u_{xx}-u_{yy}+u_y=0

**dwsmith** · December 29th 2010, 07:53 PM

Using exponential substitution:

$\exp{(rx+sy)}$

$u_{xx}-u_{yy}+u_y=0$

$\exp{(rx+sy)}(r^2-s^2+s)=0$

$r=\pm\sqrt{s^2-s}$

$u=\exp{(\pm x\sqrt{s^2-s}+ys)}$
$\ \ s<img src=$ -\infty,0]\cup [1,\infty)" alt=" \ \ s

-\infty,0]\cup [1,\infty)" />

Can this be simplified more or is there a more tactful way to represent this?

**dwsmith** · December 29th 2010, 09:11 PM

The answer is this

$\displaystyle U=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}*\sin{(x\sqrt{\lambda})}$

$\displaystyle V=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}*\cos{(x\sqrt{\lambda})}$

Lambda is any constant.

I am not sure how this is the solution.

**Sudharaka** · December 29th 2010, 10:34 PM

Originally Posted by dwsmith

The answer is this

$\displaystyle U=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}*\sin{(x\sqrt{\lambda})}$

$\displaystyle V=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}*\cos{(x\sqrt{\lambda})}$

Lambda is any constant.

I am not sure how this is the solution.

Dear dwsmith,

Therefore the general solution would be,

$u=Ae^{(x\sqrt{s^2-s}+ys)}+Be^{(-x\sqrt{s^2-s}+ys)}$

Substitute, $\sqrt{s^2-s}=\sqrt{-\lambda}$

$s=\frac{1\pm\sqrt{1-4\lambda}}{2}$

$u=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}\left(Ae^{ix\sqrt{\lambda}}+ Be^{-ix\sqrt{\lambda}}\right)$

$u=\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}\left[(A+B)\cos x\sqrt{\lambda}+i(A-B)\sin x\sqrt{\lambda}\right)$

Take A+B=F and i(A-B)=G,

$u=F\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}\cos x\sqrt{\lambda}+G\exp{\left(\frac{1\pm\sqrt{1-4\lambda}}{2}y\right)}\sin x\sqrt{\lambda}\right)$

Hope you understood.

**dwsmith** · December 29th 2010, 10:39 PM

Originally Posted by Sudharaka

$s=\frac{1\pm\sqrt{1-4\lambda}}{2}$

How would you know to let s = that expression without knowing the answer?

**dwsmith** · December 29th 2010, 10:41 PM

Originally Posted by Sudharaka

Substitute, $\sqrt{s^2-s}=\sqrt{-\lambda}$

Why is this negative lambda?

**Sudharaka** · December 30th 2010, 04:39 AM

Originally Posted by dwsmith

Why is this negative lambda?

Dear dwsmith,

I have taken $\sqrt{s^2-s}=\sqrt{-\lambda}$ so that, I get required value for s;

$s^2-s=-\lambda\Rightarrow s^2-s+\lambda=0\Rightarrow{s=\frac{1\pm\sqrt{1-4\lambda}}{2}}$

If I didnt know the answer I would not present it as I had done in post #3. I would have given the answer as,

$u=Ae^{(x\sqrt{s^2-s}+ys)}+Be^{(-x\sqrt{s^2-s}+ys)}$

But since you had given thes answers in the book and told that you are not sure how they have been obtained I had shown that both anwers are equivelent. Does this clarify your doubts?

**dwsmith** · December 30th 2010, 12:33 PM

Now I am solving the same equation using separation of variables. However, I have hit a snag.

$u(x,y)=\varphi(x)\psi(y)$

$\varphi''(x)\psi(y)-\varphi(x)\psi''(y)+\varphi(x)\psi'(y)=0$

$\displaystyle \frac{\varphi''(x)\psi(y)-\varphi(x)\psi''(y)+\varphi(x)\psi'(y)}{\varphi(x) \psi(y)}=0\Rightarrow \frac{\varphi''(x)}{\varphi(x)}-\frac{\psi''(y)}{\psi(y)}+\frac{\psi'(y)}{\psi(y)} =0$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=\frac{\psi''(y)-\psi'(y)}{\psi(y)}$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda \ \ \frac{\psi''(y)-\psi'(y)}{\psi(y)}=\lambda$

$\displaystyle \varphi''(x)-\lambda\varphi(x)=0\Rightarrow m^2-\lambda=0$

$\displaystyle \psi''(y)-\psi'(y)-\lambda\psi(y)=0\Rightarrow n^2-n-\lambda=0$

The first equation is suppose to yield $s^2+\lambda$ , and the second equation needs to be of the form $s^2-s+\lambda$ in order to obtain the same solutions from the exponential method.

I haven't come across this in the book but can I just set both equations equal to $-\lambda\mbox{?}$

**Sudharaka** · December 30th 2010, 04:15 PM

Originally Posted by dwsmith

Now I am solving the same equation using separation of variables. However, I have hit a snag.

$u(x,y)=\varphi(x)\psi(y)$

$\varphi''(x)\psi(y)-\varphi(x)\psi''(y)+\varphi(x)\psi'(y)=0$

$\displaystyle \frac{\varphi''(x)\psi(y)-\varphi(x)\psi''(y)+\varphi(x)\psi'(y)}{\varphi(x) \psi(y)}=0\Rightarrow \frac{\varphi''(x)}{\varphi(x)}-\frac{\psi''(y)}{\psi(y)}+\frac{\psi'(y)}{\psi(y)} =0$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=\frac{\psi''(y)-\psi'(y)}{\psi(y)}$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda \ \ \frac{\psi''(y)-\psi'(y)}{\psi(y)}=\lambda$

$\displaystyle \varphi''(x)-\lambda\varphi(x)=0\Rightarrow m^2-\lambda=0$

$\displaystyle \psi''(y)-\psi'(y)-\lambda\psi(y)=0\Rightarrow n^2-n-\lambda=0$

The first equation is suppose to yield $s^2+\lambda$ , and the second equation needs to be of the form $s^2-s+\lambda$ in order to obtain the same solutions from the exponential method.

I haven't come across this in the book but can I just set both equations equal to $-\lambda\mbox{?}$

Dear dwsmith,

Yes you can. What is the reasoning behind setting $\displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda ~and~ \frac{\psi''(y)-\psi'(y)}{\psi(y)}=\lambda$ ?? It is that the two parts are independent of y and x respectively. So that if they are equal to each other there must be a common constant to which both are equal. In this case you had taken it to be $\lambda$ . But you can take the constant as you like it to be. For example instead of $-\lambda$ you can take $\alpha$ if you dont like taking a the negative sign. Then there exists $\lambda$ such that $\alpha=-\lambda$ . Substitute for $\alpha$ and you will get the same answer.

Hope you got the idea. I am a non native speaker of english. So if you find anything that confuses you please do not hesitate to ask me.

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