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The function is $f(x,y,z)=e^{ax+by}\cos(5z)$. I am supposed to find a and b given that it satisfies Laplace equation: $f_{xx}+f_{yy}+f_{zz}=0$

So, I found each derivative and replaced it back into the equation:

$\cos(5z)e^{ax+by}(a^2+b^2-25)=0$

So, all I can say is that $a^2+b^2=25$ which doesn't solve the problem. Am I missing something?

John
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  • why not consider $cos(5*z)=0$?if it is related to $a,b$ only say that equation is all circle with radius $5$ and center $0,0$ – dato datuashvili Nov 10 '13 at 09:04
  • https://www.google.ge/search?q=x%5E2%2By%5E2%3D25&tbm=isch&tbo=u&source=univ&sa=X&ei=u0x_UrCKL6e60QWI14HQDQ&ved=0CCUQsAQ&biw=1280&bih=619#facrc=_&imgrc=KoBKJ1kyG9EjRM%3A%3BxuXVPENRxRGUmM%3Bhttp%253A%252F%252Fwww.mathman.biz%252Fimages%252Fx2%252B4y2%253D25.jpg%3Bhttp%253A%252F%252Fwww.mathman.biz%252Fhtml%252Fgraphx2%252Bky2%253D25.html%3B2550%3B3509 – dato datuashvili Nov 10 '13 at 09:07
  • If we consider that cos(5z) is equal to zero then aren't we changing the initial function to zero? – John Nov 10 '13 at 09:15
  • first of all we are evaluating $cos(5*z)=0$ after derivative,so we are just trying to find ciritical points,but because they have asked for $a$ and $b$,it is not necessary in this time,sorry i have misread question,for the equation $x^2+y^2=25$ – dato datuashvili Nov 10 '13 at 09:20
  • points could be (3,4),(4,3),(-3,-4),(-3,4),(-4,3),(4,-3),(-4,-3),(0,5),(0,-5),(5,0),(-5,0) – dato datuashvili Nov 10 '13 at 09:21
  • So, the solution has multiple values--all of which that lie on the circle with radius 5 and center 0,0. – John Nov 10 '13 at 09:22
  • see update comment,it has finite solution in real numbers – dato datuashvili Nov 10 '13 at 09:23
  • How can there be a finite-number of solutions to $a^2+b^2=25$. Are there any restrictions I am missing? – John Nov 10 '13 at 09:25
  • ,look $a^2+b^2=25$ is all $a,b$ which satisfy this equation,they meant real numbers i think,so you have to take all integer which satisfy this equation,there of course exist complex numbers too,but you dont need it here,just can you tell me all numbers which sastisfy $x^2+y^2=25$? – dato datuashvili Nov 10 '13 at 09:30
  • I understand what you mean. You already stated all integer solutions. Thanks – John Nov 10 '13 at 09:34
  • see my answer,please – dato datuashvili Nov 10 '13 at 09:41
  • see please answer on your previous question – dato datuashvili Nov 10 '13 at 10:17

1 Answers1

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$f(x,y,z)=e^{ax+by}\cos(5z)$. let us find derivatives seperately

1.$f_{x}=a*e^{ax+by}\cos(5z)$

2.$f_{xx}=a^2*e^{ax+by}\cos(5z)$

3.$f_{yy}=b^2*e^{ax+by}\cos(5z)$

4.$f_{z}=-5*e^{ax+by}\sin(5z)$

5.$f_{zz}=-25*e^{ax+by}\cos(5z)$

not if we arrange similar terms we get the same as you got it

$\cos(5z)e^{ax+by}(a^2+b^2-25)=0$

clearly solution is all $a,b$ which satisfy this solution and they are

$(-3,-4)$,

$(-3,4)$,

$(-4,-3)$,

$(-4,3)$,

$(3,4)$,

$(4,3)$,

$(-5,0)$,

$(0,-5)$,

$(0,5)$,

$(5,0)$ that is all.good lucks

dato datuashvili
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  • Of course, as we said previosuly this is only if $a,b \in \mathbb{Z}$ – John Nov 10 '13 at 09:44
  • exactly,you are right,as a geometry yes that is all point which lies in circle with radius $5$ and center $(0,0)$ – dato datuashvili Nov 10 '13 at 09:45
  • I would really appreciate it if you could take a look at this problem as well: http://math.stackexchange.com/questions/550047/directional-and-partial-derivatives. Its been days now and no one has been able to answer it . I was wondering if you know how to tackle it? – John Nov 10 '13 at 09:48
  • that is chain rule – dato datuashvili Nov 10 '13 at 09:52
  • http://en.wikipedia.org/wiki/Chain_rule – dato datuashvili Nov 10 '13 at 09:53
  • so now ask again and that that if partial derivative of $x$ with respect to $z$ is something ,they how can i find partial derivative of $z$ with respect to $x$ – dato datuashvili Nov 10 '13 at 09:54
  • Can you show me hows its done using the chain rule? – John Nov 10 '13 at 09:54
  • sure suppose that f is function of u,which itself is function of x,then $f_{x}=f_{u}*u_{x}$ – dato datuashvili Nov 10 '13 at 09:55
  • If that is so then what is the significance of the gradient vector(not having any zero elements and the function being equal to 100? – John Nov 10 '13 at 09:56
  • it somehow will help you,because gradient vector is sum of its partial derivatives,also because this function is equal constant,its gradient equal to zero,so it somehow should be related to each other,sorry i am bussy now and cant go deeply in problem,try to repost this question again – dato datuashvili Nov 10 '13 at 09:59
  • Thanks for your help though :) – John Nov 10 '13 at 10:02
  • try it,it will help you and consider also chain rule principles,can you accept also my answer,just for increasing reputation :D,thanks my friend,by the way i will return this question and will post answer if nobody will answer it – dato datuashvili Nov 10 '13 at 10:03
  • i found answer already, i will pot it – dato datuashvili Nov 10 '13 at 10:11