1. ## u_{xx}-u_{yy}+u_y=0

Using exponential substitution:

$\displaystyle \exp{(rx+sy)}$

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

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

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

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

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

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

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

Lambda is any constant.

I am not sure how this is the solution.

3. Originally Posted by dwsmith

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

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

Lambda is any constant.

I am not sure how this is the solution.
Dear dwsmith,

Therefore the general solution would be,

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

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

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

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

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

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

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

Hope you understood.

4. Originally Posted by Sudharaka

$\displaystyle s=\frac{1\pm\sqrt{1-4\lambda}}{2}$
How would you know to let s = that expression without knowing the answer?

5. Originally Posted by Sudharaka

Substitute, $\displaystyle \sqrt{s^2-s}=\sqrt{-\lambda}$
Why is this negative lambda?

6. Originally Posted by dwsmith
Why is this negative lambda?
Dear dwsmith,

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

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

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,

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

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?

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

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

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

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

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

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

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

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

The first equation is suppose to yield $\displaystyle s^2+\lambda$, and the second equation needs to be of the form $\displaystyle s^2-s+\lambda$ 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 $\displaystyle -\lambda\mbox{?}$

8. Originally Posted by dwsmith
Now I am solving the same equation using separation of variables. However, I have hit a snag.

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

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

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

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

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

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

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

The first equation is suppose to yield $\displaystyle s^2+\lambda$, and the second equation needs to be of the form $\displaystyle s^2-s+\lambda$ 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 $\displaystyle -\lambda\mbox{?}$
Dear dwsmith,

Yes you can. What is the reasoning behind setting $\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda ~and~ \frac{\psi''(y)-\psi'(y)}{\psi(y)}=\lambda$ ?? 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 $\displaystyle \lambda$. But you can take the constant as you like it to be. For example instead of $\displaystyle -\lambda$ you can take $\displaystyle \alpha$ if you dont like taking a the negative sign. Then there exists $\displaystyle \lambda$ such that $\displaystyle \alpha=-\lambda$. Substitute for $\displaystyle \alpha$ 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.