Heat equation

The heat equation is a Partial Differential equation that explains how temperature behaves in space and time.

About

The heat equation can be written on many forms:

\frac{\partial u}{\partial t} = α \frac{\partial ^{2} u}{\partial x ^{2}}

Where $u$ is the temperature as a function of time and position. And $α$ is a diffusivity constant. It can also be written on the form

\overset{u}{˙} = α △ u

Where $△ = \nabla \cdot \nabla = \nabla^{2}$ is known as the Laplace operator.

In one dimension $u$ is a function dependent on time $t$ and the position along a line $x$ .

\frac{\partial u}{\partial t} (t, x) = α \frac{\partial ^{2} u}{\partial x ^{2}} (t, x)

For each added dimension, an extra variable is added. In two dimensions, the equation changes to the following:

\frac{\partial u}{\partial t} (t, x, y) = α (\frac{\partial ^{2} u}{\partial x ^{2}} (t, x, y) + \frac{\partial ^{2} u}{\partial y ^{2}} (t, x, y))

Derivation

Heat equation in two dimensions

\frac{\partial u}{\partial t} (t, x, y) = α (\frac{\partial ^{2} u}{\partial x ^{2}} (t, x, y) + \frac{\partial ^{2} u}{\partial y ^{2}} (t, x, y))

Solving

Before solving the heat equation, we need a few parameters.

Lets assume we have a plate with height $H$ and length $L$ , that we want to track the temperature over.

We will need to know of the temperature distribution at $t = 0$ . This is referred to as an initial condition. In this case, we can set the initial condition to a generic function.

u (0, x, y) = f (x, y)

We also need a boundary condition. This gives information about how the temperature should act on the boundaries. In this case, lets assume we have isolated edges, so that the temperature on the edges can not be changed:

u^{'} (t, 0, y) = u^{'} (t, L, y) = u^{'} (t, x, 0) = u^{'} (t, x, H) = 0

To start of, we assume that we can use separation of variables:

u (t, x, y) = T (t) \cdot X (x) \cdot Y (y)

We then plug this into the heat equation

\frac{d T}{d t} (t) \cdot X (x) \cdot Y (y) = α (T (t) \cdot \frac{d ^{2} X}{d x ^{2}} (x) \cdot Y (y) + T (t) \cdot X (x) \frac{d ^{2} Y}{d y ^{2}} (y))

If we then divide both sides by $α \cdot T (t) \cdot X (x) \cdot Y (y)$ :

\frac{1}{α} \frac{T ^{'} ( t )}{T ( t )} = \frac{X ^{''} ( x )}{X ( x )} + \frac{Y ^{''} ( y )}{Y ( y )}

If you stare at this for a while, you notice that each side of this equation contains functions that are dependent on each independent variable. This means that if the value of $t$ changes, $x$ and $y$ will remain the same, and the expression on the right side of the equals sign will stay the same. For this equation to then be fulfilled, both sides has to stay constant:

\frac{1}{α} \frac{T ^{'} ( t )}{T ( t )} = \frac{X ^{''} ( x )}{X ( x )} + \frac{Y ^{''} ( y )}{Y ( y )} = K_{1}

Moving around on this equation to separate the functions dependent on $x$ and $y$ gives:

\frac{X ^{''} ( x )}{X ( x )} = K_{1} - \frac{Y ^{''} ( y )}{Y ( y )} = K_{2}

All of this can now be separated into three equations:

\frac{1}{α} \frac{T ^{'} ( t )}{T ( t )} = K_{1} ⟹ T^{'} (t) = α K_{1} T (t) (1)

\frac{X ^{''} ( x )}{X ( x )} = K_{2} ⟹ X^{''} (x) - K_{2} X (x) = 0 (2)

Y^{''} (y) - (K_{1} - K_{2}) Y (y) = 0 (3)

We will start by solving equation (2):

X^{''} (x) - K_{2} X (x) = 0

We then guess that the solution is the following:

X (x) = C e^{λ x}

Plugging this in and solving for $λ$ gives the actual solution;

C λ^{2} e^{λ x} - K_{2} e^{λ x} = 0

λ = \pm K_{2}

The solution is then given on the following form:

X (x) = C_{1} e^{K_{2} x} + C_{2} e^{- K_{2} x}

When we have come to this stage, we three different situations depending on $K_{2}$ :

Lets see what happens when $K_{2} = 0$

X (x) = C_{1} + C_{2} ⟹ X^{'} (x) = 0

This is not an interesting solution, as it states that $X$ is just a constant.

$K_{2} > 0$

X (x) = C_{1} e^{K_{2} x} + C_{2} e^{- K_{2} x}

Using the boundary condition.

X^{'} (x) = C_{1} K_{2} e^{K_{2} x} - C_{2} K_{2} e^{- K_{2} x}

X^{'} (0) = 0 = C_{1} - C_{2} ⟹ C_{1} = C_{2}

X^{'} (L) = C_{1} K_{2} e^{K_{2} L} - C_{2} K_{2} e^{- K_{2} L} = C_{1} - C_{2}

X^{'} (0) = X^{'} (L) = 0 ⟹ C_{1} = C_{2} = 0

This is not an interesting solution.

$K_{2} < 0$

X (x) = C_{1} e^{i - K_{2} x} + C_{2} e^{- i - K_{2} x}

By using Eulers formula we end up with the following:

X (x) = C_{1} (cos (- K_{2} x) + i sin (- K_{2} x)) + C_{2} (cos (- - K_{2} x) + i sin (- - K_{2} x))

X (x) = C_{1} (cos (- K_{2} x) + i sin (- K_{2} x)) + C_{2} (cos (- K_{2} x) - i sin (- K_{2} x))

X (x) = (C_{1} + C_{2}) cos (- K_{2} x) + (C_{1} i - C_{2} i) sin (- K_{2} x)

We now merge $(C_{1} + C_{2})$ and $(C_{1} i - C_{2} i)$ into two complex constants $A$ and $B$ :

X (x) = A cos (- K_{2} x) + B sin (- K_{2} x)

Now we want to use the boundary condition again.

X^{'} (x) = - A - K_{2} sin (- K_{2} x) + B - K_{2} cos (- K_{2} x)

X^{'} (0) = 0 = B - K_{2} ⟹^{- K_{2} \neq = 0} B = 0

$B = 0$ , this means that we only need the $A$ for the next check.

X^{'} (L) = 0 = - A - K_{2} sin (- K_{2} L)

We are not interested in the solution where $A = 0$ , $- K_{2}$ can never be zero as $K_{2}$ is a negative number. Therefore, we need to find a solution where the sine function gives zero.

sin - K_{2} L = 0 ⟹ - K_{2} L = nπ ⟹ - K_{2} = \frac{nπ}{L}

Where $n \in N$ .

X (x) = A_{n} cos (\frac{nπ}{L} x) (4)

We add a subscript $n$ to the constant in the equation as the constant might change based on what the number $n$ is. We have now found a solution for equation (2).

Next we can find a solution for equation (3):

Y^{''} (y) - (K_{1} - K_{2}) Y (y) = 0

This equation is very similar to equation (2). Finding the solution will be identical up until we get to adding the boundary condition:

Y (y) = A cos (- (K_{1} - K_{2}) y) + B sin (- (K_{1} - K_{2}) y)

TODO:

Legg inn informasjon om å løse varmelikningen med fourier transform,
- Ender opp med varmelikningens fundamentalløsning: løsning av varmelikningen over hele R uten rand og initialkrav.

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Heat equation

About

Derivation

Heat equation in two dimensions

Solving

TODO:

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