$u(x, y) = ax^3 + bxy$, here $a$ and $b$ are real-valued constants. Determine $a$ and $b$ so that the function $u$ is harmonic.

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$u(x, y) = ax^3 + bxy$, here $a$ and $b$ are real-valued constants. Determine $a$ and $b$ so that the function $u$ is harmonic.



I really don't know. Since when I derive, '$y$' just disappear.




complex-analysis harmonic-functions




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    please consider using Latex or MathJax to format your answer.
    – Ahmad Bazzi
    Aug 3 at 3:10














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enter image description here



a



$u(x, y) = ax^3 + bxy$, here $a$ and $b$ are real-valued constants. Determine $a$ and $b$ so that the function $u$ is harmonic.



I really don't know. Since when I derive, '$y$' just disappear.




complex-analysis harmonic-functions




share|cite|improve this question

















  • 2




    please consider using Latex or MathJax to format your answer.
    – Ahmad Bazzi
    Aug 3 at 3:10












up vote
0
down vote

favorite









up vote
0
down vote

favorite











enter image description here



a



$u(x, y) = ax^3 + bxy$, here $a$ and $b$ are real-valued constants. Determine $a$ and $b$ so that the function $u$ is harmonic.



I really don't know. Since when I derive, '$y$' just disappear.




complex-analysis harmonic-functions




share|cite|improve this question













enter image description here



a



$u(x, y) = ax^3 + bxy$, here $a$ and $b$ are real-valued constants. Determine $a$ and $b$ so that the function $u$ is harmonic.



I really don't know. Since when I derive, '$y$' just disappear.





complex-analysis harmonic-functions





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 3:30









user529760

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asked Aug 3 at 3:06









Carlos

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  • 2




    please consider using Latex or MathJax to format your answer.
    – Ahmad Bazzi
    Aug 3 at 3:10












  • 2




    please consider using Latex or MathJax to format your answer.
    – Ahmad Bazzi
    Aug 3 at 3:10







2




2




please consider using Latex or MathJax to format your answer.
– Ahmad Bazzi
Aug 3 at 3:10




please consider using Latex or MathJax to format your answer.
– Ahmad Bazzi
Aug 3 at 3:10










1 Answer
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I really don't know. Since when I derive, $y$ just disappear.




The second derivative $fracpartial^2 upartial y^2$ is, indeed, $0$. Nothing wrong with that! Just make sure that the same is true for $fracpartial^2 upartial x^2$ and you’re good to go.






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    very likely that there was a typo, the question, as it ought to have been typed, involved $u(x,y) = a x^3 + bxy^2$
    – Will Jagy
    Aug 3 at 4:09










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active

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up vote
2
down vote














I really don't know. Since when I derive, $y$ just disappear.




The second derivative $fracpartial^2 upartial y^2$ is, indeed, $0$. Nothing wrong with that! Just make sure that the same is true for $fracpartial^2 upartial x^2$ and you’re good to go.






share|cite|improve this answer

















  • 2




    very likely that there was a typo, the question, as it ought to have been typed, involved $u(x,y) = a x^3 + bxy^2$
    – Will Jagy
    Aug 3 at 4:09














up vote
2
down vote














I really don't know. Since when I derive, $y$ just disappear.




The second derivative $fracpartial^2 upartial y^2$ is, indeed, $0$. Nothing wrong with that! Just make sure that the same is true for $fracpartial^2 upartial x^2$ and you’re good to go.






share|cite|improve this answer

















  • 2




    very likely that there was a typo, the question, as it ought to have been typed, involved $u(x,y) = a x^3 + bxy^2$
    – Will Jagy
    Aug 3 at 4:09












up vote
2
down vote










up vote
2
down vote










I really don't know. Since when I derive, $y$ just disappear.




The second derivative $fracpartial^2 upartial y^2$ is, indeed, $0$. Nothing wrong with that! Just make sure that the same is true for $fracpartial^2 upartial x^2$ and you’re good to go.






share|cite|improve this answer














I really don't know. Since when I derive, $y$ just disappear.




The second derivative $fracpartial^2 upartial y^2$ is, indeed, $0$. Nothing wrong with that! Just make sure that the same is true for $fracpartial^2 upartial x^2$ and you’re good to go.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 3 at 3:35









Alon Amit

10.2k3665




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  • 2




    very likely that there was a typo, the question, as it ought to have been typed, involved $u(x,y) = a x^3 + bxy^2$
    – Will Jagy
    Aug 3 at 4:09












  • 2




    very likely that there was a typo, the question, as it ought to have been typed, involved $u(x,y) = a x^3 + bxy^2$
    – Will Jagy
    Aug 3 at 4:09







2




2




very likely that there was a typo, the question, as it ought to have been typed, involved $u(x,y) = a x^3 + bxy^2$
– Will Jagy
Aug 3 at 4:09




very likely that there was a typo, the question, as it ought to have been typed, involved $u(x,y) = a x^3 + bxy^2$
– Will Jagy
Aug 3 at 4:09












 

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