Question

a) A thin metal plate bounded by the 𝑥-axis and the lines 𝑥 = 0 and 𝑥 = 1 and

stretching to infinity in the 𝑦-direction has its upper and lower faces perfectly

insulated and its vertical edges and edge at infinity are maintained at the constant

temperature 0 °𝐶, while over the base temperature of 50 °𝐶 is maintained. Find

the steady state temperature 𝑢(𝑥,𝑡).

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b) If 𝑓(𝑥𝑦^2, 𝑧 − 2𝑥) = 0 then prove that 𝑥 𝑧x −1/2𝑦 𝑧𝑦 = 2 𝑥

Answer 1

b)

prove:

$x\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\frac{y}{2}=2x$

Differentiating $𝑓(𝑥𝑦^2, 𝑧 − 2𝑥) = 0$ , we get:

$y^2f_1+(\frac{\partial z}{\partial x}-2)f_2=0$

$2xyf_1+\frac{\partial z}{\partial y}f_2=0$

where f₁ and f₂ are partial derivatives of f with respect to its first input and second input. When we multiply the first equality above by 2x and the second by y:

$2x(\frac{\partial z}{\partial x}-2)=y\frac{\partial z}{\partial y}\implies x\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\frac{y}{2}=2x$

a)

solution of two dimensional heat equation $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$ is

$u(x,y)=(Acos\lambda x+Bsin \lambda x)(Ce^{\lambda y}+De^{-\lambda y})$

boundary conditions:

$u(x,\infin)=0$

$u(x,0)=50$

$u(0,y)=0$

$u(1,y)=0$

$0\le x\le 1,0\le y< \infin$

then:

$u(0,y)=A(Ce^{\lambda y}+De^{-\lambda y})=0\implies A=0$

$u(x,y)=Bsin \lambda x(Ce^{\lambda y}+De^{-\lambda y})$

$u(1,y)=Bsin \lambda (Ce^{\lambda y}+De^{-\lambda y})=0\implies \lambda=n\pi$

then:

$u(x,y)=Bsin (\pi nx)(Ce^{\pi n y}+De^{-\pi n y})$

$u(x,\infin)=Bsin (\pi nx)Ce^{\pi n y}=0\implies C=0$

$u(x,y)=B_1sin (\pi nx)e^{-\pi n y}$

where B₁ = BD

the most general solution:

$u(x,y)=\sum B_nsin (\pi nx)e^{-\pi n y}$

where

$B_n=\int^1_0 u(x,0)sin (\pi nx)=50\int^1_0 sin (\pi nx)=$

$=-\frac{50}{\pi n}cos(n\pi x)|^1_0=\frac{50}{\pi n}(1-cos(n\pi))$

for odd n:

$u(x,y)=\sum\frac{100}{\pi n}sin (\pi nx)e^{-\pi n y}$

for even n:

$u(x,y)=0$