How to Solve a Nonlinear PDE with Sinh Function?

  • #1
Adri
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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.
 
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  • #3
Setting [itex]u(x,y) = X(x)Y(y)[/itex] gives [tex]
\frac{X^2Y'^2}{a^2x^2} - X'^2Y^2 = 1.[/tex] You can separate this by taking [tex]
\frac{X^2}{x^2} = X'^2 = k^2[/tex] for some constant [itex]k[/itex]. This has solution [itex]X = \pm kx[/itex]. This leaves [tex]
k^2(\frac{Y'^2}{a^2} - Y^2) = 1[/tex] to which [itex]Y(y) = \pm\frac{1}{k}\sinh ay[/itex] is a standard solution. This gives [tex]u = \pm x \sinh ay.[/tex]

EDIT: Better is to start from [tex]
\frac{X^2}{x^2X'^2} \frac{Y'^2}{a^2} - Y^2 = \frac{1}{X'^2}.[/tex] Making this independent of [itex]x[/itex] requires [itex]X'^2 = k^2[/itex] and [itex]X^2 = C^2x^2X'^2[/itex] for some constants [itex]k[/itex], [itex]C[/itex]. This has solution [itex]X = \pm kCx = \pm \lambda x[/itex]. Then once again [tex]
\frac{Y'^2}{a^2} - Y^2 = \frac1{\lambda^2}.[/tex]
 
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  • #4
Adri said:
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.
 
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  • #5
Slightly simpler is to set [itex]f = e^g[/itex], so that [tex]
u_y = ax \cosh g \quad u_x = \sinh g[/tex] with [tex]
(g_y - a)\cosh g = ax g_x \sinh g.[/tex] Now with [itex]g_x = 0[/itex] we get [itex]g = ay + c[/itex] which leads to [tex]u = x\sinh(ay + c).[/tex] With [itex]g_y = 0[/itex] we have [tex]\cosh g = \frac kx[/tex] and [tex]
u = aky \pm \int \sqrt{ \frac{k^2}{x^2} - 1 }\,dx.[/tex] By the method of characteristics we can obtain the general solution as [tex]
\begin{split}
x &= D(\eta)\operatorname{sech}(a\zeta + B(\eta)) \\
y &= \zeta + C(\eta) \\
g &= a \zeta + B(\eta) \end{split}[/tex] and [itex]u[/itex] can be found from [tex] \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}[/tex] which yields [tex]
u = x \sinh (a \zeta + B(\eta)) + \int aC'(\eta)D(\eta)\,d\eta.[/tex]
 
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