Partial Differential Equations III/IV

[6pt] Exercise Sheet 2: Solutions

1. Burger’s equation as a model of traffic flow. We will rewrite the following PDE in the form of Burger’s equation:

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

   By expanding the second term on the left-hand side we obtain

   $${\rho }_t+v_m{\rho }_x\left(1-\frac{\rho }{{\rho }_m}\right)-\frac{v_m}{{\rho }_m}\rho {\rho }_x=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{\rho }_t+\left(v_m-\frac{2v_m}{{\rho }_m}\rho \right){\rho }_x=0.$$

   This suggest that we should define $u(x,t)=v_m-\frac{2v_m}{{\rho }_m}\rho (x,t)$. Then $\rho =\frac{{\rho }_m}{2v_m}(v_m-u)$ and

   ${\rho }_t+\left(v_m-{\displaystyle \frac{2v_m}{{\rho }_m}}\rho \right){\rho }_x=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}-{\displaystyle \frac{{\rho }_m}{2v_m}}{u}_t-{\displaystyle \frac{{\rho }_m}{2v_m}}u{u}_x=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{u}_t+u{u}_x=0.$

   Therefore $u$ satisfies Burger’s equation, as required.

2. Revision of ODEs. $$

   • Linear first-order ODEs. Multiplying the ODE by the integrating factor $e^{A(\tau )}$ gives

     $e^A\dot{x}+e^{A}ax=e^{A}b$ $⟺\mathit{\hspace{1em}}{\displaystyle \frac{d}{d\tau }}\left(e^{A}x\right)=e^{A}b$

     $⟺\mathit{\hspace{1em}}{\displaystyle {\int }_0^{\tau }}{\displaystyle \frac{d}{d\tau }}\left(e^{A}x\right)𝑑t={\displaystyle {\int }_0^{\tau }}{e}^{A(t)}b(t)𝑑t$

     $⟺\mathit{\hspace{1em}}{e}^{A(\tau )}x(\tau )-e^{A(0)}x(0)={\displaystyle {\int }_0^{\tau }}{e}^{A(t)}b(t)𝑑t$

     $⟺\mathit{\hspace{1em}}x(\tau )=e^{A(0)-A(\tau )}{x}_0+e^{-A(\tau )}{\displaystyle {\int }_0^{\tau }}{e}^{A(t)}b(t)𝑑t.$

     as required.

   • Separable ODEs. Since $f(x(\tau ))\ne 0$, we can divide the ODE by $f(x(\tau ))$ to obtain

     $\dot{x}(\tau )=f(x(\tau ))g(\tau )$ $⟺\mathit{\hspace{1em}}{\displaystyle \frac{\dot{x}(\tau )}{f(x(\tau ))}}=g(\tau )$

     $⟺\mathit{\hspace{1em}}{\displaystyle {\int }_0^{\tau }}{\displaystyle \frac{\dot{x}(t)}{f(x(t))}}𝑑t={\displaystyle {\int }_0^{\tau }}g(t)𝑑t$

     $⟺\mathit{\hspace{1em}}{\displaystyle {\int }_{x(0)}^{x(\tau )}}{\displaystyle \frac{1}{f(X)}}dX={\displaystyle {\int }_0^{\tau }}g(t)dt\mathit{\hspace{1em}}(\text{change variables: }X=x(t))$

     $⟺\mathit{\hspace{1em}}{\displaystyle {\int }_{x(0)}^{x(\tau )}}\dot{h}(X)𝑑X={\displaystyle {\int }_0^{\tau }}g(t)𝑑t$

     $⟺\mathit{\hspace{1em}}h(x(\tau ))-h(x(0))={\displaystyle {\int }_0^{\tau }}g(t)𝑑t$

     $⟺\mathit{\hspace{1em}}h(x(\tau ))=h(x_0)+{\displaystyle {\int }_0^{\tau }}g(t)𝑑t.$

     Observe that $h^{\prime }(x(\tau ))=1/f(x(\tau ))>0$ by assumption. Therefore $h$ is invertible in a neighbourhood of $x(\tau )$, and so

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

     as required.

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

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

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

   The PDE has the form

   $$a_1(x,y,u(x,y))u_x(x,y)+a_2(x,y,u(x,y))u_y(x,y)=b(x,y,u(x,y))$$

   with

   $$a_1(x,y,z)=1,a_2(x,y,z)=1,b(x,y,z)=1-z.$$

   The value of $u$ is prescribed on the line $\mathrm{\Gamma }=ℝ\times \{0\}$, which we can parametrise by $\gamma (s)=(x_0(s),y_0(s))=(s,0)$, $s\in ℝ$. Define

   $$z_0(s)=x_0^2(s)=s^2.$$

   Step 1: We need to solve

   ${\tilde{x}}_{\tau }$ $=a_1(\tilde{x},\tilde{y},\tilde{z})=1,$ (1)

   ${\tilde{y}}_{\tau }$ $=a_2(\tilde{x},\tilde{y},\tilde{z})=1,$ (2)

   ${\tilde{z}}_{\tau }$ $=b(\tilde{x},\tilde{y},\tilde{z})=1-\tilde{z},$ (3)

   subject to the initial conditions

   $\tilde{x}(0,s)$ $=x_0(s)=s,$ (4)

   $\tilde{y}(0,s)$ $=y_0(s)=0,$ (5)

   $\tilde{z}(0,s)$ $=z_0(s)=s^2.$ (6)

   Equations (1), (4) imply that

   $$\tilde{x}(\tau ,s)=\tau +s.$$

   Equations (2), (5) imply that

   $$\tilde{y}(\tau ,s)=\tau .$$

   Multiply equation (3) by the integrating factor

   $$\exp \left\{\int 1𝑑\tau \right\}=e^{\tau }$$

   to obtain

   $$e^{\tau }{\tilde{z}}_{\tau }+e^{\tau }\tilde{z}=e^{\tau }\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\frac{\partial }{\partial \tau }(e^{\tau }\tilde{z})=e^{\tau }.$$

   Integrating from $0$ to $\tau$ gives

   $$e^{\tau }\tilde{z}(\tau ,s)-e^0\tilde{z}(0,s)={\int }_0^{\tau }{e}^t𝑑t\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{e}^{\tau }\tilde{z}(\tau ,s)-s^2=e^{\tau }-1\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\tilde{z}(\tau ,s)=1+e^{-\tau }(s^2-1).$$

   Step 2: We need to invert the map $(\tau ,s)\mapsto (\tilde{x}(\tau ,s),\tilde{y}(\tau ,s))=(\tau +s,\tau )$. Setting $(x,y)=(\tau +s,\tau )$ and solving for $\tau$ and $s$ in terms of $x$ and $y$ gives $\tau =y$, $s=x-\tau =x-y$. Therefore

   $$\tilde{\tau }(x,y)=y,\tilde{s}(x,y)=x-y.$$

   Step 3: Finally,

   $$u(x,y)=\tilde{z}(\tilde{\tau }(x,y),\tilde{s}(x,y))=\boxed{1+e^{-y}({(x-y)}^2-1)}$$

   It is easy to check that $u$ satisfies the Cauchy problem.

   Plotting the characteristics: The characteristics are the curves

   $$\tau \mapsto (\tilde{x}(\tau ,s),\tilde{y}(\tau ,s))=(\tau +s,\tau )=(s,0)+\tau (1,1),$$

   which are in fact straight lines. We can write these lines in nonparametric form as $y=x-s$. Some representative characteristics are plotted below.

   

4. The method of characteristics for a linear PDE. We solve the linear first-order PDE

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

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

   where $a>0$ is a constant.

   The PDE has the form

   $$a_1(x,y,u)u_x+a_2(x,y,u)u_y=b(x,y,u)$$

   with

   $$a_1(x,y,z)=y,a_2(x,y,z)=-x,b(x,y,z)=0.$$

   The value of $u$ is prescribed on the line $\mathrm{\Gamma }=\{0\}\times (-a,a)$, which we can parametrise by $\gamma (s)=(x_0(s),y_0(s))=(0,s)$, $s\in (-a,a)$. Define

   $$z_0(s)=u(0,y_0(s))=\sqrt{a^2-s^2}.$$

   Step 1: We need to solve

   ${\tilde{x}}_{\tau }$ $=\tilde{y},$ (7)

   ${\tilde{y}}_{\tau }$ $=-\tilde{x},$ (8)

   ${\tilde{z}}_{\tau }$ $=0,$ (9)

   subject to the initial conditions

   $\tilde{x}(0,s)$ $=x_0(s)=0,$ (10)

   $\tilde{y}(0,s)$ $=y_0(s)=s,$ (11)

   $\tilde{z}(0,s)$ $=z_0(s)=\sqrt{a^2-s^2}.$ (12)

   The system of ODEs for $\tilde{x}$ and $\tilde{y}$ decouples from the ODE for $\tilde{z}$. Let $\widetilde{𝒙}=(\tilde{x},\tilde{y})$. Multiply equation (7) by $\tilde{x}$, equation (8) by $\tilde{y}$, and add the resulting equations together to get

   $$\tilde{x}{\tilde{x}}_{\tau }+\tilde{y}{\tilde{y}}_{\tau }=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\frac{\partial }{\partial \tau }({\tilde{x}}^2+{\tilde{y}}^2)=0.$$

   Therefore $|\widetilde{𝒙}|$ is independent of $\tau$ and $|\widetilde{𝒙}(\tau ,s)|=|\widetilde{𝒙}(0,s)|=|({\tilde{x}}_0(s),{\tilde{y}}_0(s))|=s$. Since $|\widetilde{𝒙}|=s$ is constant in $\tau$, we seek a solution of the form

   $$\tilde{x}(\tau ,s)=s\cos \theta (\tau ,s),\tilde{y}(\tau ,s)=s\sin \theta (\tau ,s),$$

   where $\theta (\tau ,s)$ is to be determined. Substituting these expressions into equations (7) and (8) gives

   $-s{\theta }_{\tau }\sin \theta =s\sin \theta ,$

   $s{\theta }_{\tau }\cos \theta =-s\cos \theta .$

   It follows that ${\theta }_{\tau }=-1$ and $\theta (\tau ,s)=-\tau +{\theta }_0(s)$. The initial conditions $\tilde{x}(0,s)=0$, ${\tilde{y}}_0(s)=s$ imply that ${\theta }_0=\pi /2$. Therefore

   $\tilde{x}(\tau ,s)$ $=s\cos \left({\displaystyle \frac{\pi }{2}}-\tau \right),$

   $\tilde{y}(\tau ,s)$ $=s\sin \left({\displaystyle \frac{\pi }{2}}-\tau \right).$

   Remark: An alternative way of finding $\tilde{x}$ and $\tilde{y}$ is to differentiate equation (7) with respect to $\tau$ to get ${\tilde{x}}_{\tau \tau }={\tilde{y}}_{\tau }=-\tilde{x}$. Now solve this second-order ODE for $\tilde{x}$.

   Equations (9) and (12) imply that

   $$\tilde{z}(\tau ,s)=\sqrt{a^2-s^2}.$$

   Step 2: If $(\tilde{x},\tilde{y})=(s\cos (\pi /2-\tau ),s\sin (\pi /2-\tau ))$, then ${\tilde{x}}^2+{\tilde{y}}^2=s^2$. It follows that

   $$\tilde{s}{(x,y)}^2=x^2+y^2.$$

   Since $\tilde{z}$ is independent of $\tau$, we do not need to know $\tilde{\tau }$ in order to find $u$.

   Step 3: Define

   $$u(x,y)=\tilde{z}(\tilde{\tau }(x,y),\tilde{s}(x,y))=\sqrt{a^2-\tilde{s}{(x,y)}^2}=\boxed{\sqrt{a^2-x^2-y^2}}$$

   It is easy to check that $u$ satisfies the initial value problem.

5. The method of characteristics for a semilinear PDE. We solve the semilinear first-order PDE

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

   subject to the condition $u(0,y)=y^2$ for all $y\in ℝ$. The PDE has the form

   $$a_1(x,y,u)u_x+a_2(x,y,u)u_y=b(x,y,u)$$

   with

   $$a_1(x,y,z)=-1,a_2(x,y,z)=x^2,b(x,y,z)=x^2{z}^2\cos y.$$

   %

   The value of $u$ is prescribed on the line $\mathrm{\Gamma }=\{0\}\times ℝ$, which we can parametrise by $\gamma (s)=(x_0(s),y_0(s))=(0,s)$, $s\in ℝ$. Define

   $$z_0(s)=u(x_0(s),y_0(s))=u(0,s)=s^2.$$

   Now we use the Method of Characteristics:

   Step 1: We need to solve

   ${\tilde{x}}_{\tau }$ $=-1,$ (13)

   ${\tilde{y}}_{\tau }$ $={\tilde{x}}^2,$ (14)

   ${\tilde{z}}_{\tau }$ $={\tilde{x}}^2{\tilde{z}}^2\cos \tilde{y},$ (15)

   subject to the initial conditions

   $\tilde{x}(0,s)$ $=0,$ (16)

   $\tilde{y}(0,s)$ $=s,$ (17)

   $\tilde{z}(0,s)$ $=s^2.$ (18)

   Equations (13), (16) imply that

   $$\tilde{x}(\tau ,s)=-\tau .$$

   Substituting this into equation (14) gives

   $${\tilde{y}}_{\tau }={\tau }^2.$$ (19)

   Equations (19), (17) imply that

   $$\tilde{y}(\tau ,s)=\frac{1}{3}{\tau }^3+s.$$

   Substituting this into equation (18) gives

   $${\tilde{z}}_{\tau }={\tau }^2\cos \left(\frac{1}{3}{\tau }^3+s\right){\tilde{z}}^2.$$

   We solve this using the method of separation:

   ${\tilde{z}}_{\tau }={\tau }^2\cos \left(\frac{1}{3}{\tau }^3+s\right){\tilde{z}}^2$ $⟺\mathit{\hspace{1em}}{\displaystyle {\int }_0^{\tau }}{\displaystyle \frac{1}{{\tilde{z}}^2(t,s)}}{\tilde{z}}_{\tau }(t,s)𝑑t={\displaystyle {\int }_0^{\tau }}{t}^2\cos \left(\frac{1}{3}{t}^3+s\right)𝑑t$

   $⟺\mathit{\hspace{1em}}{\displaystyle {\int }_{\tilde{z}(0,s)}^{\tilde{z}(\tau ,s)}}{\displaystyle \frac{1}{Z^2}}𝑑Z={\sin \left(\frac{1}{3}{t}^3+s\right)|}_0^{\tau }$

   $⟺\mathit{\hspace{1em}}{\displaystyle \frac{1}{\tilde{z}(0,s)}}-{\displaystyle \frac{1}{\tilde{z}(\tau ,s)}}=\sin \left(\frac{1}{3}{\tau }^3+s\right)-\sin s$

   $⟺\mathit{\hspace{1em}}\tilde{z}(\tau ,s)={\displaystyle \frac{1}{\frac{1}{s^2}-\sin \left(\frac{1}{3}{\tau }^3+s\right)+\sin s}}$

   $⟺\mathit{\hspace{1em}}\tilde{z}(\tau ,s)={\displaystyle \frac{s^2}{1-s^2\left[\sin \left(\frac{1}{3}{\tau }^3+s\right)-\sin s\right]}}$

   since $\tilde{z}(0,s)=s^2$.

   Let us underline that the definition of $\tilde{z}$ is meaningful, one for such values of $\tau ,s$ for which

   $$1-s^2\left[\sin \left(\frac{1}{3}{\tau }^3+s\right)-\sin s\right]\ne 0.$$

   Let us discuss for which values of $\tau ,s$ can we have

   $$\sin ({\tau }^3/6)\cos (s+{\tau }^3/6)=\frac{1}{2s^2}.$$

   Let us set $\tau :=-{(6s)}^{\frac{1}{3}}$, then the previous equality reduces to

   $$\sin (s)=-\frac{1}{2s^2},$$

   which clearly has infinitely many solutions, since $-\frac{1}{2s^2}\in [-1/2,0)$ as $|s|\ge 1$. So for these values of $\tau ,s$ for instance, $\tilde{z}$ is not well-defined.

   Step 2: We need to invert the map $(\tau ,s)\mapsto (-\tau ,\frac{1}{3}{\tau }^3+s)$. Setting $(x,y)=(-\tau ,\frac{1}{3}{\tau }^3+s)$ and solving for $\tau$ and $s$ in terms of $x$ and $y$ gives $\tau =-x$, $s=y-\frac{1}{3}{\tau }^3=y+\frac{1}{3}{x}^3$. Therefore

   $\tilde{\tau }(x,y)$ $=-x,$

   $\tilde{s}(x,y)$ $=y+\frac{1}{3}{x}^3.$

   Step 3: Finally,

   $$u(x,y)=\tilde{z}(\tilde{\tau }(x,y),\tilde{s}(x,y))=\boxed{\frac{{(y+\frac{1}{3}{x}^3)}^2}{1+{(y+\frac{1}{3}{x}^3)}^2\left[\sin (y+\frac{1}{3}{x}^3)-\sin y\right]}}$$

   By the same arguments as at the end of Step 1, this definition of $u$ is meaningful only for points, where the denominator is nonzero. As we have seen there, there are infinitely many points, where this is not the case.

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

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

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

   The PDE has the form

   $$a_1(x,t,u)u_x+a_2(x,t,u)u_t=b(x,t,u)$$

   with

   $$a_1(x,t,z)=t+z,a_2(x,t,z)=t,b(x,t,z)=x-t.$$

   The value of $u$ is prescribed on the line $\mathrm{\Gamma }=ℝ\times \{1\}$, which we can parametrise by $\gamma (s)=(x_0(s),t_0(s))=(s,1)$, $s\in ℝ$. Define

   $$z_0(s)=u(x_0(s),t_0(s))=1+t_0(s)=1+s.$$

   Now we use the Method of Characteristics:

   Step 1: We need to solve

   ${\tilde{x}}_{\tau }$ $=\tilde{t}+\tilde{z},$ (20)

   ${\tilde{t}}_{\tau }$ $=\tilde{t},$ (21)

   ${\tilde{z}}_{\tau }$ $=\tilde{x}-\tilde{t},$ (22)

   subject to the initial conditions

   $\tilde{x}(0,s)$ $=s,$ (23)

   $\tilde{t}(0,s)$ $=1,$ (24)

   $\tilde{z}(0,s)$ $=1+s.$ (25)

   Equations (21) and (24) imply that

   $$\tilde{t}(\tau ,s)=e^{\tau }.$$

   Differentiating equation (20) with respect to $\tau$ gives

   $${\tilde{x}}_{\tau \tau }={\tilde{t}}_{\tau }+{\tilde{z}}_{\tau }=\tilde{t}+\tilde{x}-\tilde{t}=\tilde{x}$$

   by equations (21) and (22). The linear second-order ODE ${\tilde{x}}_{\tau \tau }=\tilde{x}$ has solution

   $$\tilde{x}(\tau ,s)=A{e}^{\tau }+B{e}^{-\tau },$$

   where $A$ and $B$ are constants determined by the initial conditions $\tilde{x}(0,s)=s$ and

   $${\tilde{x}}_{\tau }(0,s)\stackrel{(\mathit{\text{<a href="\#S0.E20" title="(20) ‣ item 6" class="ltx\_ref ltx\_font\_italic" style="font-size:70\%;"><span class="ltx\_text ltx\_ref\_tag">20</span></a>}})}{=}\tilde{t}(0,s)+\tilde{z}(0,s)=2+s.$$

   The initial conditions give

   $A+B$ $=s,$

   $A-B$ $=2+s.$

   Therefore $A=1+s$ and $B=-1$ and

   $$\tilde{x}(\tau ,s)=(1+s)e^{\tau }-e^{-\tau }.$$

   Substituting this into equation (22) yields

   $${\tilde{z}}_{\tau }=(1+s)e^{\tau }-e^{-\tau }-e^{\tau }=s{e}^{\tau }-e^{-\tau }.$$

   By integrating we find that

   $$\tilde{z}(\tau ,s)=\tilde{z}(0,s)+{\int }_0^{\tau }{\tilde{z}}_{\tau }(t,s)𝑑t=\tilde{z}(0,s)+{\int }_0^{\tau }(s{e}^t-e^{-t})𝑑t=1+s+s(e^{\tau }-1)+(e^{-\tau }-1)=s{e}^{\tau }+e^{-\tau }.$$

   Step 2: We need to invert the map $(\tau ,s)\mapsto (\tilde{x}(\tau ,s),\tilde{t}(\tau ,s))=((1+s)e^{\tau }-e^{-\tau },e^{\tau })$. Setting $(x,t)=((1+s)e^{\tau }-e^{-\tau },e^{\tau })$ and solving for $\tau$ and $s$ in terms of $x$ and $t$ gives

   $$\tau =\log t,s=e^{-\tau }(x+e^{-\tau })-1=\frac{1}{t}\left(x+\frac{1}{t}\right)-1.$$

   Therefore

   $$\tilde{\tau }(x,t)=\log t,\tilde{s}(x,t)=\frac{1}{t}\left(x+\frac{1}{t}\right)-1.$$

   Step 3: We have discovered that

   $u(x,t)$ $=\tilde{z}(\tilde{\tau }(x,t),\tilde{s}(x,t))=\tilde{s}(x,t)e^{\tilde{\tau }(x,t)}+e^{-\tilde{\tau }(x,t)}$

   $=\left({\displaystyle \frac{1}{t}}\left(x+{\displaystyle \frac{1}{t}}\right)-1\right)t+{\displaystyle \frac{1}{t}}=\boxed{x+{\displaystyle \frac{2}{t}}-t}$

   Let’s verify that $u$ satisfies the Cauchy problem:

   $$(t+u)u_x+t{u}_t=\left(x+\frac{2}{t}\right)+t\left(-\frac{2}{t^2}-1\right)=x-t\mathit{\hspace{1em}}\mathrm{✓}$$

   $$u(x,1)=x+1\mathit{\hspace{1em}}\mathrm{✓}$$

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 $u(x_1,x_1)=\frac{1}{2}{x}_1$. The PDE has the form

   $$a_1(x_1,x_2,u)u_{x_1}+a_2(x_1,x_2,u)u_{x_2}=b(x_1,x_2,u)$$

   with

   $$a_1(x_1,x_2,z)=z,a_2(x_1,x_2,z)=1,b(x,t,z)=1.$$

   The value of $u$ is prescribed on the line $\mathrm{\Gamma }=\{(x_1,x_1):x_1\in ℝ\}$, which we can parametrise by $\gamma (s)=(x_1{}_0(s),x_2{}_0(s))=(s,s)$, $s\in ℝ$. Define

   $$z_0(s)=u(x_1{}_0(s),x_2{}_0(s))=u(s,s)=\frac{1}{2}s.$$

   Now we use the Method of Characteristics:

   Step 1: We need to solve

   ${\displaystyle \frac{\partial {\tilde{x}}_1}{\partial \tau }}$ $=\tilde{z},$ (26)

   ${\displaystyle \frac{\partial {\tilde{x}}_2}{\partial \tau }}$ $=1,$ (27)

   ${\displaystyle \frac{\partial \tilde{z}}{\partial \tau }}$ $=1,$ (28)

   subject to the initial conditions

   ${\tilde{x}}_1(0,s)$ $=s,$ (29)

   ${\tilde{x}}_2(0,s)$ $=s,$ (30)

   $\tilde{z}(0,s)$ $={\displaystyle \frac{1}{2}}s.$ (31)

   Equations (27), (30) imply that

   $${\tilde{x}}_2(\tau ,s)=\tau +s,$$

   and equations (28), (31) imply that

   $$\tilde{z}(\tau ,s)=\tau +\frac{1}{2}s.$$

   Substituting this into equation (26) yields

   $$\frac{\partial {\tilde{x}}_1}{\partial \tau }=\tau +\frac{1}{2}s.$$

   Integrating gives

   $${\tilde{x}}_1(\tau ,s)={\tilde{x}}_1(0,s)+{\int }_0^{\tau }\left(t+\frac{1}{2}s\right)𝑑t=s+\frac{1}{2}{\tau }^2+\frac{1}{2}s\tau =s+\frac{1}{2}\tau (\tau +s).$$

   Step 2: We need to invert the map $(\tau ,s)\mapsto ({\tilde{x}}_1(\tau ,s),{\tilde{x}}_2(\tau ,s))=(s+\frac{1}{2}\tau (\tau +s),\tau +s)$. Let us notice that this is not possible in neighborhoods of points $(\tau ,s)$ for which $\tau +s=2$.

   We set $(x_1,x_2)=(s+\frac{1}{2}\tau (\tau +s),\tau +s)$ and try to solve for $\tau$ and $s$ in terms of $x_1$ and $x_2$. We get $\tau =x_2-s$ and

   $$x_1=s+\frac{1}{2}\tau (\tau +s)=s+\frac{1}{2}(x_2-s)x_2\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}s=\frac{x_1-\frac{1}{2}{x}_2^2}{1-\frac{1}{2}{x}_2}=\frac{2x_1-x_2^2}{2-x_2}.$$

   Therefore

   $$\tau =x_2-\frac{2x_1-x_2^2}{2-x_2}=\frac{2(x_2-x_1)}{2-x_2}$$

   and

   $$\tilde{\tau }(x_1,x_2)=\frac{2(x_2-x_1)}{2-x_2}\mathit{\hspace{1em}\hspace{1em}}\tilde{s}(x_1,x_2)=\frac{2x_1-x_2^2}{2-x_2}.$$

   We notice that these terms are not meaningful for $x_2=2$. This is corresponding exactly to the case where $\tau +s=2$ (since $x_2=\tau +s$).

   Step 3: Finally

   $u(x_1,x_2)$ $=\tilde{z}(\tilde{\tau }(x_1,x_2),\tilde{s}(x_1,x_2))=\tilde{\tau }(x_1,x_2)+{\displaystyle \frac{1}{2}}\tilde{s}(x_1,x_2)$

   $={\displaystyle \frac{2(x_2-x_1)}{2-x_2}}+{\displaystyle \frac{2x_1-x_2^2}{2(2-x_2)}}$

   $=\boxed{{\displaystyle \frac{4x_2-2x_1-x_2^2}{2(2-x_2)}}}$

   This expression is meaningful only when $x_2\ne 0$. If $(x_1,x_2)=(2,2)$, then the value of the solution is given by the value of the Cauchy data, for any other point $(x_1,2)$ the solution is not defined.

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

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

   $u(x,0)=g(x)$ $\text{for }x\in ℝ.$

   This equation has the form $a_1{u}_x+a_2{u}_t=b$ with

   $$a_1(x,t,z)=c,a_2(x,t,z)=1,b(x,t,z)=0.$$

   The value of $u$ is prescribed on the line $\mathrm{\Gamma }=ℝ\times \{0\}$. We can parametrise $\mathrm{\Gamma }$ by $\gamma :ℝ\to \mathrm{\Gamma }$, $\gamma (s)=(x_0(s),y_0(s))=(\lambda s,0)$, $\lambda >0$. Define

   $$z_0(s)=u_0(x_0(s),t_0(s))=g(x_0(s))=g(\lambda s).$$

   Now we use the Method of Characteristics:

   Step 1: We need to solve

   ${\tilde{x}}_{\tau }$ $=c,$ (32)

   ${\tilde{t}}_{\tau }$ $=1,$ (33)

   ${\tilde{z}}_{\tau }$ $=0,$ (34)

   subject to the initial conditions

   $\tilde{x}(0,s)$ $=\lambda s,$ (35)

   $\tilde{t}(0,s)$ $=0,$ (36)

   $\tilde{z}(0,s)$ $=g(\lambda s).$ (37)

   Equation (32) implies that

   $$\tilde{x}(\tau ,s)=c\tau +A(s),$$

   where $A$ is to be determined. Equation (35) implies that $A(s)=\lambda s$. Therefore

   $$\tilde{x}(\tau ,s)=c\tau +\lambda s.$$

   Similarly, equations (33) and (36) imply that

   $$\tilde{t}(\tau ,s)=\tau ,$$

   and equations (34) and (37) imply that

   $$\tilde{z}(\tau ,s)=g(\lambda s).$$

   Step 2: Now we need to invert the function $(\tilde{x}(\tau ,s),\tilde{t}(\tau ,s))=(c\tau +\lambda s,\tau )$. Setting $(x,t)=(c\tau +\lambda s,\tau )$ and solving for $\tau$ and $s$ in terms of $x$ and $t$ gives $\tau =t$ and $s=(x-c\tau )/\lambda =(x-ct)/\lambda$. Therefore

   $$\tilde{\tau }(x,t)=t,\tilde{s}(x,t)=\frac{x-ct}{\lambda }.$$

   Step 3: Finally,

   $$u(x,t):=\tilde{z}(\tilde{\tau }(x,t),\tilde{s}(x,t))=g(\lambda \tilde{s}(x,t))=g(x-ct)$$

   as required.

9. Nonexistence due to a characteristic point.

   • We can parametrise $\partial \mathrm{\Omega }$ by $\gamma (\theta )=(\cos \theta ,\sin \theta )$, $\theta \in [0,2\pi )$. The point $𝒑$ corresponds to parameter $\theta =\frac{\pi }{4}$ since $\gamma (\frac{\pi }{4})=(\cos \frac{\pi }{4},\sin \frac{\pi }{4})=𝒑$. The vector $𝒏(\theta )=(\cos \theta ,\sin \theta )$ is the unit outward-pointing normal to $\partial \mathrm{\Omega }$ at point $\gamma (\theta )$. We have

     $$𝒂(\cos \theta ,\sin \theta )\cdot 𝒏(\theta )=(-1,1)\cdot (\cos \theta ,\sin \theta )=\sin \theta -\cos \theta .$$

     In particular, taking $\theta =\frac{\pi }{4}$ gives

     $$𝒂(\cos \frac{\pi }{4},\sin \frac{\pi }{4})\cdot 𝒏\left(\frac{\pi }{4}\right)=\sin \frac{\pi }{4}-\cos \frac{\pi }{4}=0.$$

     Therefore the vector field $𝒂$ is tangent to $\partial \mathrm{\Omega }$ at $𝒑$ and hence $𝒑$ is a characteristic point. All the other points of $\partial \mathrm{\Omega }$ are noncharacteristic except for the point $-𝒑$.

   • Suppose for contradiction that there exists a function $u:U\to ℝ$ satisfying

     $𝒂\cdot \nabla u=0$ $\text{in }U,$ (38)

     $u=g$ $\text{on }U\cap \partial \mathrm{\Omega },$ (39)

     where $U\subset {ℝ}^2$ is an open set containing $𝒑$. Chose $\varphi \in (0,\frac{\pi }{4})$ sufficiently small so that the line joining ${𝒑}_1:=\gamma (\frac{\pi }{4}-\varphi )$ to ${𝒑}_2:=\gamma (\frac{\pi }{4}+\varphi )$ lies in $U$. This line is tangent to the constant vector field $𝒂$ since

     ${𝒑}_2-{𝒑}_1$ $=\left(\begin{array}{c}\hfill \cos \left(\frac{\pi }{4}+\varphi \right)-\cos \left(\frac{\pi }{4}-\varphi \right)\hfill \\ \hfill \sin \left(\frac{\pi }{4}+\varphi \right)-\sin \left(\frac{\pi }{4}-\varphi \right)\hfill \end{array}\right)=\left(\begin{array}{c}\hfill -\sqrt{2}\sin \varphi \hfill \\ \hfill \sqrt{2}\sin \varphi \hfill \end{array}\right)=\sqrt{2}\sin \varphi \left(\begin{array}{c}\hfill -1\hfill \\ \hfill 1\hfill \end{array}\right)=\sqrt{2}\sin \varphi 𝒂.$

     Since $𝒂\cdot \nabla u=0$, then $u$ is constant in direction $𝒂$, and hence $u({𝒑}_1)=u({𝒑}_2)$. In case this doesn’t convince you, a more algebraic argument is as follows:

     $u({𝒑}_2)-u({𝒑}_1)$ $={\displaystyle {\int }_0^1}{\displaystyle \frac{\partial }{\partial t}}u(t{𝒑}_2+(1-t){𝒑}_1)𝑑t$ (Fundamental Theorem of Calculus)

     $={\displaystyle {\int }_0^1}\nabla u(t{𝒑}_2+(1-t){𝒑}_1)\cdot ({𝒑}_2-{𝒑}_1)𝑑t$ (Chain Rule)

     $=\sqrt{2}\sin \varphi {\displaystyle {\int }_0^1}\underset{=0}{\underbrace{\nabla u(t{𝒑}_2+(1-t){𝒑}_1)\cdot 𝒂}}𝑑t$ (equation (38))

     $=0,$

     which again shows that $u({𝒑}_1)=u({𝒑}_2)$. This contradicts the Cauchy condition (39) since

     $$u({𝒑}_1)=g({𝒑}_1)=\cos \left(\frac{\pi }{4}-\varphi \right)\ne \cos \left(\frac{\pi }{4}+\varphi \right)=g({𝒑}_2)=u({𝒑}_2).$$

     (Sketch the graph of the cosine function to convince yourself of this, and recall that $\varphi \in (0,\frac{\pi }{4})$.)

     Finally, the Cauchy problem has a global solution $u:\overline{\mathrm{\Omega }}\to ℝ$ for any continuously differentiable function $g$ satisfying

     $$g\left(\gamma \left(\frac{\pi }{4}-\varphi \right)\right)=g\left(\gamma \left(\frac{\pi }{4}+\varphi \right)\right)\mathit{\hspace{1em}}\forall \varphi \in [0,\pi ].$$

     For example, we could take $g(\cos \theta ,\sin \theta )=\sin (2\theta )$. Another possibility is to simply take $g(x,y)=x+y$, in which case the solution will be $u(x,y)=x+y$.

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

    $u_x(x,y)+u_y(x,y)=1$ $\text{for }(x,y)\in {ℝ}^2,$ (40)

    $u(x,x)=x$ $\text{for }x\in ℝ.$

    • The PDE has the form

      $$a_1(x,y,u(x,y))u_x(x,y)+a_2(x,y,u(x,y))u_y(x,y)=b(x,y,u(x,y))$$

      with

      $$a_1(x,y,z)=1,a_2(x,y,z)=1,b(x,y,z)=1.$$

      Let $𝒂=(a_1,a_2)$. The value of $u$ is prescribed on the Cauchy curve $\mathrm{\Gamma }=\{(x,x):x\in ℝ\}$, which can be parametrised by $\gamma (s)=(x_0(s),y_0(s))=(s,s)$, $s\in ℝ$. We can write the Cauchy condition as $u(\gamma (s))=z_0(s)$ with $z_0(s)=x_0(s)=s$. The vector ${\gamma }_s(s)=(1,1)$ is tangent to $\mathrm{\Gamma }$ at the point $\gamma (s)$. Therefore $𝒏(s)=(1,-1)$ is normal to $\mathrm{\Gamma }$ at each point and

      $$𝒂(\gamma (s),z_0(s))\cdot 𝒏(s)=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 1\hfill \end{array}\right)\cdot \left(\begin{array}{c}\hfill 1\hfill \\ \hfill -1\hfill \end{array}\right)=0$$

      for all $s\in ℝ$. Therefore no point on the Cauchy curve is noncharacteristic.

      Alternative proof: We have

      $$a_1(x_0(s),y_0(s),z_0(s))x_0^{\prime }(s)-a_2(x_0(s),y_0(s),z_0(s))y_0^{\prime }(s)=1-1=0$$

      for all $s\in ℝ$, and again we see that no point on the Cauchy curve is noncharacteristic.

      The vector field $𝒂(x,y,z)=(1,1)$ is constant and is tangent to the Cauchy curve $\mathrm{\Gamma }$ at every point. Therefore the flow of the vector field $𝒂(x,y,u(x,y))$ through $\mathrm{\Gamma }$ lies along $\mathrm{\Gamma }$ and the characteristics coincide with $\mathrm{\Gamma }$, as shown in the figure below. The vector field $𝒂$ is also plotted.

      

    • We 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 $f(0)=0$. As above, the PDE has the form

      $$a_1(x,y,u(x,y))u_x(x,y)+a_2(x,y,u(x,y))u_y(x,y)=b(x,y,u(x,y))$$

      with

      $$a_1(x,y,z)=1,a_2(x,y,z)=1,b(x,y,z)=1.$$

      The Cauchy curve is $\mathrm{\Gamma }=ℝ\times \{0\}$, which we can parametrise by $\gamma (s)=(x_0(s),y_0(s))=(s,0)$, $s\in ℝ$. Let $z_0(s)=f(x_0(s))=f(s)$.

      Step 1: We need to solve

      ${\tilde{x}}_{\tau }$ $=a_1(\tilde{x},\tilde{y},\tilde{z})=1,$ (41)

      ${\tilde{y}}_{\tau }$ $=a_2(\tilde{x},\tilde{y},\tilde{z})=1,$ (42)

      ${\tilde{z}}_{\tau }$ $=b(\tilde{x},\tilde{y},\tilde{z})=1,$ (43)

      subject to the initial conditions

      $\tilde{x}(0,s)$ $=x_0(s)=s,$ (44)

      $\tilde{y}(0,s)$ $=y_0(s)=0,$ (45)

      $\tilde{z}(0,s)$ $=z_0(s)=f(s).$ (46)

      Equations (41), (44) imply that

      $$\tilde{x}(\tau ,s)=\tau +s.$$

      Equations (42), (45) imply that

      $$\tilde{y}(\tau ,s)=\tau .$$

      Equations (43), (46) imply that

      $$\tilde{z}(\tau ,s)=\tau +f(s).$$

      Step 2: We need to invert the map $(\tau ,s)\mapsto (\tilde{x}(\tau ,s),\tilde{y}(\tau ,s))=(\tau +s,\tau )$. Setting $(x,y)=(\tau +s,\tau )$ and solving for $\tau$ and $s$ in terms of $x$ and $y$ gives $\tau =y$, $s=x-\tau =x-y$. Therefore

      $$\tilde{\tau }(x,y)=y,\tilde{s}(x,y)=x-y.$$

      Step 3: Define

      $$u(x,y):=\tilde{z}(\tilde{\tau }(x,y),\tilde{s}(x,y))=\tilde{\tau }(x,y)+f(\tilde{s}(x,y))=y+f(x-y).$$

      Observe that

      $$u_x+u_y=f^{\prime }(x-y)+1-f^{\prime }(x-y)=1,u(x,x)=x+f(x-x)=x+f(0)=x.$$

      Therefore $u$ satisfies the Cauchy problem (40) for all differentiable functions $f$ with $f(0)=0$. Hence the Cauchy problem (40) has infinitely many solutions.

    • Suppose that $u$ satisfies the Cauchy problem

      $u_x(x,y)+u_y(x,y)=1$ $\mathrm{\hspace{1em}}\text{for }(x,y)\in {ℝ}^2,$ (47)

      $u(x,x)=1$ $\mathrm{\hspace{1em}}\text{for }x\in ℝ.$ (48)

      Differentiating equation (48) gives

      $$u_x(x,x)+u_y(x,x)=0\mathit{\hspace{1em}}\forall x\in ℝ.$$

      But this contradicts equation (47), which states that

      $$u_x(x,x)+u_y(x,x)=1\mathit{\hspace{1em}}\forall x\in ℝ.$$

11. Obstacles to global existence.

    • Let $u:\overline{\mathrm{\Omega }}\setminus B_{\epsilon }(\mathrm{𝟎})\to ℝ$ satisfy

      $𝒂\cdot \nabla u=0$ $\text{in }\mathrm{\Omega }\setminus B_{\epsilon }(\mathrm{𝟎}),$

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

      Here we derive the solution without using the method of characteristics. Since $0=𝒂\cdot \nabla u=-𝒙\cdot \nabla u$, then $u$ is constant in the radial direction. If this isn’t obvious to you, then argue as follows: For all $r>0$, $\theta \in ℝ$,

      ${\displaystyle \frac{\partial }{\partial r}}u(r𝒆(\theta ))$ $=\nabla u(r𝒆(\theta ))\cdot 𝒆(\theta )$ (Chain Rule)

      $=\nabla u(𝒙)\cdot {\displaystyle \frac{𝒙}{|𝒙|}}$ $(𝒙:=r𝒆(\theta ))$

      $=-{\displaystyle \frac{1}{|𝒙|}}𝒂\cdot \nabla u(𝒙)$ $\text{(since }𝒂=-𝒙)$

      $=0,$

      which means that $u$ is constant in the radial direction. Therefore

      $$u(r𝒆(\theta ))=u(1\cdot 𝒆(\theta ))=u(𝒆(\theta ))=g(𝒆(\theta ))$$

      as required.

      Let $g$ be a non-constant function. Assume for contradiction that $u:\overline{\mathrm{\Omega }}\to ℝ$ is a global solution. By the calculation above $\frac{\partial }{\partial r}u(r𝒆(\theta ))=0$ for all $r>0$. Therefore

      $$0={\int }_0^1\frac{\partial }{\partial r}u(r𝒆(\theta ))𝑑r=u(1\cdot 𝒆(\theta ))-u(0\cdot 𝒆(\theta ))=g(𝒆(\theta ))-u(\mathrm{𝟎}).$$

      Therefore $g(𝒆(\theta ))=u(\mathrm{𝟎})$ for all $\theta \in ℝ$, which is a contradiction since $g$ is non-constant.

      If $g$ is a constant function, then the global solution is clearly $u=g$.

    • Let $\mathrm{\Gamma }=\{0\}\times ℝ$ and let $u:\overline{\mathrm{\Omega }}\to ℝ$ satisfy

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

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

      Since $𝒂(x,y)=(1,0)$, the PDE has the form

      $$a_1(x,y,u(x,y))u_x(x,y)+a_2(x,y,u(x,y))u_y(x,y)=b(x,y,u(x,y))$$

      with

      $$a_1(x,y,z)=1,a_2(x,y,z)=0,b(x,y,z)=1.$$

      We can parametrise $\mathrm{\Gamma }$ by $\gamma (s)=(x_0(s),y_0(s))=(0,s)$, $s\in ℝ$. Define

      $$z_0(s)=u(\gamma (s))=u_0(x_0(s),y_0(s))=y_0(s)=s.$$

      Step 1: We need to solve

      ${\tilde{x}}_{\tau }$ $=a_1(\tilde{x},\tilde{y},\tilde{z})=1,$ (49)

      ${\tilde{y}}_{\tau }$ $=a_2(\tilde{x},\tilde{y},\tilde{z})=0,$ (50)

      ${\tilde{z}}_{\tau }$ $=b(\tilde{x},\tilde{y},\tilde{z})=1,$ (51)

      subject to the initial conditions

      $\tilde{x}(0,s)$ $=x_0(s)=0,$ (52)

      $\tilde{y}(0,s)$ $=y_0(s)=s,$ (53)

      $\tilde{z}(0,s)$ $=z_0(s)=s.$ (54)

      Equations (49), (52) imply that

      $$\tilde{x}(\tau ,s)=\tau .$$

      Equations (50), (53) imply that

      $$\tilde{y}(\tau ,s)=s.$$

      Equations (51), (54) imply that

      $$\tilde{z}(\tau ,s)=\tau +s.$$

      Step 2: We need to invert the map $(\tau ,s)\mapsto (\tilde{x}(\tau ,s),\tilde{y}(\tau ,s))=(\tau ,s)$. Clearly the inverse map is given by

      $$\tilde{\tau }(x,y)=x,\tilde{s}(x,y)=y.$$

      Step 3: Finally,

      $$u(x,y)=\tilde{z}(\tilde{\tau }(x,y),\tilde{s}(x,y))=\tilde{\tau }(x,y)+\tilde{s}(x,y)=x+y,$$

      as required. (It’s actually easier to simply integrate the PDE with respect to $x$, as below, instead of using the method of characteristics.)

      Now consider the case $\mathrm{\Gamma }=\partial \mathrm{\Omega }=(\{0\}\times ℝ)\cup (\{1\}\times ℝ)$. A quick sketch of the vector field shows that each characteristic intersects $\mathrm{\Gamma }$ twice, at $(0,y)$ and $(1,y)$ for some $y\in ℝ$. Let $u:([0,a]\cup [b,1])\times ℝ\to ℝ$ be a local solution, which satisfies

      $u_x(x,y)=1$

      $u(0,y)=u(1,y)=y$ $\text{for }y\in ℝ.$

      Let $x\in [0,a]$. Then

      $$u(x,y)=u(0,y)+{\int }_0^x{u}_x(z,y)𝑑z=y+{\int }_0^{x}1𝑑z=y+x,$$

      as required. Similarly, for $x\in [b,1]$,

      $$u(x,y)=u(1,y)-{\int }_x^1{u}_x(z,y)𝑑z=y-{\int }_x^{1}1𝑑z=y-(1-x)=x+y-1,$$

      as required.

      Assume for contraction that $u:\overline{\mathrm{\Omega }}\to ℝ$ is a global solution. Then

      $$1={\int }_0^{1}1𝑑z={\int }_0^1{u}_x(z,y)𝑑z=u(1,y)-u(0,y)=u_0(1,y)-u_0(0,y)=y-y=0.$$

      Contradiction! From this calculation we see that a necessary condition for a global solution to exist is $u_0(1,y)-u_0(0,y)=1$. In this case the global solution is $u(x,y)=x+y$.

12. Geometric interpretation of the method of characteristics. Let $𝒏=\nabla F$. Then $𝒏$ is normal to the solution surface and

    $𝒏\cdot 𝒇$ $=\nabla F\cdot (a_1,a_2,b)$

    $=(u_x,u_y,-1)\cdot (a_1,a_2,b)$

    $=a_1{u}_x+a_2{u}_y-b$

    $=0.$

    Therefore $𝒇$ is orthogonal to $𝒏$ at every point of the solution surface. Hence $𝒇$ is tangent to the solution surface, as required.

13. The method of characteristics for PDEs in $n$ independent variables. The transport equation in $n$ spatial variables is

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

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

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

    $$𝒂(x,t,u(𝒙,t))\cdot {\nabla }_{(𝒙,t)}u(𝒙,t)=b(𝒙,t,u(𝒙,t)),$$

    where ${\nabla }_{(𝒙,t)}$ is the gradient with respect to $(𝒙,t)$, i.e., ${\nabla }_{(𝒙,t)}=(\nabla ,{\partial }_t)$, and

    $$𝒂(𝒙,t,z)=\left[\begin{array}{c}\hfill 𝒄\hfill \\ \hfill 1\hfill \end{array}\right],b(𝒙,t,z)=0.$$

    The value of $u$ is prescribed on the $n$–dimensional hypersurface $\mathrm{\Gamma }={ℝ}^n\times \{0\}$, which we can parametrise by $\gamma (𝒔)=({𝒙}_0(𝒔),t_0(𝒔))=(𝒔,0)$, $𝒔\in {ℝ}^n$. Define

    $$z_0(𝒔)=u(\gamma (𝒔))=u(𝒔,0)=g(𝒔).$$

    Now we apply the Method of Characteristics:

    Step 1. We need to solve

    ${\widetilde{𝒙}}_{\tau }$ $=𝒄,$

    ${\tilde{t}}_{\tau }$ $=1,$

    ${\tilde{z}}_{\tau }$ $=0,$

    subject to the initial conditions

    $\widetilde{𝒙}(0,𝒔)$ $=𝒔,$

    $\tilde{t}(0,𝒔)$ $=0,$

    $\tilde{z}(0,𝒔)$ $=g(𝒔).$

    The solution is

    $\widetilde{𝒙}(\tau ,𝒔)$ $=𝒄\tau +𝒔,$

    $\tilde{t}(\tau ,𝒔)$ $=\tau ,$

    $\tilde{z}(\tau ,𝒔)$ $=g(𝒔).$

    Step 2. We need to invert the map $(\tau ,𝒔)\mapsto (\widetilde{𝒙}(\tau ,𝒔),\tilde{t}(\tau ,𝒔))=(𝒄\tau +𝒔,\tau )$. Setting $(𝒙,t)=(𝒄\tau +𝒔,\tau )$ and solving for $\tau$ and $𝒔$ in terms of $𝒙$ and $t$ gives $\tau =t$, $𝒔=𝒙-𝒄\tau =𝒙-𝒄t$. Therefore

    $\tilde{\tau }(𝒙,t)$ $=t,$

    $\widetilde{𝒔}(𝒙,t)$ $=𝒙-𝒄t.$

    Step 3. Finally,

    $$u(𝒙,t):=\tilde{z}(\tilde{\tau }(𝒙,t),\widetilde{𝒔}(𝒙,t))=g(𝒙-𝒄t),$$

    as required.

14. Uniqueness theorem for ODEs.

    • 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

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

      Taking the dot product of this equation with ${𝒙}_1-{𝒙}_2$ gives

      $$({\dot{𝒙}}_1-{\dot{𝒙}}_2)\cdot ({𝒙}_1-{𝒙}_2)=(𝒇({𝒙}_1)-𝒇({𝒙}_2))\cdot ({𝒙}_1-{𝒙}_2)\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\frac{d}{d\tau }\frac{1}{2}{|{𝒙}_1-{𝒙}_2|}^2=(𝒇({𝒙}_1)-𝒇({𝒙}_2))\cdot ({𝒙}_1-{𝒙}_2).$$

      Therefore by the Cauchy-Schwarz inequality

      ${\displaystyle \frac{d}{d\tau }}{\displaystyle \frac{1}{2}}{|{𝒙}_1-{𝒙}_2|}^2\le |𝒇({𝒙}_1)-𝒇({𝒙}_2)||{𝒙}_1-{𝒙}_2|\le L{|{𝒙}_1-{𝒙}_2|}^2$

      since $𝒇$ is Lipschitz continuous. Define $E(\tau )={|{𝒙}_1(\tau )-{𝒙}_2(\tau )|}^2$. We have shown that

      $$\dot{E}\le 2LE.$$

      Rearranging and multiplying by the integrating factor $e^{-2L\tau }$ gives

      $$e^{-2L\tau }\dot{E}(\tau )-2L{e}^{-2L\tau }E\le 0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\frac{d}{d\tau }(e^{-2L\tau }E(\tau ))\le 0.$$

      Integrating over the interval $[0,\tau ]$ yields

      $$e^{-2L\tau }E(\tau )-E(0)\le 0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}E(\tau )\le e^{2L\tau }E(0).$$

      But ${𝒙}_1(0)={𝒙}_2(0)={𝒙}_0$. Therefore $E(0)=0$ and hence $E(\tau )=0$ for all $\tau \in ℝ$. Therefore ${𝒙}_1(\tau )={𝒙}_2(\tau )$ for all $\tau \in ℝ$, as required.

    • Let $f\in C^1([a,b])$. By the Mean Value Theorem, there exists $c\in (a,b)$ such that

      $$f(y)-f(x)=f^{\prime }(c)(y-x).$$

      Therefore, for all $x,y\in [a,b]$,

      $$|f(y)-f(x)|=|f^{\prime }(c)||y-x|\le \underset{[a,b]}{\max }|f^{\prime }||y-x|=L|y-x|$$

      with $L={\max }_{[a,b]}|f^{\prime }|$.

15. Nonuniqueness for ODEs. Consider the separable ODE

    $\dot{x}(\tau )$ $={|x(\tau )|}^{1/2},$

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

    Exactly the same method we used in Example 2.3 leads to the solution $x:(-\mathrm{\infty },2)\to ℝ$,

    $$x(\tau )=-\frac{1}{4}{(2-\tau )}^2.$$

    At $\tau =2$, $x=0$, where $f(x)={|x|}^{1/2}$ is not continuously differentiable. Let $T>2$. For any $c\in (0,T-2)$, $x_c:(-\mathrm{\infty },T)\to ℝ$ is a solution, where

16. The Eikonal equation.

    • If $𝒙\in \partial \mathrm{\Omega }$, then $|𝒙|=1$ and hence $u(𝒙)=0$ for $𝒙\in \partial \mathrm{\Omega }$. Recall that $\nabla |𝒙|=𝒙/|𝒙|$ for $𝒙\ne \mathrm{𝟎}$. Therefore

      $$\nabla u=\nabla (1-|𝒙|)=-\frac{𝒙}{|𝒙|}\mathit{\hspace{1em}}⟹\mathit{\hspace{1em}}|\nabla u|=1$$

      for $𝒙\ne 0$, as required.

    • Clearly $u=0$ when $x=\pm 1$, $y=\pm 1$. Therefore $u(𝒙)=0$ if $𝒙\in \partial \mathrm{\Omega }$. We compute

      Therefore $|\nabla u|=1$ in $\mathrm{\Omega }$, except along the lines $y=\pm x$, as required.

    • 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 $𝒙\in \partial \mathrm{\Omega }$. Assume that $u$ is 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$(𝒙,\partial \mathrm{\Omega })=|𝒙-{𝒑}_{*}|$. Then

      ${\displaystyle \frac{u(𝒙+h𝒗)-u(𝒙)}{h}}$ $={\displaystyle \frac{{\min }_{𝒒\in \partial \mathrm{\Omega }}|𝒙+h𝒗-𝒒|-{\min }_{𝒑\in \partial \mathrm{\Omega }}|𝒙-𝒑|}{h}}$

      $={\displaystyle \frac{{\min }_{𝒒\in \partial \mathrm{\Omega }}|𝒙+h𝒗-𝒒|-|𝒙-{𝒑}_{*}|}{h}}$

      $\le {\displaystyle \frac{|𝒙+h𝒗-{𝒑}_{*}|-|𝒙-{𝒑}_{*}|}{h}}.$ (55)

      Observe that

      $$\underset{h\to 0}{lim}\frac{|𝒙+h𝒗-{𝒑}_{*}|-|𝒙-{𝒑}_{*}|}{h}=\nabla |𝒙+{𝒑}_{*}|\cdot 𝒗=\frac{𝒙-{𝒑}_{*}}{|𝒙-{𝒑}_{*}|}\cdot 𝒗.$$

      Therefore taking the limit $h\to 0$ in equation (55) gives

      $$\nabla u(𝒙)\cdot 𝒗\le \frac{𝒙-{𝒑}_{*}}{|𝒙-{𝒑}_{*}|}\cdot 𝒗.$$ (56)

      There was nothing special about our choice of $𝒗$, and so this expression also holds with $𝒗$ replaced by $-𝒗$:

      $$\nabla u(𝒙)\cdot (-𝒗)\le \frac{𝒙-{𝒑}_{*}}{|𝒙-{𝒑}_{*}|}\cdot (-𝒗)\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\nabla u(𝒙)\cdot 𝒗\ge \frac{𝒙-{𝒑}_{*}}{|𝒙-{𝒑}_{*}|}\cdot 𝒗.$$ (57)

      Combining (56) and (57) yields

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

      But this holds for all $𝒗\in {ℝ}^2$. Therefore

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

      and $|\nabla u(𝒙)|=1$, as required.