Question

Verify that the equations

i) z =sqrt (2x + a )+ sqrt(2y + b) and

ii)z^2+u=2(1+l ^x)(x+ly)

are both complete integrals of the PDEz=1/p+1/q .

Also show that the complete integral

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

solution (i)

Answer 1

Question (1i)

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

Then,

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

Question (ii)

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

Then,

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

Question 2

Let suppose that $b=b(a;\lambda,\mu)$ . Then,

$$F(x,y,z,a,b)=\sqrt{2x+a}+\sqrt{2y+b}-z=0\longrightarrow\\[0.3cm] \frac{\partial F}{\partial a}=\frac{1}{2\sqrt{2x+a}}+\frac{1}{2\sqrt{2y+b}}\cdot\frac{db}{da}=0\longrightarrow\\[0.3cm] \frac{1}{2\sqrt{2x+a}}=-\frac{1}{2\sqrt{2y+b}}\cdot\frac{db}{da}\longrightarrow\\[0.3cm] 2\sqrt{2y+b}=-2\sqrt{2x+a}\cdot\frac{db}{da}\longrightarrow\\[0.3cm] \boxed{\sqrt{2y+b}=-\sqrt{2x+a}\cdot\frac{db}{da}}$$

Let

$$b(a)=-\frac{a}{\lambda}+C\longrightarrow\frac{db}{da}=-\frac{1}{\lambda}$$

Then,

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

Consider the found function along with

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

Then,

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

Conclusion,

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

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