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I would like to solve the following partial differential equation

$$\begin{cases} u_{x}^{2}+u_{y}^{2} & =\dfrac{(1+\sin u)^{2}}{4a^{2}},\\ u|\varGamma & =0, \end{cases}$$

where $a\in\mathbb{R}-\{0\}$. The equation can be rewritten to

$$F(x,y,z,p,q)=p^{2}+q^{2}-\frac{(1+\sin z)^{2}}{4a^{2}}=0.$$

The characteristic equation is given by

\begin{align*} \frac{dx}{ds}(r,s) & =F_{p}=2p,\\ \frac{dy}{ds}(r,s) & =F_{q}=2q,\\ \frac{dz}{ds}(r,s) & =pF_{p}+qF_{q}=2p^{2}+2q^{2},\\ \frac{dp}{ds}(r,s) & =-F_{x}-pF_{z}=\frac{p}{2a^{2}}\cos z(1+\sin z),\\ \frac{dq}{ds}(r,s) & =-F_{y}-qF_{z}=\frac{q}{2a^{2}}\cos z(1+\sin z). \end{align*}

The parametrization $\Gamma$ is

\begin{align*} x(r,0) & =c\cos r,\\ x(r,0) & =c\sin r,\\ z(r,0) & =c^{2}(\cos^{2}r+\sin^{2}r)-\frac{(1+\sin z)^{2}}{4a^{2}}=0, \end{align*}

where $c=(1+\sin z)/(2a)$. We search for functions $\psi_{1}(r)$, $\psi_{2}(r)$ satisfying

\begin{align*} F(\gamma_{1}(r),\gamma_{2}(r),\phi(r),\psi_{1}(r),\psi_{2}(r)) & =0,\\ \phi^{\prime}(r) & =\psi_{1}(r)\gamma_{1}^{\prime}(r)+\psi_{2}(r)\gamma_{2}^{\prime}(r). \end{align*}

Hence

$$ \psi_{1}^{2}+\psi_{2}^{2}=c^{2},$$

and

$$0=\psi_{1}(r)c(-\sin r)c+\psi_{2}(r)c\cos r,$$

with the solutions

\begin{align*} (1)\psi_{1}(r) & =c\cos r\qquad\psi_{2}(r)=c\sin r,\\ (2)\psi_{1}(r) & =-c\cos r\qquad\psi_{2}(r)=-c\sin r. \end{align*}

For the first solution are

\begin{align*} p(r,s) & =c\cos r,\\ q(r,s) & =c\sin r, \end{align*}

and

\begin{align*} x(r,s) & =2cs\cos r+c\cos r=c\cos r(2s+1),\\ y(r,s) & =2cs\sin r+c\sin r=c\sin r(2s+1),\\ z(r,s) & =2cs(\cos^{2}r+\sin^{2}r)=2cs. \end{align*}

Then,

$$x^{2}+y^{2}=c^{2}(2s+1)=(z+c)^{2}.$$

Because $u(x,y)=z(r,s),$ then

\begin{align*} \left(u+c\right)^{2} & =x^{2}+y^{2},\\ u+\frac{1+\sin u}{2a} & =\sqrt{x^{2}+y^{2}}, \end{align*}

which represents the transcendental equation for $u$, but the solution is wrong. The correct result is

$$ u=\frac{\pi}{2}-2\arctan\frac{\sqrt{x^{2}+y^{2}}}{2a}.$$

Could I ask for your help with the solution of this PDE? Thank you very much...

Updated question:

In my opinion, the problem of the last solution was the substitution, because $c=f(z)$. Let us put

$$ c(z)=\sqrt{\frac{(1+\sin z)^{2}}{4a^{2}}}=\pm(1+\sin z)/(2a).$$

We parametrize curve $\Gamma$ by $r$ such that $\Gamma=\{c(r)\cos r,c(r)\sin r\}$, on $\Gamma$ is $u|\varGamma=0$.

\begin{align*} x(r,0) & =c(r)\cos r=\frac{1+\sin r}{2a}\cos r,\\ x(r,0) & =c(r)\sin r=\frac{1+\sin r}{2a}\sin r,\\ z(r,0) & =\frac{(1+\sin r)^{2}}{4a^{2}}(\cos^{2}r+\sin^{2}r)-\frac{(1+\sin r)^{2}}{4a^{2}}=0, \end{align*}

We search for functions $\psi_{1}(r)$, $\psi_{2}(r)$ satisfying

\begin{align*} F(\gamma_{1}(r),\gamma_{2}(r),\phi(r),\psi_{1}(r),\psi_{2}(r)) & =0,\\ \phi^{\prime}(r) & =\psi_{1}(r)\gamma_{1}^{\prime}(r)+\psi_{2}(r)\gamma_{2}^{\prime}(r). \end{align*}

Hence $$ \psi_{1}^{2}+\psi_{2}^{2}=c^{2},$$ and $$0=-\psi_{1}(r)\frac{\sin r+2\sin^{2}r-1}{2a}+\psi_{2}(r)\frac{\sin2r+\cos r}{2a}.$$

However, I am not sure, if I am not completely wrong with my assumptions and substitutions.

Thank for your help.

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  • $\begingroup$ What is the boundary condition? Without boundary condition they are an infinity of solutions and we cannot say which solution among them is what you call "the correct result". $\endgroup$ – JJacquelin Feb 24 '18 at 9:15
  • $\begingroup$ @JJacquelin: The boundary condition has been added,thanks for your comment. $\endgroup$ – justik Feb 24 '18 at 10:20
  • $\begingroup$ But they are still several typos ($x$ instead of $y$, $\:p$ instead of $q$, in several places and a power $2$ missing elsewhere). Also, what is $\Gamma$, where is it defined ? $\endgroup$ – JJacquelin Feb 24 '18 at 10:32
  • $\begingroup$ @JJacquelin: Thanks for your careful reading and corrections. The parametrization has been updated, some typos have been corrected.... $\endgroup$ – justik Feb 24 '18 at 11:10
  • $\begingroup$ I am afraid about a possible confusion between the boundary and the parametric curve $\Gamma$. $\Gamma$ is clearly defined, but the boundary condition is not yet defined. Of course, if no boundary condition is specified, you are allowed to define the same curve for the boundary as for the parametrisation. Why do you think that it is the boundary condition which was used in the textbook? If it is not the same, your result will be different from the result in the textbook, even if both results are correct. So, you cannot compare your result and the result from the textbook. $\endgroup$ – JJacquelin Feb 24 '18 at 11:21
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They are many typos in your calculus ($x$ instead of $y$, $\:p$ instead of $q$, in several places and a power $2$ missing elsewhere) which makes difficult to follow what is correct or not.

HINT : it seems that the initial mistake is here:

enter image description here

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  • $\begingroup$ Thanks for your comment. Taking into account $c=f(z)$ I am not sure whether the last solution was correct. See the updated question, please. Thanks. $\endgroup$ – justik Feb 24 '18 at 10:13

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