Dear Jeong, Thanks for testing my example, and pointing to the solution using \dm (which I didn’t know…). I think the correct behaviour of math alignments should avoid using additional commands such as \dm. We’ll see what Hans and Mikael say. Best regards: OK

On 2 Nov 2022, at 12:53, Jeong Dal via ntg-context wrote:

Dear Otared,

...

I think there is an issue with the size of math fonts in \frac when used in math alignments, as it is is shown in the following example.

I confirm what you said.

However, if I use \dm{} instead of \displaystyle as following, then the output is correct.

\placeformula \startformula \startalignedcases \NC - \Delta u \NC = 0 \NC \qquad\text{in }\, \Omega \NR \NC \dm{\frac{\partial u}{\partial {\bi n}}} \NC = V u + \phi \NC \qquad\text{on }\, \partial\Omega. \NR \stopalignedcases \stopformula or this (with\type{\displaystyle} added) \placeformula \startformula \startalignedcases \NC - \Delta u \NC = 0 \NC \qquad\text{in }\, \Omega \NR \NC \dm{\frac{\partial u}{\partial {\bi n}}} \NC = V u + \phi \NC \qquad\text{on }\, \partial\Omega. \NR \stopalignedcases \stopformula

In this example, it is OK to use \dm inside of \startformula … \stopformula. I am not sure that it is what Hans and Mikael want.

Thanks,

Best regards,

Dalyoung

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Otared Kavian e-mail: otared@gmail.com Phone: +33 6 88 26 70 95

Hi Mikael, Thanks for your reply. Indeed with your hint the issue disappears. I’ll change my definition… Best regards: Otared

On 2 Nov 2022, at 21:44, Mikael Sundqvist via ntg-context wrote: […] I think in this case it is better to use a mathsimplealign instead of a matrix.

\definemathsimplealign[pdeproblem][ left={\startmathfenced[cases]}, right=\stopmathfenced, align={1:right,2:left,3:left}, strut=yes, ]

\definemathsimplealign[collected][ left={\startmathfenced[sesac]}, right=\stopmathfenced, align={1:right,2:left,3:left}, strut=yes, ]

\starttext

\placeformula \startformula \startpdeproblem \NC - \Delta u \NC = 0 \NC \qquad\text{in } \Omega \NR \NC \frac{\partial u}{\partial {\bi n}} \NC = V u + \phi \NC \qquad\text{on } \partial\Omega \NR \stoppdeproblem \stopformula

\placeformula \startformula \startcollected \NC - \Delta u \NC = 0 \NC \qquad\text{in } \Omega \NR \NC \frac{\partial u}{\partial {\bi n}} \NC = V u + \phi \NC \qquad\text{on } \partial\Omega \NR \stopcollected \stopformula

\stoptext

Best, Mikael ___________________________________________________________________________________ If your question is of interest to others as well, please add an entry to the Wiki!

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Otared Kavian e-mail: otared@gmail.com Phone: +33 6 88 26 70 95