Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
I have tried to fix up your equation. –  Aryabhata Jun 29 '11 at 19:13
1  
Try looking for a solution independent of y. This reduces your problem to a fairly easy ODE. –  gfes Jun 30 '11 at 1:06
    
Are there some conditions on constsants? The Laplace equation is one thing and the wave equation is another. –  Andrew Jul 1 '11 at 17:15
    
@gfes: a solution independent of $y$ is not compatible with the boundary conditions. The first boundary condition requires $v(x,0) = 0$. If $v$ is independent of $y$ then $v$ is 0 every where. Which is incompatible with the equation. –  Willie Wong Jul 1 '11 at 17:15
    
@anks: This ought to be an amazing question, but you introduce an awful I.C. $y(0,y)=0$ makes this question impossible to solve in most cases and becomes uninteresting. Maybe you are better to modify either the I.C.s or the inhomogeneous term so that this question becomes possible to solve, otherwise I can only provide this disappointing answer to you. –  doraemonpaul Aug 30 '12 at 12:05
show 1 more comment

1 Answer 1

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. $v(0,y)=0$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.