Partial Differential Equations III & V

[6pt] Exercise Sheet 2

Lecturer: Alpár R. Mészáros

1. Burger’s equation as a model of traffic flow. In Example 2.1 we introduced the following PDE as a simple model of traffic flow:

   $${\rho }_t+{(\rho v_m(1-\rho /{\rho }_m))}_x=0$$ (1)

   where $\rho (x,t)$ is the density of the traffic at point $x\in ℝ$ at time $t$, ${\rho }_m>0$ is a constant representing the maximum traffic density ($\rho ={\rho }_m$ if the vehicles are bumper-to-bumper), and $v_m>0$ is the speed limit. Show that this PDE can be converted into Burger’s equation with a suitable change of variables.
   Hint: Given $\rho$ satisfying (1), you need to define a function $u$ in terms of $\rho$ such that $u_t+u{u}_x=0$. You can guess the form of $u$ by expanding the term ${(\rho v_m(1-\rho /{\rho }_m))}_x$ of (1).

2. Revision of ODEs. In this exercise we recall how to solve first-order linear and separable ODEs.

   • Linear first-order ODEs. Let $x\in C^1(ℝ)$ satisfy the first-order linear ODE

     $\dot{x}(\tau )+a(\tau )x(\tau )$ $=b(\tau ),\tau \in ℝ,$

     $x(0)$ $=x_0,$

     where $a,b\in C^1(ℝ)$. Let $A$ be a primitive of $a$, i.e., let $A$ satisfy $\dot{A}(\tau )=a(\tau )$. This is sometimes denoted as $A(\tau )=\int a(\tau )𝑑\tau$. Show that

     $$x(\tau )=e^{A(0)-A(\tau )}{x}_0+e^{-A(\tau )}{\int }_0^{\tau }{e}^{A(t)}b(t)𝑑t.$$

     Hint: Multiply the ODE by the integrating factor $e^{A(\tau )}$ and observe that $e^A\dot{x}+e^{A}ax=\frac{d}{d\tau }\left(e^{A}x\right)$.

   • Separable ODEs. Let $I\subseteq ℝ$ be an open interval containing $0$ and let $x\in C^1(I)$ satisfy the separable ODE

     $\dot{x}(\tau )$ $=f(x(\tau ))g(\tau ),\tau \in I,$

     $x(0)$ $=x_0,$

     where $f,g\in C^1(ℝ)$. Let $h$ be a primitive of $1/f$, i.e., let $h$ satisfy $\dot{h}(\tau )=1/f(\tau )$. Assume that $f(x(\tau ))\ne 0$ for all $\tau \in I$. Show that

     $$x(\tau )=h^{-1}\left(h(x_0)+{\int }_0^{\tau }g(t)𝑑t\right).$$

     Hint: See Example 2.4 in the lecture notes.

   Remark: It is not worth memorising these formulas. It is easier to simply derive them whenever needed.

3. The method of characteristics for a linear PDE. Solve the linear first-order PDE

   $u_x+u_y+u=1$ $\mathrm{\hspace{1em}}\text{for }(x,y)\in {ℝ}^2,$

   $u=x^2$ $\mathrm{\hspace{1em}}\text{for }(x,y)\in ℝ\times \{0\}.$

   Plot the characteristics $\tau \mapsto (x(\tau ,s),y(\tau ,s))$ for some representative values of $s$.

4. The method of characteristics for a linear PDE. Solve the Cauchy problem

   $y{u}_x-x{u}_y=0$

   $u(0,y)=\sqrt{a^2-y^2}$ $\mathrm{\hspace{1em}}\text{for }y\in (-a,a),$

   where $a>0$ is a constant.

5. The method of characteristics for a semilinear PDE. Use the method of characteristics to find the solution $u$ of the semilinear first-order PDE

   $x^2{u}_y-u_x=x^2{u}^2\cos y$

   that satisfies the condition $u(0,y)=y^2$ for all $y\in ℝ$. This appeared on the PDEs exam in May 2015.

6. The method of characteristics for a quasilinear PDE. Solve the quasilinear first-order PDE

   $(t+u)u_x+t{u}_t=x-t$ $\mathrm{\hspace{1em}}\text{for }(x,t)\in ℝ\times (0,\mathrm{\infty }),$

   $u(x,1)=1+x$ $\mathrm{\hspace{1em}}\text{for }x\in ℝ.$

   Verify that the answer you obtain using the method of characteristics does indeed satisfy the PDE.

7. The method of characteristics for a quasilinear PDE. Find $u(x_1,x_2)$ satisfying the quasilinear PDE

   $u{u}_{x_1}+u_{x_2}=1$

   and the Cauchy condition $u(x_1,x_1)=\frac{1}{2}{x}_1$. This question is from L.C. Evans (1998) Partial Differential Equations, American Mathematical Society.

8. The method of characteristics is independent of the parametrisation of $\mathrm{\Gamma }$. Consider the transport equation

   $u_t+c{u}_x=0$ $\text{for }(x,t)\in ℝ\times (0,\mathrm{\infty }),$

   $u(x,0)=g(x)$ $\text{for }x\in ℝ.$

   Let $\mathrm{\Gamma }=ℝ\times \{0\}$ be the Cauchy curve. Show that the method of characteristics produces the solution $u(x,t)=g(x-ct)$ for all parametrisations of $\mathrm{\Gamma }$ of the form $\gamma :ℝ\to \mathrm{\Gamma }$, $\gamma (s)=(\lambda s,0)$, $\lambda >0$. (In Example 2.10 we considered the case $\lambda =1$.) With more work it can be shown that the method of characteristics is independent of the choice of parametrisation of $\mathrm{\Gamma }$.

9. Nonexistence due to a characteristic point. Let $𝒂:{ℝ}^2\to {ℝ}^2$ be the vector field $𝒂(x,y)=(-1,1)$. Let $\mathrm{\Omega }=B_1(\mathrm{𝟎})$ be the open unit ball in ${ℝ}^2$ centred at the origin. Consider the linear PDE

   $𝒂\cdot \nabla u=0$ $\text{in }\mathrm{\Omega },$

   $u=g$ $\text{on }\partial \mathrm{\Omega },$

   where $g:\partial \mathrm{\Omega }\to ℝ$ is defined by $g(x,y)=x$.

   • Show that the point $𝒑=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\in \partial \mathrm{\Omega }$ is characteristic, i.e., is not noncharacteristic. Are any other points of $\partial \mathrm{\Omega }$ characteristic?

   • Prove that there does not exist any function $u:U\to ℝ$, where $U\subset {ℝ}^2$ is an open set containing $𝒑$, satisfying

     $𝒂\cdot \nabla u=0$ $\text{in }U,$

     $u=g$ $\text{on }U\cap \partial \mathrm{\Omega }.$

     Give an example of a nonconstant function $g$ for which there is a solution.

10. Nonuniqueness or nonexistence due to a characteristic Cauchy curve. Consider the PDE

    $u_x(x,y)+u_y(x,y)=1$ $\text{for }(x,y)\in {ℝ}^2,$ (2)

    $u(x,x)=x$ $\text{for }x\in ℝ.$

    Let $\mathrm{\Gamma }=\{(x,x):x\in ℝ\}$ denote the Cauchy curve where the Cauchy data is prescribed.

    • Show that every point of $\mathrm{\Gamma }$ is characteristic, i.e., that no point is noncharacteristic. Sketch the Cauchy curve $\mathrm{\Gamma }$ and the characteristics.

    • Show that the Cauchy problem (2) has infinitely many solutions.
      Hint: Use the method of characteristics to solve the PDE

      $u_x(x,y)+u_y(x,y)=1$ $\mathrm{\hspace{1em}}\text{for }(x,y)\in {ℝ}^2,$

      $u(x,0)=f(x)$ $\mathrm{\hspace{1em}}\text{for }x\in ℝ,$

      where $f:ℝ\to ℝ$ is an arbitrary differentiable function satisfying $f(0)=0$. Verify that your solution $u$ satisfies the Cauchy problem (2) for all $f$.

    • Now consider the same PDE but with different Cauchy data:

      $u_x(x,y)+u_y(x,y)=1$ $\mathrm{\hspace{1em}}\text{for }(x,y)\in {ℝ}^2,$

      $u(x,x)=1$ $\mathrm{\hspace{1em}}\text{for }x\in ℝ.$

      As above, the Cauchy data is characteristic everywhere. Show that this time, however, the Cauchy problem has no solution.
      Hint: Differentiate the equation $u(x,x)=1$.

    More generally, it can be shown that the Cauchy problem (2) has no solution with Cauchy data $u(x,x)=g(x)$, $x\in ℝ$, unless $g(x)=x+c$ for some constant $c\in ℝ$. In summary, if the Cauchy data is characteristic everywhere, then the Cauchy problem can have no solution or infinitely many solutions. This example is taken from Y. Pinchover & J. Rubinstein (2005) An Introduction to Partial Differential Equations, Cambridge University Press.

11. Obstacles to global existence. Even if the Cauchy data is noncharacteristic at every point, a Cauchy problem might not have a global solution, i.e., a solution for all values of the independent variables. This could be because the solution blows up (see Example 2.18) or because the characteristics intersect (see Chapter 3). In this exercise we look at two others obstacles to global existence.

    • Vanishing vector fields. Let $𝒂:{ℝ}^2\to {ℝ}^2$ be the vector field $𝒂(x,y)=-(x,y)$. Let $\mathrm{\Omega }=B_1(\mathrm{𝟎})$ be the open unit ball in ${ℝ}^2$ centred at the origin. Consider the linear PDE

      $𝒂\cdot \nabla u=0$ $\text{in }\mathrm{\Omega },$

      $u=g$ $\text{on }\partial \mathrm{\Omega }.$

      Observe that $𝒂$ vanishes at $\mathrm{𝟎}\in \mathrm{\Omega }$. Sketch $𝒂$ and $\mathrm{\Omega }$. Let $𝒆(\theta )=(\cos (\theta ),\sin (\theta ))$ and let $\epsilon >0$. By using the method of characteristics (or otherwise), derive the solution $u:\overline{\mathrm{\Omega }}\setminus B_{\epsilon }(\mathrm{𝟎})\to ℝ$ given in polar coordinates by

      $$u(r𝒆(\theta ))=g(𝒆(\theta )),r\in [\epsilon ,1],\theta \in [0,2\pi ).$$

      This can also be written as $u(𝒙)=g\left(\frac{𝒙}{|𝒙|}\right)$, $\epsilon \le |𝒙|\le 1$.

      Assume that $g$ is not a constant function. Prove that the PDE does not have a global solution $u:\overline{\mathrm{\Omega }}\to ℝ$. In the special case where $g$ is a constant function, write down the global solution.

    • Characteristics intersecting the Cauchy curve multiple times. Let $𝒂:{ℝ}^2\to {ℝ}^2$ be the vector field $𝒂(x,y)=(1,0)$. Let $\mathrm{\Omega }=(0,1)\times ℝ$. Consider the linear PDE

      $𝒂\cdot \nabla u=1$ $\text{in }\mathrm{\Omega },$

      $u=u_0$ $\text{on }\mathrm{\Gamma }.$

      Define $u_0(x,y)=y$. First consider the case $\mathrm{\Gamma }=\{0\}\times ℝ$. Using the method of characteristics (or otherwise) derive the solution $u:\overline{\mathrm{\Omega }}\to ℝ$, $u(x,y)=x+y$. Sketch $\mathrm{\Omega }$, $\mathrm{\Gamma }$, and the characteristics.

      Now consider the case $\mathrm{\Gamma }=\partial \mathrm{\Omega }=(\{0\}\times ℝ)\cup (\{1\}\times ℝ)$. Sketch $\mathrm{\Omega }$, $\mathrm{\Gamma }$, and the characteristics again. How many times does each characteristic intersect $\mathrm{\Gamma }$? Let . Derive the local solution $u:([0,a]\cup [b,1])\times ℝ\to ℝ$,

      (It is simpler to integrate the PDE with respect to $x$ than to use the method of characteristics.) Prove that the PDE does not have a global solution $u:\overline{\mathrm{\Omega }}\to ℝ$. Finally, suppose that $u_0$ is arbitrary, and give a condition on $u_0$ such that a global solution exists.

      For another example where the characteristics intersect the Cauchy curve more than once, revisit Q5(ii) from Exercise Sheet 1.

12. Geometric interpretation of the method of characteristics. Let $u:{ℝ}^2\to ℝ$ satisfy

    $a_1(x,y,u)u_x+a_2(x,y,u)u_y$ $=b(x,y,u)\mathit{\hspace{1em}}\text{in }{ℝ}^2,$

    $u$ $=u_0\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}}\text{on }\mathrm{\Gamma },$

    where $\mathrm{\Gamma }\subset {ℝ}^2$ is a simple smooth curve. The graph of $u$ can be written in the form

    $$\{(x,y,z)\in {ℝ}^3:F(x,y,z)=0\}$$

    where $F(x,y,z)=u(x,y)-z$. This is called the solution surface. Recall that $\nabla F$ is normal to the surface. Show that the vector field $𝒇=(a_1,a_2,b)$ is tangent to the solution surface. This means that the flow of the vector field $𝒇$ through any point on the surface stays on the surface. In particular, since the curve

    $$\tilde{\mathrm{\Gamma }}=\{(x,y,u_0(x,y)):(x,y)\in \mathrm{\Gamma }\}$$

    lies in the solution surface, then the flow of the vector field $𝒇$ through $\tilde{\mathrm{\Gamma }}$ generates a portion of the solution surface (provided that $𝒇$ is not tangent to $\tilde{\mathrm{\Gamma }}$). Step 1 of the method of characteristics is exactly to find the flow of $𝒇$ through $\tilde{\mathrm{\Gamma }}$.

13. The method of characteristics for PDEs in $n$ independent variables. Use the method of characteristics for quasilinear first-order PDEs in $n$ independent variables (see Section 2.4 of the lecture notes) to derive the solution $u(𝒙,t)=g(𝒙-𝒄t)$ of the transport equation

    $u_t+𝒄\cdot \nabla u=0$ $\text{for }(𝒙,t)\in {ℝ}^n\times (0,\mathrm{\infty }),$

    $u(𝒙,0)=g(𝒙)$ $\text{for }𝒙\in ℝ,$

    where $𝒄\in {ℝ}^n$ is a constant vector and $u:{ℝ}^n\times [0,\mathrm{\infty })\to ℝ$.

14. Uniqueness theorem for ODEs. In this problem we prove the easy part of the well-posedness theorem for ODEs.

    • Let $𝒇:{ℝ}^n\to {ℝ}^n$ be Lipschitz continuous, i.e., suppose that there exists a constant $L>0$ such that

      $$|𝒇(𝒚)-𝒇(𝒙)|\le L|𝒚-𝒙|\mathit{\hspace{1em}}\forall 𝒙,𝒚\in {ℝ}^n.$$

      Let ${𝒙}_0\in {ℝ}^n$. Show that the ODE

      $\dot{𝒙}(\tau )$ $=𝒇(𝒙(\tau )),\tau \in ℝ,$

      $𝒙(0)$ $={𝒙}_0,$

      has at most one solution.
      Hint: Let ${𝒙}_1,{𝒙}_2$ be solutions, ${\dot{𝒙}}_1=𝒇({𝒙}_1)$, ${\dot{𝒙}}_2=𝒇({𝒙}_2)$, ${𝒙}_1(0)={𝒙}_2(0)={𝒙}_0$. Then

      $${\dot{𝒙}}_1-{\dot{𝒙}}_2=𝒇({𝒙}_1)-𝒇({𝒙}_2).$$

      Take the dot product of this equation with ${𝒙}_1-{𝒙}_2$. Show that

      $$\frac{d}{d\tau }\frac{1}{2}{|{𝒙}_1-{𝒙}_2|}^2\le L{|{𝒙}_1-{𝒙}_2|}^2.$$

      Define $E(\tau )={|{𝒙}_1(\tau )-{𝒙}_2(\tau )|}^2$. Prove that $E(\tau )=0$ for all $\tau$.

    • In Theorem 2.5 (the well-posedness theorem for ODEs) we took $𝒇$ to be continuously differentiable rather than Lipschitz continuous. In this exercise we show that a continuously differentiable function on a compact set is Lipschitz continuous. For simplicity we restrict our attention to the case $n=1$. Let $f\in C^1([a,b])$. Show that $f$ is Lipschitz continuous, i.e., show that there exists $L>0$ such that

      $$|f(y)-f(x)|\le L|x-y|\mathit{\hspace{1em}}\forall x,y\in [a,b].$$

      Hint: Use the Mean Value Theorem.

15. Nonuniqueness for ODEs. Consider the separable ODE

    $\dot{x}(\tau )$ $=f(x(\tau )),$

    $x(0)$ $=-1,$

    for $f(x)={|x|}^{1/2}$. Since $f$ is continuously differentiable in a neighbourhood of $x=-1$, the well-posedness theory for ODEs ensures that the ODE has a unique solution $x:(-\epsilon ,\epsilon )\to ℝ$ for some $\epsilon >0$. Solve this ODE to find an explicit formula for the solution on the time interval $(-\mathrm{\infty },2)$. What happens at time $\tau =2\mathrm{?}$ Write down infinitely many solutions $x:(-\mathrm{\infty },T)\to ℝ$ for any $T>2$. (Hint: See Example 2.3.) Like Example 2.4, this is an example where global existence and uniqueness fails. In Example 2.4 it is because the solution blows up in finite time (the solution reaches the boundary of the domain of $f$). In this example it is because $x(\tau )$ reaches the point $x=0$ where $f$ is not continuously differentiable.

16. The Eikonal equation. The Eikonal equation is an example of a fully nonlinear first-order PDE. The method of characteristics can be extended to treat fully nonlinear first-order PDEs. In this exercise you are simply asked to verify that given functions satisfy boundary value problems of the form

    $|\nabla u(𝒙)|=1$ $\text{for }𝒙\in \mathrm{\Omega },$

    $u(𝒙)=0$ $\text{for }𝒙\in \partial \mathrm{\Omega },$

    where $\mathrm{\Omega }\subset {ℝ}^2$ is open and bounded, and $u:\overline{\mathrm{\Omega }}\to ℝ$.

    • Let $\mathrm{\Omega }=B_1(\mathrm{𝟎})$ be the open unit disc centred at the origin. Show that $u(𝒙)=1-|𝒙|$ satisfies the Eikonal equation for all $𝒙\in B_1(\mathrm{𝟎})$ except for $𝒙=\mathrm{𝟎}$, where $u$ is not differentiable.

    • Let $\mathrm{\Omega }=(-1,1)\times (-1,1)$ be an open square. Show that the tent function

      satisfies the Eikonal equation for all $𝒙\in (-1,1)\times (-1,1)$ except along the lines $y=\pm x$, where $u$ is not differentiable.

    • Parts (i) and (ii) are a special case of the following: Define $u(𝒙)$ to be the distance from $𝒙$ to the boundary of $\mathrm{\Omega }$, i.e.,

      $$u(𝒙)=\mathrm{dist}(𝒙,\partial \mathrm{\Omega })=\underset{𝒑\in \partial \mathrm{\Omega }}{\min }|𝒙-𝒑|.$$

      Clearly $u(𝒙)=0$ when $𝒙\in \partial \mathrm{\Omega }$. Show that if $u$ is continuously differentiable in a neighbourhood of $𝒙$, then $|\nabla u(𝒙)|=1$.

      Hints: Let $u$ be continuously differentiable in a neighbourhood of $𝒙$. Then for all $𝒗\in {ℝ}^2$

      $$\nabla u(𝒙)\cdot 𝒗=\underset{h\to 0}{lim}\frac{u(𝒙+h𝒗)-u(𝒙)}{h}.$$

      Let ${𝒑}_{*}$ satisfy dist$(𝒙,\partial \mathrm{\Omega })=|𝒙-{𝒑}_{*}|$. Show that

      $$\frac{u(𝒙+h𝒗)-u(𝒙)}{h}\le \frac{|𝒙+h𝒗-{𝒑}_{*}|-|𝒙-{𝒑}_{*}|}{h}.$$

      Therefore deduce that

      $$\nabla u(𝒙)\cdot 𝒗\le \frac{𝒙-{𝒑}_{*}}{|𝒙-{𝒑}_{*}|}\cdot 𝒗.$$

      (Use that $\nabla |𝒙|=𝒙/|𝒙|$.) There was nothing special about our choice of $𝒗$, and so this expression also holds with $𝒗$ replaced by $-𝒗$. Use this to show that

      $$\nabla u(𝒙)\cdot 𝒗=\frac{𝒙-{𝒑}_{*}}{|𝒙-{𝒑}_{*}|}\cdot 𝒗.$$

      Since this holds for all $𝒗\in {ℝ}^2$, conclude that $|\nabla u(𝒙)|=1$.