In a study of heat transfer, we find that for a solid rod, there is a relationship between the second derivative of the temperature with respect to position along the rod and the first with respect to time. (A linear temperature change with position would imply as much heat flowing into a region as out. so the temperature there would not change with time).

$\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{T}\left(x,\tau \right)}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\mathbf{\beta }\frac{\mathbf{\partial }\mathbf{T}\left(x,\tau \right)}{\mathbf{\partial }\mathbf{\tau }}\mathbf{\delta x}$

(a) Separate variables this assume a solution that is a product of a function of xand a function of tplug it in then divide by it, obtain two ordinary differential equations.

(b) consider a fairly simple, if somewhat unrealistic case suppose the temperature is 0 atx=0and, and x=1 positive in between, write down the simplest function of xthat (1) fits these conditions and (2) obey the differential equation involving x.Does your choice determine the value, including sign of some constant ?

(c) Obtain the full$\mathbf{T}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}$for this case.

Step by step solution

01

Formula for Heat Equation.

We know that:

In a study of heat transfer, we find that for a solid rod, there is a relationship between the second derivative of the temperature with respect to position along the rod and the first with respect to time.

Heat equation:

$\frac{{\partial }^2\mathrm{T}(\mathrm{x},\mathrm{t})}{\partial {\mathrm{x}}^2}=\mathrm{b}\frac{\partial \mathrm{T}(\mathrm{x},\mathrm{t})}{\partial \mathrm{t}}$

02

Separating the Variables

Let

$\begin{array}{rcl}& & \mathrm{T}(\mathrm{x},\mathrm{t})\; =\mathrm{X(}\mathrm{x}).\mathrm{T}(\mathrm{t})\\ & & =\mathrm{T}\frac{{\partial }^{\mathrm{2}}\mathrm{X}}{\partial {\mathrm{x}}^{\mathrm{2}}}=\mathrm{bX}\frac{\partial \mathrm{T}}{\partial \mathrm{t}}\\ & =& \frac{\mathrm{1}}{\mathrm{X}}\frac{{\partial }^{\mathrm{2}}\mathrm{X}}{\partial {\mathrm{x}}^{\mathrm{2}}}=\frac{\mathrm{b}}{\mathrm{T}}\frac{\partial \mathrm{T}}{\partial \mathrm{t}}={\mathrm{a}}^{\mathrm{2}}\end{array}$

Then

03

Assigning the Value

Then $\frac{\mathrm{1}}{\mathrm{X}}\frac{{\partial }^{\mathrm{2}}\mathrm{X}}{\partial {\mathrm{x}}^{\mathrm{2}}}={\mathrm{a}}^{\mathrm{2}}$and $\frac{\mathrm{b}}{\mathrm{T}}\frac{\partial \mathrm{T}}{\partial \mathrm{t}}={\mathrm{a}}^{\mathrm{2}}$

$\frac{{\partial }^{\mathrm{2}}\mathrm{X}}{\partial {\mathrm{x}}^{\mathrm{2}}}-{\mathrm{a}}^{\mathrm{2}}\mathrm{X}=\; 0$And $\frac{\mathrm{1}}{\mathrm{T}}\frac{\partial \mathrm{T}}{\partial \mathrm{t}}=\frac{{\mathrm{a}}^{\mathrm{2}}}{\mathrm{b}}$

$\mathrm{X}={\mathrm{Ae}}^{-\mathrm{ax}}+{\mathrm{Be}}^{\mathrm{ax}}$And $\frac{\partial \mathrm{T}}{\mathrm{T}}=\frac{{\mathrm{a}}^{\mathrm{2}}}{\mathrm{b}}\partial \mathrm{t}$

$\mathrm{X}={\mathrm{Ae}}^{-\mathrm{ax}}+{\mathrm{Be}}^{\mathrm{ax}}$And $\mathrm{lnT}=\frac{{\mathrm{a}}^{\mathrm{2}}\mathrm{t}}{\mathrm{b}}$

As$\text{x}\to \infty$so $\text{B = 0}\Rightarrow {\text{X = Ae}}^{\text{- ax}}$ and $\mathrm{T}=\exp \left(\frac{{\mathrm{a}}^{\mathrm{2}}\mathrm{t}}{\mathrm{b}}\right)$

Therefore .$\mathrm{T}(\mathrm{x},\mathrm{t})\; =\mathrm{Aexp}(\; -\mathrm{ax})\exp \left(\frac{{\mathrm{a}}^{\mathrm{2}}\mathrm{t}}{\mathrm{b}}\right)$

04

Finding X as Constant.

Suppose at $\mathrm{x}=\; 0\&\mathrm{x}=\mathrm{L}\Rightarrow \mathrm{T}=\; 0$Then

$\begin{array}{rcl}& & \mathrm{T}(0,\mathrm{t})\; =\mathrm{T(}\mathrm{L},\mathrm{t})\; =\mathrm{Aexp}\left(\frac{{\mathrm{a}}^{\mathrm{2}}\mathrm{t}}{\mathrm{b}}\right)=\mathrm{Aexp}(\; -\mathrm{aL})\exp \left(\frac{{\mathrm{a}}^{\mathrm{2}}\mathrm{t}}{\mathrm{b}}\right)\\ & \Rightarrow & \exp (\; -\mathrm{aL})\; =\; 1\\ & \Rightarrow & \mathrm{a}=\; 0\end{array}$

So X = A

Therefore, X=A is some constant.

05

Full Temperature

For this case $\mathrm{T}(\mathrm{x},\mathrm{t})\; =\mathrm{A}$ , temperature is also constant throughout the rod.

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