12.2 Real (scalar) $1^{st}$ order PDEs

Example 12.2.1.label We consider first the quasilinear case

\begin{align*}\sums{n }{i=1}a_{i}(x,u)\pd_{i}u=b(x,u)\end{align*}

where $x=(x_{1},...,x_{n})$. We consider the Cauchy problem $u|_{S}=\p$ for $S$ hypersurface in $\R^{n}$ and $\p$ is given (prescribes the graph of $u$ over $S$). Let $S^{*}=\curl{(x,\p(x):x\in S)}$ then this problem is reducible to an ODEs problem. Rewrite the graph of $u$ as $F=u(x)-y=0$ then we rewrite the original problem as

\begin{align*}\vec{A}\cd\nabla F=0&&\vec{A}=(a_{1},...,a_{n},b),\nabla F=(\pd_{1}u,...,\pd_{n}u,-1)\end{align*}

Thus $\vec{A}$ must be everywhere tangent to graph $F=0$. This suggests constructing graph of $u$ by ”flowing off” $S^{*}$ along integral curves of $\vec{A}$ for this to work it would be sensible to require that $\vec{A}$ to never be tangent to $S^{*}$.

Theorem 12.2.2 (Local Existence and Uniqueness of $1^{st}$ order Quasilinear PDE).label For the Cauchy problem above such that $(a_{1},...,a_{n})$ is not tangent to $S$ at any $x\in S$. Then for any sufficiently small neighborhood $\OM$ of $S$ in $\R^{n}$, there exists a unique solution to the Cauchy problem.

Proof. Let us express non-tangency condition. Suppose $S$ is parametrized by

\begin{align*}U\ni s=(s_{1},...,s_{n-1})\mto (g_{1}(s),...,g_{n}(s))&&\det\begin{pmatrix}\pder{g_1}{s_1}&\cds&\pder{g_1}{s_{n-1}}&a_{1}(\OBC{g(s)}{x\in S},\OBC{\p(g(s))}{y(x)})\\ \vdots&&\vdots\\ \pder{g_n}{s_1}&\cds&\pder{g_n}{s_{n-1}}&a_{n}(g(s),\p((g(s))))\end{pmatrix}\neq 0\end{align*}

$\fall s\in U$. If we were in the linear case then this $\det\neq 0$ condition is equivalent to $(a_{1},...,a_{n})$ is non-characteristic.

We now turn this into ODE system

\begin{align*}\der{x_j}{t}=a_{j}(x,y)&&\der{y}{t}=b(x,y)\end{align*}

with initial conditions

\begin{align*}x_{j}(s,0)=g_{j}(s)&&y(s,0)=\p(g(s))\end{align*}

By $\exists!$ for ODE (Cauchy-Lipschitz Picard) $\exists!$ solution

\begin{align*}(s_{1},...,s_{n-1},t)\mto (x_{1}(s,t),...,x_{n}(s,t))\end{align*}

which is $C^{1}$ in both $s$ (parametrizes the initial data) and $t$ (flow parameter). By IFT, $\exists C^{1}$ inverse map

\begin{align*}s=s(x)&&t=t(x)\end{align*}

in a small neighborhood $\OM$ of any point in $S$. Then we may define $u(x)\define y(s(x),t(s))$. If $x\in S$ then we have $u(x)=\p(s)$ hence the Cauchy condition is satisfied. It remains to show that $u(x)$ solves the PDE. We have

\begin{align*}\sums{n }{j=1}a_{j}\pd_{j}u&=\sums{n }{j=1}a_{j} \parens{\sums{n-1 }{k=1}\pder{u}{s_k}\pder{s_k}{x_j}+\pder{u}{t}\pder{t}{x_j}}\\&=\sums{n-1}{k=1}\pder{u}{s_k}\sums{n }{j=1}\UBC{a_j}{\pder{x_j}{t}}\pder{s_k}{x_j}+\pder{u}{t}\sums{n}{j=1}\UBC{a_j}{\pder{x_j}{t}}\pder{t}{x_j}\\&=\sums{n-1}{k=1}\pder{u}{s_k}\sums{n }{j=1}\UBC{\pder{x_j}{t}\pder{s_k}{x_j}}{\pder{s_k}{t}=0}+\pder{u}{t}\sums{n }{j=1}\UBC{\pder{x_j}{t}\pder{t}{x_j}}{\pder{t}{t}=1}\\&=\pder{u}{t}=b\end{align*}

The integral curve of ODE system is called characteristic curve$\square$

Example 12.2.3 (Linear Example).label

(1)
Consider the linear PDE $x_{1}\pd_{1}u+2x_{2}\pd_{2}u+\pd_{3}u=3u$ and $u=\p(x_{1},x_{2})$ on $S$ given by $x_{3}=0$ then
\begin{align*}\der{x_1}{t}=x_{1}&&\der{x_2}{t}=2x_{2}&&\der{x_3}{t}=1&&\der{u}{t}=3u\end{align*}

then
\begin{align*}(x_{1},x_{2},x_{3},u)|_{t=0}=(s_{1},s_{2},0,\p(x_{1},x_{2}))\end{align*}

so
\begin{align*}x_{1}=s_{1}e^{t}&&x_{2}=s_{2}e^{2}t&&x_{3}=t+s_{3}&&u=s_{4}e^{3t}\end{align*}

By initial condition $s_{3}=0,s_{4}=\p(x_{1},x_{2})$. Express $t,s$ as funtions of $x$ we have $t=x_{3},s_{1}=e^{-x_3}x_{1},s_{2}=e^{-2x_3}x_{2}$ thus
\begin{align*}u=\p(e^{-x_3}x_{1},e^{-2x_3}x_{2})e^{3x_3}\end{align*}

We check the Cauchy condition
\begin{align*}u|_{x_3=0}=\p(x_{1},x_{2})\cd 1\end{align*}

and check the PDE is satisfied
\begin{align*}x_{1}[\p_{,1}e^{-x_3}]+2x_{2}[\p_{,2}e^{-2x_3}]+[-x_{1}\p_{,1}e^{-x_3}+\p_{,2}(-2x_{2})e^{-2x_3}]e^{3x_3}+\p 3e^{3x_3}=3u\end{align*}
(2)
$u\pd_{1}u+\pd_{2}u=1$ with $S: x_{1}=x_{2}=s$ for $0<s<1$ and $u|_{S}=\frac{1}{2}s$ so
\begin{align*}\der{x_1}{t}=u&&\der{x_2}{t}=1&&\der{u}{t}=1\end{align*}

then
\begin{align*}(x_{1},x_{2},y)|_{t=0}=(s,s,\frac{1}{2}s)\end{align*}

and
\begin{align*}\det\begin{pmatrix}\pder{x_1}{s}&a_{1}(s,s,\frac{1}{2}s)\\ \pder{x_2}{s}&a_{2}(s,s,\frac{1}{2}s)\end{pmatrix}=\det\begin{pmatrix}1&\frac{1}{2}s\\ 1&1\end{pmatrix}=1-\frac{1}{2}s\neq 0\end{align*}

Solving for $x_{1},x_{2},u$ we have
\begin{align*}x_{2}(s,t)=t+s&&u(s,t)=t+\frac{1}{2}s&&x_{1}(s,t)=\frac{1}{2}t^{2}+\frac{1}{2}st+s\end{align*}

Collecting terms we have solve for $s$ and $t$
\begin{align*}x_{2}-x_{1}=-\frac{1}{2}t(-2+\UBC{t+s}{x_2})\implies t=\frac{2(x_{2}-x_{1})}{2-x_{2}}\end{align*}

and
\begin{align*}x_{2}=t+s\implies s=x_{2}-t=x_{2}-\frac{2(x_{2}-x_{1})}{2-x_{2}}=\frac{-x_{2}^{2}+2x_{1}}{2-x_{2}}\end{align*}

Lastly
\begin{align*}u(x)=t(x)+\frac{1}{2}s(x)=\frac{4x_{2}-2x_{1}-x_{2}^{2}}{2(2-x_{2})}\end{align*}

We check that the Cauchy condition are satisfied on $x_{1}=x_{2}=s$
\begin{align*}u=\frac{4s-2s-s^{2}}{2(2-s)}=\frac{2s-s^{2}}{2(2-s)}=\frac{1}{2}s\end{align*}

Check that $u$ solves the PDE
\begin{align*}\pd_{1}u=\frac{-2}{2(2-x_{2})}=\frac{1}{x_{2}-2}&&\pd_{2}u=\frac{1}{2}\frac{(4-2x_{2})(2-x_{2})-(4x_{2}-2x_{1}-x_{2}^{2})(-1)}{(2-x_{2})^{2}}=\frac{1}{2}\frac{2(2-x_{2})^{2}+4x_{2}-2x_{1}-x_{2}^{2}}{(2-x_{2})^{2}}\end{align*}

so
\begin{align*}u\pd_{1}u+\pd_{2}u=-\frac{4x_{2}-2_{x}1-x_{2}^{2}}{2(2-x_{2})^{2}}+\frac{2(2-x_{2})^{2}+4x_{2}-x_{2}^{2}-2x_{1}}{2(2-x_{2}^{2})}=1\end{align*}

We note that at $s=2$ then the solution to pde we wrote isn’t defined since the IFT step breaks down. Indeed $s=2$ the $\det$ step gives $0$ hence not invertible.

Example 12.2.4 (Nonlinear $1^{st}$ Order PDE).label Finally we consider the Cauchy problem for a fully nonlinear $1^{th}$ order PDE.

\begin{align*}F(x_{1},...,x_{n},u,\pd_{1}u,...,\pd_{n}u)=0&&u|_{S}=\p\end{align*}

We assume that $F\in C^{2}$. We let $\xi_{i}\define \pd_{i}u$ and want $u$ such that image of the map

\begin{align*}x\mto (x,u(x),\xi(x))\end{align*}

lies in $F=0$. We note $x_{1},...,x_{n},u,\xi_{1},...,\xi_{n}$ is $2n+1$ dimensional and denoted $J^{1}(\R^{n},\R)$ the space of $1$-jets of maps $\R^{n}\to \R$ and $(x,u(x),\pd u(x))$ is a solution if and only if the graph is contained in $\curl{F=0}$. One can consider the PDE as a hypersurface in the space of $1$-jets.

By analogy with the quasilinear case Theorem 12.2.2, we write the Cauchy characteristic equation as follows

\begin{align*}\der{x_i}{t}=\pder{F}{\xi_i},1\leq i\leq n\end{align*}

to be augmented by ODEs for $y\geq \xi_{i}$

\begin{align*}\der{y}{t}=\sums{n}{i=1}\pder{y}{x_i}\der{x_i}{t}=\sums{n }{i=1}\pder{y}{x_i}\pder{F}{\xi_i}\end{align*}

thus we have

\begin{align*}\der{y}{t}=\sums{n }{i=1}\xi_{i}\pder{F}{\xi_i}\end{align*}

Furthermore

\begin{align*}\der{\xi_j}{t}&=\sums{n }{k=1}\pder{\xi_j}{x_k}\der{x_k}{t}=\sums{n }{k=1}\pder{x_j}{x_k}\pder{F}{\xi_k}\\&=\sums{n }{k=1}\pder{^2u}{x_j\pd x_k}\pder{F}{\xi_k}=\sums{n }{k=1}\pder{^2u}{x_k\pd x_j}\pder{F}{\xi_k}\\&=\sums{n }{k=1}\pder{\xi_k}{x_j}\pder{F}{\xi_k}\end{align*}

thus

\begin{align*}\pder{F}{x_j}+\pder{F}{u}\UBC{\pder{u}{x_j}}{\xi_j}+\UBC{\sums{n }{k=1}\pder{F}{\xi_k}\pder{x_k}{\xi_j}}{=-\der{\xi_j}{t}}=0\end{align*}

So the characteristic equations are

\begin{align*}\der{x_i}{t}&=\pder{F}{\xi_i}(x,y,\xi)\\ \der{y}{t}&=\sums{n }{i=1}\xi_{i}\pder{F}{\xi_i}(x,y,\xi)\\ \der{\xi_i}{t}&=\parens{-\pder{F}{x_i}-\xi_i\pder{F}{y}}(x,y,\xi)\end{align*}

We remark that $\pder{F}{u}=0$ then the characteristic equation describes Hamiltonian flow in $(x,\xi)$. The solutions of this ODE system is called characteristic strips. What about initial conditions for the characteristic system of ODEs?

\begin{align*}x(s,0)&=g(s)\\ y(s,0)&=\p(g(s))\end{align*}

To get conditions for $\xi$ we note that on $S$

\begin{align*}\pder{y}{s_j}=\pder{\p(g(s))}{s_j}=\sums{n }{k=1}\pder{y}{x_k}\pder{g_k}{s_j}=\sums{n }{k=1}\xi_{k}\pder{g_k}{s_j}&&F(g(s),\p(g(s)),\xi_{j}(g(s)))=0&&\end{align*}

for $1\leq j\leq n-1$. We need

\begin{align*}\det\begin{pmatrix}\pder{g_1}{s_1}&\cds&\pder{g_1}{s_{n-1}}&\pder{F}{\xi_1}\\ \vdots&&\vdots\\ \pder{g_n}{s_1}&\cds&\pder{g_n}{s_{n-1}}&\pder{F}{\xi_n}\end{pmatrix}\neq 0\end{align*}

so here the ambient space also include $\xi$ and thus more things can go wrong than just tangency. The first thing that can go wrong (for $\xi$) is related the $1$-jets the second (for $y$) comes from graph of function.

\begin{align*}(s_{1},...,s_{n-1},t)\bij (x_{1},...,x_{n})&&u(x)\define y(s(x),t(x))\end{align*}

Example 12.2.5 (Nonlinear $1^{st}$ Order Example).label Consider

\begin{align*}\begin{cases}\pd_{1}u\pd_{2}u=u\\ u(0,x_{2})=(x_{2})^{2}\end{cases}\end{align*}

So $F(x_{1},x_{2},u,\xi_{1},\xi_{2})=\xi_{1}\xi_{2}-u$ and characteristic equations for this example is given by

\begin{align*}\begin{cases}\der{x_1}{t}=\xi_{1}&\der{x_2}{t}=\xi_{1}\\ \der{y}{t}=2\xi_{1}\xi_{2}\\ \der{\xi_1}{t}=\xi_{1}&\der{\xi_2}{t}=\xi_{2}\end{cases}\end{align*}

\begin{align*}\xi_{1}=\xi_{1}^{o}e^{t}&&\xi_{2}=\xi_{2}^{o}e^{t}\\ x_{1}=\xi_{2}^{o}e^{t}+x_{1}^{o}&&x_{2}=\xi_{1}^{o}e^{t}+x_{2}^{o}&&y=\xi_{1}^{o}\xi_{2}^{o}e^{2t}+u^{o}\end{align*}

Impose Cauchy data for characteristic strips

\begin{align*}x_{1}|_{t=0}&=0\iff\xi_{2}^{o}+x_{1}^{o}=0\implies x_{1}(t,s)=\xi_{2}^{o}(e^{t}-1)\\ x_{2}|_{t=0}&=s\iff\xi_{1}^{o}+x_{2}^{o}=s\implies x_{2}(t,s)=\xi_{1}^{o}(e^{t}-1)+s\\ u|_{t=0}&=s^{2}\iff \xi_{1}^{o}\xi_{2}^{o}+u^{o}=s^{2}\implies y(t,s)=\xi_{1}^{o}\xi_{2}^{o}(e^{2t}-1)+s^{2}\\ \xi_{2}(0,x_{2})&=\pder{u}{x_2}(0,x_{2})=2x_{2}\\ \implies \xi_{2}|_{t=0}&=2s\implies \xi_{2}(t,s)=2se^{t}\end{align*}

Use PDE to solve for the $\xi_{1}$

\begin{align*}\xi_{1}\xi_{2}-u=0\implies \xi_{1}(0,s)\xi_{2}(0,s)-u(0,s)=\xi_{1}|_{t=0}2s-s^{2}=0\implies \xi_{1}|_{t=0}=\frac{s}{2}\end{align*}

but here we divided by $s$ which is not allowed when $s=0$. Here we see $(g_{1}(s),g_{2}(s))=(0,s)$. This is where $\det\begin{pmatrix}0&2s\\1&\frac{s}{2}\end{pmatrix}=-2s$ which is $0$ when $s=0$.

Finally we carry out the inversion $(s,t)\mto x$. Rewrite the characteristic strips

\begin{align*}x_{1}(t,s)&=2s(e^{t}-1)\\ x_{2}(t,s)&=\frac{s}{2}(e^{t}-1)+s=\frac{s}{2}(e^{t}+1)\\ y(t,s)&=\frac{s}{2}2s(e^{2t}-1)+s^{2}=s^{2}e^{2t}\\ \xi_{1}(t,s)&=\frac{s}{2}e^{t}\\ \xi_{2}(t,s)&=2se^{t}\end{align*}

Inverting

\begin{align*}x_{1}-4x_{2}&=-4s\implies s=\frac{4x_{2}-x_{1}}{4}\\ x_{1}+4x_{2}&=4se^{t}\implies e^{t}=\frac{x_{1}+4x_{2}}{4x_{2}-x_{1}}\end{align*}

\begin{align*}u(x_{1},x_{2})&=y(s(x_{1},x_{2}),t(x_{1},x_{2}))\\&=\parens{s(x_1,x_2)}^{2}e^{2t(x_1,x_2)}\\&=\parens{\frac{4x_{2}-x_{1}}{4}}^{2} \parens{\frac{x_{1}+4x_{2}}{4x_{2}-x_{1}}}^{2}\\&=\frac{(x_{1}+4x_{2})^{2}}{16}\end{align*}

Check Cauchy data

\begin{align*}u(0,x_{2})=\frac{(4x_{2})^{2}}{16}=(x_{2})^{2}\end{align*}

Check PDE

\begin{align*}u_{1}=\frac{1}{8}(x_{1}+4x_{2})&&u_{2}=\frac{1}{8}4(x_{1}+4x_{2})\implies u_{1}u_{2}=u\end{align*}

Direct References

circleTheorem 12.2.2: Local Existence and Uniqueness of $1^{st}$ order Quasilinear PDE

The Eigenspace

12.2 Real (scalar) $1^{st}$ order PDEs

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Direct References

The Eigenspace

Direct References