Question

The temperature distribution θ(x,t) along an insulated metal rod of length L is described by the differential equation ∂x 2∂ 2θ= 1/D ∂t/∂θ(00) where D≠0 is a constant. The rod is held at a fixed temperature of 0 ∘C at one end and is insulated at the other end, which gives rise to the boundary conditions ∂x/∂θ=0 when x=0 for t>0 together with θ=0 when x=L for t>0. The initial temperature distribution in the rodis given by θ(x,0)=0.3cos(7Lπx)(0≤x≤L).

(a) Use the method of separation of variables, with θ(x,t)=X(x)T(t), to show that the function X(x) satisfies the differential equation X ′′−μX=0 for some constant μ. Write down the corresponding differential equation that T(t) must satisfy.
(b) Find the two boundary conditions that X(x) must satisfy.
(c) Suppose that μ<0, so μ=−k²
for some k>0. In this case the general solution of equation ( ∗) is X(x)=Acos(kx)+Bsin(kx). Find the non-trivial solutions of equation (∗) that satisfy the boundary conditions, stating clearly what values k is allowed to take.
(d) Show that the function f(x,y)=exp(−Dk²t)cos(kx) satisfies the given partial differential equation for any constant k.

Answer 1

The method of separation of variables is used to solve the given partial differential equation. X(x) satisfies the differential equation X'' - μX = 0, while T(t) satisfies the corresponding differential equation obtained by dividing both sides by D. X(x) must satisfy the boundary conditions ∂x/∂θ = 0 when x = 0 for t > 0 and θ = 0 when x = L for t > 0. The non-trivial solutions of the differential equation satisfying the boundary conditions are X(x) = Acos(kx) and X(x) = Bsin(kx), where k can take any positive value.

Step-by-step explanation:

To solve the differential equation using the method of separation of variables, we assume that the temperature distribution can be expressed as a product of two functions: θ(x,t) = X(x)T(t). Substituting this into the given equation, we get X'' - μX = 0, where μ is a constant. The corresponding differential equation for T(t) is found by dividing both sides of the equation by D.

The two boundary conditions that X(x) must satisfy are ∂x/∂θ = 0 when x = 0 for t > 0 and θ = 0 when x = L for t > 0.

For the case where μ < 0, we have μ = -k^2 for some k > 0. The general solution of equation (*) is X(x) = Acos(kx) + Bsin(kx). The non-trivial solutions that satisfy the boundary conditions are X(x) = Acos(kx) and X(x) = Bsin(kx), where k can take any positive value.

The function f(x,t) = exp(-Dk^2t)cos(kx) satisfies the given partial differential equation for any constant k.