Answer to Question #199533 in Differential Equations for Rajkumar

Question

Verify that the equations

i) z = √(2x + a) + √(2y+ b)

ii) z² + 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 the one parameter sub-system obtained by taking b = -( a/λ ) - ( μ/1+λ ) in the solution i).

Answer 1

1(i)

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

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,

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

1(ii)

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

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,

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

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,

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

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

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

Substituting the found relationship in the second equation :

$z^ 2 +μ=2(1+ \dfrac{1}{λ} ​ )(x+λy)⟶$

$(\dfrac{1+λ}{λ})^2(2x+a)+ \mu =2\times \dfrac{1+ \lambda}{λ}(x+ λy)| \div \dfrac{1}{(1+ \lambda)^2}$

$\dfrac{2x+a}{ \lambda^2 }+ \dfrac{ \mu}{(1+ \lambda ^2)} =\dfrac{2(x+ \lambda y)}{\lambda(1+ \lambda)}$

$\dfrac{2x+a}{\lambda^2}+\dfrac{\mu}{(1+\lambda)^2}=\dfrac{2(x+\lambda y)}{\lambda (1+ \lambda)} \longrightarrow$

$\dfrac{2x+a}{\lambda^2} + \dfrac{\mu}{1+ \lambda^2}= \dfrac{2y}{1+ \lambda}+ \dfrac{2x}{\lambda(1+\lambda)} \longrightarrow$

$\dfrac{2x+a}{\lambda^2} + \dfrac{\mu}{1+ \lambda^2}= \dfrac{ \dfrac{2x+a}{\lambda^2}-b}{1+ \lambda} + \dfrac{2x}{\lambda(1+ \lambda)} \longrightarrow$

On further solving it will become:

$\dfrac{2x+a+2x \lambda +a \lambda-2x \lambda-2x-2a}{\lambda^2(1+\lambda)} = \dfrac{\mu}{(1+ \lambda)^2}= - \dfrac{b}{(1+ \lambda)}$

$[ \dfrac{a \lambda}{\lambda^2(1+\lambda)}+ \dfrac{\mu}{(1+\lambda)^2}=- \dfrac{b}{(1+ \lambda)}] \times[-(1+ \lambda)]$

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

Thus,

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

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