User:Eml5526.s11.team6.deshpande/hwk4

=Problem 4.3 Finding weak form for heat conduction problem=

Given
Given the strong form for the heat conduction problem in a circular plane:

with the essential boundary conditions and natural boundary condition

where R is the total radius of the plate, s is the heat source per unit length along the plate radius. T is the temperature and k is the conductivity. Assume that k, s and R are given.

Find
a. Construct the weak form for the above strong form.

b. Use quadratic trial (candidate) solution of the form $$T= \alpha_0+\alpha_1 r+\alpha_2 r^2$$ and weight function of the same form to obtain a solution of the weak form.

c. Solve the differential equation with the boundary conditions and show that the temperature distribution along the radius is given by

Solution
a. Weak form can be found as follows, with weighted function

Now, multiply eq. 4.3.1 by $$ \displaystyle w $$ and integrate between 0 to R.

Use integration by parts for the first term in eq. 4.3.6

Using eq. (4.3.3) and eq. (4.3.5), we get

Put eq. (4.3.8) in eq. (4.3.7), we get the weak form as follows

b. Now, consider the trial solution

By given essential boundary condition eq. (4.3.2) and by natural boundary condition eq. (4.3.3) Therefore, we get Put this eq. (4.3.14) in eq. (4.3.10), we get Hence, By eq. (4.3.2) and eq. (4.3.16) we get,

Taking arbitrary function $$ \displaystyle w $$ such that $$\displaystyle \beta_1=0$$, by comparing with eq. (4.3.15) Put this in weak form eq. (4.3.9), we get Integrating, factoring out $$\beta_0,\beta_2 $$ and rearranging the terms gives As the above must hold for arbitrary weight functions, it must hold for arbitrary $$\displaystyle\beta_0 $$ and $$\displaystyle\beta_2 $$. Therefore, the coefficients of $$\displaystyle\beta_0 $$ and $$\displaystyle\beta_2 $$ must vanish, which gives the following algebraic equation in $$\displaystyle\alpha_0 $$ and $$\displaystyle\alpha_2 $$: Solving above, second row gives Therefore, From eq. (4.3.18)

c. Therefore, the solution is Hence, the solution for the given differential equation is