QUESTION 2: Solve the initial boundary value problem Tt...

QUESTION 2: Solve the initial boundary value problem T_t (x,t) = alpha T_{xx} (x,t) , x > 0, t > 0 T (x,0) = 0 , x > 0 -KT_x (0,t) = Phi_0 , lim_{x -> infinity} T (x,t) = 0 , t > 0

Transcript

00:01 We have to solve the initial boundary value problem and it is given as t of x, t is equal to alpha t double x of x, t where x is greater than 0 and t is greater than 0.

00:18 Then t of x, 0 it is equal to 0 whenever x is greater than 0.

00:24 Then minus k t of x, 0, t is equal to phi of 0 and limit extending to infinity, t of x, t is equal to 0 when t is greater than 0.

00:41 This was given to us and this can be written as this was t1.

00:47 So, here we can write it as del t by del x, this del t by del small t, this will be equal to alpha into del 2 of t by del x square where x is greater than 0 and t is greater than 0.

01:06 With the initial condition t of x, 0 it is equal to 0 where x is greater than 0 and these are the boundary conditions which we are provided here.

01:18 This is nothing but a partial differential equation.

01:21 This is a pde that is partial differential equation.

01:26 This is a heat equation with the source term given by alpha and the given boundaries and initial conditions.

01:34 Solving this kind of problem typically involves using the separation of variables.

01:39 Here we use the separation of variables and fourier series technique.

01:45 So, here step first is the separation of variables.

01:48 We will assume that t of x, t can be represented as a product of functions of x and t.

01:54 So, it is xx into t of t.

01:58 Substitute this into the partial differential equation.

02:02 So, x of x into t dash of t will be equal to alpha into x double dash of x and multiplied by t over t.

02:13 Then dividing both sides by this alpha and x of x and t of t, we get here t dash of t divided by alpha t of t is equal to x double dash of x divided by x of x.

02:33 This is equal to minus lambda where lambda is a separation constant.

02:38 Now, we will solve for the time part.

02:42 So, we will move into the step number 2 that is the time part.

02:47 The equation for t over t is a simple first order ordinary differential equation.

02:54 Therefore, t dash of t divided by alpha t of t, this is equal to minus lambda means this is equal to this term and integrating both sides with respect to t, we get over integration 1 by alpha log of that is natural log of t over t.

03:16 This is equal to minus lambda t plus c1 as a constant.

03:22 Now, we will solve for this t over t.

03:24 This t over t is equal to this alpha will multiply on that side that is alpha multiplied by minus lambda t, this minus lambda t plus c1...

Question

QUESTION 2: Solve the initial boundary value problem T_t (x,t) = alpha T_{xx} (x,t) , x > 0, t > 0 T (x,0) = 0 , x > 0 -KT_x (0,t) = Phi_0 , lim_{x -> infinity} T (x,t) = 0 , t > 0

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