2.1 Boundary conditions both Dirichlet and Homogeneous

2.1.1 Example 1

2.1.2 Example 2

2.1.3 Example 3

2.2 Boundary conditions both Dirichlet, one homogeneous other end nonHomogeneous

2.3 Boundary conditions both Neumann and homogeneous

2.4 Boundary conditions one end Neumann, the other nonHomogeneous Dirichlet

2.5 Boundary conditions at both ends homogeneous Neumann (insulated)

2.6 Boundary conditions at one end non-homogeneous Neumannm other end homogeneous Dirichlet

2.1.1 Example 1

2.1.2 Example 2

2.1.3 Example 3

2.2 Boundary conditions both Dirichlet, one homogeneous other end nonHomogeneous

2.3 Boundary conditions both Neumann and homogeneous

2.4 Boundary conditions one end Neumann, the other nonHomogeneous Dirichlet

2.5 Boundary conditions at both ends homogeneous Neumann (insulated)

2.6 Boundary conditions at one end non-homogeneous Neumannm other end homogeneous Dirichlet

Solve

Boundary conditions

Initial conditions

Solution.

A bar length cm with insulated sides is initially at degrees. starting at , the ends are held at zero degree. Find the temperature distribution in the bar at time

Boundary conditions

Initial conditions

Solution

Let

Substituting this into the PDE gives

Dividing by

Therefore

Where is some constant. Hence we obtain two ODE’s to solve. They are

And

case

From first boundary conditions, , the above becomes

Hence . From second boundary conditions , or , therefore trivial solution and is not an eigenvalue.

case

The solution is

From first B.C. and the solution becomes . From second B.C.

For non-trivial solution, we want or

Therefore

Now we solve the time ODE . This has solution

Therefore the solution becomes

To find , we use orthoginality. At , we are given . Hence at the solution becomes

Multiplying both sides by and integrating

Therefore

Therefore, the solution is

Here is an animation.

Problem

Solve with boundary conditions and initial conditions

Solution

Let . By separation of variables we obtain the following two ODEs. The spatial ODE is

And the time ODE

| (2) |

We can quickly see that is only possible case. Hence solution to (1) is

| (2A) |

Applying boundary conditions gives

Applying boundary conditions gives

| (4) |

(4),(3) give the system

| (5) |

For non-trivial solution we want

But , hence or for . Therefore

Each eigenvalue has associated eigenvector. For , (5) becomes

First equation gives . Hence and can be anything. Hence first eigenfunction from (2A) is

For (5) becomes

First equation gives . Hence and can be anything. Hence second eigenfunction from (2A) is

We continue this way and find that for the eigenfunctions are

And for the eigenfunctions are

Therefore the spatial solution is

The solution to the time domain ODE is , therefore the complete solution is

To find , we apply orthogonality. At . For case

To find , we apply orthogonality. At . For case

Hence the solution is

To verify

Example 1 here is solution for which satisfies boundary conditions. Using this we find

And

Hence analytical solution (6) is

The source code for Manipulate

Animation of the above

Example 2 which satisfies boundary conditions. Using this we find

And

Hence analytical solution (6) is

The source code for Manipulate

Animation of the above

Example 3

which satisfies boundary conditions. Using this we find

And

Hence analytical solution (6) is

The source code for Manipulate

Animation of the above

A bar length

Boundary conditions

Initial conditions

Solution,

A bar length

Boundary conditions, insulated

Initial conditions

Solution,

A bar length

Boundary conditions, left end only insulated

Initial conditions

Solution,

See my solution for exam 1, Math 322. UW, Fall 2016, problem 3.

Solve

with,

And initial conditions

Solution

Solution is not unique, since there is unknown . To find , the solvability condition for with Neumann B.C. is used, which says that total flux must be zero at steady state, which is the case here, since flux is zero as given (insulated). Since solvability condition is satisfied, then energy is conserved. Solution at from above is

Since energy is conserved then comparing it to energy at initial conditions

But , hence

Hence the final solution is

Using , here is animation for 3.5 seconds

This 1D heat PDE has one end with boundary condition that is time dependent.

Now let

Be some reference temperature distribution that only needs to satisfy the boundary conditions given. i.e. . Then the difference temperature distribution is

Since satisfy the nonhomogeneous B.C’s, then satisfies the homogeneous B.C which is . With the initial conditions

Since has homogeneous B.C. it can be easily solved. The solution can be found by separation of variables to be

Where eigenvalue . is now found from initial conditions.

Multiplying both sides by integrating and changing the order of integration and summation on RHS gives

By orthogonality of the above reduces to one term

Hence the solution is

But , therefore the final solution is

Here is animation using Mathematica, for for 2 seconds.