# How to eliminate Q(x,y) in system of two PDE

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)

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

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

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

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.

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yes, but I need to eliminate Q and all derivatives of function Q, how to do that? – George Jun 4 '12 at 21:20
you just need to solve this linear system (i used the uniform notation $i=0$ corresponds to $Q$) – Valentin Jun 4 '12 at 21:22
@ Valentin Thanks, right answer! – George Jun 4 '12 at 21:25
@ Valentin in second equation should be derivatives x and y? – George Jun 4 '12 at 21:37
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 – Valentin Jun 4 '12 at 21:46

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)

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

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@ Robert Israel Is it possible to do this in $Mathematica$? – George Jun 5 '12 at 8:15
@George: I'm not a Mathematica user, so I don't know. – Robert Israel Jun 5 '12 at 18:14