Aerospace Engineering (AE)Page 9 of 35Organizing Institute: IIT KanpurQ.13 Consider the one-dimensional wave equation0u ut x + = forx− ,0t .For an initial condition2( ,0) xu x e−= , the solution at1t = is(A)2( 1)( ,1) xu x e− −=(B)1( ,1)u x e−=(C)2( ,1) xu x e−=(D)2( 1)( ,1) xu x e− +=Q.14 A two-dimensional potential flow solution for flow past an airfoil has a streamlinepattern as shown in the figure. Which of the following conditions is additionallyrequired to satisfy the Kutta condition?(A) Addition of a source of strength0Q (B) Addition of a source of strength0Q (C) Addition of a circulation of strength0 (counter-clockwise)(D) Addition of a circulation of strength0 (clockwise
Question
Aerospace Engineering (AE)Page 9 of 35Organizing Institute: IIT KanpurQ.13 Consider the one-dimensional wave equation0u ut x + = forx− ,0t .For an initial condition2( ,0) xu x e−= , the solution at1t = is(A)2( 1)( ,1) xu x e− −=(B)1( ,1)u x e−=(C)2( ,1) xu x e−=(D)2( 1)( ,1) xu x e− +=Q.14 A two-dimensional potential flow solution for flow past an airfoil has a streamlinepattern as shown in the figure. Which of the following conditions is additionallyrequired to satisfy the Kutta condition?(A) Addition of a source of strength0Q (B) Addition of a source of strength0Q (C) Addition of a circulation of strength0 (counter-clockwise)(D) Addition of a circulation of strength0 (clockwise
Solution
To solve question 13, we are given the one-dimensional wave equation:
∂²u/∂t² = c² ∂²u/∂x²
where c is a constant. We are also given the initial condition:
u(x,0) = 2x e^(-x²)
To find the solution at t = 1, we need to solve the wave equation with the given initial condition.
The general solution to the one-dimensional wave equation is given by:
u(x,t) = f(x - ct) + g(x + ct)
where f and g are arbitrary functions.
Substituting the initial condition into the general solution, we have:
2x e^(-x²) = f(x) + g(x)
To find f(x) and g(x), we can differentiate both sides of the equation with respect to x.
Differentiating the left side with respect to x gives:
2e^(-x²) - 4x²e^(-x²)
Differentiating the right side with respect to x gives:
f'(x) + g'(x)
Setting these two expressions equal to each other, we have:
2e^(-x²) - 4x²e^(-x²) = f'(x) + g'(x)
To simplify the equation, we can assume that f'(x) = g'(x) = 0, which means f(x) and g(x) are constants.
Integrating both sides of the equation, we have:
∫(2e^(-x²) - 4x²e^(-x²)) dx = ∫(f'(x) + g'(x)) dx
Simplifying the integrals, we get:
2∫e^(-x²) dx - 4∫x²e^(-x²) dx = f(x) + g(x)
The integral of e^(-x²) is not expressible in terms of elementary functions, but it can be approximated using numerical methods. Similarly, the integral of x²e^(-x²) is not expressible in terms of elementary functions.
Therefore, the solution at t = 1 cannot be determined analytically. The correct answer is (E) The solution cannot be determined with the given information.
Moving on to question 14, we are given a two-dimensional potential flow solution for flow past an airfoil. The streamline pattern is shown in the figure.
To satisfy the Kutta condition, we need to add a circulation of strength Γ.
The correct answer is (C) Addition of a circulation of strength Γ > 0 (counter-clockwise).
Similar Questions
We have seen that the one-dimensional wave equation has lots of possible solutions. In fact there is a whole vector subspace of solutions (you may like to prove this). We can narrow down the range of possible solutions by adding extra conditions that the function must satisfy. These are usually called initial conditions. For example, suppose that u(x,t)𝑢(𝑥,𝑡) satisfies the one-dimensional wave equation:∂2u∂t2−c2∂2u∂x2=0∂2𝑢∂𝑡2−𝑐2∂2𝑢∂𝑥2=0 .In addition, suppose u(x,t)𝑢(𝑥,𝑡) satisfies the initial conditions:u(x,0)=f(x)𝑢(𝑥,0)=𝑓(𝑥) and ∂u∂t(x,0)=g(x)∂𝑢∂𝑡(𝑥,0)=𝑔(𝑥) .In 1746, Jean-Baptiste le Rond d'Alembert discovered that there is only one possible solution which satisfies these conditions:u(x,t)=12(f(x−ct)+f(x+ct))+12c∫x−ctx+ctg(s)ds.𝑢(𝑥,𝑡)=12(𝑓(𝑥−𝑐𝑡)+𝑓(𝑥+𝑐𝑡))+12𝑐∫𝑥−𝑐𝑡𝑥+𝑐𝑡𝑔(𝑠)𝑑𝑠. Jean-Baptise le Rond d'Alembert (1717 -1783)For example, if we have the wave equation with c=1𝑐=1 :∂2u∂t2−∂2u∂x2=0∂2𝑢∂𝑡2−∂2𝑢∂𝑥2=0 and initial conditions f(x)=sin(x)𝑓(𝑥)=sin(𝑥) and g(x)=cos(x)𝑔(𝑥)=cos(𝑥) then∫x−ctx+ctg(s)ds=∫x−tx+tcos(s)ds=∫𝑥−𝑐𝑡𝑥+𝑐𝑡𝑔(𝑠)𝑑𝑠=∫𝑥−𝑡𝑥+𝑡cos(𝑠)𝑑𝑠= .So by d'Alembert's formula, the solution isu(x,t)=𝑢(𝑥,𝑡)= .
The function u(x,t)=ei(x+ct)𝑢(𝑥,𝑡)=𝑒𝑖(𝑥+𝑐𝑡) is another solution to the wave equation∂2u∂t2−c2∂2u∂x2=0∂2𝑢∂𝑡2−𝑐2∂2𝑢∂𝑥2=0 Let's investiage the case when c=2+5i𝑐=2+5𝑖 . ThenThe real part of u𝑢 is The imaginary part of u𝑢 is .So the function oscillates in the complex plane, but these oscillations are dampened by the exponential term. As time goes onlimt→∞u(x,t)=lim𝑡→∞𝑢(𝑥,𝑡)= .
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