: $v(x,0)=0,v(0,y)=0$

i tried separation method but can't get particular solution using this initial conditions

sorry guys,for error

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    @anks: According to http://math.stackexchange.com/posts/48496/revisions, the coefficients of the PDE has been changed by you, but the fact comes out to that the former version is solvable while the latter version is not solvable in most cases. Does the PDE really $a\dfrac{\partial^2 v}{\partial x^2}+b\dfrac{\partial v}{\partial x}+c\dfrac{\partial^2 v}{\partial y^2}=10x^2+9x+6$ or really $ax^2\dfrac{\partial^2 v}{\partial x^2}+bx\dfrac{\partial v}{\partial x}+c\dfrac{\partial^2 v}{\partial y^2}=10x^2+9x+6$ ?2013-04-10

1 Answers 1

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Case $1$: $a=b=0$ and $c\neq0$

Then $c\dfrac{\partial^2v}{\partial y^2}=10x^2+9x+6$

$\dfrac{\partial^2v}{\partial y^2}=\dfrac{10x^2+9x+6}{c}$

$\dfrac{\partial v}{\partial y}=\dfrac{(10x^2+9x+6)y}{c}+C_1(x)$

$v(x,y)=\dfrac{(10x^2+9x+6)y^2}{2c}+C_1(x)y+C_2(x)$

$v(x,0)=0$ :

$C_2(x)=0$

$\therefore v(x,y)=\dfrac{(10x^2+9x+6)y^2}{2c}+C_1(x)y$

$v(0,y)=0$ :

$\dfrac{6y^2}{2c}+C_1(0)y=0$

$C_1(0)=-\dfrac{3y}{c}$

$\therefore v(x,y)=\dfrac{(10x^2+9x+6)y^2}{2c}+C_1(x)y$ , where $C_1(x)$ is any function satisfy $C_1(0)=-\dfrac{3y}{c}$

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

Then $bx\dfrac{\partial v}{\partial x}=10x^2+9x+6$

$\dfrac{\partial v}{\partial x}=\dfrac{10x}{b}+\dfrac{9}{b}+\dfrac{6}{bx}$

$v(x,y)=\dfrac{5x^2}{b}+\dfrac{9x}{b}+\dfrac{6\ln x}{b}+C(y)$

But $x$ cannot substitute $0$ as it will becomes undefined

$\therefore$ There is no solution when we force to satisfy the I.C. $v(0,y)=0$

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

Then $ax^2\dfrac{\partial^2v}{\partial x^2}+bx\dfrac{\partial v}{\partial x}=10x^2+9x+6$

$\dfrac{\partial^2v}{\partial x^2}+\dfrac{b}{ax}\dfrac{\partial v}{\partial x}=\dfrac{10}{a}+\dfrac{9}{ax}+\dfrac{6}{ax^2}$

I.F. $=e^{\int\frac{b}{ax}dx}=e^{\frac{b}{a}\ln x}=e^{\ln x^{\frac{b}{a}}}=x^{\frac{b}{a}}$

$\therefore\dfrac{\partial}{\partial x}\left(x^{\frac{b}{a}}\dfrac{\partial v}{\partial x}\right)=\dfrac{10x^{\frac{b}{a}}}{a}+\dfrac{9x^{\frac{b}{a}-1}}{a}+\dfrac{6x^{\frac{b}{a}-2}}{a}$

$\begin{cases}x^{\frac{b}{a}}\dfrac{\partial v}{\partial x}=\dfrac{10x^{\frac{b}{a}+1}}{a+b}+\dfrac{9x^{\frac{b}{a}}}{b}+\dfrac{6x^{\frac{b}{a}-1}}{b-a}+C_1(y)&\text{when}\dfrac{b}{a}\neq-1,0,1\\\dfrac{1}{x}\dfrac{\partial v}{\partial x}=\dfrac{10\ln x}{a}-\dfrac{9}{ax}-\dfrac{3}{ax^2}+C_1(y)&\text{when}\dfrac{b}{a}=-1\\\dfrac{\partial v}{\partial x}=\dfrac{10x}{a}+\dfrac{9\ln x}{a}-\dfrac{6}{ax}+C_1(y)&\text{when}\dfrac{b}{a}=0\\x\dfrac{\partial v}{\partial x}=\dfrac{5x^2}{a}+\dfrac{9x}{a}+\dfrac{6\ln x}{a}+C_1(y)&\text{when}\dfrac{b}{a}=1\end{cases}$

$\dfrac{\partial v}{\partial x}=\begin{cases}\dfrac{10x}{a+b}+\dfrac{9}{b}+\dfrac{6}{(b-a)x}+C_1(y)x^{-\frac{b}{a}}&\text{when}\dfrac{b}{a}\neq-1,0,1\\\dfrac{10x\ln x}{a}-\dfrac{9}{a}-\dfrac{3}{ax}+C_1(y)x&\text{when}\dfrac{b}{a}=-1\\\dfrac{10x}{a}+\dfrac{9\ln x}{a}-\dfrac{6}{ax}+C_1(y)&\text{when}\dfrac{b}{a}=0\\\dfrac{5x}{a}+\dfrac{9}{a}+\dfrac{6\ln x}{ax}+\dfrac{C_1(y)}{x}&\text{when}\dfrac{b}{a}=1\end{cases}$

$v(x,y)=\begin{cases}\dfrac{5x^2}{a+b}+\dfrac{9x}{b}+\dfrac{6\ln x}{b-a}+\dfrac{C_1(y)ax^{1-\frac{b}{a}}}{a-b}+C_2(y)&\text{when}\dfrac{b}{a}\neq-1,0,1\\\dfrac{5x^2(2\ln x-1)}{2a}-\dfrac{9x}{a}-\dfrac{3\ln x}{a}+\dfrac{C_1(y)x^2}{2}+C_2(y)&\text{when}\dfrac{b}{a}=-1\\\dfrac{5x^2}{a}+\dfrac{9x(\ln x-1)}{a}-\dfrac{6\ln x}{a}+C_1(y)x+C_2(y)&\text{when}\dfrac{b}{a}=0\\\dfrac{5x^2}{2a}+\dfrac{9x}{a}+\dfrac{3(\ln x)^2}{a}+C_1(y)\ln x+C_2(y)&\text{when}\dfrac{b}{a}=1\end{cases}$

But $x$ cannot substitute $0$ as it will becomes undefined

$\therefore$ There is no solution when we force to satisfy the I.C. $v(0,y)=0$

Case $4$: $a=0$ and $b\neq0$ and $c\neq0$

Then $bx\dfrac{\partial v}{\partial x}+c\dfrac{\partial^2v}{\partial y^2}=10x^2+9x+6$

It is possible that the subsititution $v(x,y)=v_c(x,y)+v_p(x,y)$ can make the inhomogeneous linear PDE becomes homogeneous linear PDE if $v_p(x,y)$ can be found.

For this question, the form of $v_p(x,y)$ is not difficult to guess, just the particular solution of the ODE $bx\dfrac{dv}{dx}=10x^2+9x+6$

But with reference to case $2$ , this ODE is no solution when we force to satisfy the I.C. $v(0,y)=0$

$\therefore$ There is no solution when we force to satisfy the I.C. $v(0,y)=0$

Case $5$: $a\neq0$ and $b\neq0$ and $c\neq0$

Then $ax^2\dfrac{\partial^2v}{\partial x^2}+bx\dfrac{\partial v}{\partial x}+c\dfrac{\partial^2v}{\partial y^2}=10x^2+9x+6$

It is possible that the subsititution $v(x,y)=v_c(x,y)+v_p(x,y)$ can make the inhomogeneous linear PDE becomes homogeneous linear PDE if $v_p(x,y)$ can be found.

For this question, the form of $v_p(x,y)$ is not difficult to guess, just the particular solution of the ODE $ax^2\dfrac{d^2v}{dx^2}+bx\dfrac{dv}{dx}=10x^2+9x+6$

But with reference to case $3$ , this ODE is no solution when we force to satisfy the I.C. $v(0,y)=0$

$\therefore$ There is no solution when we force to satisfy the I.C. $v(0,y)=0$