# Partial Differential Equation

 Question 1
$A 5 \mathrm{~cm}$ long metal rod $A B$ with initially at a uniform temperature of $T_{0}{ }^{\circ} \mathrm{C}$. Thereafter, temperature at both the ends are maintained at $0^{\circ} \mathrm{C}$. Neglecting the heat transfer from the lateral surface of the rod, the heat transfer in the rod is governed by the one-dimensional diffusion equation $\frac{\partial T}{\partial t}=D \frac{\partial^{2} T}{\partial x^{2}}$, where $D$ is the thermal diffusively of the metal, given as $1.0 \mathrm{~cm}^{2} / \mathrm{s}$.
The temperature distribution in the rod is obtained as
$T(x, t)=\sum_{n=13,5 \ldots .}^{\infty} C_{n} \sin \frac{n \pi x}{5} e^{-\beta n^{2} t}$ where $x$ is in $\mathrm{cm}$ measured from $A$ to $B$ with $x=0$ at $A, t$ is $s, C_{n}$ are constants in ${ }^{\circ} \mathrm{C}, \mathrm{T}$ is in ${ }^{\circ} \mathrm{C}$ and $\beta$ is in $\mathrm{s}^{-1}$.

The value of $\beta$ (in $\mathrm{s}^{-1}$, rounded off to three decimal places) is ___
 A 0.395 B 0.125 C 0.254 D 0.685
GATE CE 2023 SET-2   Engineering Mathematics
Question 1 Explanation:
Given: Diffusion equation:
$\frac{\partial T}{\partial t}=\mathrm{D} \frac{\partial^{2} \mathrm{~T}}{\partial \mathrm{x}^{2}}$
And also,
$D=1$
$\mathrm{T}(0, \mathrm{t})=0$
$\mathrm{T}(5,5)=0$
$T(x, 0)=T_{0}$
We know, solution of equation is given by,
$T(x, t)=\left(C_{1} \cos \alpha x+C_{2} \sin \alpha x\right) C_{3} e^{-\alpha^{2} t} \;\;\; ...(1)$

For $\mathrm{T}(0, t)=0$
$\left(C_{1}+0\right) C_{3} e^{-\alpha^{2} t}=0$
$\Rightarrow \mathrm{C}_{1}=0$

For $\mathrm{T}(\mathrm{s}, \mathrm{t})=0$
$\left(C_{2} \sin \alpha 5\right) C_{3} \mathrm{e}^{-25 t}=0$
$\Rightarrow \sin 5 \alpha=0$
$\Rightarrow \alpha=\frac{\mathrm{n} \pi}{5}$

Now, from equation (1)
$T(x, t)=C_{2} C_{3} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2}}{25} t}$
$=b_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2} t}{25}}$

Generalize solution
$T(x, t)=\sum_{n=1}^{\infty} b_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\frac{n^{2} \pi^{2} t}{25}}$
For $\mathrm{T}(\mathrm{x}, 0)=\mathrm{T}_{0}$
$\Rightarrow T_{0}=\sum_{n=1}^{\infty} b_{n} \sin \left(\frac{n \pi x}{5}\right)$

From fourier series,
$b_{n}=0$ for even value of $n$
$\therefore \mathrm{T}(\mathrm{x}, \mathrm{t})=\sum_{\mathrm{n}=13,5 \ldots .}^{\infty} \mathrm{b}_{\mathrm{n}} \sin \left(\frac{\mathrm{n} \pi \mathrm{x}}{5}\right) \mathrm{e}^{-\frac{\mathrm{n}^{2} \pi^{2} \mathrm{t}}{25}}$

Given: $T(x, t)=\sum_{n=13,5 \ldots .}^{\infty} C_{n} \sin \left(\frac{n \pi x}{5}\right) e^{-\beta n^{2} t}$
On comparison;
$\beta=\frac{\pi^{2}}{25}=0.395 \mathrm{sec}^{-1}$
 Question 2
The steady-state temperature distribution in a square plate $A B C D$ is governed by the 2-dimensional Laplace equation. The side $A B$ is kept at a temperature of $100{ }^{\circ} \mathrm{C}$ and the other three sides are kept at a temperature of $0{ }^{\circ} \mathrm{C}$. Ignoring the effect of discontinuities in the boundary conditions at the corners, the steady-state temperature at the center of the plate is obtained as $T_0 ^{\circ} \mathrm{C}$. Due to symmetry, the steady-state temperature at the center will be same $\left(\mathrm{T_0}{ }^{\circ} \mathrm{C}\right)$, when any one side of the square is kept at a temperature of $100{ }^{\circ} \mathrm{C}$ and the remaining three sides are kept at a temperature of $0{ }^{\circ} \mathrm{C}$. Using the principle of superposition, the value of $T_0$ is ___ (rounded off to two decimal places).
 A 25.35 B 14.25 C 19.92 D 28.25
GATE CE 2023 SET-2   Engineering Mathematics
Question 2 Explanation:
We know, Laplace equation in two dimensional on a unit square with Dirichlet boundary condition :
$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0 \lt (x, y) \lt 1$
This Laplace equation is called harmonic function. Solution of Laplace equation,
$u(x, y)=\left(C_{1} \cos \beta x+C_{2} \sin \beta x\right)\left(C_{3} e^{\text {By }}+C_{4} e^{-\beta y}\right)$
Let square $A B C D$ is unit square.
Now, given
\begin{aligned} & u(x, 0)=0 \\ & u(0, y)=0 \\ & u(1, y)=0 \\ & u(x, 1)=100 \end{aligned}

For $u(x, 0)=0$
$\quad 0=\left(C_{1} \cos \beta x+0\right)(C_3+C_4)$
$\Rightarrow \quad \mathrm{C}_{3}=-\mathrm{C}_{4}$

For $u(0, y)=0$;
$0=\left(C_{1}+0\right)\left(C_{3} e^{\beta y}+C_{4} e^{-\beta y}\right)$
$\Rightarrow \quad \mathrm{C}_{1}=0$

Now,
\begin{aligned} u(x, y) & =0+C_{2} \sin \beta x\left(C_{3} e^{\beta y}+C_{4} e^{-\beta y}\right) \\ & =a_{x} \sin \beta x\left(e^{\beta y}-e^{-\beta y}\right) \quad ...(1) \end{aligned}
$\left[\because \mathrm{C}_{3}=-\mathrm{C}_{4}\right.$ and Let $\left.\mathrm{C}_{2} \mathrm{C}_{3}=\mathrm{a}_{\mathrm{x}}\right]$

For $u(1, y)=0$
$0=a_{x} \sin \beta x\left(e^{\beta y}-e^{\beta y}\right)$
$\Rightarrow \quad \beta=\mathrm{n} \pi$

For $u(x, 1)=100$
$a_{x} \sin n\pi x\left(e^{n\pi}-e^{-n \pi}\right)=100$
$a_{x}=\frac{100}{\sin n\pi x\left(e^{n\pi}-e^{-n \pi}\right)}$

From eqn. (1), we get
$\begin{gathered} u(x, y)=\sum_{n=1}^{\infty }\frac{100}{\sin n \pi x\left(e^{n\pi}-e^{-n\pi}\right)} \times \sin n\pi x\left(e^{n \pi y}-e^{- n\pi y}\right) \\ =\sum_{n=1}^{\infty }\frac{100}{\left(e^{n\pi}-e^{-n \pi}\right)}\left[e^{n\pi y}-e^{-n\pi y}\right] \end{gathered}$
Now, at mid point $(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right)$
$u\left(\frac{1}{2}, \frac{1}{2}\right)=T_{0} \text { [given] }$
\begin{aligned} T_0&=\sum_{n=1}^{\infty }\frac{100}{\left(\mathrm{e}^{n\pi}-\mathrm{e}^{-n\pi}\right)}\left[\mathrm{e}^{n\pi / 2}-\mathrm{e}^{-n\pi / 2}\right] \\ &=100\left [ \left ( \frac{\mathrm{e}^{\pi / 2}-\mathrm{e}^{-\pi / 2}}{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}} \right ) +\left ( \frac{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}}{\mathrm{e}^{2\pi}-\mathrm{e}^{-2\pi}} \right )+...\right ]\\ &=100\left [ \frac{1}{(\mathrm{e}^{\pi / 2}-\mathrm{e}^{-\pi / 2})}+\frac{1}{\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}}+\frac{1}{\mathrm{e}^{3\pi / 2}-\mathrm{e}^{-3\pi / 2}} +...(neglect) \right ]\\ &=100[0.199+0.043+0.008]\\ &\approx 25^{\circ}C \end{aligned}

 Question 3
The function $f(x, y)$ satisfies the Laplace equation
$\triangledown ^2f(x,y)=0$
on a circular domain of radius $r = 1$ with its center at point P with coordinates $x = 0, y = 0$. The value of this function on the circular boundary of this domain is equal to 3.
The numerical value of $f(0, 0)$ is:
 A 0 B 2 C 3 D 1
GATE CE 2022 SET-2   Engineering Mathematics
Question 3 Explanation:
According to given condition given function f(x,y) is nothing but constant function i.e. f(x,y)=3 because this is the only function whose value is 3 at any point on the boundary of unit circle and it is also satisfying Laplace equation, so
f(0,0)=3
 Question 4
The Fourier cosine series of a function is given by:
$f(x)=\sum_{n=0}^{\infty }f_n\cos nx$
For $f(x)=\cos ^4x$, the numerical value of $f_4+f_5$ is ______ . (round off to three decimal places)
 A 2.255 B 0.652 C 0.125 D 1.585
GATE CE 2022 SET-1   Engineering Mathematics
Question 4 Explanation:
\begin{aligned} \cos ^4 x&=\cos ^2 x\cos ^2 x\\ &=\left ( \frac{1+\cos 2x}{2} \right )\left ( \frac{1+\cos 2x}{2} \right )\\ &=\frac{1}{4}(1+2 \cos 2x+ \cos ^2 2x)\\ &=\frac{1}{4}(1+2 \cos 2x+ \frac{(1+\cos 4x)}{2})\\ &=\frac{1}{4}+\frac{\cos 2x}{2}+\frac{1}{8}+\frac{\cos 4x}{8}\\ &=\frac{3}{8}+\frac{\cos 2x}{2}+\frac{\cos 4x}{8}\\ f_4&=\frac{1}{8}=0.125\\ f_5&=0\\ \rightarrow f_4+f_5&=0.125 \end{aligned}
 Question 5
Consider the following expression:
$z=\sin(y+it)+\cos(y-it)$
where $z, y,$ and $t$ are variables, and $i=\sqrt{-1}$ is a complex number. The partial differential equation derived from the above expression is
 A $\frac{\partial^2 z}{\partial t^2}+\frac{\partial^2 z}{\partial y^2}=0$ B $\frac{\partial^2 z}{\partial t^2}-\frac{\partial^2 z}{\partial y^2}=0$ C $\frac{\partial z}{\partial t}-i\frac{\partial z}{\partial y}=0$ D $\frac{\partial z}{\partial t}+i\frac{\partial z}{\partial y}=0$
GATE CE 2022 SET-1   Engineering Mathematics
Question 5 Explanation:
\begin{aligned} z&=\sin(y+it)+ \cos (y-it)\\ \frac{\partial z}{\partial y}&=\cos (y+it)-\sin (y-it)\\ \frac{\partial ^2 z}{\partial ^2 y^2}&=-\sin(y+it)- \cos (y-it)\\ \frac{\partial ^2 z}{\partial ^2 y^2}&=-z \;\;...(i)\\ \frac{\partial z}{\partial t}&=i \cos (y+it)+i\sin (y-it)\\ \frac{\partial ^2 z}{\partial ^2 t^2}&=+\sin(y+it)+ \cos (y-it)\\ \frac{\partial ^2 z}{\partial ^2 t^2}&=z\;\;...(ii)\\ &\text{Adding (i) and (ii)}\\ &\frac{\partial ^2 z}{\partial ^2 y^2}+\frac{\partial ^2 z}{\partial ^2 t^2}=0 \end{aligned}

There are 5 questions to complete.