# pde satisfy with conditions

• Jun 21st 2008, 01:25 PM
jescavez
pde satisfy with conditions
Hello everybody.

I am new and I have one new question :)

i am trying these question, i must to show:
$y=log((x-a)^2+(z-b)^2))$

satisfies

$\frac{\partial^2y}{\partial{x^2}}+\frac{\partial^2 {y}}{\partial{z^2}}=0$

except at (a,b)

I differentiated twice for z and for x (chain rule and quotient) and put in equation to get:
$\frac{2x+2y+4(x-a)^2+4(z-b)^2}{((x-a)^2 +(z-b)^2)^2}$

I think it is ok and it does not work for a, b (as it goes to infinity) but is it enough to do this and say equation is satisfied? sorry if is obvious question! or have i done calculation wrong?
Thank you
• Jun 22nd 2008, 12:42 AM
flyingsquirrel
Hello
Quote:

Originally Posted by jescavez
[...]
or have i done calculation wrong?
[...]

Yes, I think you made a mistake when you differentiated $(x-a)^2$ with respect to $x$ and $(z-b)^2$ with respect to $z$.

Remember the chain rule : $(u\circ v)'=v'\times u'\circ v$. Here, $v(x)=x-a$ $\implies v'(x)=1$ and $u(x)=x^2$ $\implies u'(x)=2x$ :

$\frac{\partial }{\partial x}\left[(x-a)^2 \right]=v'(x)\times u'\circ v(x)= 1\cdot 2(x-a)=2(x-a)\neq 2x$

The same goes for $\frac{\partial }{\partial z}\left[(z-b)^2 \right]$.

Does it help ?
• Jun 22nd 2008, 02:20 PM
jescavez
Hello flying squirrel

thank you for the reply, yes I get same for $\frac{\partial }{\partial x}$ term but i need differentiate twice. then i multiply by $\frac{\partial{y}}{\partial{u}}=\frac{1}{((x-a)^2+(z-b)^2)}$ where $u=(x-a)^2+(z-b)^2$
and i have $\frac{2(x-a)}{(x-a)^2+(z-b)^2)}$ for dy/dx. i quotient rule this and have (i am correcting now as i notice a mistake before) $\frac{\partial^2{y}}{\partial{x^2}}=\frac{2((x-a)^2+(z-b)^2)+4(x-a)^2}{((x-a)^2+(z-b)^2)^2}$
do you think this is right now (i do same for z). so 2x term (and 4(x-a)^2) is from second derivative. so finish formula is:
$\frac{4((x-a)^2+(z-b)^2)+4(x-a)^2+4(z-b)^2}{((x-a)^2+(z-b)^2)^2}$
i think is right now but now for x=a and z=b it not fail.
what do you think?
• Jun 22nd 2008, 02:40 PM
flyingsquirrel
Quote:

Originally Posted by jescavez
Hello flying squirrel

thank you for the reply, yes I get same for $\frac{\partial }{\partial x}$ term but i need differentiate twice. then i multiply by $\frac{\partial{y}}{\partial{u}}=\frac{1}{((x-a)^2+(z-b)^2)}$ where $u=(x-a)^2+(z-b)^2$
and i have $\frac{2(x-a)}{(x-a)^2+(z-b)^2)}$ for dy/dx. i quotient rule this and have (i am correcting now as i notice a mistake before) $\frac{\partial^2{y}}{\partial{x^2}}=\frac{2((x-a)^2+(z-b)^2)+4(x-a)^2}{((x-a)^2+(z-b)^2)^2}$
do you think this is right now (i do same for z). so 2x term (and 4(x-a)^2) is from second derivative. so finish formula is:
$\frac{4((x-a)^2+(z-b)^2)+4(x-a)^2+4(z-b)^2}{((x-a)^2+(z-b)^2)^2}$
i think is right now but now for x=a and z=b it not fail.
what do you think?

I think that you forgot that the quotient rule tells us that $\left(\frac{f}{g}\right)'=\frac{f'g{\color{red}-}g'f}{g^2}$. :D

With this minus sign $\frac{4((x-a)^2+(z-b)^2)+4(x-a)^2+4(z-b)^2}{((x-a)^2+(z-b)^2)^2}$ becomes $\frac{4((x-a)^2+(z-b)^2){\color{red}-}4(x-a)^2{\color{red}-}4(z-b)^2}{((x-a)^2+(z-b)^2)^2}=0$ as expected.
• Jun 22nd 2008, 02:41 PM
Mathstud28
Quote:

Originally Posted by flyingsquirrel
I thinks that you forgot that the quotient rule tells us that $({\color{red}\bold{uv}})'=\frac{u'v{\color{red}-}v'u}{v^2}$. :D

With this minus sign $\frac{4((x-a)^2+(z-b)^2)+4(x-a)^2+4(z-b)^2}{((x-a)^2+(z-b)^2)^2}$ becomes $\frac{4((x-a)^2+(z-b)^2){\color{red}-}4(x-a)^2{\color{red}-}4(z-b)^2}{((x-a)^2+(z-b)^2)^2}=0$ as expected.

..
• Jun 23rd 2008, 01:16 PM
jescavez
Thank you flying squirrel and maths stud, I see now.

How do I do the thank you for a post?
• Jun 23rd 2008, 01:33 PM
flyingsquirrel
Quote:

Originally Posted by jescavez
Thank you flying squirrel and maths stud, I see now.

How do I do the thank you for a post?

Click on the button http://www.mathhelpforum.com/math-he...ost_thanks.gif of the post you want to... thank. :D (it's at the bottom right of the post)
• Jun 23rd 2008, 04:08 PM
jescavez
Ok. am i being blind i just see quote button no thanks button.
• Jun 24th 2008, 09:03 AM
flyingsquirrel
Quote:

Originally Posted by jescavez
Ok. am i being blind i just see quote button no thanks button.

The "thanks" button does not appear on your posts since you can't thank yourself (:D): try to find it on someone else's posts.
• Jun 24th 2008, 10:31 AM
jescavez
I was not logged in! thank you now :-)