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I need to solve the following equation $$-\frac{\partial S(x,y,t)}{\partial t}=ax^2+bx\frac{\partial S(x,y,t)}{\partial x}+c\Big[\frac{\partial S(x,y,t)}{\partial x}\Big]^2+\frac{1}{x-\alpha}+\frac{1}{\beta-x}$$ for the function $S(x,y,t)$ approximately. Here, $a$, $b$, $c$, $\alpha$, and $\beta$ are given constants.

  • Naturally, we can achieve this by Taylor series approximation, i.e.,

    $S(x,y,t)=c_1+c_2x+c_3y+c_4x^2+c_5xy+c_6y^2+\cdots$, $\frac{1}{x-\alpha}$ and $\frac{1}{\beta-x}$ can also be expanded as Taylor polynomials. After substitution into the equation in the beginning, by collecting terms with same variables $x^iy^j$ and letting their coefficients equal to zero, we solve the time-varying $c_1$, $c_2$, $\dots$, which means we solve the function $S(x,y,t)$ approximately.

  • Now notice the fractional form of the equation in the beginning, I guess a rational form of $S(x,y,t)$, i.e.,

    $S(x,y,t)=\frac{c_1+c_2x+c_3y+c_4x^2+c_5xy+c_6y^2+\cdots}{(x-\alpha)(\beta-x)}$, and expect it can give more accurate solutions. But it seems not work well.

So my question: Is there any good form to approximate the function $S(x,y,t)$, except the conventional Taylor series expansion? Thanks in advance!

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  • $\begingroup$ Is it correct that $y$ does not appear in the PDE? $\endgroup$ – Antonio Vargas Jul 10 '14 at 9:43
  • $\begingroup$ @Antonio Vargas Yes, it is correct. $\endgroup$ – user143763 Jul 11 '14 at 0:54
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Hint:

$-\dfrac{\partial S}{\partial t}=ax^2+bx\dfrac{\partial S}{\partial x}+c\left(\dfrac{\partial S}{\partial x}\right)^2+\dfrac{1}{x-\alpha}+\dfrac{1}{\beta-x}$

$-\dfrac{\partial^2S}{\partial x\partial t}=2ax+bx\dfrac{\partial^2S}{\partial x^2}+b\dfrac{\partial S}{\partial x}+2c\dfrac{\partial S}{\partial x}\dfrac{\partial^2S}{\partial x^2}-\dfrac{1}{(x-\alpha)^2}+\dfrac{1}{(x-\beta)^2}$

Let $u=\dfrac{\partial S}{\partial x}$ ,

Then $-\dfrac{\partial u}{\partial t}=2ax+bx\dfrac{\partial u}{\partial x}+bu+2cu\dfrac{\partial u}{\partial x}-\dfrac{1}{(x-\alpha)^2}+\dfrac{1}{(x-\beta)^2}$

$\dfrac{\partial u}{\partial t}+(2cu+bx)\dfrac{\partial u}{\partial x}=-bu-2ax+\dfrac{1}{(x-\alpha)^2}-\dfrac{1}{(x-\beta)^2}$

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