MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Separation should work. Is this homework? – András Bátkai Jul 9 '11 at 16:58
no,this is not homework – anksh11 Jul 9 '11 at 17:04
The question to begin here would be perhaps for which values of $a$, $b$, $c\ $ a solution exists. For example, if $a=b=0\ $ then $u(x,y)=5x^2y^2/c + 3xy^2/c + 9 y^2/(2c) + C_2(x) y + C_1(x)\ $ and $u(0,y)=\frac{9 y^2}{2 c}+ C_2(0) y + C_1(0)\ $, so constants $C_1(0)$, $C_2(0)$ cannot be choosen to get $u(0,y)=0$. – Andrew Jul 10 '11 at 7:01

A particular solution of the pde (obtained with Maple's help) is $v(x,y) = \frac{5 x^2}{a+b} + \frac{9x}{b} + \frac{6 \ln x}{a-b}$.

share|cite|improve this answer
Or, if you don't want a singularity at $x=0$, $\frac{5x^2}{a+b} + \frac{9x}{b} + \frac{3y^2}{c}$. – Robert Israel Jul 10 '11 at 16:01
thanks man, i had tried mathematica without success but not maple ,but how you get second solution,means how we can exclude singularity while giving initial conditions/etc on maple? – anksh11 Jul 10 '11 at 18:18
but second solution doesn't fullfill initial conditions – anksh11 Jul 10 '11 at 18:23
Typically, when solving a nonhomogeneous boundary value problem, you take a particular solution $v_p$ of the pde (without regard to boundary conditions), and then look for a solution $v_h$ of the homogeneous pde such that $v = v_h + v_p$ satisfies the boundary conditions. Thus in your case, taking $v_p(x,y) = \frac{5x^2}{a+b} + \frac{9x}{b} + \frac{3y^2}{c}$, the boundary conditions for $v_h$ would be $v_h(x,0) = - v_p(x,0) = - \frac{5x^2}{a+b} - \frac{9x}{b}$ and $v_h(0,y) = -v_p(0,y) = - \frac{3y^2}{c}$. – Robert Israel Jul 10 '11 at 19:52
However, you're likely to have trouble with your problem because the pde is singular at $x=0$. – Robert Israel Jul 10 '11 at 20:00

Your Answer


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.