Partial Differential Equations III & V
Problem Class 8 Solution

Exercise 1 (Properties of the fundamental solution to the Heat equation). Let \[\Phi(x,t) = \frac{1}{\left( 4\pi kt\right)^{\frac{n}{2}}}e^{-\frac{\left\lvert x\right\rvert^2}{4kt}},\] where \(k>0\) is given, be the fundamental solution to the Heat equation.

  1. Show that for any \(p\geq 1\) \[\left\lVert\Phi(x,t)\right\rVert_{L^p\left( \mathbb{R}^n\right)}=\frac{\left( 4\pi kt\right)^{\frac{n(1-p)}{2p}}}{p^{\frac{n}{2p}}}\] You may use the fact that \(\int_{\mathbb{R}^n}e^{-\frac{\left\lvert y\right\rvert^2}{2}}dy=\left( 2\pi\right)^{\frac{n}{2}}\).

    In particular, \[\int_{\mathbb{R}^n} \Phi(x,t)dx=\left\lVert\Phi\left( \cdot,t\right)\right\rVert_{L^1\left( \mathbb{R}^n\right)}=1\] for all \(t>0\).

  2. Young’s convolution inequality states that if \(f\in L^p\left( \mathbb{R}^n\right)\) and \(g\in L^q\left( \mathbb{R}^n\right)\) where \(p,q\in [1,\infty]\) are such that \[\frac{1}{p}+\frac{1}{q}=1+\frac{1}{r}\] for some \(r\in [1,\infty]\) then \(f\ast g \in L^r\left( \mathbb{R}^n\right)\) and \[\left\lVert f\ast g\right\rVert_{L^r\left( \mathbb{R}^n\right)} \leq \left\lVert f\right\rVert_{L^p\left( \mathbb{R}^n\right)}\left\lVert g\right\rVert_{L^q\left( \mathbb{R}^n\right)}.\] Use this to show that for any \(g\in C_c\left( \mathbb{R}^n\right)\) we have that for any \(r\geq 1\) and any \(1\leq p\leq r\) \[\left\lVert\Phi(\cdot,t)\ast g\right\rVert_{L^r} \leq \frac{C_{p,n,k}\left\lVert g\right\rVert_{L^{\frac{rp}{rp+p-r}}\left( \mathbb{R}^n\right)}}{t^{\frac{n(p-1)}{2p}}}\] where \(C_{p,n,k}\) is an explicit constant that depends only on \(p\), \(n\), and \(k\).

Solution:

  1. \[\left\lVert\Phi(x,t)\right\rVert_{L^p\left( \mathbb{R}^n\right)}^p = \frac{1}{\left( 4\pi kt\right)^{\frac{pn}{2}}}\int_{\mathbb{R}^n}e^{-\frac{p\left\lvert x\right\rvert^2}{4kt}}dx\] \[\underset{y=\sqrt{\frac{p}{2kt}}x}{=} \frac{1}{\left( 4\pi kt\right)^{\frac{pn}{2}}} \left( \frac{2kt}{p}\right)^{\frac{n}{2}}\int_{\mathbb{R}^n}e^{-\frac{\left\lvert y\right\rvert^2}{2}}dy =\frac{\left( 4\pi kt\right)^{\frac{n(1-p)}{2}}}{p^{\frac{n}{2}}},\] from which the result follows.

  2. This follows from the fact that if1 \[\frac{1}{p}+\frac{1}{q}=1+\frac{1}{r},\] then \(q=\frac{rp}{rp+p-r}\), part \(\eqref{item:L_p_norm}\), and Young’s convolution inequality. Indeed, \[\left\lVert\Phi(\cdot,t)\ast g\right\rVert_{L^r} \leq \left\lVert\Phi\right\rVert_{L^p\left( \mathbb{R}^n\right)}\left\lVert g\right\rVert_{L^{\frac{rp}{rp+p-r}}\left( \mathbb{R}^n\right)}=\frac{\left( 4\pi kt\right)^{\frac{n(p-1)}{2p}}}{p^{\frac{n}{2p}}}\left\lVert g\right\rVert_{L^{\frac{rp}{rp+p-r}}\left( \mathbb{R}^n\right)}.\]

Exercise 2 (Bonus – additional bounds on \(u=\Phi\ast g\)). Consider the solution to the heat equation \[\nonumber \begin{split} &u_t(x,t)-k\Delta u(x,t) = 0,\qquad x\in\mathbb{R}^n,\;t>0,\\ &u(x,0)=g(x), \qquad\qquad\qquad \; x\in\mathbb{R}^n, \end{split}\] where \(k>0\) and \(g\in L^1\left( \mathbb{R}^n\right)\), given by \[u(x,t) = \int_{\mathbb{R}^n}\Phi(x-y,t)g(y)dy\] with \[\Phi(x,t) = \frac{1}{\left( 4\pi kt\right)^{\frac{n}{2}}}e^{-\frac{\left\lvert x\right\rvert^2}{4kt}}.\]

  1. Show that if \(g\in L^\infty\left( \mathbb{R}^n\right)\) then so is \(u\). Moreover \[\left\lVert u\right\rVert_{L^\infty\left( \mathbb{R}^n\times (0,\infty)\right)} \leq \left\lVert g\right\rVert_{L^\infty\left( \mathbb{R}^n\right)}.\]

  2. If \(g\in L^2\left( \mathbb{R}^n\right)\) one can show that \(u(\cdot,t)\in L^2\left( \mathbb{R}^n\right)\) for all \(t>0\) (follows from the previous exercise!). Show that for any \(t>0\) \[\left\lVert u\left( \cdot,t\right)\right\rVert_{L^2\left( \mathbb{R}^n\right)} \leq \left\lVert g\right\rVert_{L^2\left( \mathbb{R}^n\right)}.\]

Hint: You may use the following: \[\Phi\geq0,\qquad \text{and}\qquad\int_{\mathbb{R}^n}\Phi(z,t)dz=1,\;\;\forall t>0.\] \[\widehat{\Phi}(\xi,t) = \frac{1}{\left( 2\pi\right)^{\frac{n}{2}}}e^{-kt\left\lvert\xi\right\rvert^2}.\]

Solution:

  1. We have that for any \(x\in\mathbb{R}^n\) and \(t>0\) \[\left\lvert u(x,t)\right\rvert \leq \int_{\mathbb{R}^n}\left\lvert\Phi(x-y,t)\right\rvert\left\lvert g(y)\right\rvert dy \leq \left\lVert g\right\rVert_{L^\infty\left( \mathbb{R}^n\right)}\int_{\mathbb{R}^n}\Phi(x-y,t)dy\] \[\underset{\tiny{\begin{tabular}{c} $z=x-y$ \\ $dz=\left\lvert\left( -1\right)^n\right\rvert dy$ \end{tabular}}}{=}\left\lVert g\right\rVert_{L^\infty\left( \mathbb{R}^n\right)}\int_{\mathbb{R}^n}\Phi(z,t)dz=\left\lVert g\right\rVert_{L^\infty\left( \mathbb{R}^n\right)}.\] Consequently \[\left\lVert u\right\rVert_{L^\infty\left( \mathbb{R}^n\times (0,\infty)\right)} \leq \left\lVert g\right\rVert_{L^\infty\left( \mathbb{R}^n\right)}.\]

  2. Passing to the Fourier transform in the spatial variable we find that \[\widehat{u}\left( \xi,t\right) = \widehat{\Phi (\cdot,t)\ast g}(\xi) = \left( 2\pi\right)^{\frac{n}{2}}\widehat\Phi\left( \xi,t\right)\widehat{g}(\xi)=\widehat{g}\left( \xi\right)e^{-kt\left\lvert\xi\right\rvert^2}.\] Using to Plancherel’s identity we find that \[\left\lVert u\left( \cdot,t\right)\right\rVert_{L^2\left( \mathbb{R}^n\right)}^2 = \left\lVert\widehat{u}\left( \cdot,t\right)\right\rVert_{L^2\left( \mathbb{R}^n\right)}^2 = \int_{\mathbb{R}^n}\left\lvert\widehat{u}\left( \xi,t\right)\right\rvert^2d\xi\] \[=\int_{\mathbb{R}^n}\left\lvert g\left( \xi\right)\right\rvert^2e^{-2kt\left\lvert\xi\right\rvert^2}d\xi \leq \int_{\mathbb{R}^n}\left\lvert g\left( \xi\right)\right\rvert^2d\xi = \left\lVert\widehat{g}\right\rVert_{L^2\left( \mathbb{R}^n\right)}^2 = \left\lVert g\right\rVert_{L^2\left( \mathbb{R}^n\right)}^2,\] giving us the desired result.

Exercise 3 (The energy method: Uniqueness for the heat equation in a time dependent domain). Let \(k>0\), \(T>0\) be given. Let \(a,b:[0,T]\to \mathbb{R}\) be smooth functions such that \(a(t)<b(t)\) for all \(t\in [0,T]\). Let \(U\subset \mathbb{R}\times (0,T]\) be the non-cylindrical domain \[U=\left\lbrace (x,t)\in \mathbb{R}\times (0,T]\;|\;a(t)<x<b(t)\right\rbrace.\] Consider the heat equation \[\nonumber \left\{ \begin{array}{ll} u_t-ku_{xx} =f(x,t)& \left( x,t\right)\in U,\\ u(a(t),t)=g_1(t) & t\in [0,T],\\ u(b(t),t)=g_2(t) & t\in [0,T],\\ u(x,0)=u_0(x) & x\in \left( a(0),b(0)\right). \end{array} \right.\] Use the energy method to prove that the equation has at most one smooth solution.

Solution:
Assuming there exist two smooth solution to the equation, \(u_1\) and \(u_2\), we start by defining \(w=u_1-u_2\). The linearity of the equation implies that \(w\) solves the equation \[\nonumber \left\{ \begin{array}{ll} w_t-kw_{xx} =0& \left( x,t\right)\in U,\\ w(a(t),t)=0 & t\in [0,T],\\ w(b(t),t)=0 & t\in [0,T],\\ w(x,0)=0 & x\in \left( a(0),b(0)\right). \end{array} \right.\] As is common with the energy method, we will multiply our equation by a function and integrating by parts. In this case (though not always!) it will be \(w\). We have that \[\int_{a(t)}^{b(t)}w_t\left( x,t\right) w\left( x,t\right) dx = k\int_{a(t)}^{b(t)}w\left( x,t\right) w_{xx}\left( x,t\right) dx.\] Since \[\int_{a(t)}^{b(t)}w\left( x,t\right) w_{xx}\left( x,t\right) dx = \underbrace{w\left( b(t),t\right)}_{=0}w_x\left( b(t),t\right)-\underbrace{w\left( a(t),t\right)}_{=0}w_x\left( a(t),t\right)\] \[-\int_{a(t)}^{b(t)} w^2_x\left( x,t\right) dx = - \int_{a(t)}^{b(t)} w^2_x\left( x,t\right) dx\] and \(w_t w = \frac{1}{2}\partial_t \left( w^2\right)\) we find that \[\int_{a(t)}^{b(t)}\partial_t \left( w^2\right)\left( x,t\right) dx = -2k\int_{a(t)}^{b(t)} w^2_x\left( x,t\right) dx.\] We would like to write \(\int_{a(t)}^{b(t)}\partial_t \left( w^2\right)dx\) as a full time derivative. This is not immediately clear as the boundaries of the integration also depend on \(t\). We notice, however, that \[\frac{d}{dt} \int_{a(t)}^{b(t)}w^2\left( x,t\right) dx = \int_{a(t)}^{b(t)}\partial_t \left( w^2\right)\left( x,t\right)dx + \underbrace{w^2\left( b(t),t\right)}_{=0}b^\prime(t)\] \[- \underbrace{w^2\left( a(t),t\right)}_{=0}a^\prime(t)= \int_{a(t)}^{b(t)}\partial_t \left( w^2\right)\left( x,t\right)dx,\] which is justified since all the functions are smooth. We conclude that our equation can be written as \[\frac{d}{dt} \int_{a(t)}^{b(t)}w^2\left( x,t\right) dx = -2k\int_{a(t)}^{b(t)} w^2_x\left( x,t\right) dx \leq 0,\] which implies that the energy \[E(t) = \frac{1}{2}\int_{a(t)}^{b(t)}w^2\left( x,t\right)dx\] is non-increasing. Consequently \[0\leq E(t) \leq E(0) = \int_{a(0)}^{b(0)}w^2\left( x,0\right)dx =0\] from which we conclude that, since \(w\) is continuous, \(w\left( x,t\right)=0\) on \(U\) or \(u_1\equiv u_2\).
Remark: We could have started by guessing (or being given) the energy and finding its derivative. Indeed, defining \[E(t) = \frac{1}{2}\int_{a(t)}^{b(t)}w^2\left( x,t\right)dx\] we find that (again, all the functions are smooth) \[\frac{d}{dt}E(t) = \frac{1}{2}\int_{a(t)}^{b(t)}\partial_t \left( w^2\right)\left( x,t\right)dx + \underbrace{\frac{w^2\left( b(t),t\right)b^\prime(t)}{2}}_{=0}\] \[-\underbrace{\frac{w^2\left( a(t),t\right)a^\prime(t)}{2}}_{=0}=\int_{a(t)}^{b(t)}w\left( x,t\right)w_t\left( x,t\right)dx\] \[= k\int_{a(t)}^{b(t)}w\left( x,t\right) w_{xx}\left( x,t\right) dx=k\underbrace{w\left( b(t),t\right)}_{=0}w_x\left( b(t),t\right)-k\underbrace{w\left( a(t),t\right)}_{=0}w_x\left( a(t),t\right)\] \[-k\int_{a(t)}^{b(t)} w^2_x\left( x,t\right) dx = - k\int_{a(t)}^{b(t)} w^2_x\left( x,t\right) dx \leq 0.\]

Exercise 4 (Grönwall’s inequality). Another important inequality in the study of PDEs, in particular in the study of long time behaviour of solutions, is the so-called Grönwall’s inequality which state that if \(y:[0,T]\to \mathbb{R}\) is continuous and differentiable on \((0,T)\), and if there exists \(\lambda \in \mathbb{R}\) such that \[y^\prime (t) \leq \lambda y(t)\] then we have that \[y(t) \leq y(0) e^{\lambda t}.\] Prove Grönwall’s inequality.
Remark: The above inequality can be generalised to show that if \[y^\prime (t) \leq \lambda (t)y(t)\] then \(y(t) \leq y(0)e^{\int_{0}^t \lambda(s)ds}\). There are additional important Grönwall inequalities which we won’t mention at this point.

Solution:
We note that we can write our inequality as \[y^\prime (t) - \lambda y(t) \leq 0.\] Had we had equality, we would have used the integrating factor \(e^{\int \left( -\lambda \right)dt } = e^{-\lambda t}\). This motivates us to define \(z(t)=e^{-\lambda t}y(t)\). We find that \[z^\prime (t) = e^{-\lambda t}\left( y^\prime (t) -\lambda y(t)\right) \leq 0.\] Since \(z(t)\) is differentiable on \((0,T)\) and continuous on \([0,T]\) (as \(y(t)\) is) we conclude that \(z(t)\) must be non-increasing on \([0,T]\). Consequently, for any \(t\in [0,T]\) we have that \(z(t)\leq z(0)=y(0)\) which implies \[y(t) = z(t)e^{\lambda t}\leq y(0)e^{\lambda t}.\]

Exercise 5. Let \(T>0\) be given and define \[\Omega_T= \left( a,b\right)\times (0,T]\] for a given \(-\infty<a<b<\infty\).

  1. Show that there exists at most one solution \(u\in C_1^2\left( \Omega_T\right)\cap C\left( \overline{\Omega_T}\right)\) to the problem \[\begin{equation} \label{eq:heat} \left\{ \begin{array}{ll} u_t-u_{xx} =1& \left( x,t\right)\in \Omega_T,\\ u=0 & \left( x,t\right)\in \Gamma_T, \end{array} \right. \end{equation}\] where \(\Gamma_T = (a,b)\times \left\lbrace 0\right\rbrace \cup \left\lbrace a\right\rbrace\times [0,T] \cup \left\lbrace b\right\rbrace\times [0,T]\).

  2. Assume that \(u\) is a \(C_1^2\left( \Omega_T\right)\cap C\left( \overline{\Omega_T}\right)\) solution to \(\eqref{eq:heat}\). Show that for any \(\left( x,t\right)\in \Omega_T\) \[0\leq u(x,t) \leq t.\]

Solution:

  1. Assuming that there are two \(C_1^2\left( \Omega_T\right)\cap C\left( \overline{\Omega_T}\right)\) solutions, \(u_1\) and \(u_2\), we define \(w=u_1-u_2\) which is also a function in \(C_1^2\left( \Omega_T\right)\cap C\left( \overline{\Omega_T}\right)\) and satisfies the equation \[\nonumber \left\{ \begin{array}{ll} w_t-w_{xx} =0& \left( x,t\right)\in \Omega_T,\\ u=0 & \left( x,t\right)\in \Gamma_T, \end{array} \right.\] Using the weak maximum and weak minimum principles for the heat equation we find that \[\max_{\overline{\Omega_T}}w(x,t) = \max_{\Gamma_T}w(x,t) =0\] and \[\min_{\overline{\Omega_T}}w(x,t) = \min_{\Gamma_T}w(x,t) =0\] which implies that \(w\equiv 0\) on \(\overline{\Omega_T}\), or equivalently that \(u_1\equiv u_2\).

  2. We see that \(u\) satisfies \[u_t-u_{xx} =1 > 0\] and consequently, according to the weak minimum principle, \[\min_{\overline{\Omega_T}}u(x,t) = \min_{\Gamma_T}u(x,t) =0\] showing that \(u(x,t) \geq 0\) on \(\Omega_T\).

    We can’t use the weak maximum principle on \(u\) but we notice that the second inequality that we’d like to show, \(u(x,t) \leq t\), can be rewritten as \(u(x,t) -t \leq 0\). This motivates us to define \(w(x,t)=u(x,t)-t\). We find that \(w\in C_1^2\left( \Omega_T\right)\cap C\left( \overline{\Omega_T}\right)\) and it satisfies \[w_t-w_{xx} = u_t -1 -u_{xx}=0\] in \(\Omega_T\) and \[w = 0-t =-t\] on \(\Gamma_T\). Using the weak maximum principle principle for \(w\) we find that \[\max_{\overline{\Omega_T}}w(x,t) = \max_{\Gamma_T}w(x,t) =\max_{\Gamma_T}\left( -t\right)\leq 0\] which implies that for any \(x\in \Omega_T\) \[u(x,t) = w(x,t)+t \leq t.\]

  1. Note that this inequality automatically implies that \(p,q\leq r\) if \(p,q,r\in [1,\infty]\).↩︎