Analytical Solution of the Poisson's Equation for One-Dimensional Domains

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Latest revision as of 09:08, 4 November 2009

General solution using the Heat Transfer example

Consider the heat transfer without convection effects along the following bar:

Remember that the conduction phenomena refers to "the transfer of thermal energy from a region of higher temperature to a region of lower temperature through direct molecular communication within a medium or between mediums in direct physical contact without a flow of the material medium".

As boundary conditions, the temperature is fixed at the beginning of the bar, and the heat flow is given at the end of the bar:

$\varphi = \bar \varphi \mid_{x=x_0}$

$q = \bar q \mid_{x=x_L}$

Taking a differential length and by establishing the balance of heat flows, it can be written:

outflow $(q + dq) - (q) \frac{ }{ }$ inflow $= 0 \frac{ }{ }$

$dq = 0 \frac{ }{ }$

Using the Fourier law[1]:

$q = - k \cdot \frac{d \varphi}{dx}$

Therefore:

$dq = \frac{dq}{dx} dx = - \frac{d}{dx} \left(k \cdot \frac{d \varphi}{dx} \right) dx$

$\frac{d}{dx} \left(k \cdot \frac{d \varphi}{dx} \right) = 0$

If an external heat source is considered, the balance of heat flows becomes:

$(q + dq) - (q) - Q \cdot dx = 0 \frac{ }{ }$

$\frac{dq}{dx} - Q = 0$

$\frac{d}{dx} \left(k \cdot \frac{d \varphi}{dx} \right) + Q = 0$

Rewritting the boundary conditions:

$\varphi - \bar \varphi =0$ en $x=x_0 \frac{ }{ }$

$q - \bar q = - \left( k \cdot \frac{d \varphi}{dx} + \bar q \right) = 0$ en $x=x_L\frac{ }{ }$

Generic Solution:

1 General solution using the Heat Transfer example
2 The simplest case: homogeneous medium, no sources
- 2.1 Case 1. With Dirichlet boundary conditions
- 2.2 Case 2. With Dirichlet and Neumann boundary conditions
3 Heterogeneous medium, no sources
- 3.1 Case 3. With Dirichlet boundary conditions
- 3.2 Case 4. With Dirichlet and Neumann boundary conditions
4 With a source, homogeneous medium

The simplest case: homogeneous medium, no sources

Case 1. With Dirichlet boundary conditions

sources

medium

boundary

conditions

no sources

homogeneous

Dirichlet

$Q = 0 \frac{ }{ }$

$k(x) = cte \frac{ }{ }$

$\varphi (x_0)= \varphi_0$

$\varphi (x_L)= \varphi_L$

To obtain the constants cte₁ and cte₂:

$\varphi (x_0)= \frac{cte_1}{k} x_0 + cte_2 = \varphi_0$

$\varphi (x_L)= \frac{cte_1}{k} x_L + cte_2 = \varphi_L$

with the result of:

$cte_1 = k \cdot \frac{\varphi_L - \varphi_0}{x_L - x_0}$

$cte_2 = \varphi_0 - \frac{cte_1}{k} x_0 = \frac{\varphi_0 x_L - \varphi_L x_0}{x_L - x_0}$

that is:

Specific example:

parameters

sources

medium

b.c.

geometry

$Q = 0 \frac{ }{ }$

$k = 5 \frac{ }{ }$

$\varphi_0 = 3$

$\varphi_L = 7$

$x_0 = 0 \frac{ }{ }$

$x_L = 10 \frac{ }{ }$

results

unknow

gradient

flow

$\varphi (x_0)= 3$

$\varphi (x_L)= 7$

$\frac{d \varphi (x)}{dx} \bigg|_{\forall x} = 0.4$

$k \cdot \frac{d \varphi (x)}{dx} \bigg|_{\forall x} = 2$

To modify the parameters, edit this Matlab code

Case 2. With Dirichlet and Neumann boundary conditions

sources

medium

boundary

conditions

no sources

homogeneous

Dirichlet and Neumann

$Q = 0 \frac{ }{ }$

$k(x) = cte \frac{ }{ }$

$\varphi (x_0)= \varphi_0$

$k \cdot \nabla \varphi (x) \mid_{x=x_L}= q_L$

$cte_1 = q_L \frac{}{}$

$cte_2 = \varphi_0 - \frac{q_L}{k} x_0$

that is:

note that:

$\varphi (x_L) = \frac{q_L}{k} (x_L - x_0) + \varphi_0$

Specific example:

parameters

sources

medium

b.c.

geometry

$Q = 0 \frac{ }{ }$

$k = 5 \frac{ }{ }$

$\varphi_0 = 3$

$q_L = 2 \frac{ }{ }$

$x_0 = 0 \frac{ }{ }$

$x_L = 10 \frac{ }{ }$

results

unknow

gradient

flow

$\varphi (x_0)= 3$

$\varphi (x_L)= 7$

$\frac{d \varphi (x)}{dx} \bigg|_{\forall x} = 0.4$

$k \cdot \frac{d \varphi (x)}{dx} \bigg|_{\forall x} = 2$

To modify the parameters, edit this Matlab code

Heterogeneous medium, no sources

Case 3. With Dirichlet boundary conditions

sources	medium	boundary conditions
no sources	heterogeneous	Dirichlet
$Q = 0 \frac{ }{ }$	$k(x) = \begin{cases} k_1 & 0 \leqslant x \leqslant x_1 \\ k_2 & x_1 \leqslant x \leqslant x_L \end{cases}$	$\varphi (x_0)= \varphi_0$ $\varphi (x_L)= \varphi_L$

$\varphi (x) = \begin{cases} \frac{cte_1^1}{k_1} x + cte_2^1 & 0 \leqslant x \leqslant x_1 \\ \\ \frac{cte_1^2}{k_2} x + cte_2^2 & x_1 \leqslant x \leqslant x_L \end{cases}$

To obtain the constants cte₁ and cte₂ in both intervals:

$k_1 \nabla \varphi (x) \mid_{x=x_1} = k_2 \nabla \varphi (x) \mid_{x=x_1}$

therefore cte₁ is the same in both intervals.

$\varphi (x_0)= \frac{cte_1}{k_1} x_0 + cte_2^1 = \varphi_0$ and $\varphi (x_L)= \frac{cte_1}{k_2} x_L + cte_2^2 = \varphi_L$

$\varphi (x_1)= \frac{cte_1}{k_1} x_1 + cte_2^1 = \frac{cte_1}{k_2} x_1 + cte_2^2$

$\begin{bmatrix} \frac{x_0}{k_1} & 1 & 0 \\ \frac{x_L}{k_2} & 0 & 1 \\ \left( \frac{1}{k_1} - \frac{1}{k_2} \right) x_1 & 1 & -1 \end{bmatrix} \begin{bmatrix} \frac{}{} cte_1 \\ \frac{}{} cte_2^1 \\ \frac{}{} cte_2^2 \\ \end{bmatrix} = \begin{bmatrix} \varphi_0 \\ \varphi_L \\ \frac{}{} 0 \\ \end{bmatrix}$

with the result of:

$cte_1 = \frac{k_1 k_2 (\varphi_L - \varphi_0)}{k_1 (x_L - x_1) + k_2 (x_1 - x_0)}$

$cte_2^1 = \frac{k_1 \varphi_0 (x_L - x_1 ) + k_2 (\varphi_0 x_1 - \varphi_L x_0)}{k_1 (x_L - x_1) + k_2 (x_1 - x_0)}$

$cte_2^2 = \frac{k_1 (\varphi_0 x_L - \varphi_L x_1 ) + k_2 \varphi_L ( x_1 - x_0)}{k_1 (x_L - x_1) + k_2 (x_1 - x_0)}$

note that if k₁=k₂, the case 3 becomes the case 1.

Specific example:

parameters

sources

medium

b.c.

geometry

$Q = 0 \frac{ }{ }$

$k(x) = \begin{cases} k_1 & x_0 \leqslant x \leqslant x_1 \\ k_2 & x_1 \leqslant x \leqslant x_L \end{cases}$

$\varphi_0 = 3$

$\varphi_L = 7 \frac{ }{ }$

$x_0 = 0 \frac{ }{ }$

$x_1 = 9 \frac{ }{ }$

$x_L = 10 \frac{ }{ }$

results

unknow

gradient

flow

$\varphi (x_0)= 3$

$\varphi (x_L)= 7$

$\frac{d \varphi (x)}{dx} \bigg|_{x_0 \leqslant x \leqslant x_1} = 0.4$

$\frac{d \varphi (x)}{dx} \bigg|_{x_1 \leqslant x \leqslant x_L} = 1$

$k \cdot \frac{d \varphi (x)}{dx} \bigg|_{\forall x} = 5$

To modify the parameters, edit this Matlab code

Case 4. With Dirichlet and Neumann boundary conditions

sources

medium

boundary

conditions

no sources

heterogeneous

Dirichlet and Neumann

$Q = 0 \frac{ }{ }$

$k(x) = \begin{cases} k_1 & 0 \leqslant x \leqslant x_1 \\ k_2 & x_1 \leqslant x \leqslant x_L \end{cases}$

$\varphi (x_0)= \varphi_0$

$k \cdot \nabla \varphi (x) \mid_{x=x_L}= q_L$

$\varphi (x) = \begin{cases} \frac{cte_1^1}{k_1} x + cte_2^1 & 0 \leqslant x \leqslant x_1 \\ \\ \frac{cte_1^2}{k_2} x + cte_2^2 & x_1 \leqslant x \leqslant x_L \end{cases}$

To obtain the constants cte₁ and cte₂ in both intervals:

$k_1 \nabla \varphi (x) \mid_{x=x_1} = k_2 \nabla \varphi (x) \mid_{x=x_1} = q_L = cte_1$

$\varphi (x_0)= \frac{cte_1}{k_1} x_0 + cte_2^1 = \varphi_0$

$\varphi (x_1)= \frac{cte_1}{k_1} x_1 + cte_2^1 = \frac{cte_1}{k_2} x_1 + cte_2^2$

$\begin{bmatrix} \frac{}{} 1 & 0 \\ \frac{}{} 1 & -1 \end{bmatrix} \begin{bmatrix} \frac{}{} cte_2^1 \\ \frac{}{} cte_2^2 \end{bmatrix} = \begin{bmatrix} \varphi_0 - \frac{cte_1 x_0}{k_1}\\ \frac{cte_1 x_1}{k_2} - \frac{cte_1 x_1}{k_1} \end{bmatrix}$

with the result of:

$cte_1 = q_L \frac{}{}$

$cte_2^1 = \varphi_0 - \frac{q_L x_0}{k_1}$

$cte_2^2 = \varphi_0 + \frac{q_L}{k_1} \left( x_1 \frac{k_2 - k_1}{k_2} - x_0 \right)$

note that if k₁=k₂, the case 4 becomes the case 2.

Specific example:

parameters

sources

medium

b.c.

geometry

$Q = 0 \frac{ }{ }$

$k(x) = \begin{cases} k_1 & x_0 \leqslant x \leqslant x_1 \\ k_2 & x_1 \leqslant x \leqslant x_L \end{cases}$

$\varphi_0 = 3$

$q_L = 5 \frac{ }{ }$

$x_0 = 0 \frac{ }{ }$

$x_1 = 9 \frac{ }{ }$

$x_L = 10 \frac{ }{ }$

results

unknow

gradient

flow

$\varphi (x_0)= 3$

$\varphi (x_L)= 7$

$\frac{d \varphi (x)}{dx} \bigg|_{x_0 \leqslant x \leqslant x_1} = 0.4$

$\frac{d \varphi (x)}{dx} \bigg|_{x_1 \leqslant x \leqslant x_L} = 1$

$k \cdot \frac{d \varphi (x)}{dx} \bigg|_{\forall x} = 5$

To modify the parameters, edit this Matlab code

With a source, homogeneous medium

Case 5. Constant source over the whole domain, Dirichlet boundary conditions

sources	medium	boundary conditions
constant sources	homogeneous	Dirichlet
$Q = Q_0 \frac{ }{ }$	$k(x) = cte \frac{ }{ }$	$\varphi (x_0)= \varphi_0$ $\varphi (x_L)= \varphi_L$

$\varphi (x) = - \frac{Q}{2 k} x^2 + \frac{cte_1}{k} x + cte_2$

$k \nabla \varphi (x) = - Q x + cte_1$

To obtain the constants cte₁ and cte₂:

$\varphi (x_0) = \varphi_0 = - \frac{Q}{2 k} x_0^2 + \frac{cte_1}{k} x_0 + cte_2$

$\varphi (x_L) = \varphi_L = - \frac{Q}{2 k} x_L^2 + \frac{cte_1}{k} x_L + cte_2$

$\begin{bmatrix} \frac{x_0}{k} & 1 \\ \\ \frac{x_L}{k} & 1 \end{bmatrix} \begin{bmatrix} \frac{}{} cte_1 \\ \\ \frac{}{} cte_2 \end{bmatrix} = \begin{bmatrix} \varphi_0 + \frac{Q}{2 k} x_0^2\\ \\ \varphi_L + \frac{Q}{2 k} x_L^2 \end{bmatrix}$

with the result of:

$\vartriangle = \frac{1}{k} (x_0 - x_L)$

$cte_1 = \frac{\left[ (\varphi_0 - \varphi_L) + \frac{Q}{2 k} (x_0^2 - x_L^2) \right] }{\vartriangle}$

$cte_2 = \frac{\left[ (x_0 \varphi_L - x_L \varphi_0) + \frac{Q}{2 k} (x_0 x_L^2 - x_L x_0^2) \right] }{k \vartriangle}$

note that if Q = 0, the case 5 becomes the case 1.

Specific example:

parameters

sources

medium

b.c.

geometry

$Q_0 = 2\frac{ }{ }$

$k = 5 \frac{ }{ }$

$\varphi_0 = 3$

$\varphi_L = 7 \frac{ }{ }$

$x_0 = 0 \frac{ }{ }$

$x_L = 10 \frac{ }{ }$

results

unknow

gradient

flow

$\varphi (x_0)= 3$

$\varphi (x_L)= 7$

$\frac{d \varphi (x)}{dx} \bigg|_{x=x_0} = 2.4$

$\frac{d \varphi (x)}{dx} \bigg|_{x=x_L} = -1.6$

$k \cdot \frac{d \varphi (x)}{dx} \bigg|_{x=x_0} = 12$

$k \cdot \frac{d \varphi (x)}{dx} \bigg|_{x=x_L} = -8$

To modify the parameters, edit this Matlab code

Case 6. A constant source over a part of the domain, Dirichlet boundary conditions

sources	medium	boundary conditions
constant source	homogeneous	Dirichlet
$Q(x) = \begin{cases} Q_0 & x_0 \leqslant x \leqslant x_f \\ 0 & x_f \leqslant x \leqslant x_L \end{cases}$	$k(x) = cte \frac{ }{ }$	$\varphi (x_0)= \varphi_0$ $\varphi (x_L)= \varphi_L$

$\varphi (x) = \begin{cases} - \frac{Q_0}{2 k} x^2 + \frac{cte_1^1}{k} x + cte_2^1 & 0 \leqslant x \leqslant x_f \\ \\ \frac{cte_1^2}{k} x + cte_2^2 & x_f \leqslant x \leqslant x_L \end{cases}$

$k \nabla \varphi (x) = \begin{cases} - Q_0 x + cte_1^1 & 0 \leqslant x \leqslant x_f \\ \\ cte_1^2 & x_f \leqslant x \leqslant x_L \end{cases}$

To obtain the constants cte₁ and cte₂ in both intervals:

interval 1:

$\varphi (x_0) = - \frac{Q_0}{2 k} x_0^2 + \frac{cte_1^1}{k} x_0 + cte_2^1 = \varphi_0$

$\varphi (x_f) = - \frac{Q_0}{2 k} x_f^2 + \frac{cte_1^1}{k} x_f + cte_2^1$

$k \nabla \varphi (x) \mid_{x=x_f} = - Q_0 x_f + cte_1^1$

interval 2:

$\varphi (x_L) = \frac{cte_1^2}{k} x_L + cte_2^2 = \varphi_L$

$\varphi (x_f) = \frac{cte_1^2}{k} x_f + cte_2^2$

$k \nabla \varphi (x) \mid_{x=x_f} = cte_1^2$

with the result of:

$cte_1^1 = \frac{k}{x_L - x_0} \left[ (\varphi_L - \varphi_0) - \frac{Q_0}{2k} (x_0^2 - 2 x_L x_f + x_f^2) \right]$

$cte_2^1 = \frac{\varphi_0 x_L - \varphi_L x_0}{x_L - x_0} - \frac{Q_0}{2 k} \frac{x_0}{x_L - x_0} (2 x_L x_f + x_0 x_L - x_f^2)$

$cte_1^2 = - Q_0 x_f + cte_1^1 = \frac{k}{x_L - x_0} \left[ (\varphi_L - \varphi_0) - \frac{Q_0}{2k} (x_0^2 - 2 x_f x_0 + x_f^2) \right]$

$cte_2^2 = cte_2^1 + \frac{1}{2} \frac{Q_0 x_f^2}{k}$

note that if Q₀ = 0, the case 6 becomes the case 1.

Specific example:

parameters

sources

medium

b.c.

geometry

$Q(x) = \begin{cases} Q_0 = 2 & x_0 \leqslant x \leqslant x_f \\ 0 & x_f \leqslant x \leqslant x_L \end{cases}$

$k = 5 \frac{ }{ }$

$\varphi_0 = 3$

$\varphi_L = 7 \frac{ }{ }$

$x_0 = 0 \frac{ }{ }$

$x_f = 4 \frac{ }{ }$

$x_L = 10 \frac{ }{ }$

results

unknow

gradient

flow

$\varphi (x_0)= 3$

$\varphi (x_L)= 7$

$\frac{d \varphi (x)}{dx} \bigg|_{x=x_0} = 1.68$

$\frac{d \varphi (x)}{dx} \bigg|_{x=x_L} = 0.08$

$k \cdot \frac{d \varphi (x)}{dx} \bigg|_{x=x_0} = 8.4$

$k \cdot \frac{d \varphi (x)}{dx} \bigg|_{x=x_L} = 0.4$

To modify the parameters, edit this Matlab code

Case 7. Constant source over the whole domain, Dirichlet and Neumann boundary conditions

sources

medium

boundary

conditions

constant sources

homogeneous

Dirichlet and Neumann

$Q = Q_0 \frac{ }{ }$

$k(x) = cte \frac{ }{ }$

$\varphi (x_0)= \varphi_0$

$k \cdot \nabla \varphi (x) \mid_{x=x_L}= q_L$

$\varphi (x) = - \frac{Q}{2 k} x^2 + \frac{cte_1}{k} x + cte_2$

$k \nabla \varphi (x) = - Q x + cte_1$

To obtain the constants cte₁ and cte₂:

$\varphi (x_0) = \varphi_0 = - \frac{Q}{2 k} x_0^2 + \frac{cte_1}{k} x_0 + cte_2$

$k \nabla \varphi (x) |_{x_L} = - Q x_L + cte_1 = q_L$

with the result of:

$cte_1 = q_L + Q x_L \,$

$cte_2 = - \frac{Q}{2 k} x^2 + (q_L - Q x_L) \frac{x}{L} + \varphi_0$

$\varphi (x) = \frac{Q}{2 k} (x_0^2 - x^2) + \frac{q_L + Q x_L}{k} (x-x_0) + \varphi_0$

$k \nabla \varphi (x) = Q (x_L - x) + q_L$

note that if Q = 0, the case 5 becomes the case 2.

Specific example:

parameters

sources

medium

b.c.

geometry

$Q_0 = 2\frac{ }{ }$

$k = 5 \frac{ }{ }$

$\varphi_0 = 3$

$q_L = 2 \,$

$x_0 = 0 \,$

$x_L = 10 \,$

results

unknow

gradient

flow

$\varphi (x_0)= 3$

$\varphi (x_L)= 27$

$\frac{d \varphi (x)}{dx} \bigg|_{x=x_0} = 4.4$

$\frac{d \varphi (x)}{dx} \bigg|_{x=x_L} = 0.4$

$k \cdot \frac{d \varphi (x)}{dx} \bigg|_{x=x_0} = 22$

$k \cdot \frac{d \varphi (x)}{dx} \bigg|_{x=x_L} = 2$

To modify the parameters, edit this Matlab code

Back to Poisson's_Equation main page

Revision as of 16:06, 4 December 2008 (view source) JMora (Talk \| contribs) (→Case 6. A constant source over a part of the domain, Dirichlet boundary conditions) ← Older edit		Latest revision as of 09:08, 4 November 2009 (view source) JMora (Talk \| contribs) (→Case 7. Constant source over the whole domain, Dirichlet and Neumann boundary conditions)
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−	[[Category:~~Theory~~]]	+	[[Category:Poisson's Equation]]

Analytical Solution of the Poisson's Equation for One-Dimensional Domains

Latest revision as of 09:08, 4 November 2009

General solution using the Heat Transfer example

Contents

The simplest case: homogeneous medium, no sources

Case 1. With Dirichlet boundary conditions

Case 2. With Dirichlet and Neumann boundary conditions

Heterogeneous medium, no sources

Case 3. With Dirichlet boundary conditions

Case 4. With Dirichlet and Neumann boundary conditions

With a source, homogeneous medium

Case 5. Constant source over the whole domain, Dirichlet boundary conditions

Case 6. A constant source over a part of the domain, Dirichlet boundary conditions

Case 7. Constant source over the whole domain, Dirichlet and Neumann boundary conditions

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