Explained: "The Laplace equation ∂2∂x2f+∂2∂y2f=0"

diferira7c

Answered question

2021-12-10

The Laplace equation

\frac{\partial^{2}}{\partial x^{2}} f + \frac{\partial^{2}}{\partial y^{2}} f = 0

Chanell Sanborn

Beginner2021-12-11Added 41 answers

Step 1
The steady sate temparature distribution is a one diamensional slab of thermal conductivity

50 \frac{w}{m} \times k

and thickness 50 mm is found to be

T = a + b x^{2}

Where

a = 2000 c, b = 20000 \frac{c}{m^{2}}

T is in degrees celsius and x in meters.
Given:

f (x, y) = \ln {x^{2} + y^{2}}

Now,

\frac{\partial f}{\partial x} = \frac{1}{\sqrt{x^{2} + y^{2}}} \times \frac{2 x}{2 \sqrt{x^{2} + y^{2}}} = \frac{x}{x^{2} + y^{2}}

\frac{\partial^{2} f}{\partial x^{2}} = \frac{x^{2} + y^{2} - 2 x^{2}}{{(x^{2} + y^{2})}^{2}} = \frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}}

\frac{\partial f}{\partial y} = \frac{1}{\sqrt{x^{2} + y^{2}}} \times \frac{2 y}{2 \sqrt{x^{2} + y^{2}}} = \frac{y}{{(\sqrt{x^{2} + y^{2}})}^{2}} = \frac{y}{x^{2} + y^{2}}

\frac{\partial^{2} f}{\partial y^{2}} = \frac{x^{2} + y^{2} - 2 y^{2}}{{(x^{2} + y^{2})}^{2}} = \frac{x^{2} - y^{2}}{{(x^{2} + y^{2})}^{2}}

t h e r a f or e \frac{p a r t i l^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} = \frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}} + {\frac{x^{2} - y^{2}}{x^{2} + y^{2}}}^{2}}

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Toni Scott

Beginner2021-12-12Added 32 answers

It is equal to zero as

\frac{\partial^{2} f}{{d x}^{2}} + \frac{\partial^{2} f}{{d y}^{2}} = (4 x^{2} + 4 y^{2}) \frac{\partial^{2} f}{d u^{2}} + (4 x^{2} + 4 y^{2}) \frac{\partial^{2} f}{d v^{2}}

= (4 x^{2} + 4 y^{2}) (\frac{\partial^{2} f}{d u^{2}} + \frac{d^{2} f}{d v^{2}})

= 0

as f(u,v) is harmonic.

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