How to Solve a Nonlinear PDE with Sinh Function?

[Adri]

The PDE is $$ \frac{1}{a^2 x^2} (u_y)^2 - (u_x)^2 =1$$ I know the solution, its $u=x senh(ay)$, but I dont know how I can get it. I've tried variable separation and method of characteristics but they dont seem to work.

 

[jedishrfu]

If you look at how the sinh() function is defined using $(e^x - e^{-x})/2$ that might give you a clue to using $e^x$ in some manner.

https://en.wikipedia.org/wiki/Hyperbolic_functions

https://onlinehw.math.ksu.edu/math340book/chap4/euler.php

 

[pasmith]

Setting $u(x,y) = X(x)Y(y)$ gives $$\frac{X^2Y'^2}{a^2x^2} - X'^2Y^2 = 1.$$ You can separate this by taking $$\frac{X^2}{x^2} = X'^2 = k^2$$ for some constant $k$. This has solution $X = \pm kx$. This leaves $$k^2(\frac{Y'^2}{a^2} - Y^2) = 1$$ to which $Y(y) = \pm\frac{1}{k}\sinh ay$ is a standard solution. This gives $$u = \pm x \sinh ay.$$

EDIT: Better is to start from $$\frac{X^2}{x^2X'^2} \frac{Y'^2}{a^2} - Y^2 = \frac{1}{X'^2}.$$ Making this independent of $x$ requires $X'^2 = k^2$ and $X^2 = C^2x^2X'^2$ for some constants $k$, $C$. This has solution $X = \pm kCx = \pm \lambda x$. Then once again $$\frac{Y'^2}{a^2} - Y^2 = \frac1{\lambda^2}.$$

 

[renormalize]

The PDE is $$ \frac{1}{a^2 x^2} (u_y)^2 - (u_x)^2 =1$$ I know the solution, its $u=x senh(ay)$, but I dont know how I can get it. I've tried variable separation and method of characteristics but they dont seem to work.

Let me expand a bit on the approach of @pasmith to find some solutions that are more general.
Your PDE can be written in the factored form $\left(\frac{u_{y}}{ax}+u_{x}\right)\left(\frac{u_{y}}{ax}-u_{x}\right)=1$, or equivalently:$$\left(\frac{u_{y}}{ax}+u_{x}\right)=f,\;\left(\frac{u_{y}}{ax}-u_{x}\right)=f^{-1}\tag{1a,b}$$where $f$ is a function to be determined. These simultaneous equations can be solved for $u_x,u_y$:$$u_{x}=\frac{1}{2}\left(f-f^{-1}\right),\;u_{y}=\frac{ax}{2}\left(f+f^{-1}\right)\tag{2a,b}$$Differentiating (2a) w.r.t to $y$ and (2b) w.r.t. $x$, and equating the results, yields a PDE for $f$:$$\left(f^{2}+1\right)\left(af-f_{y}\right)+ax\left(f^{2}-1\right)f_{x}=0\tag{3}$$Any solution $f\left(x,y\right)$ of eq.(3) can be substituted into eqs.(2) to yield a solution $u\left(x,y\right)$ of your PDE by quadratures:$$u\left(x,y\right)=\frac{1}{2}\int dx\left(f-f^{-1}\right)=\frac{ax}{2}\int dy\left(f+f^{-1}\right)\tag{4}$$Of course, eq.(3) is not obviously any easier to solve in general than your original PDE. But it is tractable for the specific case that $f$ depends on a single variable, since eq.(3) then becomes a first-order ODE.
For example, set $f\left(x,y\right)=f\left(y\right)$ in eq.(3) to find $\left(f^{2}+1\right)\left(af-f_{y}\right)=0$, with solution $f\left(y\right)=k_{y}e^{ay}$, where $k_y$ is an arbitrary constant of integration. Using (4), one solution of your PDE is therefore:$$u\left(x,y\right)=\frac{x}{2}\left(k_{y}e^{ay}-k_{y}^{-1}e^{-ay}\right)\tag{5a}$$(Note that putting $k_y=1$ reduces this to the specific $x\sinh\left(ay\right)$ solution you gave in your post.) Alternatively, setting $f\left(x,y\right)=f\left(x\right)$ in eq.(3) gives $\left(f^{2}+1\right)f+x\left(f^{2}-1\right)f_{x}=0$ which is solved by $f\left(x\right)=\frac{1}{2k_{x}x}\left(1\pm\sqrt{1-4k_{x}^{2}x^{2}}\right)$. Then (4) gives the solution:$$u\left(x,y\right)=\frac{1}{2k_{x}}\left(ay\pm\left(\sqrt{1-4k_{x}^{2}x^{2}}-\tanh^{-1}\left(\sqrt{1-4k_{x}^{2}x^{2}}\right)\right)\right)\tag{5b}$$In summary, eqs.(5) represent two distinct solutions of your PDE, each of which involves an arbitrary constant of integration.

 

[pasmith]

Slightly simpler is to set $f = e^g$, so that $$u_y = ax \cosh g \quad u_x = \sinh g$$ with $$(g_y - a)\cosh g = ax g_x \sinh g.$$ Now with $g_x = 0$ we get $g = ay + c$ which leads to $$u = x\sinh(ay + c).$$ With $g_y = 0$ we have $$\cosh g = \frac kx$$ and $$u = aky \pm \int \sqrt{ \frac{k^2}{x^2} - 1 }\,dx.$$ By the method of characteristics we can obtain the general solution as $$\begin{split} x &= D(\eta)\operatorname{sech}(a\zeta + B(\eta)) \\ y &= \zeta + C(\eta) \\ g &= a \zeta + B(\eta) \end{split}$$ and $u$ can be found from $$\begin{split} u_\zeta &= x_\zeta \sinh g + ax y_\zeta \cosh g \\ &= -aD \tanh^2 g + a D \\ u_\eta &= x_\eta \sinh g + ax y_\eta \cosh g \\ &= -DB' \tanh^2 g + D' \tanh g +aDC'.\end{split}$$ which yields $$u = x \sinh (a \zeta + B(\eta)) + \int aC'(\eta)D(\eta)\,d\eta.$$