Complex Variables and Partial Differential Equations (3130005)

Question-5b-OR

BE | Semester-3 Summer-2020 | 27-10-2020

Q5) (b)

Let,

u (x, t) = X (x) \cdot T (t) = X \cdot T

\frac{\partial u}{\partial x} = X^{'} \cdot T & \frac{\partial u}{\partial t} = X \cdot T^{'}

Now,

2 \frac{\partial u}{\partial x} = \frac{\partial u}{\partial t} + u

\Rightarrow 2 X^{'} \cdot T = X \cdot T^{'} + X \cdot T

Dividing by X∙Y both the sides,

\Rightarrow 2 \frac{X^{'} \cdot T}{X \cdot T} = \frac{X \cdot T^{'}}{X \cdot T} + \frac{X \cdot T}{X \cdot T}

\Rightarrow 2 \frac{X^{'}}{X} = \frac{T^{'}}{T} + 1

\Rightarrow 2 \frac{X^{'}}{X} = \frac{T^{'}}{T} + 1 = k

∎ 2 \frac{X^{'}}{X} = k \Rightarrow \frac{1}{X} dX = \frac{k}{2} dx \Rightarrow \log X = \frac{kx}{2} + c \Rightarrow X = c_{1} e^{\frac{kx}{2}}

∎ \frac{T^{'}}{T} + 1 = k \Rightarrow \frac{1}{T} dT = (k - 1) dt \Rightarrow \log T = (k - 1) t + c \Rightarrow T = c_{2} e^{(k - 1) t}

Solution:

u (x, t) = X \cdot T = (c_{1} e^{\frac{kx}{2}}) (c_{2} e^{(k - 1) t}) - - - - - [1]

Now,

u (x, 0) = 4 e^{- 3 x}

[1]

, We have

u (x, 0) = (c_{1} e^{\frac{kx}{2}}) (c_{2} e^{(k - 1) 0})

\Rightarrow 4 e^{- 3 x} = c_{1} c_{2} e^{\frac{kx}{2}}

\Rightarrow c_{1} c_{2} = 4 & \frac{k}{2} = - 3 \Rightarrow c_{1} c_{2} = 4 & k = - 6

Solution:

u (x, t) = 4 e^{- 3 x} e^{- 7 t} = 4 e^{- 3 x - 7 t}