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$$ d_1u_{xx} +d_2 u_{yy} = -2 $$ I need to solve this given PDE. I tried to solve it using change of variables. The variables are $$ \xi = y-ax \ , \ \ \ \eta = y+ax$$ where $$a = \sqrt{-d_1d_2} $$ After the change of variable $(x,y)$ into $(\xi,\eta)$ , my equation looks like this, $$u_{\xi\xi}(1-d_1d_2) + 2 u_{\xi\eta} (1+d_1d_2) + u_{\eta\eta}(1-d_1d_2) = -2 $$

which seems more complicated than the original equation. Did i do something wrong? When I tried with example equations I was able to solve them without any problem. Can someone please give me a clue on how to solve this one. Thanks in advance.

dexter
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2 Answers2

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The problem is easier than you were trying to make it. Let $$ \xi = \frac{x}{\sqrt{d_1}}, \eta = \frac{y}{\sqrt{d_2}} $$ then the equation is $$ \nabla^2 u = -2 $$ and the solution to that is any solution to $$ \nabla^2 u = -0 $$ (all of which are of the form $u = pxy + ax+by+c$ if you want the equation to hold everywhere on the $(x,y)$ plane) plus a particular solution to $$ \nabla^2 u = -2 $$ which has the family of solutions $$ u = m \xi^2 -(2+m) \eta^2 $$ for any real $m$.

So your answer should be $$ u = \frac{m}{d_1}x^2 - \frac{2+m}{d_2} y^2 + pxy + ax + by + c $$

If, on the other hand, the problem is a boundary value problem one some closed subsset of the plane, there are further solutions to the homogneuous part $\nabla^2 u = 0$ which can be thought of as potentials of an arbitrary charge distribution.

Mark Fischler
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  • thanks a lot. Am new to mathematics,so am having little trouble in understanding few steps in this. I'll try to learn about deriving particular solution. thanks again. – dexter Aug 11 '14 at 09:33
  • can you please explain me how to solve the canonical form or suggest me some material regarding it ? – dexter Aug 11 '14 at 18:16
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Just for completeness,

the discriminant of your PDE is given by:

$$B^2 - 4AC = 0 - 4 d_1 d_2, \quad A = d_1, B = 0, C=d_2, $$ which can be positive, negative or zero depending on the values of $d_i$. Since, a priori, we don't known anything of the coefficients $d_i$, we have that the characteristics of your PDE can be found from the following equation:

$$d_1 \xi_{x}^2 + d_2 \xi_{y}^2 = 0, \tag{1}$$ (and the same for the other characteristic, $\eta$) which can be solved for:

a) $d_{1,2} \neq 0$: which yields: $\xi_x/\xi_y = \pm \sqrt{d_2/d_1} = - dy/dx$, giving the characteristics curves: $$y + \alpha x = \xi, \quad y - \alpha x = \eta, \quad \alpha = \sqrt{d_2/d_1}. $$ After using the chain rule, we obtain the original equation written in terms of the new variables:

$$u_{\xi\xi} + u_{\eta \eta} = \frac{1}{d_2}, \tag{2}$$

which is an elliptic PDE (Poisson's eq.) you can solve following @Mark Fischler's answer.

b) $d_1 = 0, d_2 \neq 0$: we would have $d_2 u_{yy} = -2$, whose general solution is:

$$u(x,y) = - \frac{y^2}{d_2} + y f(x) + g(x), \tag{3}$$ being $f$ and $g$ arbitrary functions of $x$.

c) $d_2 = 0, d_1 \neq 0$: reasoning as we have done above, we would get:

$$u(x,y) = - \frac{x^2}{d_1} + x f(y) + g(y). \tag{4}$$

Hope this helps!

Cheers.

Dmoreno
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  • thanks you so much. It really helped me to understand @Mark Fischler's answer. – dexter Aug 11 '14 at 09:35
  • Happy to help, @dexter! – Dmoreno Aug 11 '14 at 09:36
  • $ \frac{(dy)^2}{d2} = \frac{(dx)^2}{d1} $ is this a correct way to derive $ \xi , \eta$ ? – dexter Aug 11 '14 at 09:37
  • Of course this equation comes from eq. (1). The two different solutions correspond to $\xi$ and $\eta$, respectively. If you are interested in knowing where do eq. (1) or the fact $\xi_x /\xi_y = -dy/dx$ come from, just tell me. – Dmoreno Aug 11 '14 at 09:40
  • for first order we use $\frac{dx}{coefficient} = \frac{dy}{coefficient} $, i thought eqn(1) comes from similar kind of eqn. Can you explain me how to get it? – dexter Aug 11 '14 at 09:51
  • For 1st order linear (or quasilinear) PDEs such as $a(x,y) z_x + b(x,y) z_y = F(x,y,u)$ we often use the method of characteristics which reads: $dx/a = dy/b = dz/F$. Generally speaking, this is not applicable to this 2nd order PDEs where you may proceed as it has been described by Mark and me. Eq.(1) comes from the transformation $(x,y)\to (\xi,\eta)$. If you have any more doubts, don't hesitate to open a chat! – Dmoreno Aug 11 '14 at 10:03
  • i need some more clarification. But am not allowed to chat. it says i have nly 16 reputations. :( – dexter Aug 11 '14 at 10:52
  • i found this one, http://www.mathstats.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf – dexter Aug 11 '14 at 11:02
  • I can't access your URL either. Ok, let's try to do this quickly: 1st order PDEs $\to$ Lagrange-Charpit method (which is a generalization of characteristics (some links: http://en.wikipedia.org/wiki/Method_of_characteristics, http://math.stackexchange.com/questions/787003/pde-characteristic-equations-remembering-them/787675#787675), 2nd order PDEs: proceed as described above (here's another answer of mine (http://math.stackexchange.com/questions/891815/solving-the-pde-u-tt2u-txu-xx-2c/891889#891889) – Dmoreno Aug 11 '14 at 11:28
  • your previous answers explained it so nicely. – dexter Aug 11 '14 at 16:54
  • can you give suggest me some material to understand solving the elliptic PDE? – dexter Aug 11 '14 at 17:53
  • Thanks for the compliment @dexter! Unfortunately, I have no available on-line material for elliptic PDEs in particular right now, just my personal notes. However, this seems to be a good reference for partial differential equations from the point of view of operators: L.V. Hörmander, "The analysis of linear partial differential operators" , 1 , Springer (1983). Also: R. Haberman, "Applied Partial Differential Equations". Cheers! – Dmoreno Aug 12 '14 at 09:54
  • I got those books, thanks a lot @Dmoreno – dexter Aug 13 '14 at 16:36