Partial Differential Equations III & V
Revision Class Solution

Solution:

1. Rewriting the equation as \[\nonumber \left\{ \begin{array}{ll} 0\;\partial_{x_1}u(x_1,x_2)+x_1^2\partial_{x_2} u (x_1,x_2) = 5, & (x_1,x_2)\in\mathbb{R}^2,\\ u(x_1,0) = 3, & x_1\in\mathbb{R}. \end{array} \right.\] we see that the leading vector field is \(\vec{a}\left( x_1,x_2\right)=\left( 0,x_1^2\right)\), the Cauchy curve is the \(x_1\) axis given by \[\{\gamma(s):=(s,0):s\in\mathbb{R}\}\subset\mathbb{R}^2,\] and the Cauchy datum is \(u\left( s,0\right)=3\).

2. As \(\gamma'(s) = \left( 1,0\right)\) we see that the condition for being non-characteristic is \[0\not=\vec{a}\left( \gamma(s)\right)\cdot \left( 0,-1\right)=-s^2,\] which shows that all the points on the Cauchy curve, except \((0,0)\), are non-characteristic.

3. The ODE system is given by \[\left\{ \begin{array}{l} \partial_\tau x_1(\tau,s) = 0,\\ \partial_\tau x_2(\tau,s) = x_1^2(\tau,s),\\ \partial_\tau z(\tau,s) = 5 \end{array} \right.\] with the initial datum \[x_1(0,s) = s,\ \ \ x_2(0,s)=0,\ \ z(0,s)=3.\] The solution is given by \[x_1(\tau,s)=x_1(0,s)=s,\] \[z(\tau,s) = 5\tau + z(0,s) = 3+5\tau,\] and since \(\partial_\tau x_2(\tau,s) = x_1^2(\tau,s)=s^2\) we have that \[x_2(\tau,s)=x_2(0,s)+s^2\tau =s^2\tau.\]

4. For a fixed \(s\in \mathbb{R}\setminus \left\lbrace 0\right\rbrace\) we find that the characteristic curves are given by \[\tau\mapsto (x_1(\tau,s),x_2(\tau,s)) = (s,s^2 \tau).\] These are lines parallel to the \(x_2\)-axis.

5. In order to find the solution we need to invert the map \(\left( \tau,s\right)\to \left( x_1,x_2\right)\). We have that \(x_1=s\) and \[\tau =\frac{x_2}{s^2} = \frac{x_2}{x_1^2}.\] Note that \(x_1\not=0\) since we do not consider the point \((0,0)\). We conclude that \[u(x_1,x_2) = z\left( \tau(x_1,x_2),s(x_1,x_2)\right) = 3+ 5\frac{x_2}{x_1^2}.\] The domain of this function is \(\mathbb{R}^2\setminus\{(0,x_2): x_2\in\mathbb{R}\}.\)

Solution:

1. We recall that for any two \(C^2\) functions \(u,v\) \[v\Delta u = v\mathrm{div}\left( \nabla u\right) =\mathrm{div}\left( v\nabla u\right) - \nabla v\cdot \nabla u.\] Using the above with the divergence theorem we find that \[\int_{\Omega} v\Delta u d x = \int_{\Omega}\mathrm{div}\left( v\nabla u\right) dx - \int_{\Omega}\nabla v\cdot \nabla udx\] \[=\int_{\partial \Omega }v\nabla u \cdot \vec{n}dS - \int_{\Omega}\nabla v\cdot \nabla udx\] \[=\int_{\partial \Omega }v\partial_n u dS - \int_{\Omega}\nabla v\cdot \nabla udx,\] which is the desired result.

2. Using the formula we found in a) with \(v=1\) and \(u\) we find that \[0+0 = \int_{\partial \Omega}1\;\partial_n u dS.\]

3. Using the formula we found in a) with \(v=u\) we find that \[0 + \int_{\Omega}\left\lvert\nabla u\right\rvert^2dx = \int_{\partial \Omega}u\partial_n u dS.\] Consequently, \[\int_{\partial \Omega}u\partial_n u dS=\int_{\Omega}\left\lvert\nabla u\right\rvert^2dx\geq 0.\]

4. Since \(u\) is harmonic the formula we found in a) states that \[\int_{\Omega}\nabla u \cdot \nabla v dx = \int_{\partial \Omega}v\partial_n u dS.\] Replacing \(u\) with \(v\) and using the fact that \(v\) is harmonic gives \[\int_{\Omega}\nabla v \cdot \nabla u dx = \int_{\partial \Omega}u\partial_n v dS.\] Subtracting these two equations gives the desired result.

Solution:

1. A symmetric matrix is positive semi-definite if and only if all its eigenvalues are non-negative. It is positive definite if and only if all its eigenvalues are positive. For a \(2\times 2\) symmetric matrix \(A\) with eigenvalues \(\lambda_1\) and \(\lambda_2\) we know that \[\mathrm{det}(A) = \lambda_1 \lambda_2,\qquad \mathrm{tr}\left( A\right)=\lambda_1+\lambda_2.\] Consequently \(A\) is positive semi-definite if and only if \(\mathrm{det}(A) \geq 0\) and \(\mathrm{tr}(A) \geq 0\) and is positive definite if and only if \(\mathrm{det}(A) > 0\) and \(\mathrm{tr}(A) >0\).

   In our case \(\mathrm{det}(A)=1-\alpha^2\) and \(\mathrm{tr}(A)=2>0\). We conclude that if \(\left\lvert\alpha\right\rvert<1\) the matrix is positive definite and if \(\left\lvert\alpha\right\rvert\leq 1\) the matrix is positive semi-definite.

2. Since \(\left\lvert\alpha\right\rvert<1\) our operator \(\mathcal{L}^\alpha\) is an elliptic operator. Consequently, any classical solution \(u\) to \[\mathcal{L}^\alpha u =0\] is both a super and sub-solution to our equation, which implies that it satisfies the weak minimum and maximum principles. As a result, since \(u\) is continuous on the closure of the domain and as such must attain a minimum and maximum on it, \(u\) attains both its minimum and maximum on \(\partial\Omega\).

3. We have that \[\mathcal{L}^1 u (x_1,x_2) = -\Delta u(x_1,x_2) - 2\partial_{x_1x_2}u(x_1,x_2)\] for any \(u\in C^2\left( \Omega\right)\). If \(u(x_1,x_2) = c_1(x_1^2+x_2^2) - c_2x_1x_2\) solves the equation \(\mathcal{L}^1 u =0\) we must have that \[0 = -2c_1-2c_1 -2 \left( -c_2\right) = 2c_2-4c_1.\] In other words \[u(x_1,x_2) = c_1\left( x_1^2+x_2^2 -2x_1x_2\right) = c_1\left( x_1-x_2\right)^2.\]

4. Following the hint, we see that if we choose \(c_1>0\) we have that \(u(x_1,x_2)\geq 0\) for all \(x_1,x_2\). Choosing any open and bounded set that intersects the line \(x_1=x_2\) will give us an interior point \(\left( \hat{x}_1,\hat{x}_1\right)\) for which we find that \[u\left( \hat{x}_1,\hat{x}_1\right)=0,\] i.e. \(u\) attains a global minimum in an interior point but \(u\) is not constant.

Solution:

1. Lax’s entropy condition for weak solutions to the Cauchy problem is that every shock to the problem \(\left( \sigma(t),t\right)\) satisfies \[f'\left( u_r\right) < \dot{\sigma} < f'(u_l)\] where \(u_\ell\) and \(u_r\) are the left and right limits of the solution across the shock. Since \(f\) is strongly convex this condition is equivalent to \(u_r<u_\ell\).

2. We have that \[\partial_t u^\varepsilon(x,t)=-\frac{\alpha}{\varepsilon}v'\left( \frac{x-\alpha t}{\varepsilon}\right),\] \[\partial_x u^\varepsilon(x,t)=\frac{1}{\varepsilon}v'\left( \frac{x-\alpha t}{\varepsilon}\right),\] \[\partial_{xx} u^\varepsilon(x,t)=\frac{1}{\varepsilon^2}v''\left( \frac{x-\alpha t}{\varepsilon}\right).\] Plugging this back in \(\eqref{eq:q6-2}\) gives us \[-\frac{\alpha}{\varepsilon}v'\left( \frac{x-\alpha t}{\varepsilon}\right)+\frac{1}{\varepsilon}f'\left( v\left( \frac{x-\alpha t}{\varepsilon}\right)\right)v'\left( \frac{x-\alpha t}{\varepsilon}\right) - \frac{1}{\varepsilon}v''\left( \frac{x-\alpha t}{\varepsilon}\right)=0.\] As \(\frac{x-\alpha t}{\varepsilon}\) ranges over all points in \(\mathbb{R}\) the ODE that \(v\) must satisfy is \[v''(s)=-\alpha v'(s) + f'(v(s))v'(s) ,\qquad s\in \mathbb{R}.\]

3. We have that \[\lim_{\varepsilon\to 0}u^{\varepsilon}(x,t) = \left\{ \begin{array}{ll} u_\ell, & x<\alpha t,\\ u_r, & x>\alpha t. \end{array} \right.\]

4. We notice that \[-\alpha v'(s) + f'(v(s))v'(s) = \frac{d}{ds}\left( -\alpha v(s) + f(v(s))\right)\] and consequently we can integrate the ODE of \(v\) to get that \[v'(s) = -\alpha v(s) + f(v(s)) + C,\] for any \(s\in\mathbb{R}\). Taking \(s\) to \(\infty\) in the above we find that \[0 = -\alpha u_r + f\left( u_r\right)+C\] Similarly, taking \(s\) to \(-\infty\) yields \[0 = -\alpha u_{\ell} + f\left( u_{\ell}\right)+C.\] Subtracting the two equations we find that \[\alpha \left( u_r-u_{\ell}\right) = f\left( u_r\right)-f\left( u_{\ell}\right).\]

5. We notice that for any \(x\not=\alpha t\) \[u(x,t)=\lim_{\varepsilon\to 0}u^{\varepsilon}(x,t) = \left\{ \begin{array}{ll} u_\ell, & x<\alpha t,\\ u_r, & x>\alpha t. \end{array} \right.\] solves \(\eqref{eq:q6-1}\) with the correct initial conditions. Moreover, on the shock curve \(\sigma(t)=\alpha t\) we have that \[\dot{\sigma} = \alpha = \frac{f\left( u_r\right)-f\left( u_{\ell}\right)}{u_r-u_{\ell}}.\] The mean value theorem implies that we can find \(c\) between \(u_r\) and \(u_{\ell}\) such that \[\frac{f\left( u_r\right)-f\left( u_{\ell}\right)}{u_r-u_{\ell}}=f'(c)\] Since \(f\) is strictly convex, \(f'\) is increasing and we conclude that \[f'(u_r) < f'(c) < f'\left( u_{\ell}\right)\] as we assumed that \(u_r<u_{\ell}\). We conclude that the entropy condition is satisfied.

Solution:

1. For any interval \(I\) on which \(u_0\) is continuous we get the characteristics curves are straight lines which start at \((s,0)\) with \(s\in I\) and have a slope \(u_0(s)\). In particular we find that the characteristics are \[x = s + u_0(s)t = \left\{ \begin{array}{ll} s, & s<0,\\ s+ t, & 0<s<1,\\ s+2t, & 1<s<2,\\ s+ st, & 2<s, \end{array} \right.\] As the slopes keep increasing when defined (which we could have said from the fact that \(u_0\) is non-decreasing) we see that the characteristics will not cross, and there will be empty regions which they will not reach. In other words, there will be no shocks but we will need rarefaction waves to construct a unique entropy solution.

   A sketch of the characteristics (in blue) and the rarefaction waves we’ll need to add (in red) is given below.

   

2. We have two rarefaction regions: The wedge which is between the lines \(x=0\) and \(x=t\) which meet at \((0,0)\), and the wedge between the line \(x=1+t\) and \(x=1+2t\) which meet at \((1,0)\).

   We will look for a solution of the form \(g_1\left( \frac{x}{t}\right)\) with \(g_1(0)=0\) and \(g_1(1)=1\) for the first region and \(g_2\left( \frac{x-1}{t}\right)\) with \(g_2(1)=1\) and \(g_2(2)=2\) for the second.

   For \(v(x,t)=h\left( \frac{x-x_0}{t}\right)\) to be a solution for our equation we must have that \[-\frac{x-x_0}{t^2}h'\left( \frac{x-x_0}{t}\right) + \frac{h\left( \frac{x-x_0}{t}\right)}{t}h'\left( \frac{x-x_0}{t}\right)=0,\] which is equivalent to saying that \[h'(z)\left( h(z)-z\right)=0,\] showing that \(h(z)=z\) is a possible solution. Due to our initial conditions we see that it is indeed the correct solution, i.e. \(g_1\left( \frac{x}{t}\right) = \frac{x}{t}\) and \(g_2\left( \frac{x-1}{t}\right)=\frac{x-1}{t}\).

3. Following on our above discussion we identify the following regions:
   \(x<0\): Here we can use characteristics to find that \(u(x,t) = u_0(x)=0\).
   \(0<x<t\): Here we introduced the first rarefaction wave. We find that \(u(x,t)=g_1\left( \frac{x}{t}\right)=\frac{x}{t}\).
   \(t<x<1+t\): Here we can use characteristics to find that \(u(x,t) = u_0(x-t)=1\).
   \(1+t<x<1+2t\): Here we introduced the second rarefaction wave. We find that \(u(x,t)=g_2\left( \frac{x-1}{t}\right)=\frac{x-1}{t}\).
   For \(x>1+2t\) we are in the realm of characteristics. However we have two distinct domains: \(1+2t<x<2+2t\): Here we can use characteristics to find that \(u(x,t) = u_0(x-2t)=2\).
   \(x\geq 2+2t\): Here we can use characteristics to find that \(u(x,t) = u_0\left( \frac{x}{1+t}\right)=\frac{x}{1+t}\).
   In conclusion: \[u(x,t): = \left\{ \begin{array}{ll} 0, & x\le0,\\ \frac{x}{t}, & 0<x\le t,\\ 1, & t<x\le 1+t,\\ \frac{x-1}{t}, & 1+t<x\le1+2t,\\ 2, & 1+ 2t<x\le2+2t,\\ \frac{x}{1+t}, & 2+2t\le x. \end{array} \right.\]

4. The proposed solution is piecewise continuous, so we only need to check the continuity at the boundary of the regions. For instance: \[\lim_{\left( x,t\right)\to (0,t)_{\ell}}u(x,t) = 0 = u(0,t)=\lim_{\left( x,t\right)\to (0,t)_{r}}u(x,t)\] where we have used \((0,t)_{\ell}\) and \((0,t)_{r}\) to indicate we are approaching from the region to the left of the problematic curve and its right. We can similarly check all other boundaries.

5. Since our solution is continuous, and as such has no shocks, it satisfies Lax’s entropy condition automatically.

Solution:

1. From the assumptions on \(F\) we know that there exists \(C>0\) such that \(F(x)\leq C\). Consequently, \[E[u] = \int_{\Omega}\frac12 (\Delta u(x))^2 d x - \int_{\Omega}F(u(x))d x \geq -\int_{\Omega}Cdx = -C\left\lvert\Omega\right\rvert=-c_0.\]

2. Assume that \(u\in\mathcal{V}\) is a minimiser of \(E\). Let \(\varphi\in \mathcal{V}\) be given and consider the function \[u_{\varepsilon}=u+\varepsilon\varphi.\] Since \[u_{\varepsilon}\vert_{\partial \Omega}=u\vert_{\partial \Omega}+\varepsilon\varphi\vert_{\partial \Omega}=0\] and \[\nabla u_{\varepsilon}\cdot \vec{n}\vert_{\partial \Omega}=\nabla u\cdot \vec{n}\vert_{\partial \Omega}+\varepsilon\nabla \varphi\cdot \vec{n}\vert_{\partial \Omega}=0\] we conclude that \(u_{\varepsilon}\in \mathcal{V}\) and consequently, by the assumption of minimality of \(u\), \[E[u_{\varepsilon}]\leq E[u].\] This implies that if \(E[u_{\varepsilon}]\) is differentiable with respect to \(\varepsilon\) then as \(u_0=u\) we must have that \(\frac{d}{d\varepsilon}E[u_{\varepsilon}]=0\). Since \[E[u_{\varepsilon}] = \int_{\Omega}\frac12 (\Delta u(x)+\varepsilon\Delta \varphi(x))^2 d x - \int_{\Omega}F(u(x)+\varepsilon\varphi(x))d x\] \[=\frac12\int_{\Omega} (\Delta u(x))^2dx+\varepsilon\int_{\Omega} \Delta u(x)\Delta\varphi(x)dx+ \frac{\varepsilon^2}{2}\int_{\Omega} (\Delta \varphi(x))^2- \int_{\Omega}F(u(x)+\varepsilon\varphi(x))d x\] and since \(F\) is smooth, we have that \[\frac{d}{d\varepsilon}E[u_{\varepsilon}] = \int_{\Omega} \Delta u(x)\Delta\varphi(x)dx+ \varepsilon\int_{\Omega} (\Delta \varphi(x))^2- \int_{\Omega}F'(u(x)+\varepsilon\varphi(x)) \varphi(x)d x.\] We conclude that for any \(\varphi\in\mathcal{V}\) we must have \[0=\int_{\Omega} \Delta u(x)\Delta\varphi(x)dx- \int_{\Omega}F'(u(x)) \varphi(x)d x.\]

3. To find the PDE that \(u\) must satisfy we need to rewrite our optimality condition in a way that involves only \(\varphi\) (and not its derivatives). This means that we will need to integrate by parts, which is justified by the assumption that \(u\in C^4\left( \overline{\Omega}\right)\). We have that \[\Delta u \Delta \varphi= \Delta u \mathrm{div}\left( \nabla \varphi\right) = \mathrm{div}\left( \Delta u \nabla \varphi\right) - \nabla \varphi\nabla \Delta u\] \[=\mathrm{div}\left( \Delta u \nabla \varphi\right) - \left( \mathrm{div}\left( \varphi\nabla \Delta u\right) - \varphi\mathrm{div}\left( \nabla \Delta u\right)\right)\] \[=\mathrm{div}\left( \Delta u \nabla \varphi\right) - \mathrm{div}\left( \varphi\nabla \Delta u\right) + \varphi\Delta^2 u.\] Using the divergence theorem we find that \[0=\int_{\Omega} \Delta u(x)\Delta\varphi(x)dx- \int_{\Omega}F'(u(x)) \varphi(x)d x =\int_{\partial \Omega}\Delta u (y)\nabla \varphi(y)\cdot \vec{n(y)}dS(y)\] \[-\int_{\partial \Omega}\varphi(y) \nabla \Delta u(y) \cdot \vec{n(y)}dS(y) + \int_{\Omega}\Delta^2 u(x) \varphi(x) dx- \int_{\Omega}F'(u(x)) \varphi(x)d x\] \[=\int_{\Omega}\left( \Delta^2 u(x)-F'(u(x))\right) \varphi(x)d x ,\] where we have used the fact that \(\varphi\in\mathcal{V}\). Choosing \(\varphi\in C^1_c\left( \Omega\right)\subset \mathcal{V}\) and using the fundamental lemma of the calculus of variation we conclude the \(u\) must satisfy \[\Delta ^2 u(x) = F'\left( u(x)\right),\qquad x\in\Omega.\] As \(u\in\mathcal{V}\) we have the boundary conditions \(u\vert_{\partial \Omega}=0\), \(\nabla u\cdot \vec{n}\vert_{\partial \Omega}=0\).

4. We have not discussed this topic this year.

5. Assume that we have two minimisers \(u_1\) and \(u_2\). Since \(F=0\) we must have that for any \(\varphi\in\mathcal{V}\) \[\int_{\Omega}\Delta u_1(x) \Delta\varphi(x)d x=0=\int_{\Omega}\Delta u_2(x) \Delta\varphi(x)d x.\] In particular \[\int_{\Omega}\Delta \left( u_1(x)-u_2(x)\right) \Delta\varphi(x)d x=0.\] Choosing \(\varphi=u_1-u_2\in \mathcal{V}\) we find that \[\int_{\Omega}\left( \Delta \left( u_1(x)-u_2(x)\right)\right)^2dx=0\] which implies \[\Delta\left( u_1(x)-u_2(x)\right)=0,\qquad x\in \Omega.\] This implies that \(u_1-u_2\) is harmonic and is zero on the boundary (even if the set is not connected). Consequently, \(u_1\equiv u_2\).