0
$\begingroup$

Analytical problem is simple but I am not sure is it possible in this kind of system. I will give my idea. We can apply on first equation $ \frac{\partial }{\partial x} $ and then that it is possible to substitute from second equation $ \frac{\partial Q}{\partial x} $ in first, but the problem is that we have Q on two places and coefficient in front is not the same (A and B). A, B, C, D , E and F are constants.

i need one PDE equation just with P(x,y)

$$ -A\frac{\partial ^2Q(x,y)}{\partial x^2}+B\frac{\partial ^3P(x,y)}{\partial x^3}+C Q(x,y)-C\frac{\partial P(x,y)}{\partial x}+D\frac{\partial ^2Q(x,y)}{\partial y^2}=0 $$

$$ -B\frac{\partial ^3Q(x,y)}{\partial x^3}+E\frac{\partial ^4P(x,y)}{\partial x^4}-C \frac{\partial Q(x,y)}{\partial x}+C\frac{\partial ^2P(x,y)}{\partial x^2}-F\frac{\partial ^2P(x,y)}{\partial y^2}=0 $$

Thank you in advance

$\endgroup$

2 Answers 2

2
$\begingroup$

Maple can handle this with casesplit in the PDEtools package.

des:= {-A*diff(Q(x,y),x,x)+B*diff(P(x,y),x,x,x)+C*Q(x,y) -C*diff(P(x,y),x)+D*diff(Q(x,y),y,y)=0, -B*diff(Q(x,y),x,x,x)+E*diff(P(x,y),x,x,x,x)-C*diff(Q(x,y),x) +C*diff(P(x,y),x,x)-F*diff(P(x,y),y,y)=0};

PDEtools:-casesplit(des,[Q,P]);

and the last equation returned is (after some cleaning up)

$$\left( -{B}^{2}+EA \right) {\frac {\partial ^{6}}{\partial {x}^{6}}}P \left( x,y \right) + \left( - D C-AF \right) {\frac { \partial ^{4}}{\partial {y}^{2}\partial {x}^{2}}}P \left( x,y \right) + \left( AC-CE \right) {\frac {\partial ^{4}}{\partial {x}^{4}}}P \left( x,y \right) + D F{\frac {\partial ^{4}}{ \partial {y}^{4}}}P \left( x,y \right) - D E { \frac {\partial ^{6}}{\partial {y}^{2}\partial {x}^{4}}}P \left( x,y \right) +CF{\frac {\partial ^{2}}{\partial {y}^{2}}}P \left( x,y \right) = 0 $$

$\endgroup$
2
  • $\begingroup$ @ Robert Israel Is it possible to do this in $Mathematica$? $\endgroup$
    – George
    Jun 5, 2012 at 8:15
  • $\begingroup$ @George: I'm not a Mathematica user, so I don't know. $\endgroup$ Jun 5, 2012 at 18:14
1
$\begingroup$

$$-A\frac{\partial^{2}Q}{\partial x^{2}}+B\frac{\partial^{3}P}{\partial x^{3}}+CQ-C\frac{\partial P}{\partial x}+D\frac{\partial^{2}Q}{\partial y^{2}}=0\tag{{1}}$$ $$-B\frac{\partial^{3}Q}{\partial x^{3}}+E\frac{\partial^{4}P}{\partial x^{4}}-C\frac{\partial Q}{\partial x}+C\frac{\partial^{2}P}{\partial x^{2}}-F\frac{\partial^{2}P}{\partial y^{2}}=0\tag{{2}}$$

Differentiate $(1)$ with respect to $x$ :$$-A\frac{\partial^{3}Q}{\partial x^{3}}+B\frac{\partial^{4}P}{\partial x^{4}}+C\frac{\partial Q}{\partial x}-C\frac{\partial^{2}P}{\partial x^{2}}+D\frac{\partial^{3}Q}{\partial x\partial y^{2}}=0\tag{{3}}$$

Add $(2)$ and $(3)$:$$-\left(A+B\right)\frac{\partial^{3}Q}{\partial x^{3}}+\left(E+B\right)\frac{\partial^{4}P}{\partial x^{4}}+D\frac{\partial^{3}Q}{\partial x\partial y^{2}}-F\frac{\partial^{2}P}{\partial y^{2}}=0\tag{4}$$ Differentiate $(2)$ twice with respect to $y$ :$$-B\frac{\partial^{5}Q}{\partial x^{3}\partial y^{2}}+E\frac{\partial^{6}P}{\partial x^{4}\partial y^{2}}-C\frac{\partial^{3}Q}{\partial x\partial y^{2}}+C\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}-F\frac{\partial^{4}P}{\partial y^{4}}=0\tag{{5}}$$

Differentiate $(4)$ twice with respect to $x$ $$-\left(A+B\right)\frac{\partial^{5}Q}{\partial x^{5}}+\left(E+B\right)\frac{\partial^{6}P}{\partial x^{6}}+D\frac{\partial^{5}Q}{\partial x^{3}\partial y^{2}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}=0\tag{6}$$

Differentiate $(2)$ twice with respect to $x$ :$$-B\frac{\partial^{5}Q}{\partial x^{5}}+E\frac{\partial^{6}P}{\partial x^{6}}-C\frac{\partial^{3}Q}{\partial x^{3}}+C\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}=0\tag{{7}}$$

From $(7)$:

$$\frac{\partial^{5}Q}{\partial x^{5}}=\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{6}}-C\frac{\partial^{3}Q}{\partial x^{3}}+C\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}\right)\tag{{8}}$$

Substitute into $(6)$: $$-\left(A+B\right)\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{6}}-C\frac{\partial^{3}Q}{\partial x^{3}}+C\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}\right)+\left(E+B\right)\frac{\partial^{6}P}{\partial x^{6}}+D\frac{\partial^{5}Q}{\partial x^{3}\partial y^{2}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}=0 \tag{9}$$

From $(5)$ $$\frac{\partial^{5}Q}{\partial x^{3}\partial y^{2}}=\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{4}\partial y^{2}}-C\frac{\partial^{3}Q}{\partial x\partial y^{2}}+C\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}-F\frac{\partial^{4}P}{\partial y^{4}}\right)\tag{10}$$

Substitute into $(9)$: $$-\left(A+B\right)\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{6}}-C\frac{\partial^{3}Q}{\partial x^{3}}+C\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}\right)+\left(E+B\right)\frac{\partial^{6}P}{\partial x^{6}}+D\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{4}\partial y^{2}}-C\frac{\partial^{3}Q}{\partial x\partial y^{2}}+C\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}-F\frac{\partial^{4}P}{\partial y^{4}}\right)-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}=0\tag{11}$$

From $(4)$: $$\frac{\partial^{3}Q}{\partial x\partial y^{2}}=-\frac{1}{D}\left(-\left(A+B\right)\frac{\partial^{3}Q}{\partial x^{3}}+\left(E+B\right)\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{2}P}{\partial y^{2}}\right)\tag{12}$$

Substitute into $(11)$: $$-\left(A+B\right)\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{6}}-C\frac{\partial^{3}Q}{\partial x^{3}}+C\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}\right)+\left(E+B\right)\frac{\partial^{6}P}{\partial x^{6}}+D\frac{1}{B}\left(E\frac{\partial^{6}P}{\partial x^{4}\partial y^{2}}+C\frac{1}{D}\left(-\left(A+B\right)\frac{\partial^{3}Q}{\partial x^{3}}+\left(E+B\right)\frac{\partial^{4}P}{\partial x^{4}}-F\frac{\partial^{2}P}{\partial y^{2}}\right)+C\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}-F\frac{\partial^{4}P}{\partial y^{4}}\right)-F\frac{\partial^{4}P}{\partial x^{2}\partial y^{2}}=0\tag{13}$$

Differentiate $(1)$ with respect to $x$ three times:$$-A\frac{\partial^{5}Q}{\partial x^{5}}+B\frac{\partial^{6}P}{\partial x^{6}}+C\frac{\partial^{3}Q}{\partial x^{3}}-C\frac{\partial^{4}P}{\partial x^{4}}+D\frac{\partial^{5}Q}{\partial y^{5}}=0\tag{14}$$

$(13)$ gives an expression for $$\frac{\partial^{3}Q}{\partial x^{3}}$$ . substitution into $(8)$ gives equation for $$\frac{\partial^{5}Q}{\partial x^{5}}$$ . Substituting the results in $(14)$ gives an equation in terms of P and its derivatives only.

$\endgroup$
9
  • $\begingroup$ yes, but I need to eliminate Q and all derivatives of function Q, how to do that? $\endgroup$
    – George
    Jun 4, 2012 at 21:20
  • $\begingroup$ you just need to solve this linear system (i used the uniform notation $i=0$ corresponds to $Q$) $\endgroup$
    – Valentin
    Jun 4, 2012 at 21:22
  • $\begingroup$ @ Valentin Thanks, right answer! $\endgroup$
    – George
    Jun 4, 2012 at 21:25
  • $\begingroup$ @ Valentin in second equation should be derivatives x and y? $\endgroup$
    – George
    Jun 4, 2012 at 21:37
  • $\begingroup$ yes, you are right i made a slip, so you actually might need to differentiate each a few times more by $x$ and $y$ until the number of equations matches the number of variables $\endgroup$
    – Valentin
    Jun 4, 2012 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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