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)
Expert's answer

Answer 1

i) Given

$$\ \ \ z=\sqrt{2x+a}+\sqrt{2y+b} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

Differentiate (1) partially with respect to $x$

$$\\ \ \ {\partial z \over \partial x}={2 \over 2\sqrt{2x+a}}={ 1\over \sqrt{2x+a}} \ \ \ \ \ \ \ \ \ \ \ (2)$$

Differentiate (1) partially with respect to $y$

$$\\ \ \ {\partial z \over \partial y}={2 \over 2\sqrt{2+b}}={ 1\over \sqrt{2y+b}} \ \ \ \ \ \ \ \ \ \ \ (3)$$

From (2) and (3)

$$p=\sqrt{2x+a},\ q=\sqrt{2y+b}$$

Substitutiing in (1) we get

$$z={1 \over p}+{1 \over q}$$

ii) Given

$$\ \ z^2+\mu=2(1+\lambda^{-1})(x+\lambda y) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$

Differentiate (4) partially with respect to $x$

$$2z{\partial z \over \partial x}=2(1+\lambda^{-1})$$

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ z{\partial z \over \partial x}=1+\lambda^{-1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)$$

Differentiate (4) partially with respect to $y$

$$2z{\partial z \over \partial y}=2\lambda(1+\lambda^{-1})$$

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ z{\partial z \over \partial y}=\lambda(1+\lambda^{-1}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)$$

From (5) and (6)

$$p={1 \over z}(1+\lambda^{-1}),\ q={1 \over z}\lambda(1+\lambda^{-1})$$

Then

$${1 \over p}+{1 \over q}={z \over 1+\lambda^{-1}}+{z \over \lambda(1+\lambda^{-1})}$$

$${1 \over p}+{1 \over q}=z{\lambda+1 \over \lambda(1+\lambda^{-1})}$$

$${1 \over p}+{1 \over q}=z{\lambda+1 \over \lambda+1}$$

Hence

$$z={1 \over p}+{1 \over q}$$

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

$$b=-{a \over \lambda}-{\mu \over1+\lambda}$$

Given

$$z=\sqrt{2x+a}+\sqrt{2y+b}$$

Then

$$f(x, y, z, a, b)=\sqrt{2x+a}+\sqrt{2y+b}-z=0$$

$${\partial f \over \partial a}={1 \over 2\sqrt{2x+a}}+{1 \over 2\sqrt{2y+b}}\cdot{db \over da}=0$$

$$\sqrt{2y+b}=-{db \over da}\sqrt{2x+a}$$

Let

$$b=-{a \over \lambda}+c$$

Then

$$\sqrt{2y+b}={1 \over \lambda}\sqrt{2x+a}$$

$$2y+b={2 \over \lambda^2}x+{1 \over \lambda^2}a$$

$$z=(1+{1 \over \lambda})\sqrt{2x+a}$$

$$z^2={(1+\lambda)^2 \over \lambda^2}(2x+a)$$

If

$$z^2+\mu=2(1+\lambda^{-1})(x+\lambda y)$$

Then

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

$${1+\lambda \over \lambda}(2x+a)=2x+\lambda(2y)-{\lambda \over 1+\lambda}\mu$$

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

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

$$b=-{a \over \lambda}-{\mu \over1+\lambda}$$

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

$$b=-{a \over \lambda}-{\mu \over1+\lambda}$$