Computing H1 norm numerically












1












$begingroup$


I'm solving a PDE numerically using FDM and Spectral Methods.



I understand how to compute the $L_{2}$, but I dont understand how to compute the $H_{1}$ norm.



What does the $u'$ mean in the below definition?



$E_{H^1}=||u_{h}-u_{true}||=Error_{L^2} +Big(sumlimits_{i=1}^nint_{x_{i-1}}^{x}big({u'_{h}-u'_{true}}big)^2Big)^{1/2}$






numerical-methods norm







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$endgroup$












  • $begingroup$
    $u'$ is the (spatial) derivative of $u$. This comes directly from the definition of the Sobolev $H^1$ norm: en.wikipedia.org/wiki/…
    $endgroup$
    – Erick Wong
    Feb 21 '14 at 3:40










  • $begingroup$
    haha...oops how would you actually compute that in matlab?
    $endgroup$
    – Rambi
    Feb 21 '14 at 4:05
















1












$begingroup$


I'm solving a PDE numerically using FDM and Spectral Methods.



I understand how to compute the $L_{2}$, but I dont understand how to compute the $H_{1}$ norm.



What does the $u'$ mean in the below definition?



$E_{H^1}=||u_{h}-u_{true}||=Error_{L^2} +Big(sumlimits_{i=1}^nint_{x_{i-1}}^{x}big({u'_{h}-u'_{true}}big)^2Big)^{1/2}$






numerical-methods norm







share|cite|improve this question









$endgroup$












  • $begingroup$
    $u'$ is the (spatial) derivative of $u$. This comes directly from the definition of the Sobolev $H^1$ norm: en.wikipedia.org/wiki/…
    $endgroup$
    – Erick Wong
    Feb 21 '14 at 3:40










  • $begingroup$
    haha...oops how would you actually compute that in matlab?
    $endgroup$
    – Rambi
    Feb 21 '14 at 4:05














1












1








1


1



$begingroup$


I'm solving a PDE numerically using FDM and Spectral Methods.



I understand how to compute the $L_{2}$, but I dont understand how to compute the $H_{1}$ norm.



What does the $u'$ mean in the below definition?



$E_{H^1}=||u_{h}-u_{true}||=Error_{L^2} +Big(sumlimits_{i=1}^nint_{x_{i-1}}^{x}big({u'_{h}-u'_{true}}big)^2Big)^{1/2}$






numerical-methods norm







share|cite|improve this question









$endgroup$




I'm solving a PDE numerically using FDM and Spectral Methods.



I understand how to compute the $L_{2}$, but I dont understand how to compute the $H_{1}$ norm.



What does the $u'$ mean in the below definition?



$E_{H^1}=||u_{h}-u_{true}||=Error_{L^2} +Big(sumlimits_{i=1}^nint_{x_{i-1}}^{x}big({u'_{h}-u'_{true}}big)^2Big)^{1/2}$






numerical-methods norm




numerical-methods norm






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 21 '14 at 3:24









RambiRambi

185




185












  • $begingroup$
    $u'$ is the (spatial) derivative of $u$. This comes directly from the definition of the Sobolev $H^1$ norm: en.wikipedia.org/wiki/…
    $endgroup$
    – Erick Wong
    Feb 21 '14 at 3:40










  • $begingroup$
    haha...oops how would you actually compute that in matlab?
    $endgroup$
    – Rambi
    Feb 21 '14 at 4:05


















  • $begingroup$
    $u'$ is the (spatial) derivative of $u$. This comes directly from the definition of the Sobolev $H^1$ norm: en.wikipedia.org/wiki/…
    $endgroup$
    – Erick Wong
    Feb 21 '14 at 3:40










  • $begingroup$
    haha...oops how would you actually compute that in matlab?
    $endgroup$
    – Rambi
    Feb 21 '14 at 4:05
















$begingroup$
$u'$ is the (spatial) derivative of $u$. This comes directly from the definition of the Sobolev $H^1$ norm: en.wikipedia.org/wiki/…
$endgroup$
– Erick Wong
Feb 21 '14 at 3:40




$begingroup$
$u'$ is the (spatial) derivative of $u$. This comes directly from the definition of the Sobolev $H^1$ norm: en.wikipedia.org/wiki/…
$endgroup$
– Erick Wong
Feb 21 '14 at 3:40












$begingroup$
haha...oops how would you actually compute that in matlab?
$endgroup$
– Rambi
Feb 21 '14 at 4:05




$begingroup$
haha...oops how would you actually compute that in matlab?
$endgroup$
– Rambi
Feb 21 '14 at 4:05










1 Answer
1






active

oldest

votes


















0












$begingroup$

The finite difference method computes a point-wise approximation of $u_{mathrm{true}}$. You have to do a sort of post-processing of the FDM approximation $u_h$ for which you can compute/approximate its derivative. E.g., in 1D, it is reasonable to reconstruct a $tilde{u}_h$ which is linear on each interval such that $tilde{u}_h(x_i)=u_h(x_i)$ in the point $x_i$ of the discretisation. Such a function you can already differentiate.



For computing the error, I would suggest you to compute the integrals $int_{x_{i-1}}^{x_i}(tilde{u}_h'(x)-u_{mathrm{true}}(x)')^2mathrm{d}x$ on a reasonable discretisation of each interval $[x_{i-1},x_i]$ in order to diminish the quadrature error due to the nonlinearity of $u_{mathrm{true}}$. Of course the same is true also for the $L_2$-norm.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    sounds good! Could you apply the same approach to finding $H_{1}$ numerically for a spectral method with a periodic domain?
    $endgroup$
    – Rambi
    Feb 21 '14 at 14:34










  • $begingroup$
    I'm not very familiar with the spectral method, but I guess that yes except that in that case you don't need to do any post-processing since the solution is already expressed in terms of the spectral basis (if I'm not wrong).
    $endgroup$
    – Algebraic Pavel
    Feb 21 '14 at 14:43











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

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votes









0












$begingroup$

The finite difference method computes a point-wise approximation of $u_{mathrm{true}}$. You have to do a sort of post-processing of the FDM approximation $u_h$ for which you can compute/approximate its derivative. E.g., in 1D, it is reasonable to reconstruct a $tilde{u}_h$ which is linear on each interval such that $tilde{u}_h(x_i)=u_h(x_i)$ in the point $x_i$ of the discretisation. Such a function you can already differentiate.



For computing the error, I would suggest you to compute the integrals $int_{x_{i-1}}^{x_i}(tilde{u}_h'(x)-u_{mathrm{true}}(x)')^2mathrm{d}x$ on a reasonable discretisation of each interval $[x_{i-1},x_i]$ in order to diminish the quadrature error due to the nonlinearity of $u_{mathrm{true}}$. Of course the same is true also for the $L_2$-norm.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    sounds good! Could you apply the same approach to finding $H_{1}$ numerically for a spectral method with a periodic domain?
    $endgroup$
    – Rambi
    Feb 21 '14 at 14:34










  • $begingroup$
    I'm not very familiar with the spectral method, but I guess that yes except that in that case you don't need to do any post-processing since the solution is already expressed in terms of the spectral basis (if I'm not wrong).
    $endgroup$
    – Algebraic Pavel
    Feb 21 '14 at 14:43
















0












$begingroup$

The finite difference method computes a point-wise approximation of $u_{mathrm{true}}$. You have to do a sort of post-processing of the FDM approximation $u_h$ for which you can compute/approximate its derivative. E.g., in 1D, it is reasonable to reconstruct a $tilde{u}_h$ which is linear on each interval such that $tilde{u}_h(x_i)=u_h(x_i)$ in the point $x_i$ of the discretisation. Such a function you can already differentiate.



For computing the error, I would suggest you to compute the integrals $int_{x_{i-1}}^{x_i}(tilde{u}_h'(x)-u_{mathrm{true}}(x)')^2mathrm{d}x$ on a reasonable discretisation of each interval $[x_{i-1},x_i]$ in order to diminish the quadrature error due to the nonlinearity of $u_{mathrm{true}}$. Of course the same is true also for the $L_2$-norm.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    sounds good! Could you apply the same approach to finding $H_{1}$ numerically for a spectral method with a periodic domain?
    $endgroup$
    – Rambi
    Feb 21 '14 at 14:34










  • $begingroup$
    I'm not very familiar with the spectral method, but I guess that yes except that in that case you don't need to do any post-processing since the solution is already expressed in terms of the spectral basis (if I'm not wrong).
    $endgroup$
    – Algebraic Pavel
    Feb 21 '14 at 14:43














0












0








0





$begingroup$

The finite difference method computes a point-wise approximation of $u_{mathrm{true}}$. You have to do a sort of post-processing of the FDM approximation $u_h$ for which you can compute/approximate its derivative. E.g., in 1D, it is reasonable to reconstruct a $tilde{u}_h$ which is linear on each interval such that $tilde{u}_h(x_i)=u_h(x_i)$ in the point $x_i$ of the discretisation. Such a function you can already differentiate.



For computing the error, I would suggest you to compute the integrals $int_{x_{i-1}}^{x_i}(tilde{u}_h'(x)-u_{mathrm{true}}(x)')^2mathrm{d}x$ on a reasonable discretisation of each interval $[x_{i-1},x_i]$ in order to diminish the quadrature error due to the nonlinearity of $u_{mathrm{true}}$. Of course the same is true also for the $L_2$-norm.






share|cite|improve this answer









$endgroup$



The finite difference method computes a point-wise approximation of $u_{mathrm{true}}$. You have to do a sort of post-processing of the FDM approximation $u_h$ for which you can compute/approximate its derivative. E.g., in 1D, it is reasonable to reconstruct a $tilde{u}_h$ which is linear on each interval such that $tilde{u}_h(x_i)=u_h(x_i)$ in the point $x_i$ of the discretisation. Such a function you can already differentiate.



For computing the error, I would suggest you to compute the integrals $int_{x_{i-1}}^{x_i}(tilde{u}_h'(x)-u_{mathrm{true}}(x)')^2mathrm{d}x$ on a reasonable discretisation of each interval $[x_{i-1},x_i]$ in order to diminish the quadrature error due to the nonlinearity of $u_{mathrm{true}}$. Of course the same is true also for the $L_2$-norm.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Feb 21 '14 at 13:07









Algebraic PavelAlgebraic Pavel

16.3k31840




16.3k31840












  • $begingroup$
    sounds good! Could you apply the same approach to finding $H_{1}$ numerically for a spectral method with a periodic domain?
    $endgroup$
    – Rambi
    Feb 21 '14 at 14:34










  • $begingroup$
    I'm not very familiar with the spectral method, but I guess that yes except that in that case you don't need to do any post-processing since the solution is already expressed in terms of the spectral basis (if I'm not wrong).
    $endgroup$
    – Algebraic Pavel
    Feb 21 '14 at 14:43


















  • $begingroup$
    sounds good! Could you apply the same approach to finding $H_{1}$ numerically for a spectral method with a periodic domain?
    $endgroup$
    – Rambi
    Feb 21 '14 at 14:34










  • $begingroup$
    I'm not very familiar with the spectral method, but I guess that yes except that in that case you don't need to do any post-processing since the solution is already expressed in terms of the spectral basis (if I'm not wrong).
    $endgroup$
    – Algebraic Pavel
    Feb 21 '14 at 14:43
















$begingroup$
sounds good! Could you apply the same approach to finding $H_{1}$ numerically for a spectral method with a periodic domain?
$endgroup$
– Rambi
Feb 21 '14 at 14:34




$begingroup$
sounds good! Could you apply the same approach to finding $H_{1}$ numerically for a spectral method with a periodic domain?
$endgroup$
– Rambi
Feb 21 '14 at 14:34












$begingroup$
I'm not very familiar with the spectral method, but I guess that yes except that in that case you don't need to do any post-processing since the solution is already expressed in terms of the spectral basis (if I'm not wrong).
$endgroup$
– Algebraic Pavel
Feb 21 '14 at 14:43




$begingroup$
I'm not very familiar with the spectral method, but I guess that yes except that in that case you don't need to do any post-processing since the solution is already expressed in terms of the spectral basis (if I'm not wrong).
$endgroup$
– Algebraic Pavel
Feb 21 '14 at 14:43


















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