User:EML5526.S11.Team1.HS/HW4

3.6
$$ \left. \left( wJG\frac{\text{d}\phi }{\text{d}x} \right) \right|_{x=0}^{x=l} $$

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

$$ k \frac{d}{dr} \left( r \frac{dT}{dr} \right) + rs = 0, 0 < r \leq R. $$

natural boundary condition : $$ \frac{dT}{dr} \left( r = 0 \right) = 0, $$

essential boundary condition : $$ T \left( r = R \right) = 0, $$

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.

Find
Assume $$ k, ~s~ $$ and $$ ~R~ $$ are given:

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

b. Use quadratic trial (candidate) solutions of the form $$ T = \alpha_{0} + \alpha_{1}r + \alpha_{2}r^2 $$ and weight functions 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

$$ T = \frac{s}{4k} \left(R^2 - r^2 \right). $$

a)
First we multiply the equation in the strong form by the weight function and integrate over the domain.

$$ \int_{0}^{R} wk \frac{d}{dr} \left( r \frac{dT}{dr}\right )dr + \int_{0}^{R} wrsdr = 0 ~ \forall w, $$

$$ \int_{0}^{R} k \frac{dw}{dr} \left( r \frac{dT}{dr}\right )dr = \left[wkr \frac{dT}{dr} \right]_0^R + \int_{0}^{R} wrsdr  \forall w(r)  with w(R) = 0 $$

The middle term cancels out when evaluated from $$ 0 to  R $$ which yields

$$ \int_{0}^{R} k \frac{dw}{dr} \left( r \frac{dT}{dr}\right )dr = \int_{0}^{R} wrsdr  \forall w(r)  with w(R) = 0 $$

b)
The trial solution $$ T\left ( r \right )$$ and the weight function $$ w\left ( r \right )$$ can be written as follows:

$$ T\left ( r \right ) = \alpha_{0} + \alpha_{1}r + \alpha_{2}r^{2}$$

$$ w\left ( r \right ) = \beta_{0} + \beta_{1}r + \beta_{2}r^{2} $$

Using the boundary conditions, the coefficients of the trial solutions can be solved. First the you take the derivative of the trial solution and set it to the natural boundary condition.

$$ \frac{dT}{dr} = \alpha_1 + 2 \alpha_{2}r $$

$$ \frac{dT}{dr} \left(r=0 \right) = 0 = \alpha_1 + 2 \alpha_{2}r $$

Therefore $$ \alpha_1 = 0 $$

Now use the essential boundary

$$ T \left(r = R \right) = 0 = \alpha_0 + \alpha_{2}R $$

Therefore $$ \alpha_0 = - \alpha_2 R^2 $$ which solves for our two temperature functions

T \left(r \right) = \alpha_2 \left(r^2 - R^2 \right) $$ \frac{dT}{dr} = 2 \alpha_2 r $$

Now solve for the weight functions using $$ w \left( R \right) = 0 $$

$$ w \left( R \right) = 0 = \beta_0 + \beta_1 R + \beta_2 R^2 $$

Which yields

$$ \beta_0 = -\beta_1 R -\beta_2 R^2 $$

$$ w \left(r \right) = \beta_1 \left( r -R \right) + \beta_2 \left( r^2 - R^2 \right) $$

And the derivative of the weight function is

$$ \frac{dw}{dr} = \beta_1 + 2 \beta_2 r $$

Before substitution, when the middle term is evaluated from $$ \left( 0 to R \right) $$ it goes to zero

$$ w \left( R \right) k r \frac{dT}{dr} \left( R \right) - w \left( 0 \right) k r \frac{dT}{dr} \left( 0 \right) = 0 $$

Therefore we can cancel out the middle term and substitute in the rest

$$ \int_{0}^{R} k \left( \beta_1 + 2 \beta_2 r \right) \left( r \left( 2 \alpha_2 r \right) \right )dr =  \int_{0}^{R} \left( \beta_1 \left( r -R \right) + \beta_2 \left( r^2 - R^2 \right) \right) rsdr \forall w  with w = 0 $$

Integrating we get:

$$ -k \left[ \frac{2 \alpha_2 \beta_1 r^3}{3} + 3 \alpha_2 \beta_2 r^4 \right]_0^R + s \left[ \beta_1 \left( \frac{r^3}{3} - \frac{R r^2}{2} \right) + \beta_2 \left( \frac{r^4}{4} - \frac{R^2 r^2}{2} \right) \right]_0^R = 0 $$

After evaluating the domain, the equation simplifies to:

$$ -k \alpha_2 \left( \beta_1 \left( \frac{2 R^3}{3} \right) + \beta_2 \left( R^4 \right) \right) + s \left( \beta_1 \left( - \frac{R^3}{6} \right) + \beta_2 \left( - \frac{R^4}{4} \right) \right) = 0 $$

Factoring out $$ \beta_1 $$ and $$ \beta_2 $$ we get:

$$ \beta_1\left ( 2\alpha_2 \frac{R^3}{3} + s\frac{R^3}{6} \right ) + \beta_2 \left ( k \alpha_2 R^4 + s \frac{R^4}{4} \right ) = 0 $$

Assuming that $$ \beta_1 $$ and $$ \beta_2 $$ equal zero, $$ \alpha_2 $$ can be solved:

$$ \alpha_2 = \frac{-s}{4k} $$

Now $$ \alpha_2 $$ can be substituted into $$ T \left( r \right) $$ to obtain the solution of the weak form

$$ T \left( r \right) = \frac{-s}{4k} \left( r^2 - R^2 \right) $$

c)
Using the strong form the differential equation can be solved by taking the integral:

$$ k \frac{d}{dr} \left( r \frac{dT}{dr} \right) + rs = 0, 0 < r \leq R. $$

$$ \int \frac{d}{dr} \left( r \frac{dT}{dr} \right) = \int -r \frac{s}{k} $$

$$ r \frac{dT}{dr} = - r^2 \frac{s}{2k} + c_1 $$

Using the natural boundary condition $$ \frac{dT}{dr} \left(r = 0 \right) = 0 \rightarrowc_1 = 0 $$

Integrate once again we get:

$$ T = -r^2 \frac{s}{4k} + c_2 $$

Using the essential boundary condition $$ T \left( r = R \right) = 0 $$ solves for:

$$ c_2 = \frac{s R^2}{4k} $$

substituting in $$ c_2 $$ yields the temperature distribution

$$ T = \frac{s}{4k} \left(R^2 - r^2 \right) $$

=Problem 8=

Given
Based off Problem 3.4

Given Strong Form

$$ \frac{d}{dx}\left ( 2 \frac{du}{dx} \right ) + 2x = 3 \frac{d^{2}u}{dt^{2}}  on  1<x<3 $$

$$ \left. 2 \frac{du}{dx} \right|_{x=1} = 0.1 $$

$$ u \left( x=3 \right) = 0.001 $$

$$ u^{h}\left ( x \right ) = \alpha _{0} + \alpha _{1}\left ( x-3 \right )+ \alpha _{2}\left ( x-3 \right )^2+ \alpha _{3}\left ( x-3 \right )^3 $$

and

$$ w^{h}\left ( x \right ) = \beta{0} + \beta_{1}\left ( x-3 \right )+ \beta_{2}\left ( x-3 \right )^2+ \beta_{3}\left ( x-3 \right )^3 $$

with $$ u^h \left(x =3,t \right) = sin \left(2t \right) $$

Find
1. Assume

$$ A = 1, E = 2, \bar{m} = 3 $$

Find $$ \tilde{M} $$

2. Assume

$$ n = 3 $$

$$ u^{h} \left ( p,t \right ) = g\left ( t \right ) = sin2t $$

Find $$ F \left(t \right) $$

Solution
From the strong form we can easily obtain the weak form. Find $$ u(x) $$ smooth enough and satisfy the essential boundary condition:

$$ u \left( x = 3 \right) = 0.001 $$ such that:

$$ \int_{1}^{3} 3\frac{dw}{dx} 2\frac{du}{dx}dx = -0.1 \left(w \right)_{x=3} + \int_{1}^{3} 2xwdx + \int_{1}^{3} 3 \frac{d^{2}u}{dt^{2}}dx \forall w \left(x \right)  with  w \left(3\right) = 0 $$

since $$ u^h \left(x=3,t \right) = sin \left(2t \right) $$

$$ u^{h}\left ( x = 3 \right ) = \alpha _{0} + \alpha _{1}\left ( 3-3 \right )+ \alpha _{2}\left ( 3-3 \right )^2+ \alpha _{3}\left ( 3-3 \right )^3 \rightarrow \alpha_0 = sin\left(2t \right) $$

Thus $$ \frac{d^2 u}{dt^2} = -4 sin \left(2t \right) $$

$$ w^{h}\left ( x = 3 \right ) = \beta _{0} + \beta_{1}\left ( 3-3 \right )+ \beta_{2}\left ( 3-3 \right )^2+ \beta_{3}\left ( 3-3 \right )^3 \rightarrow \beta_0 = 0 $$

From discrete weak form we get:

$$ \tilde{M} = \int_{1}^{3} \bar{m} w \left ( x \right ) \frac{d^{2}u}{dt^{2}}dx = \int_{1}^{3} 3 \left [ \beta_{1}\left ( x-3 \right )+ \beta_{2}\left ( x-3 \right )^2+ \beta_{3}\left ( x-3 \right )^3 \right ]\left ( -4 sin\left ( 2t \right ) \right )dx $$

$$ \tilde{M} = -12 sin \left(2t \right) \times \left[ -2 \beta_1 - \frac{8}{3} \beta_2 - 4 \beta_3 \right] $$