I've a little troubles in proving local uniqueness of solution for Cauchy problems concerning quasilinear PDE's. It's a little bit boring, but I tried to be as clear as possible.

Suppose $\Omega$ is an open and connected subset of $\mathbb{R}^2$ and let $a(x,y,z),b(x,y,z),c(x,y,z)$ scalar functions of class $C^1$ in $\Omega \times \mathbb{R}$. Let $I$ an open interval and $f=f(s)$, $g=g(s)$ and $h=h(s)$ be $C^1(I)$.

We want to prove local existence and uniqueness of a solution for the Cauchy problem $$\begin{cases} a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u)\\u(f(s),g(s))=h(s)\qquad s\in I\end{cases},$$ under certain conditions using method of characteristic.

Consider for each fixed $s\in I$ the autonomous system of ODE's $$\begin{cases}\frac{d}{dt}x=a(x,y,z)\\\frac{d}{dt}y=b(x,y,z)\\\frac{d}{dt}z=c(x,y,z)\end{cases}$$ with initial conditions $$\begin{cases}x(0)=f(s)\\y(0)=g(s)\\z(0)=h(s)\end{cases}.$$

Because $a,b,c$ are $C^1$, then for every $s \in I$ I find a unique maximal and global solution $x=X(s,t),y=Y(s,t),z=Z(s,t)$ defined on an open interval $J_s$ containing $0$ such that $X(s,0)=f(s)$, $Y(s,0)=g(s)$, $Z(s,0)=h(s)$.

Now consider the function $$(s,t)\longrightarrow (X(s,t),Y(s,t),Z(s,t))\qquad [1] $$ for $s \in I$, $t \in J_s$.

Fix $s_0 \in I$ and let us reason in a neighborhood of $(s_0,0)$, trying to define a good domain for $[1]$ and then talk about invertibility.

Luckily we can choose an interval $J$ independent of s (on which IVPs for characteristic equations have solutions) provided we are willing to restrict ourselves to an interval $I_0$ containing $s = s_0 $ instead of the entire interval I. Thus we may assume that the domain of the vector-valued function given in $[1]$ is $I_0 × J$. Thanks to the differentiable dependence of solutions to initial value problems for ODEs, the vector-valued function given in $[1]$ is continuously differentiable. Thus we are interested in the invertibility, near $(s, t) = (s_0,0)$, of the function $[1]$. Note that at the point $(s, t) = (s_0,0)$, the Jacobian of the function in $[1]$ is given by $$J=\left|\begin{matrix} X_s(s_0,0) &X_t(s_0,0)\\Y_s(s_0,0) &Y_t(s_0,0)\end{matrix}\right|=\left|\begin{matrix} f'(s_0)& a(f(s_0),g(s_0),h(s_0))\\g'(s_0) &b(f(s_0),g(s_0),h(s_0))\end{matrix}\right|.$$ Provided that $J\neq 0$ we can use inverse function theorem and state that exist a neighbourhood $U$ containing $(s_0,0)$ and a neighbourhood $W$ containing $(f(s_0),g(s_0))$ such that the previous map considered from $U$ to $W$ is invertible. That is, we get two functions $S,T$ defined on $W$ such that $$s=S(x,y),\,\,\,t=T(x,y).$$ If we now define $$u(x,y):=Z(S(x,y),T(x,y))$$ then $u$ solves the Cauchy problem on $W$. In fact for every $s \in I$ such that $(f(s),g(s))\in W$ we have $$u(f(s),g(s))=Z(S(f(s),g(s)),T(f(s),g(s)))=Z(S(X(s,0),Y(s,0)),T(X(s,0),Y(s,0)))=Z(s,0)=h(s)$$and it's easy to see that $u$ solves the PDE by differentiating.

So a solution exist in a neighborhood $W$ of $(f(s_0),g(s_0))$.

Now come my problems, because I want to prove that $u$ is the unique solution on $W$. Following the analytical proof given by F. John - Partial Differential Equations-Springer US (1975), suppose that $u'$ is another solution of the Cauchy problem on $W$. Let $(x',y')\in W$.

Set $s'=S(x',y')$ and consider the characteristic curve $\Gamma$ that solves $$\begin{cases}\frac{d}{dt}x=a(x,y,z)\\\frac{d}{dt}y=b(x,y,z)\\\frac{d}{dt}z=c(x,y,z)\end{cases}$$ with initial conditions $$\begin{cases}x(0)=f(s')\\y(0)=g(s')\\z(0)=h(s')\end{cases}.$$

Because $u$ and $u'$ solve the Cauchy problem, their corresponding integral surfaces both passes through the point $(f(s'),g(s'),h(s'))$ as the characteristic curve $\Gamma$ does for $t=0$. So the integral surfaces must contain the part of $\Gamma$ whose projection on $xy$ plane is contained in $W$. In particular for $t'=T(x',y')$ we have $$u'(x',y')=u'(X(s',t'),Y(s',t'))=Z(s',t')=Z(S(x',y'),T(x',y'))=u(x',y')$$ by definition of $u$.

Here is my question: even if we now that $(s',t')\in U$, who ensure me that $(s',0)\in U$ too, and so I'm sure that both integral surfaces have the point $(f(s'),g(s'),h(s'))$ in common? I mean, am I sure that if a point $(\overline{x},\overline{y})\in W$ and $s=S(\overline{x},\overline{y})$ then $(f(s),g(s))\in W$?

I think I have to consider a neighborhood of $(s_0,0)$ contained in $U$ that is a rectangle to be sure that the previous hold: in this way every selected characteristic will pass through the space initial curve $(f(s),g(s),h(s))$ and so the problem is solved. I apologies for all this words for a problem that is probably trivial and is not about PDE's!!

Thanks in advance.

  • $\begingroup$ Could you in particular clarify the sentence beginning with "Here is my question: even if we now that..."? I'm having trouble parsing it; there isn't an independent clause before "and so" for one thing. $\endgroup$ – epimorphic Oct 24 '18 at 16:44

Let $\boldsymbol{x}:=(x,y,z)$, $\boldsymbol{F}(\boldsymbol{x}% ):=(a(\boldsymbol{x}),b(\boldsymbol{x}),z(\boldsymbol{x}))$, $\boldsymbol{\gamma}(s):=(f(s),g(s),h(s))$, and $\boldsymbol{x}_{0}% :=\boldsymbol{\gamma}(s_{0})$. I assume that $\boldsymbol{F}$ is defined on an open set $U$ of $\mathbb{R}^{3}$ which contains $\boldsymbol{x}_{0}$. Since $\boldsymbol{F}$ is of class $C^{1}$, there exist a closed cube $Q(\boldsymbol{x}_{0},r)$ centered at $\boldsymbol{x}_{0}$ and of side-length $r$ which is contained in $U$ and two constants $M>0$ and $L>0$ such that \begin{align*} \Vert\boldsymbol{F}(\boldsymbol{x})\Vert & \leq M\quad\text{for all }\boldsymbol{x}\in Q(\boldsymbol{x}_{0},r),\\ \Vert\boldsymbol{F}(\boldsymbol{x}_{1})-\boldsymbol{F}(\boldsymbol{x}% _{2})\Vert & \leq L\Vert\boldsymbol{x}_{1}-\boldsymbol{x}_{2}\Vert \quad\text{for all }\boldsymbol{x}_{1},\boldsymbol{x}_{2}\in Q(\boldsymbol{x}% _{0},r). \end{align*} Since $f$, $g$, $h$ are continuous, there exist $\delta>0$ such that $$ |f(s)-f(s_{0})|\leq\frac{r}{4},\quad|g(s)-g(s_{0})|\leq\frac{r}{4}% ,\quad|h(s)-h(s_{0})|\leq\frac{r}{4}% $$ for all $s\in\lbrack s_{0}-\delta,s_{0}+\delta]$. Hence, $\boldsymbol{\gamma }(s)\in Q(\boldsymbol{x}_{0},r/2)$.

Taking $0<T\leq\min\{r/(4M),1/(2L\}\}$, it follows that for all $s\in\lbrack s_{0}-\delta,s_{0}+\delta]$ the Cauchy problem \begin{align*} \frac{d\boldsymbol{x}}{d\tau}(\tau) & =\boldsymbol{f}(\boldsymbol{x}% (\tau)),\\ \boldsymbol{x}(0) & =\boldsymbol{\gamma}(s), \end{align*} has a unique solution (this is just by Banach's fixed point theorem). We claim that $\boldsymbol{x}(\tau)\in Q(\boldsymbol{x}_{0},r)$ for all $\tau\in \lbrack-T,T]$. Indeed, $\boldsymbol{x}(0)=\boldsymbol{\gamma}(s)\in Q(\boldsymbol{x}_{0},r/2)$, and so by continuity we have that for $\tau$ very small $\boldsymbol{x}(\tau)\in Q(\boldsymbol{x}_{0},r)$. But as long as $\boldsymbol{x}(t)\in Q(\boldsymbol{x}_{0},r)$, we have that $$ \boldsymbol{x}(\tau)=\boldsymbol{x}(0)+\int_{0}^{\tau}\boldsymbol{f}% (\boldsymbol{x}(t))\,dt $$ and so \begin{align*} |x(\tau)-f(s_{0})| & \leq|x(\tau)-f(s)|+|f(s)-f(s_{0})|\\ & \leq\int_{0}^{\tau}|a(\boldsymbol{x}(t))|\,dt+|f(s)-f(s_{0})|\\ & \leq MT+\frac{r}{4}\leq\frac{r}{2}% \end{align*} by the choice of $M$, and similarly for $y$ and $z$. Hence, $\boldsymbol{x}% (\tau)\in Q(\boldsymbol{x}_{0},r)$ for all $\tau\in\lbrack-T,T]$.

Going back to your proof, take $U$ and take $W$ so small that \begin{align*} U & \subseteq[s_{0}-\delta,s_{0}+\delta]\times\lbrack-T,T],\\ W & \subseteq[f(s_{0})-r/2,f(s_{0})+r/2]\times\lbrack g(s_{0})-r/2,g(s_{0}% )+r/2]. \end{align*}

Let $\bar{u}=\bar{u}(x,y)$ be any solution of your Cauchy problem in $W$. Given $(\bar{x},\bar{y})\in W$, you have that $(s^{\prime},t^{\prime})\in U$. In particular, $s^{\prime}\in\lbrack s_{0}-\delta,s_{0}+\delta]$. Consider the following system of ODE \begin{align*} \frac{d\bar{x}}{d\tau}(\tau) & =a(\bar{x}(\tau),\bar{y}(\tau),\bar{u}(\bar {x}(\tau),\bar{y}(\tau))),\\ \frac{d\bar{y}}{d\tau}(\tau) & =b(\bar{x}(\tau),\bar{y}(\tau),\bar{u}(\bar {x}(\tau),\bar{y}(\tau))), \end{align*} with the initial conditions$$ \bar{x}(0)=f(s^{\prime}),\quad\bar{y}(0)=g(s^{\prime}). $$ Define $\bar{w}(\tau):=\bar{u}(\bar{x}(\tau),\bar{y}(\tau))$. Then as you showed \begin{align*} \frac{d\bar{w}}{d\tau}(\tau) & =\frac{\partial\bar{u}}{\partial x}(\bar {x}(\tau),\bar{y}(\tau))\frac{d\bar{x}}{d\tau}(\tau)+\frac{\partial\bar{u}% }{\partial y}(\bar{x}(\tau),\bar{y}(\tau))\frac{d\bar{y}}{d\tau}(\tau)\\ & =a(\bar{x}(\tau),\bar{y}(\tau),\bar{u}(\bar{x}(\tau),\bar{y}(\tau )))\frac{\partial\bar{u}}{\partial x}(\bar{x}(\tau),\bar{y}(\tau))+b(\bar {x}(\tau),\bar{y}(\tau),\bar{u}(\bar{x}(\tau),\bar{y}(\tau)))\frac {\partial\bar{u}}{\partial y}(\bar{x}(\tau),\bar{y}(\tau))\\ & =c(\bar{x}(\tau),\bar{y}(\tau),\bar{u}(\bar{x}(\tau),\bar{y}(\tau))) \end{align*} and $$ \bar{w}(0)=\bar{u}(\bar{x}(0),\bar{y}(0))=\bar{u}(f(s^{\prime}),g(s^{\prime }))=h(s^{\prime}). $$ Hence, $(\bar{x}(\tau),\bar{y}(\tau),\bar{w}(\tau))$ is a solution of the Cauchy problem \begin{align*} \frac{d\bar{x}}{d\tau}(\tau) & =a(\bar{x}(\tau),\bar{y}(\tau),\bar{w}% (\tau)),\\ \frac{d\bar{y}}{d\tau}(\tau) & =b(\bar{x}(\tau),\bar{y}(\tau),\bar{w}% (\tau)),\\ \frac{d\bar{w}}{d\tau}(\tau) & =c(\bar{x}(\tau),\bar{y}(\tau),\bar{w}(\tau)), \end{align*} with initial data $$ \bar{x}(0)=f(s^{\prime}),\quad\bar{y}(0)=g(s^{\prime}),\quad\bar {w}(0)=h(s^{\prime}). $$ But by uniqueness $\bar{x}(t)=X(s^{\prime},t)$, $\bar{y}(t)=Y(s^{\prime},t)$ and $\bar{w}(t)=Z(s^{\prime},t)$ for all $t$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.