User:Eml5526.s11.team6.deshpande/hwk2

=Problem 2.1= = Conductive Bar PDE, Dynamic situation.=

Find
For the heat problem, derive the following partial differential equation by performing a balance of heat:

$$ \frac{\partial}{\partial x}\left [ A(x)k(x)\frac{\partial u}{\partial x} \right ]+f(x,t)=A(x) \rho(x)c\cdot \frac{\partial u}{\partial t}\ $$

Solution
Consider the FBD of an infinitesimally small section of length dx of the 1-D conductive bar with varying material properties at a distance x from the origin. Where, A(x) - Varying cross section of the bar, k(x) - conductivity of the material, rho (x) - Mass density, u - Temperature constrained only to x- axis (1-D bar), t - Time.



Where, heat flux at $$\displaystyle x=q(x)$$,

heat flux at $$\displaystyle x+dx=q(x+dx)$$

outward heat flow through A(x)=Q(x), at n(x)=-1,

$$\displaystyle Q(x)=q(x)A(x)n(x)= -q(x)A(x)$$ $$\displaystyle (Eq. 2.1.1) $$

Similarly, outward heat flow through $$\displaystyle A(x+dx)=Q(x+dx), at n(x+dx)=1,$$

$$\displaystyle Q(x+dx)=q(x+dx)A(x+dx)n(x+dx)= q(x+dx)A(x+dx) $$ $$\displaystyle (Eq. 2.1.2) $$

Also, internal heat source per unit volume= r(x,t)

Hence, heat source per unit length= r(x,t)A(x) = f(x,t) $$\displaystyle (Eq. 2.1.3) $$

Now, balance of heat can be done as follows:

$$\displaystyle H_{1}= $$heat flow in control volume

$$\displaystyle =-[Q(x)+Q(x+dx)]$$

$$\displaystyle =-[-q(x)A(x)+q(x+dx)A(x+dx)]$$ $$\displaystyle (Eq. 2.1.4) $$

$$\displaystyle H_{2}= $$Heat by internal source r(x,t) $$\displaystyle =f(x,t)dx$$$$\displaystyle (Eq. 2.1.5) $$

$$\displaystyle H_{3}= $$heat by temperature difference $$\displaystyle=\;\; A(x)\rho (x)c\cdot\frac{\partial u}{\partial t}$$$$\displaystyle (Eq. 2.1.6) $$

Now, for balance of heat

$$\displaystyle H_{1}+H_{2}=H_{3}$$$$\displaystyle (Eq. 2.1.7) $$

Therefore, from $$\displaystyle (Eq. 2.1.4) $$, $$\displaystyle (Eq. 2.1.5) $$,  $$\displaystyle (Eq. 2.1.6) $$,  $$\displaystyle (Eq. 2.1.7) $$ we can write,

$$\displaystyle q(x)A(x)-q(x+dx)A(x+dx)+f(x,t)dx=\;\; A(x)\rho (x)c\cdot\frac{\partial u}{\partial t}$$$$\displaystyle (Eq. 2.1.8) $$

put equations $$\displaystyle (Eq. 2.1.1) $$ and $$\displaystyle (Eq. 2.1.2) $$ in $$\displaystyle (Eq. 2.1.8) $$

$$\displaystyle Q(x,t)-Q(x+dx,t)+f(x,t)dx=\;\; A(x)\rho (x)c\cdot\frac{\partial u}{\partial t}$$

Including Taylor series expansion for $$Q(x+dx,t)$$ and neglecting the higher order terms (h.o.t) yields:

$$ \cancel {Q(x,t)}-[\cancel {Q(x,t)}+\frac {\partial }{\partial x}(Q(x,t))+(h.o.t)]+f(x,t)=\;\; A(x)\rho (x)c\cdot\frac{\partial u}{\partial t}$$

Hence, $$ -\frac {\partial }{\partial x}(Q(x,t))+(h.o.t)]+f(x,t)=\;\; A(x)\rho (x)c\cdot\frac{\partial u}{\partial t}$$$$\displaystyle (Eq. 2.1.9) $$

where, $$\displaystyle Q(x,t)= -q(x)A(x)$$ and $$\displaystyle q(x)=k(x)\frac {\partial u}{\partial x}$$

Therefore we get,

=Problem 2.4=

Find
To Prove Solution provided by Wolfram Alpha

Prove Following steps and use them to get the final proof of the above eqution:

Part 1: $$ \int log(x)dx = x*log(x) - x$$ Solution provided by Wolfram Alpha using integration by parts.

Part 2: $$ \int x*log(x)dx = \frac {1}{2}*x^2* \left [ log(x) - \frac {1}{2} \right ] $$

Part 3: $$ \int \frac {x^2 dx}{1+C*x} = \frac {2*log(C*x + 1) + x*(C*x-2)*C}{2*C^3} $$

Part 4: $$ \int \frac {x^2 dx}{A+C*x} = \frac {2*log(C*x + A) + C*x*(C*x-2*A)}{2*C^3} $$

Part 5: Now refer to meeting 9 [[media:fe1.s11.mtg9.djvu|Mtg 9 (c)]] and find exact solution $$\displaystyle u(x)$$ for problem (3)

Part 6: Plot $$\displaystyle u(x)$$

Solution
Part 1

Prove Solution provided by Wolfram Alpha

using integration by parts, i.e.,

Part 2

Prove Solution provided by Wolfram Alpha

using integration by parts, (Eq. 2.4.3)

Part 3

Prove Solution provided by Wolfram Alpha

Using integration by parts, (Eq. 2.4.3)

Continuing with the integration by parts with the latter part of the RHS of (Eq. 2.4.15), Given (Eq. 2.4.7) and the substitution given in (Eq. 2.4.14),

Applying integration by parts again, we obtain the following relations:

Finally, the integral in the RHS of (Eq. 2.4.19) can be solved for and simplified using (Eq. 2.4.12) and the integration by substitution given (Eq. 2.4.14), i.e., Combining the former portion of the RHS of (Eq. 2.4.15), the former portion of the RHS of (Eq. 2.4.19), and the latter portion of the RHS of (Eq. 2.4.20), and then simplifying, the final solution to the integral in (Eq. 2.4.13) can be found to be

Part 4

Prove Solution provided by Wolfram Alpha

and the first integration by parts given in Step 3, (Eq. 2.4.15),

Continuing with the integration by parts with the latter part of the RHS of (Eq. 2.4.24),

and given the second integration by parts in Step 3, (Eq. 2.4.19)

Finally, the integral in the RHS of (Eq. 2.4.27) can be solved for and simplified using (Eq. 2.4.12) and the integration by substitution given in (Eq. 2.4.23), i.e.,

Combining the former portion of the RHS of (Eq. 2.4.24), the former portion of the RHS of (Eq. 2.4.27), the latter portion of the RHS of (Eq. 2.4.28), and then simplifying, the final solution to the integral in (Eq. 2.4.22) can be found in a similar fashion to (Eq. 2.4.20) of Step 3, i.e.,

Final Step

Given (Eq. 2.4.29) from Step 4, if A = C = 1, then

Thus, required result stated in ' Find' section Eq 2.4.1 is achieved.

Part 5

Part 6