Question #68370

Form Partial differential eq. of x^2/a^2 +y^2/b^2 +u^2/c^2 =1

Expert's answer

Answer on Question #68370 – Math – Differential Equations

Question

Form Partial differential eq. of x2/a2+y2/b2+u2/c2=1x^2/a^2 + y^2/b^2 + u^2/c^2 = 1

Solution

We have the following equation


x2a2+y2b2+u2c2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{u^2}{c^2} = 1


Note that the number of constants (a,b,c)(a, b, c) is more than the number of independent variables (x,y)(x, y). Hence the order of the resulting differential equation will be more than 1.

Differentiating (1) partially with respect to xx we get


2xa2+2uc2ux=0\frac{2x}{a^2} + \frac{2u}{c^2} \frac{\partial u}{\partial x} = 0


or


c2a2=uxux\frac{c^2}{a^2} = -\frac{u}{x} \frac{\partial u}{\partial x}


Differentiating (2) partially with respect to xx we get


1a2+1c2uxux+uc22ux2=0\frac{1}{a^2} + \frac{1}{c^2} \frac{\partial u}{\partial x} \frac{\partial u}{\partial x} + \frac{u}{c^2} \frac{\partial^2 u}{\partial x^2} = 0


or


c2a2=(ux)2u2ux2\frac{c^2}{a^2} = -\left(\frac{\partial u}{\partial x}\right)^2 - u \frac{\partial^2 u}{\partial x^2}


From (3) and (4) we get


xu2ux2+x(ux)2=uuxx u \frac{\partial^2 u}{\partial x^2} + x \left(\frac{\partial u}{\partial x}\right)^2 = u \frac{\partial u}{\partial x}


which is required differential equation.

This equation is not unique. Similarly differentiating (1) partially with respect to yy we get


2yb2+2uc2uy=0\frac{2y}{b^2} + \frac{2u}{c^2} \frac{\partial u}{\partial y} = 0


or


c2b2=uyuy\frac{c^2}{b^2} = -\frac{u}{y} \frac{\partial u}{\partial y}


and differentiating (6) partially with respect to yy we get


1b2+1c2uyuy+uc22uy2=0\frac{1}{b^2} + \frac{1}{c^2} \frac{\partial u}{\partial y} \frac{\partial u}{\partial y} + \frac{u}{c^2} \frac{\partial^2 u}{\partial y^2} = 0


or


c2b2=(uy)2u2uy2\frac{c^2}{b^2} = -\left(\frac{\partial u}{\partial y}\right)^2 - u \frac{\partial^2 u}{\partial y^2}


From (7) and (8) we get


yu2uy2+y(uy)2=uuyy u \frac{\partial^2 u}{\partial y^2} + y \left(\frac{\partial u}{\partial y}\right)^2 = u \frac{\partial u}{\partial y}


which is also a required differential equation. Equations (5) and (9) can be summed so that the resulting equation is symmetric with respect to xx and yy

xu2ux2+yu2uy2+x(ux)2+y(uy)2=uux+uuyx u \frac {\partial^ {2} u}{\partial x ^ {2}} + y u \frac {\partial^ {2} u}{\partial y ^ {2}} + x \left(\frac {\partial u}{\partial x}\right) ^ {2} + y \left(\frac {\partial u}{\partial y}\right) ^ {2} = u \frac {\partial u}{\partial x} + u \frac {\partial u}{\partial y}


Answer: The required partial differential equations are


xu2ux2+x(ux)2=uuxx u \frac {\partial^ {2} u}{\partial x ^ {2}} + x \left(\frac {\partial u}{\partial x}\right) ^ {2} = u \frac {\partial u}{\partial x}


or


yu2uy2+y(uy)2=uuyy u \frac {\partial^ {2} u}{\partial y ^ {2}} + y \left(\frac {\partial u}{\partial y}\right) ^ {2} = u \frac {\partial u}{\partial y}


or


xu2ux2+yu2uy2+x(ux)2+y(uy)2=uux+uuyx u \frac {\partial^ {2} u}{\partial x ^ {2}} + y u \frac {\partial^ {2} u}{\partial y ^ {2}} + x \left(\frac {\partial u}{\partial x}\right) ^ {2} + y \left(\frac {\partial u}{\partial y}\right) ^ {2} = u \frac {\partial u}{\partial x} + u \frac {\partial u}{\partial y}


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