# What does degrees of freedom (DOFs) mean in COMSOL Multiphysics?

 Solution Number: 875 Title: What does degrees of freedom (DOFs) mean in COMSOL Multiphysics? Platform: All Platforms Applies to: All Products Versions: All versions Categories: Solver, Physics Keywords: DOF, degree of freedom

## Problem Description

What does degrees of freedom (DOFs) mean in COMSOL Multiphysics?

## Solution

The solution time and memory requirements are strongly related to the number of degrees of freedom in the model. It is often desirable to be able to estimate the number of degrees of freedom based on the number of elements in the model.

For most physics interfaces (or application modes in 3.5a) each dependent variable is present in all nodes in the mesh. This means that the number of degrees of freedom is given by the number of nodes multiplied by the number of dependent variables. The relation between the number of nodes and the number of elements depends on the order of the elements and differs between 2D and 3D. The relation is only approximate since it depends on the ratio of the elements that lie on the boundary of the geometry. For thin geometries, where a large proportion of the elements lie on the boundary, the number of nodes per element is a bit higher.

The following are approximate relations between the number of nodes and the number of elements in 2D and 3D for Lagrange elements of different order. Quadrilateral (quad) meshes have roughly twice as many nodes as triangular meshes, and hexahedral (brick) meshes have about 6 times as many nodes as tetrahedral meshes.

### 2D

• Linear triangular elements: (#nodes) = 0.5 * (#elements)
• Linear quad elements: (#nodes) = 1 * (#elements)
• Quadratic triangular elements: (#nodes) = 2 * (#elements)
• Cubic triangular elements: (#nodes) = 4.5 * (#elements)
• Cubic quad elements: (#nodes) = 9 * (#elements)

### 3D

• Linear tetrahedral elements: (#nodes) = 0.2 * (#elements)
• Linear brick elements: (#nodes) = 1.2 * (#elements)
• Quadratic tetrahedral elements: (#nodes) = 1.4 * (#elements)
• Quadratic brick elements: (#nodes) = 8.5 * (#elements)
• Cubic tetrahedral elements: (#nodes) = 4.6 * (#elements)
• Cubic brick elements: (#nodes) = 28 * (#elements)

The total number of degrees of freedom is then given by:
(#degrees of freedom) = (#nodes) * (#dependent variables)

For more information, either search the Help (press F1 in COMSOL Multiphysics), or open and browse the electronic Documentation (press Ctrl F1 in COMSOL Multiphysics) and go to COMSOL Multiphysics Reference Manual > Studies and Solvers > Solution Utility Nodes > Compile Equations and also, The Statistics Page.

To find the Mesh statistics in COMSOL 4, right-click the Mesh node in the Model Builder and choose Statistics. The total number of degrees of freedom can be found in COMSOL 4.1 and later by right-clicking the node Study > Solver Configurations > Solver > Compile Equations and choosing Statistics.

In COMSOL 3.5a, the number of elements and the total number of degrees of freedom is shown in the Mesh Statistics dialog box.

Note that the number of degrees of freedom is not the only factor determining the memory requirements and the solution time of a problem. For more information on how to avoid running out of memory, please visit Knowledge Base entry 830

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