0
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Separation should work. Is this homework? $\endgroup$ Jul 9, 2011 at 16:58
  • $\begingroup$ no,this is not homework $\endgroup$
    – anksh11
    Jul 9, 2011 at 17:04
  • 1
    $\begingroup$ 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 $u(0,y)=0$. $\endgroup$
    – Andrew
    Jul 10, 2011 at 7:01

1 Answer 1

1
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ Or, if you don't want a singularity at $x=0$, $\frac{5x^2}{a+b} + \frac{9x}{b} + \frac{3y^2}{c}$. $\endgroup$ Jul 10, 2011 at 16:01
  • $\begingroup$ 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? $\endgroup$
    – anksh11
    Jul 10, 2011 at 18:18
  • $\begingroup$ but second solution doesn't fullfill initial conditions $\endgroup$
    – anksh11
    Jul 10, 2011 at 18:23
  • $\begingroup$ 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 $v_h(0,y) = -v_p(0,y) = - \frac{3y^2}{c}$. $\endgroup$ Jul 10, 2011 at 19:52
  • $\begingroup$ However, you're likely to have trouble with your problem because the pde is singular at $x=0$. $\endgroup$ Jul 10, 2011 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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