# Thread: pde satisfy with conditions

1. ## pde satisfy with conditions

Hello everybody.

I am new and I have one new question

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

satisfies

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

except at (a,b)

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

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

2. Hello
Originally Posted by jescavez
[...]
or have i done calculation wrong?
[...]
Yes, I think you made a mistake when you differentiated $\displaystyle (x-a)^2$ with respect to $\displaystyle x$ and $\displaystyle (z-b)^2$ with respect to $\displaystyle z$.

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

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

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

Does it help ?

3. Hello flying squirrel

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

4. Originally Posted by jescavez
Hello flying squirrel

thank you for the reply, yes I get same for $\displaystyle \frac{\partial }{\partial x}$ term but i need differentiate twice. then i multiply by $\displaystyle \frac{\partial{y}}{\partial{u}}=\frac{1}{((x-a)^2+(z-b)^2)}$ where $\displaystyle u=(x-a)^2+(z-b)^2$
and i have $\displaystyle \frac{2(x-a)}{(x-a)^2+(z-b)^2)}$ for dy/dx. i quotient rule this and have (i am correcting now as i notice a mistake before) $\displaystyle \frac{\partial^2{y}}{\partial{x^2}}=\frac{2((x-a)^2+(z-b)^2)+4(x-a)^2}{((x-a)^2+(z-b)^2)^2}$
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:
$\displaystyle \frac{4((x-a)^2+(z-b)^2)+4(x-a)^2+4(z-b)^2}{((x-a)^2+(z-b)^2)^2}$
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 $\displaystyle \left(\frac{f}{g}\right)'=\frac{f'g{\color{red}-}g'f}{g^2}$.

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

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

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

6. Thank you flying squirrel and maths stud, I see now.

How do I do the thank you for a post?

7. 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 of the post you want to... thank. (it's at the bottom right of the post)

8. Ok. am i being blind i just see quote button no thanks button.

9. 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 (): try to find it on someone else's posts.

10. I was not logged in! thank you now :-)