# How Large is the Residual Error in FEA Solutions of PDEs?

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• feynman1
In summary, a PDE is solved by finite elements and then becomes a discrete system solved by Newton iteration. Every iteration step has a residual error, which can be estimated using Taylor's Theorem. When the solution is completed, the residual error of the PDE can be compared to the residual error of the finite element discrete system. To estimate the error, one can compute a second solution on a finer mesh and use Richardson extrapolation.
feynman1
A PDE is solved by finite elements. The PDE then becomes a discrete system solved by Newton iteration. Every iteration step comes with a residual error. When the solution is completed, how far is the residual error of the PDE from the residual error of the finite element discrete system?

feynman1 said:
A PDE is solved by finite elements. The PDE then becomes a discrete system solved by Newton iteration. Every iteration step comes with a residual error. When the solution is completed, how far is the residual error of the PDE from the residual error of the finite element discrete system?
I thought the residual error minimization and trial functions come into play deriving the element equations which are then solved by Gaussian Elimination.

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bob012345 said:
I thought the residual error minimization and trial functions come into play deriving the element equations which are then solved by Gaussian Elimination.
If the discrete system is nonlinear, either iteration is needed or linearization is to be done? Which is more commonly applied in software?

There is a theoretical bound on the discretization error (the difference bewtween the actual solution and the numerical solution).

If you can't solve the resulting algebraic system exactly, then this will be a further source of error. You can estimate this using Taylor's Theorem: if you are attempting to solve $f(x) = 0$ and the actual solution is $x_{c}$, then for the approximation $x_n$ you have $|f(x_n)| \approx |f'(x_n)||x_n - x_{c}|$ so that $|x_n - x_{c}| \approx |f(x_n)|/|f'(x_n)|$. You can then continue your iteration until this error is less than the theoretical maximum discretization error.

feynman1 said:
A PDE is solved by finite elements. The PDE then becomes a discrete system solved by Newton iteration. Every iteration step comes with a residual error. When the solution is completed, how far is the residual error of the PDE from the residual error of the finite element discrete system?
You have to distinguish between the solution of the discrete equations $u_d$ and the exact solution of the continuous equations $u$.
When using an iterative solver, you start with an estimate of $u_d$ and you update this estimate at every iteration, let's call this $u_d^{i}$. So at every iteration, you compute a new estimate of your discrete solution, and the error is $e_d = |u_d^{i} - u_d|$. The residual is what you get when you substitute your current estimate $u_d^{i}$ into your discretized PDE. If your numerical method converges, your residual goes to zero. The error and the residual are related: $|e| < C\cdot|R|$, and a lot of study has gone into getting accurate estimates of this constant C. So when the residual is zero, you can be pretty sure that you have found the solution of the discretized PDE (within machine precision). What remains is the error $e = |u_d - u|$. In general, you do not know this error, although we do know that the error goes to zero when the mesh element size goes to zero, and we also know that the rate of convergence is equal to the order of the numerical method. What you can do is compute a second solution on a finer mesh and compare this to the first solution.
You now have 2 solutions, one on the coarse mesh 1: $u_{d_1}$ on the fine mesh 2: $u_{d_2}$. An important observation is that if you interpolate the solution of mesh 1 (with residual R=0) to the finer mesh 2 and recompute the residual, the residual on the new mesh is not zero anymore.
If you have 3 meshes with 3 solutions, you can use Richardson extrapolation to get an estimate of the error.

feynman1
bigfooted said:
You have to distinguish between the solution of the discrete equations $u_d$ and the exact solution of the continuous equations $u$.
When using an iterative solver, you start with an estimate of $u_d$ and you update this estimate at every iteration, let's call this $u_d^{i}$. So at every iteration, you compute a new estimate of your discrete solution, and the error is $e_d = |u_d^{i} - u_d|$. The residual is what you get when you substitute your current estimate $u_d^{i}$ into your discretized PDE. If your numerical method converges, your residual goes to zero. The error and the residual are related: $|e| < C\cdot|R|$, and a lot of study has gone into getting accurate estimates of this constant C. So when the residual is zero, you can be pretty sure that you have found the solution of the discretized PDE (within machine precision). What remains is the error $e = |u_d - u|$. In general, you do not know this error, although we do know that the error goes to zero when the mesh element size goes to zero, and we also know that the rate of convergence is equal to the order of the numerical method. What you can do is compute a second solution on a finer mesh and compare this to the first solution.
You now have 2 solutions, one on the coarse mesh 1: $u_{d_1}$ on the fine mesh 2: $u_{d_2}$. An important observation is that if you interpolate the solution of mesh 1 (with residual R=0) to the finer mesh 2 and recompute the residual, the residual on the new mesh is not zero anymore.
If you have 3 meshes with 3 solutions, you can use Richardson extrapolation to get an estimate of the error.
What drives the iterations? Changing mesh size to check for convergence?

bigfooted said:
You have to distinguish between the solution of the discrete equations $u_d$ and the exact solution of the continuous equations $u$.
When using an iterative solver, you start with an estimate of $u_d$ and you update this estimate at every iteration, let's call this $u_d^{i}$. So at every iteration, you compute a new estimate of your discrete solution, and the error is $e_d = |u_d^{i} - u_d|$. The residual is what you get when you substitute your current estimate $u_d^{i}$ into your discretized PDE. If your numerical method converges, your residual goes to zero. The error and the residual are related: $|e| < C\cdot|R|$, and a lot of study has gone into getting accurate estimates of this constant C. So when the residual is zero, you can be pretty sure that you have found the solution of the discretized PDE (within machine precision). What remains is the error $e = |u_d - u|$. In general, you do not know this error, although we do know that the error goes to zero when the mesh element size goes to zero, and we also know that the rate of convergence is equal to the order of the numerical method. What you can do is compute a second solution on a finer mesh and compare this to the first solution.
You now have 2 solutions, one on the coarse mesh 1: $u_{d_1}$ on the fine mesh 2: $u_{d_2}$. An important observation is that if you interpolate the solution of mesh 1 (with residual R=0) to the finer mesh 2 and recompute the residual, the residual on the new mesh is not zero anymore.
If you have 3 meshes with 3 solutions, you can use Richardson extrapolation to get an estimate of the error.
thanks a lot and I wonder what could be the reason if |u-ud| doesn't decrease with refined meshes? If the true solution of the PDEs is of a highly oscillatory nature in space, would the residual of the PDEs (containing spatial derivatives) be hard to be suppressed?

if the solution is time dependent, then you have to converge the residuals every time step. If the residuals do not go down anymore, this is usually because for practical problems, you have a complicated geometry with an unstructured mesh, and the cells are not perfect squares or triangles.

feynman1
bigfooted said:
if the solution is time dependent, then you have to converge the residuals every time step. If the residuals do not go down anymore, this is usually because for practical problems, you have a complicated geometry with an unstructured mesh, and the cells are not perfect squares or triangles.
what if the residual of the discrete system yielding the ud solution goes down with iterations but f(ud) doesn't (f defined in #4 so f(uex)=0 where uex is exact)?

how is the "residual of the discrete system yielding $u_d$" different from $f(u_d)=R$?

bigfooted said:
how is the "residual of the discrete system yielding $u_d$" different from $f(u_d)=R$?
I'm not sure. f is some differential operator and it takes derivatives, whereas the discrete system has no derivative. Would the gibbs phenomenon make the discrete solution so bad on highly oscillatory regions that the residual will never be small?

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feynman1 said:
I'm not sure. f is some differential operator and it takes derivatives, whereas the discrete system has no derivative. Would the gibbs phenomenon make the discrete solution so bad on highly oscillatory regions that the residual will never be small?
?

how large is the residual of a discrete solution of a PDE from 0 in order of magnitude?

## 1. What is the residual of a PDE?

The residual of a PDE (Partial Differential Equation) is the difference between the exact solution of the PDE and the solution obtained through Finite Element Analysis (FEA). It is a measure of the error in the FEA solution and is typically represented by a vector or a scalar value.

## 2. How is the residual calculated in FEA?

The residual in FEA is calculated by subtracting the FEA solution from the exact solution of the PDE at each node or element in the finite element mesh. This results in a set of equations that can be solved to obtain the residual values at each node or element.

## 3. What does a large residual value indicate?

A large residual value indicates that the FEA solution is significantly different from the exact solution of the PDE. This could be due to various factors such as mesh size, element type, boundary conditions, or the choice of numerical methods used in the FEA analysis.

## 4. How can the residual be minimized in FEA?

The residual can be minimized in FEA by refining the finite element mesh, using higher-order elements, adjusting the boundary conditions, or using more accurate numerical methods. It is also important to ensure that the FEA model accurately represents the physical problem being analyzed.

## 5. Is a small residual value always desirable in FEA?

While a small residual value is generally desired in FEA, it is not always an accurate measure of the quality of the solution. In some cases, a small residual value may be achieved by using a highly refined mesh, but this may not necessarily result in a more accurate solution. It is important to consider other factors such as convergence, stability, and physical relevance of the FEA solution.

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