MPH-001 Solved Assignment 2024 | IGNOU

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MPH-001 Solved Assignment 2024 | IGNOU Sample Solution
Expert Answer
mph-001-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

mph-001-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

PART A
  1. a) Using separation of variables, solve:
∂ u ∂ x = 2 ∂ u ∂ t + u ∂ u ∂ x = 2 ∂ u ∂ t + u (del u)/(del x)=2(del u)/(del t)+u\frac{\partial u}{\partial x}=2 \frac{\partial u}{\partial t}+u∂u∂x=2∂u∂t+u
where u ( x , 0 ) = 3 e − 3 x u ( x , 0 ) = 3 e − 3 x u(x,0)=3e^(-3x)u(x, 0)=3 e^{-3 x}u(x,0)=3e−3x.
Answer:
The PDE given is:
∂ u ∂ x = 2 ∂ u ∂ t + u ∂ u ∂ x = 2 ∂ u ∂ t + u (del u)/(del x)=2(del u)/(del t)+u\frac{\partial u}{\partial x} = 2 \frac{\partial u}{\partial t} + u∂u∂x=2∂u∂t+u
with the initial condition:
u ( x , 0 ) = 3 e − 3 x u ( x , 0 ) = 3 e − 3 x u(x,0)=3e^(-3x)u(x, 0) = 3 e^{-3 x}u(x,0)=3e−3x
To solve this PDE using separation of variables, let’s assume a solution of the form u ( x , t ) = X ( x ) T ( t ) u ( x , t ) = X ( x ) T ( t ) u(x,t)=X(x)T(t)u(x, t) = X(x)T(t)u(x,t)=X(x)T(t), where X ( x ) X ( x ) X(x)X(x)X(x) is a function of x x xxx only, and T ( t ) T ( t ) T(t)T(t)T(t) is a function of t t ttt only. Substituting this form into the PDE gives:
X ′ ( x ) T ( t ) = 2 X ( x ) T ′ ( t ) + X ( x ) T ( t ) X ′ ( x ) T ( t ) = 2 X ( x ) T ′ ( t ) + X ( x ) T ( t ) X^(‘)(x)T(t)=2X(x)T^(‘)(t)+X(x)T(t)X'(x)T(t) = 2X(x)T'(t) + X(x)T(t)X′(x)T(t)=2X(x)T′(t)+X(x)T(t)
Rearranging, we get:
X ′ ( x ) X ( x ) = 2 T ′ ( t ) T ( t ) + 1 X ′ ( x ) X ( x ) = 2 T ′ ( t ) T ( t ) + 1 (X^(‘)(x))/(X(x))=2(T^(‘)(t))/(T(t))+1\frac{X'(x)}{X(x)} = 2\frac{T'(t)}{T(t)} + 1X′(x)X(x)=2T′(t)T(t)+1
Since the left side depends only on x x xxx and the right side depends only on t t ttt, each side must be equal to a constant, which we shall call λ λ lambda\lambdaλ. Thus, we have two ordinary differential equations (ODEs):
X ′ ( x ) X ( x ) = λ and 2 T ′ ( t ) T ( t ) + 1 = λ X ′ ( x ) X ( x ) = λ and 2 T ′ ( t ) T ( t ) + 1 = λ (X^(‘)(x))/(X(x))=lambdaquad”and”quad2(T^(‘)(t))/(T(t))+1=lambda\frac{X'(x)}{X(x)} = \lambda \quad \text{and} \quad 2\frac{T'(t)}{T(t)} + 1 = \lambdaX′(x)X(x)=λand2T′(t)T(t)+1=λ
Let’s solve these ODEs separately.
For X ( x ) X ( x ) X(x)X(x)X(x), we have:
X ′ ( x ) = λ X ( x ) X ′ ( x ) = λ X ( x ) X^(‘)(x)=lambda X(x)X'(x) = \lambda X(x)X′(x)=λX(x)
And for T ( t ) T ( t ) T(t)T(t)T(t), rearranging the equation gives:
2 T ′ ( t ) = ( λ − 1 ) T ( t ) 2 T ′ ( t ) = ( λ − 1 ) T ( t ) 2T^(‘)(t)=(lambda-1)T(t)2T'(t) = (\lambda – 1)T(t)2T′(t)=(λ−1)T(t)
These lead to exponential solutions for both X ( x ) X ( x ) X(x)X(x)X(x) and T ( t ) T ( t ) T(t)T(t)T(t). Let’s proceed to find the explicit solutions for X ( x ) X ( x ) X(x)X(x)X(x) and T ( t ) T ( t ) T(t)T(t)T(t), and apply the initial condition to determine any specific constants.
The solutions to the ordinary differential equations are:
X ( x ) = C 1 e λ x X ( x ) = C 1 e λ x X(x)=C_(1)e^(lambda x)X(x) = C_1 e^{\lambda x}X(x)=C1eλx
T ( t ) = C 2 e ( λ − 1 ) t 2 T ( t ) = C 2 e ( λ − 1 ) t 2 T(t)=C_(2)e^(((lambda-1)t)/(2))T(t) = C_2 e^{\frac{(\lambda – 1)t}{2}}T(t)=C2e(λ−1)t2
Thus, our solution u ( x , t ) u ( x , t ) u(x,t)u(x, t)u(x,t) can be written as a product of X ( x ) X ( x ) X(x)X(x)X(x) and T ( t ) T ( t ) T(t)T(t)T(t):
u ( x , t ) = X ( x ) T ( t ) = C 1 C 2 e λ x e ( λ − 1 ) t 2 u ( x , t ) = X ( x ) T ( t ) = C 1 C 2 e λ x e ( λ − 1 ) t 2 u(x,t)=X(x)T(t)=C_(1)C_(2)e^(lambda x)e^(((lambda-1)t)/(2))u(x, t) = X(x)T(t) = C_1 C_2 e^{\lambda x} e^{\frac{(\lambda – 1)t}{2}}u(x,t)=X(x)T(t)=C1C2eλxe(λ−1)t2
Let’s simplify this expression:
u ( x , t ) = C e λ x + ( λ − 1 ) t 2 u ( x , t ) = C e λ x + ( λ − 1 ) t 2 u(x,t)=Ce^(lambda x+((lambda-1)t)/(2))u(x, t) = C e^{\lambda x + \frac{(\lambda – 1)t}{2}}u(x,t)=Ceλx+(λ−1)t2
where C = C 1 C 2 C = C 1 C 2 C=C_(1)C_(2)C = C_1 C_2C=C1C2 is a constant.
Now, to determine the specific value of λ λ lambda\lambdaλ and the constant C C CCC, we apply the initial condition u ( x , 0 ) = 3 e − 3 x u ( x , 0 ) = 3 e − 3 x u(x,0)=3e^(-3x)u(x, 0) = 3 e^{-3 x}u(x,0)=3e−3x. Let’s substitute t = 0 t = 0 t=0t = 0t=0 into our solution to match the initial condition.
Upon applying the initial condition and aligning our universe of equations, we’ve discovered that λ = − 3 λ = − 3 lambda=-3\lambda = -3λ=−3 and C = 3 C = 3 C=3C = 3C=3. Therefore, the particular solution to our initial value problem is:
u ( x , t ) = 3 e − 3 x + ( − 3 − 1 ) t 2 = 3 e − 3 x − 2 t u ( x , t ) = 3 e − 3 x + ( − 3 − 1 ) t 2 = 3 e − 3 x − 2 t u(x,t)=3e^(-3x+((-3-1)t)/(2))=3e^(-3x-2t)u(x, t) = 3 e^{-3x + \frac{(-3 – 1)t}{2}} = 3 e^{-3x – 2t}u(x,t)=3e−3x+(−3−1)t2=3e−3x−2t
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