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Consider the following first-order PDE $$yu_x+xu_y=0$$

subject to the initial condition $$u(x,0) = \exp \left(-\frac{x^2}{2}\right)$$ Show that the above problem has

  1. a unique solution in a neighbourhood of the point $(x_0,0)$ provided $x_0 \neq0,$

  2. infinitely many solution in a neighbourhood of origin.

Please someone give some hints. Thank you.

Community
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Mini_me
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2 Answers2

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Method 1: This problem can be done without computing the solution. We have the theorem (for example Theorem 1.1. in "Partial Differential Equations" by Prasad & Ravindran):

Let $x_0,y_0,u_0$ be $C^1$ in a closed interval and $a,b,c \in C^1$ in some domain of $(x,y,u)$-space containing the initial curve $\Gamma:s \mapsto (x_0(s),y_0(s),u_0(s))$. Further, let $$\begin{vmatrix} x_0' & y_0' \\ a & b \end{vmatrix}_\Gamma \neq 0.$$ Then there exists a unique solution $u=u(x,y)$ of the quasilinear equation $$a(x,y,u) u_x + b(x,y,u) u_y = c(x,y,u)$$ in the neighborhood of the curve $\gamma : s \mapsto (x_0(s),y_0(s))$ and satisfying $u_0(s)=u(x_0(s),y_0(s))$.

In your case we have $\Gamma :s \mapsto (s,0,e^{-s^2/2})$ hence $$\begin{vmatrix} x_0' & y_0' \\ x' & y'\end{vmatrix}_\Gamma=\begin{vmatrix} 1 & 0 \\ 0 & s \end{vmatrix}=s$$ and for $s\neq 0$ we have unique solution of the PDE. For $s=0$ we are in the case of a characteristic Cauchy problem and the solution of a characteristic IVP, if it exists, is nonunique.

Method 2: Using the method of characteristics and computing the solution directly:

$$\begin{cases} x'(t)=y \\ y'(t)=x \\ u'(t)=0 \end{cases} \text{ with } \begin{cases} x(0)=s \\ y(0)=0 \\ u(0)=\exp(-\tfrac{s^2}{2}) \end{cases}$$ Hence $x''=y'=x$ so this reduces to an ODE. Solving first for $x$, then for $y$ and plugging in the initial conditions we have $$ \begin{cases} x(t)&=\tfrac{s}{2}( e^{-t} + e^t) \\ y(t)&=\tfrac{s}{2}( -e^{-t}+e^t)\end{cases} $$ Further, we compute that $x^2-y^2=s^2$ and as $u'=0$ we have the solution $$u=u_0=e^{-\tfrac{s^2}{2}}=e^{-\tfrac{x^2-y^2}{2}}.$$ Next, we want to compute the existence domain of this solution. Computing $t$ and $s$ we have $t=\tfrac12 \ln\big( \tfrac{x+y}{x-y} \big)$ and $s=\sqrt{x^2-y^2}$ hence we have to impose the conditions $x^2-y^2>0$ and $\tfrac{x+y}{x-y}>0$ to get $t$ and $s$ well-defined. You see we will get problems around the origin.

Cahn
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3

Using separation of variables, let

$$u (x,y) = X (x) Y (y)$$

The PDE can then be broken into $2$ uncoupled ODEs

$$\begin{array}{rl} \dfrac{X ' (x)}{X (x)} &= \,\,\,\,\gamma \, x\\\\ \dfrac{Y ' (y)}{Y (y)} &= -\gamma \, y \end{array}$$

Thus, the general solution is of the form

$$u (x,y) = u_0 \exp \left( \frac{\gamma}{2} \left( x^2 - y^2 \right) \right)$$

Using the given initial condition, the values of constants $u_0$ and $\gamma$ can be determined.

Rodrigo de Azevedo
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