3
$\begingroup$

Having a look to my old PDE notes, I have come across with the following problem:

Consider the 2nd order PDE: $$ \varphi_{xx} - \varphi_{xy} = 0, \quad (x,y)\in \mathbb{R}^2, \quad \varphi = \varphi(x,y). $$

The solution of this problem is well known and can be computed in several many ways (characteristics, differential operators, common sense*, etc.):

$$\varphi(x,y) = F(y) + G(x+y),$$ where $F$ and $G$ are arbitrary functions of their respective arguments. But eventually my teacher claimed: $``$ since the coefficients of the PDE are constants, we can proceed as if we were dealing with ODEs and then guess solution of the form:

$$ \varphi(x,y) = A \exp(\alpha x+ \beta y), \quad A, \alpha, \beta \in \mathbb{C},$$

then, after substitution we come up with:

$$ (\alpha^2- \alpha \beta) \, \varphi = 0 $$

which holds for either $\varphi = 0$ (trivial solution) or $\alpha = 0$ or $\alpha = \beta$, for which we can deduce:

$$ \varphi(x,y) = A_1 \exp(\beta y) + A_2 \exp(\beta(x+y)) \color{blue}{\stackrel{\text{magic}}{=}} F(y) + G(x+y)."$$

What are the reasons behind the step before magic appears? I guess this has something to do with the fact that we can sweep on all possibles values of $\beta$ so we can therefore write something like:

$$ \int_B e^{ \beta (x+y) } \, \mathrm{d} \beta := F(x+y), $$ where $B$ is somewhat the set of all posibles values of $\beta$.

Can anybody enlight me about this? Any help will be greatly appreciated.

Cheers!

Edit: (another example)

Consider the Helmholtz $(k \neq 0)$ / Laplace $(k = 0)$ equation: $$z_{xx} + z_{yy} + k^2 z =0, \quad (x,y)\mathbb{R}^2, \quad z = z(x,y),$$ and $k$ is the wave number.

Again, assuming $z = A \exp{(\alpha x+ \beta y)}$ we arrive at: $$ (\alpha^2 + \beta^2 + k^2) z = 0, $$ which yields to $z=0$ or $\alpha = \pm \mathrm{i} \, \sqrt{\beta^2 + k^2}$. Then, magic takes part, but this time in a stronger manner, as my teacher claims:

$$ z(x,y) = \int^{\beta_2}_{\beta_1} A_1(\beta) \, e^{\beta y + \mathrm{i} \, \sqrt{\beta^2 + k^2} x} \, \mathrm{d \beta} + \int^{\beta_4}_{\beta_1} A_2(\beta) \, e^{\beta y - \mathrm{i} \, \sqrt{\beta^2 + k^2} x } \, \mathrm{d}\beta, $$

for $\beta_i$ such that $z$ satisfies the PDE. This time, no closed form of the solution was given but he points out that these integrals resemble somewhat a Fourier and Laplace transforms combination of the unknown functions $A_1$ and $A_2$. At this point, there is utter confusion.


$^*$ i.e., noting that the PDE can be rewritten as $(\varphi_x - \varphi_y)_x = 0$.

$\endgroup$
  • $\begingroup$ I'll be back within 5-6 hours so I will not be able to accept any answers or to answer comments till then. Thanks! $\endgroup$ – Dmoreno Aug 11 '14 at 14:15
2
$\begingroup$

It seems like something is missing. I mean, the answer your teacher gives is certainly a valid form of solution, but it is by no means the only one. Without boundary or initial conditions, the best you can do in this case is leave it in terms of arbitrary functions. For example, consider,

$F\left( y \right)=y,\; \; G\left( x+y \right)=\left( x+y \right)^{2}$.

A quick mental check will show that this is a valid solution to the PDE, and not in the form given by the teacher. Maybe he/she was giving an example of how you can take a solution that is not in the form $\varphi(x,y) = F(y) + G(x+y)$ and find the conditions for which it satisfies the PDE (in this case, when $\varphi=0, \alpha=\beta,$ or, $\alpha=0$.)

$\endgroup$
  • $\begingroup$ See the second example I have posted in my edit. Maybe that's the underlying idea of this approach but I still cannot see the relation. $\endgroup$ – Dmoreno Aug 12 '14 at 9:35
1
$\begingroup$

i think that its because $F(y)$ and $G(x+y)$ are arbitrary functions, since you get:

$$\varphi(x,y)=\color{red}{A_1\exp(\beta y)}+\color{blue}{A_2\exp[\beta(x+y)]}=\color{red}{F(y)}+\color{blue}{G(x+y)}$$

for the steps before, i think that because if $\varphi_1$ and $\varphi_2$ are two solutions of $\varphi_{xx}-\varphi_{xy}=0$, then $\varphi=\varphi_1+\varphi_2$ are also a solution, since $(\varphi_1+\varphi_2)_{xx}-(\varphi_1+\varphi_2)_{xy}=[(\varphi_{1})_{xx}-(\varphi_{1})_{xy}]+[(\varphi_{2})_{xx}-(\varphi_{2})_{xy}]=0$

$\endgroup$
  • $\begingroup$ Hi there, @cand. I still don't see how this two arbitrary functions came from the exponentials. Furthermore, doesn't your prove show that the PDE is linear, i.e., for any linear combination $a_1 \varphi_1 + a_2 \varphi_2$, then $\mathcal{L}[a_1 \varphi_1 + a_2 \varphi_2 ] = a_1 \mathcal{L}[\varphi_1] + a_2 \mathcal{L}[\varphi_2]$ is a solution of $\mathcal{L}[\varphi] = \varphi_{xx} - \varphi_{xy} = 0$, but nothing else? $\endgroup$ – Dmoreno Aug 11 '14 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.