Answer to Question #108014 in Differential Equations for Nikesh gautam pandit ji

Question

Verify that the equations
i) z = √(2x+a) + √(2y+b) and
ii) z^2 + µ = 2(1+λ^-1) (x+λy)
are both complete integrals of the PDE .
z = 1/p + 1/q Also show that the complete integral

(ii) is the envelope of one parameter sub-system obtained by taking
b = -a/λ - µ/1+λ in the solution (i)

Answer 1

QUESTION (1i)

$$z=\sqrt{2x+a}+\sqrt{2y+p}\longrightarrow\\[0.3cm] p=\frac{\partial z}{\partial x}=\frac{2}{2\sqrt{2x+a}}\equiv\frac{1}{\sqrt{2x+a}}\\[0.3cm] q=\frac{\partial z}{\partial y}=\frac{2}{2\sqrt{2y+b}}\equiv\frac{1}{\sqrt{2y+b}}\\[0.3cm]$$

Then,

$$\frac{1}{p}+\frac{1}{q}=\frac{1}{\displaystyle\frac{1}{\sqrt{2x+a}}}+\frac{1}{\displaystyle\frac{1}{\sqrt{2y+b}}}\longrightarrow\\[0.3cm] \frac{1}{p}+\frac{1}{q}=\sqrt{2x+a}+\sqrt{2y+b}\equiv z$$

Conclusion,

$$\boxed{z=\sqrt{2x+a}+\sqrt{2y+p}\to\frac{1}{p}+\frac{1}{q}=z}$$

QUESTION (1ii)

$$z^2+\mu=2\left(1+\lambda^{-1}\right)(x+\lambda y)\longrightarrow\\[0.3cm] 2z\cdot\frac{\partial z}{\partial x}=2\left(1+\lambda^{-1}\right)\longrightarrow\\[0.3cm] p=\frac{\partial z}{\partial x}=\frac{1+\lambda^{-1}}{z}=\frac{\lambda+1}{z\lambda}\\[0.3cm] 2z\cdot\frac{\partial z}{\partial y}=2\left(1+\lambda^{-1}\right)\lambda\longrightarrow\\[0.3cm] q=\frac{\partial z}{\partial y}=\frac{\lambda+1}{z}\\[0.3cm]$$

Then,

$$\frac{1}{p}+\frac{1}{q}=\frac{1}{\displaystyle\frac{\lambda+1}{z\lambda}}+\frac{1}{\displaystyle\frac{\lambda+1}{z}}=\frac{z\lambda}{\lambda+1}+\frac{z}{\lambda+1}\\[0.3cm] \frac{1}{p}+\frac{1}{q}=\frac{z\left(\lambda+1\right)}{\lambda+1}\equiv z$$

Conclusion,

$$\boxed{z^2+\mu=2\left(1+\lambda^{-1}\right)(x+\lambda y)\longrightarrow\frac{1}{p}+\frac{1}{q}=z}$$

QUESTION 2

Suppose that $b=b(a;\lambda;\mu)$ , then

$$\frac{\partial }{\partial a}\left|\sqrt{2x+a}+\sqrt{2y+b}-z=0\right.\longrightarrow\\[0.3cm] \frac{1}{2\sqrt{2x+a}}+\frac{1}{2\sqrt{2y+b}}\cdot\frac{\partial b}{\partial a}=0\longrightarrow\\[0.3cm] \boxed{\sqrt{2y+b}=-\sqrt{2x+a}\cdot\frac{\partial b}{\partial a}}$$

Suppose that

$$b(a;\lambda;\mu)=-\frac{a}{\lambda}+C\longrightarrow\frac{\partial b}{\partial a}=-\frac{1}{\lambda}$$

Then,

$$\sqrt{2y+b}=-\sqrt{2x+a}\cdot\left(-\frac{1}{\lambda}\right)\longrightarrow\\[0.3cm] \boxed{\sqrt{2y+b}=\frac{1}{\lambda}\cdot\sqrt{2x+a}}\\[0.3cm] \left[\sqrt{2y+b}\right]^2=\left[\frac{1}{\lambda}\cdot\sqrt{2x+a}\right]^2\longrightarrow\\[0.3cm] \boxed{2y=\frac{2x+a}{\lambda^2}-b}$$

Substituting the found relationship in the first equation and express $z^2$ :

$$z=\sqrt{2x+a}+\sqrt{2y+b}\longrightarrow\\[0.3cm] z=\sqrt{2x+a}+\frac{1}{\lambda}\cdot\sqrt{2x+a}\longrightarrow\\[0.3cm] \boxed{z^2=\left(1+\frac{1}{\lambda}\right)^2\cdot(2x+a)}$$

Substituting the found relationship in the second equation :

$$z^2+\mu=2\left(1+\frac{1}{\lambda}\right)(x+\lambda y)\longrightarrow\\[0.3cm] \left.\left(\frac{1+\lambda}{\lambda}\right)^2(2x+a)+\mu=2\cdot\frac{1+\lambda}{\lambda}(x+\lambda y)\right|\div\frac{1}{(1+\lambda)^2}\\[0.3cm] \frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}=\frac{2(x+\lambda y)}{\lambda(1+\lambda)}\longrightarrow\\[0.3cm] \frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}=\frac{2y}{(1+\lambda)}+\frac{2x}{\lambda(1+\lambda)}\longrightarrow\\[0.3cm] \frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}=\frac{\displaystyle\frac{2x+a}{\lambda^2}-b}{(1+\lambda)}+\frac{2x}{\lambda(1+\lambda)}\longrightarrow\\[0.3cm] \frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}=\frac{(2x+a)-\lambda^2b}{\lambda^2(1+\lambda)}+\frac{2x}{\lambda(1+\lambda)}\longrightarrow\\[0.3cm] \frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}=\frac{(2x+a)}{\lambda^2(1+\lambda)}-\frac{\lambda^2b}{\lambda^2(1+\lambda)}+\frac{2x}{\lambda(1+\lambda)}\\[0.3cm] \frac{2x+a}{\lambda^2}+\frac{\mu}{(1+\lambda)^2}-\frac{2x}{\lambda(1+\lambda)}-\frac{(2x+a)}{\lambda^2(1+\lambda)}=-\frac{b}{(1+\lambda)}\\[0.3cm] \frac{(1+\lambda)(2x+a)-2x\lambda-(2x+a)}{\lambda^2(1+\lambda)}+\frac{\mu}{(1+\lambda)^2}=-\frac{b}{(1+\lambda)}\\[0.3cm] \frac{2x+a+2x\lambda+a\lambda-2x\lambda-2x-a}{\lambda^2(1+\lambda)}+\frac{\mu}{(1+\lambda)^2}=-\frac{b}{(1+\lambda)}\\[0.3cm] \left.\frac{a\lambda}{\lambda^2(1+\lambda)}+\frac{\mu}{(1+\lambda)^2}=-\frac{b}{(1+\lambda)}\right|\times-(1+\lambda)\\[0.5cm] \boxed{b=-\frac{a}{\lambda}-\frac{\mu}{(1+\lambda)}}$$

Conclusion,

The complete integral (1ii) is the envelope of one parameter sub-system obtained by taking

$$b=-\frac{a}{\lambda}-\frac{\mu}{(1+\lambda)}$$