Toronto Math Forum

$u$ can be written as a sum of 3 homogeneous harmonic polynomials.
\begin{equation}
u = -\frac{1}{5}(z^3-3zx^2) -\frac{1}{5}(z^3-3zy^2) + \frac{2}{5}z
\end{equation}

We look for solutions such that $u_{tt}-u_{xx}=u(1-2u^2)=0$.
$u_{tt}-u_{xx}=0$ implies that $u(x,t) = f(x \pm t)$ for some function $f$.
$u(1-2u^2)=0$ implies that either $u = 0$ or $u = \pm \frac{1}{\sqrt 2}$.

A kink may be described by
\begin{equation}
u(x,t) = \left\{\begin{array}{21}
 &\pm \frac{1}{\sqrt 2} \qquad & x \ge t\\
 & 0 & x < t \end{array} \right.
\end{equation}

A soliton may be described by
\begin{equation}
u(x,t) = \left\{\begin{array}{21}
 &\pm \frac{1}{\sqrt 2} \qquad & x = t\\
 & 0 & x \neq t \end{array} \right.
\end{equation}

Substitute the partial derivatives of $u(x,y)=X(x)Y(y)$ into the pde and divide by $X^{(2)}Y^{(2)}$.

\begin{equation}
\frac{\frac{X^{(4)}}{X^{(2)}}}{\frac{Y^{(2)}}{Y}} + 2 + \frac{\frac{Y^{(4)}}{Y^{(2)}}}{\frac{X^{(2)}}{X}}=0
\end{equation}

One set of solutions can be obtained by assuming the following:
\begin{equation}
\frac{X^{(4)}}{X^{(2)}} = \frac{X^{(2)}}{X} = \omega^2 \qquad \frac{Y^{(4)}}{Y^{(2)}} = \frac{Y^{(2)}}{Y} = -\omega^2
\end{equation}

where $\omega > 0$. In this case, the solution is as follows:
\begin{equation}
u(x,y)=X(x)Y(y)=(A\cosh\omega x + B\sinh\omega x) (C\cos\omega y + D\sin\omega y)
\end{equation}

Similarly, another solution is as follows:
\begin{equation}
u(x,y)=X(x)Y(y)=(A\cos\omega x + B\sin\omega x) (C\cosh\omega y + D\sinh\omega y)
\end{equation}

Let $y = x - L$. The problem can be written as follows.

\begin{equation}
u_{tt} + Ku_{yyyy} = 0
\end{equation}

\begin{equation}
u(-L,t) = u_y(-L,t) = u(L,t) = u_y(L,t) = 0 \label{BC}
\end{equation}

Let $u(y,t) = Y(y)T(t)$ and apply separation of  variables.
\begin{equation}
\frac{Y^{(4)}(y)}{Y(y)} = \omega^4 \label{y}
\end{equation}

The general solution to (\ref{y}) is
\begin{equation}
Y(y) = A\cosh(\omega y) + B\sinh(\omega y) + C\cos(\omega y) + D\sin(\omega y)
\end{equation}

Boundary conditions (\ref{BC}) imply

\begin{align}
Y(-L) &= A\cosh(\omega L) - B\sinh(\omega L) + C\cos(\omega L) - D\sin(\omega L) = 0\\
Y(L) &= A\cosh(\omega L) + B\sinh(\omega L) + C\cos(\omega L) + D\sin(\omega L) = 0 \\
Y'(-L) &= -A\sinh(\omega L) + B\cosh(\omega L) + C\sin(\omega L) + D\cos(\omega L) = 0 \\
Y'(L) &= A\sinh(\omega L) + B\cosh(\omega L) - C\sin(\omega L) + D\cos(\omega L) = 0
\end{align}

This $4\times4$ system is equivalent to two $2\times2$ systems:

\begin{equation}
 \left[\begin{array}{cc} \sinh(\omega L) & -\sin(\omega L) \\ \cosh(\omega L) & \cos(\omega L) \end{array} \right] \left[\begin{array}{cc} A \\ C \end{array} \right] \equiv \Delta_2 \left[\begin{array}{cc} A \\ C \end{array} \right] = 0
\end{equation}

\begin{equation}
 \left[\begin{array}{cc} \sinh(\omega L) & \sin(\omega L) \\ \cosh(\omega L) & \cos(\omega L) \end{array} \right] \left[\begin{array}{cc} B \\ D \end{array} \right] \equiv \Delta_1 \left[\begin{array}{cc} B \\ D \end{array} \right] = 0
\end{equation}

For any given $\omega \in {\rm I\!R}$, $\Delta_1=\Delta_2=0$ iff $\omega = 0$. Since $0$ is not an eigenvalue, every eigenvalue corresponds to either an even or an odd eigenfunction. For odd eigenfunctions, the coefficients $[B \space D]'$ belong to the null space of $\Delta_1$. For even eigenfunctions, the coefficients $[A \space C]'$ belong to the null space of $\Delta_2$. Because both of these 2 null spaces are 1-dimensional, eigenvalues are simple. All eigenfunctions corresponding to the same eigenvalue are proportional.

A general odd eigenfunction can be written as follows:

\begin{equation}
Y(y) = B\sin(\omega_{odd} L)\sinh(\omega_{odd} y) - B\sinh(\omega_{odd} L) \sin(\omega_{odd} y)
\end{equation}

where $\omega_{odd}$ is a solution to $\tanh(\omega L) = \tan(\omega L)$.


A general even eigenfunction can be written as follows:

\begin{equation}
Y(y) = A\cos(\omega_{even} L)\cosh(\omega_{even} y) - A\cosh(\omega_{even} L) \cos(\omega_{even} y)
\end{equation}

where $\omega_{even}$ is a solution to $\tanh(\omega L) = - \tan(\omega L)$.

Thanks Prof. I have revised the solution.

Transform the problem into the first quadrant of the characteristic coordinates $(\xi,\eta)$.
\begin{align}
-4c^2\tilde{u}_{\xi \eta} &= \tilde{f}(\xi, \eta) && \xi > 0, \eta > 0 \label{a}  \\
\tilde{u}|_{\xi=0} &= \tilde{g} \left( -\frac{\eta}{2c} \right) && \eta > 0  \\
\tilde{u}|_{\eta=0} &= \tilde{h} \left( \frac{\xi}{2c} \right)   && \xi > 0
\end{align}
The solution to $(\ref{a})$ is as follows.
\begin{align}
\tilde{u}(\xi,\eta)= -\frac{1}{4c^2} \int_0 ^\xi
\int_0 ^\eta \tilde{f}(\xi',\eta' )\,d\eta' d\xi' + \psi(\eta) + \phi(\xi)
\end{align}
The domain of dependence is a rectangle defined as $\tilde{R}(\xi,\eta) = \{ (\xi',\eta') \vert 0< \xi' < \xi,\, 0< \eta' < \eta\}$.

Assume $\psi(0) = \phi(0) = \frac{1}{2}g(0) = \frac{1}{2}h(0)$.
\begin{align}
\phi(\xi) &= h \left( \frac{\xi}{2c} \right) - \frac{1}{2}h(0)\\
\psi(\eta) &= g \left( -\frac{\eta}{2c} \right) - \frac{1}{2}g(0)
\end{align}

Translate back to the $(x,t)$ coordinates. Use the hint provided and the fact that the Jacobian is equal to $2c$.
\begin{align}
u(x,t) = -\frac{1}{2c}\iint _{R(x,t)} f(x',t')\,dx'dt' + h \left( \frac{x+ct}{2c} \right) + g \left( -\frac{x-ct}{2c} \right) - h(0)
\end{align}
where $R(x,t)=\{ (x',t'):\, 0< x'-ct' < x-ct,\, 0< x'+ct' < x+ct\}$.

I think so. You probably will have 2 equations for the positive and negative square roots.

Problem 5a
$$yu_x-xu_y=x\\
\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x}\\
x^2+y^2 = C\\
u = -y + \phi(x^2+y^2)$$

Problem 5b
$$yu_x-xu_y=x^2\\
\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x^2}\\
x^2+y^2 = C\\
\left\{\begin{array}{2}
x = r\cos\theta\\
y = r\sin\theta\\
\end{array}\right.\\
du = -xdy = -r^2\cos^2\theta d\theta\\
u = - \frac{r^2}{2} \left(\theta + \frac{1}{2}\sin2\theta\right) + \psi(r)$$

Problem 5c. I think there is a typo in equation (14). The coefficient on $u_x$ should be $y$. Yes indeed. V.I.
$$yu_x+xu_y=x\\
\frac{dx}{y} = \frac{dy}{x} = \frac{du}{x}\\
x^2-y^2 = C\\
u = y + \zeta(x^2-y^2)$$

Problem 5d
$$yu_x+xu_y=x^2\\
\frac{dx}{y} = \frac{dy}{x} = \frac{du}{x^2}\\
x^2-y^2 = C,\qquad -\infty < C < \infty$$

For $0 < C < \infty$, let $C = r^2$.
$$\left\{\begin{array}{2}x = r\sec\theta\\
y = r\tan\theta\\
\end{array}\right.\\
du = xdy = r^2\sec^3\theta d\theta\\
u =  \frac{r^2}{2} \left(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|\right) + \psi(r)$$

For $-\infty < C < 0$, let $C = -r^2$.
$$\left\{\begin{array}{2}
x = r\tan\theta\\
y = r\sec\theta\\
\end{array}\right.\\
du = xdy = r^2\tan^2\theta\sec\theta d\theta = r^2\left(\sec^3\theta - \sec\theta\right)d\theta\\
u =  \frac{r^2}{2} \left(\sec\theta \tan\theta - \ln|\sec\theta+\tan\theta|\right) + \phi(r)$$

At ${x = \pm y}$, $u$ is undefined and discontinuous.