## Thursday, July 6, 2017

### Unizor - Partial Differential Equations - Heat Solution

Notes to a video lecture on http://www.unizor.com

Partial Differential Equations
Solution to Heat Equation

Our purpose is to solve the heat equation that describes the dynamics of temperature distribution within a thin rod and looks like this:
T(t,x)/t = a²·∂²T(t,x)/
where
t - time,
x - distance of a point on a thin rod from its edge,
T(t,x) - temperature at time t of a point on a rod at distance x from its edge,
a - constant that depends on physical characteristics of a rod.

This is a partial differential equation for function T(t,x) describing the distribution of temperature T within a thin rod at time t at location with X-coordinate x, assuming it's positioned along the X-axis with the left edge at x=0.

First of all, let's state that this distribution of temperature depends not only on the equation itself, which reflects dynamics of heat movement within a rod, but also depends on the initial conditions - what was the temperature at different locations of a rod at time t=0, described by function T(x,0).

Another type of condition that might be necessary to take into account is so-called boundary condition. It plays an important role in cases when edges of a rod are maintained at certain (not necessarily fixed) temperature all the time, which can be described by functions T(0,t) and T(L,t) (here L is the length of a rod).

Now we are ready to solve the equation.
Let's find a solution to our partial differential equation in a form
T(t,x) = f(t)·g(x)
We don't know if we will succeed on this way, but it's worth to try, since this relatively simple approach might bring us to solution. If it does, great. If it does not, at least, we tried.

Then
T(t,x)/t = f'(t)·g(x)
∂²T(t,x)/x² = f(t)·g''(x)
Now our heat equation looks like
f'(t)·g(x) = a²·f(t)·g''(x)
or
f'(t)/f(t) = a²·g''(x)/g(x)
Now we have a peculiar situation, when a function of t is equal to a function of x. The only possibility of this might be if both functions are constants that are equal to each other.
Therefore,
f'(t)/f(t) = A and
a²·g''(x)/g(x) = A
where A - some unknown constant that might be determined using initial conditions.

These are two ordinary differential equations, both types were addressed in the previous lectures. Let's get to their solutions.

The equation for f(t) is solved as follows:
df/f = A·dt
Integrating both sides results in
ln|f| = A·t + C
(where C - any constant)
|f(t)| = eA·t+C
or, considering C is any constant,
f(t) = C·eA·t

It's easy to check that this is a solution since
f'(t) = C·A·eA·t and
f'(t)/f(t) = A

The equation for g(x) is a linear equation of the second order, considered in a prior lecture about oscillation of a spring.
We can represent it as
g''(x) − (A/a²)·g(x) = 0
We will look for its solution in a form
g(x) = eλ·x
where λ - any complex number.
Since
g''(x) = λ²·eλ·x
we come to a simple quadratic equation for λ:
λ² − A/a² = 0
with solutions
λ = ±(1/a)·√A
This results in solutions for g(x):
g(x) = C1·e(1/a)·√A·x + C2·e−(1/a)·√A·x

General solution to a heat equation can now be represented as
T(t,x) = f(t)·g(x) =
= C·eA·t·(C1·e(1/a)·√A·x + C2·e−(1/a)·√A·x)

where CC1 and C2 are unknown constants, that can be defined only if additional conditions (initial conditions or boundary conditions) are given.

What remains is analysis of our solution for a purpose to get only real component of it, ignoring imaginary part, and apply initial condition on distribution of temperatures at time t=0, like T(0,x)=u(x), and boundary conditions (if applicable) T(t,0)=T0 and T(t,L)=TL in order to determine unknown constants. These are purely technical issues and lie outside the scope of this course.