# What is the thermal equation solution

## Thermal equation

How can heat changes be mapped mathematically? In this article we will show you how to solve thermal conduction equations using the separation approach.

- One-dimensional heat conduction equationin the text
- Solve the separated DGLin the text
- in the text
- Solution of the second boundary conditionin the text

### One-dimensional heat conduction equation

The heat conduction equation states how the temperature in a material or gas changes over time. The diffusion equation is also described in this way. It indicates how the concentration of a substance changes in space and time because the substance diffuses. The one-dimensional heat equation, or diffusion equation, looks like this. It's a parabolic differential equation.

This is where we put the product approach

a.

On the left we can pull out of the lead and on the right side .

We write it a little more clearly,

separate the terms

and set it equal to the constant lambda.

We do this because both sides only depend on one variable. They can only be the same if they are constant.

### Solve the separated DGL

which we solve separately from each other. Now let's look at the initial and boundary conditions.

From these you can derive the initial and boundary conditions for the functions T and X:

We start by solving the equation that depends on t.

Here an exponential approach leads to the goal.

And then it goes on with the second equation dependent on x. It is an ordinary second order differential equation

Here we introduce the characteristic polynomial for on

### Storm Liouville Problem

Depending on the value of Lambda, the solution looks different. There is a so-called Sturm-Liouville problem that you can solve with case distinction. You have to consider three cases:

The first case is that lambda is greater than zero.

Then you get purely real values for lambda.

And your solution is a combination of two exponential functions. We use the boundary conditions.

results in one, so the sum plus remains. So it turns out that equal is.

### Solution of the second boundary condition

The second boundary condition leads to the trivial solution equal equals zero, since the expression in brackets is not equal to zero. The second case is that equals zero.

With the double zero at zero the result is the following X,

in which you insert the boundary conditions.

The first constraint implies that must be zero.

The second boundary condition leads to c_Eins equal to zero. So here too we only get the trivial solution. The third and last case is lambda less than zero.

The eigenvalues are imaginary numbers. In order to have a better overview, you can define alpha, so that applies

The result is the following form of the solution,

in which we use the boundary conditions again.

results in zero.

can be chosen arbitrarily as long as alpha is a natural number n.

This finally results in a nontrivial solution, even several, since you can use different n. You can write the total solution as a sum.

Since we are now know,

we can also insert it in the approach to T.

The overall solution u looks like this:

Now we put in the initial condition

By comparing coefficients you now know that and all the others are.

Your final solution looks like this:

Make sure that there is no n, , or C occurs more. Because you should have determined all of them beforehand. Now you know how to solve the heat conduction equation and how to deal with a Sturm-Liouville eigenvalue problem.

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