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Exercise 8.3

Show that [tex] u(x, y) = x \sqrt{xy} [/tex] satisfies the equation
[tex]\[ x u_x - y u_y = u \][/tex]
subject to
[tex]\[ u(y, y) = y^2, \quad y \geq 0 \][/tex]


Sagot :

To demonstrate that the function [tex]\( u(x, y) = x \sqrt{x y} \)[/tex] satisfies the equation [tex]\( x u_x - y u_y = u \)[/tex], we need to perform the following steps. Let's do this step-by-step:

1. Find the partial derivatives [tex]\( u_x \)[/tex] and [tex]\( u_y \)[/tex]:
- Partial derivative with respect to [tex]\( x \)[/tex]:
[tex]\[ u(x, y) = x \sqrt{x y} \][/tex]
To find [tex]\( u_x \)[/tex], use the product rule and the chain rule:
[tex]\[ u_x = \frac{\partial}{\partial x} (x \cdot \sqrt{x y}) \][/tex]

First, recognize that [tex]\( \sqrt{x y} = (x y)^{1/2} \)[/tex].

Then,
[tex]\[ u(x, y) = x \cdot (x y)^{1/2} \][/tex]

Now, apply the product rule:
[tex]\[ u_x = \frac{\partial x}{\partial x} \cdot (x y)^{1/2} + x \cdot \frac{\partial (x y)^{1/2}}{\partial x} \][/tex]

We know [tex]\( \frac{\partial x}{\partial x} = 1 \)[/tex]:
[tex]\[ u_x = (x y)^{1/2} + x \cdot \frac{\partial (x y)^{1/2}}{\partial x} \][/tex]

Using the chain rule for [tex]\( (x y)^{1/2} \)[/tex]:
[tex]\[ \frac{\partial (x y)^{1/2}}{\partial x} = \frac{1}{2} (x y)^{-1/2} \cdot y = \frac{y}{2 \sqrt{x y}} \][/tex]

Thus,
[tex]\[ u_x = \sqrt{x y} + x \cdot \frac{y}{2 \sqrt{x y}} = \sqrt{x y} + \frac{x y}{2 \sqrt{x y}} = \sqrt{x y} + \frac{1}{2} \sqrt{x y} = \frac{3}{2} \sqrt{x y} \][/tex]

2. Partial derivative with respect to [tex]\( y \)[/tex]:
[tex]\[ u(x, y) = x \sqrt{x y} \][/tex]

To find [tex]\( u_y \)[/tex]:
[tex]\[ u_y = \frac{\partial}{\partial y} (x \cdot \sqrt{x y}) \][/tex]

Recognize [tex]\( \sqrt{x y} = (x y)^{1/2} \)[/tex]:
[tex]\[ u_y = x \cdot \frac{\partial (x y)^{1/2}}{\partial y} \][/tex]

Using the chain rule for [tex]\( (x y)^{1/2} \)[/tex]:
[tex]\[ \frac{\partial (x y)^{1/2}}{\partial y} = \frac{1}{2} (x y)^{-1/2} \cdot x = \frac{x}{2 \sqrt{x y}} \][/tex]

Thus,
[tex]\[ u_y = x \cdot \frac{x}{2 \sqrt{x y}} = \frac{x^2}{2 \sqrt{x y}} \][/tex]

Simplify further:
[tex]\[ u_y = \frac{x^2}{2 \sqrt{x y}} \][/tex]

3. Verify [tex]\( x u_x - y u_y = u \)[/tex]:
- Substitute [tex]\( u_x \)[/tex] and [tex]\( u_y \)[/tex] into the equation:
[tex]\[ x u_x - y u_y = x \left( \frac{3}{2} \sqrt{x y} \right) - y \left( \frac{x^2}{2 \sqrt{x y}} \right) \][/tex]

Simplify each term:
[tex]\[ x u_x = x \cdot \frac{3}{2} \sqrt{x y} = \frac{3}{2} x \sqrt{x y} \][/tex]
[tex]\[ y u_y = y \cdot \frac{x^2}{2 \sqrt{x y}} = \frac{1}{2} y \cdot \frac{x^2}{\sqrt{x y}} = \frac{1}{2} \frac{yx^2}{\sqrt{x y}} = \frac{1}{2} x^2 \sqrt{\frac{y}{x}} y = \frac{1}{2} x \sqrt{x y} \][/tex]

Therefore,
[tex]\[ x u_x - y u_y = \frac{3}{2} x \sqrt{x y} - \frac{1}{2} x \sqrt{x y} = \sqrt{x y} \][/tex]

Recognize that
[tex]\[ u = x \sqrt{x y} \][/tex]

Hence,
[tex]\( x u_x - y u_y = u \)[/tex]

Thus, we have successfully shown that [tex]\( x u_x - y u_y = u \)[/tex].

Lastly, let's check the condition [tex]\( u(y, y) = y^2 \)[/tex]:
[tex]\[ u(y, y) = y \sqrt{y \cdot y} = y \sqrt{y^2} = y \cdot y = y^2 \][/tex]

Therefore, the function [tex]\( u(x, y) = x \sqrt{x y} \)[/tex] satisfies both the differential equation [tex]\( x u_x - y u_y = u \)[/tex] and the condition [tex]\( u(y, y) = y^2 \)[/tex] for [tex]\( y \geq 0 \)[/tex].