Section A
Let us consider the following Cauchy problem associated to a first order PDE \[\begin{equation} \label{eq:q1} \left\{ \begin{array}{ll} x_2\partial_{x_1}u(x_1,x_2) = 1, & (x_1,x_2)\in\mathbb{R}^2,\\ u(0,x_2) = 0, & x_2\in\mathbb{R}. \end{array} \right. \end{equation}\]
Identify the leading vector field, the Cauchy data and the Cauchy curve.
Are the points on the Cauchy curve characteristic or non-characteristic? Justify your answer.
Using the method of characteristics, solve the problem in \(\eqref{eq:q1}\). Give the domain of definition of the solution.
Solution:
The leading vector field \(\vec{a}:\mathbb{R}^2\to\mathbb{R}^2\) is given by \(\vec{a}(x_1,x_2)=(x_2,0)\), the Cauchy datum is \(u(0,x_2) = 0\) and the Cauchy curve is given by \(\{\gamma(s):=(0,s):s\in\mathbb{R}\}\subset\mathbb{R}^2\), which is in fact the \(x_2\)-axis.
We have \(\gamma'(s) = (0,1)\), and we have \(\vec{a}(x_1,x_2)\cdot\gamma'(s)=0\), so the vector field is always orthogonal to the Cauchy curve. The only ‘problematic’ point is the origin, as \(\vec{a}(0,0)\) vanishes, and therefore all points on the Cauchy curve, except \((0,0)\), are non-characteristic. The point \((0,0)\) is characteristic and we expect the solutions to have problems at this point.
We have the system of ODEs (where \(\tau\in\mathbb{R}\) stands for the artificial time parameter and \(s\in\mathbb{R}\) is used for the parametrisation of the Cauchy curve) \[\left\{ \begin{array}{l} \partial_\tau x_1(\tau,s) = x_2(\tau,s),\\ \partial_\tau x_2(\tau,s) = 0,\\ \partial_\tau z(\tau,s) = 1 \end{array} \right.\] equipped with the boundary conditions \[x_1(0,s) = 0,\ \ \ x_2(0,s)=s,\ \ z(0,s)=0.\] From here, we directly find \[x_1(\tau,s) = \tau s,\ \ x_2(\tau,s) = s,\ \ z(\tau,s) = \tau.\] For a generic point \((X_1,X_2) \in\mathbb{R}^d\), we find \(s=X_2\) and \(\tau = X_1/X_2\), provided \(X_2\neq 0\). Therefore the solution becomes \(u(x_1,x_2)=x_1/x_2\) and the domain of this function is \(\mathbb{R}^2\setminus\{(x_1,0): x_1\in\mathbb{R}\}.\) We notice that since \(\vec{a}(0,0)=(0,0)\), no information can be transported along the \(x_1\)-axis.
Consider the Cauchy problem for Burgers’ equation \[\begin{equation} \label{eq:q2} \left\{ \begin{array}{ll} \partial_{t}u(x,t)+\frac12\partial_x(u^2(x,t)) = 0, & (x,t)\in\mathbb{R}\times (0,+\infty),\\ u(x,0) = u_0(x), & x\in\mathbb{R}, \end{array} \right. \end{equation}\] where \(u_0:\mathbb{R}\to\mathbb{R}\) is given.
Let \(u_0(x)=\frac17 x^{7}\). Show that \(\eqref{eq:q2}\) has a global in time classical solution.
Let \(u_0(x) = \sin(x)\). Write down the definition of the critical time \(t_c\) (until when we can guarantee the existence of a classical solution to \(\eqref{eq:q2}\)) associated to this initial datum. Show that \(t_c\le 1\).
Solution: Some general fact used in both questions below. In the case of Burgers’ equation, we have that the wave speed is given by \(c(u) = u\) and the characteristics are given in the \((x,t)\)-(half)plane (\(t\ge 0\)) by \[x = s+u_0(s)t,\] where \(s\in\mathbb{R}\) parametrises the characteristics.
If \(u_0\in C^1(\mathbb{R})\), the global existence of a classical solution to Burgers’ boils down to the global invertibility of the flow map \(s\mapsto s+u_0(s)t\), for every \(t>0\). A sufficient condition for this is \(1+u_0'(s)t>0\) for all \(t>0\) and for all \(s\in\mathbb{R}\). As \(u_0\) is increasing, we have \(1+u_0'(s)t = 1 + t s^6>0\) for all \(t>0\) and all \(s\in\mathbb{R}\). This implies the global in time existence of a classical solution.
The critical time is defined as the greatest \(t>0\) such that we can guarantee the global invertibility of the flow map. This is formally defined as \[t_c:=\inf\left\{-\frac{1}{u_0'(s)} : s\in I \right\},\] where \(I=\{s\in\mathbb{R}: u_0'(s)<0\}.\)
In this case \(u_0'(s) = \cos(s)\) and so \[I=\cup_{k\in\mathbb{Z}} (\pi/2+2k\pi,3\pi/2+2k\pi).\] Therefore, in order to find information about \(t_c\), we would need to optimise the function \(h(s) = -\cos(s)^{-1}\) on the interval \((\pi/2,3\pi/2)\) (notice that this function is \(2\pi\)-periodic, so it is enough to consider this interval). We have clearly \[\begin{equation} \label{eq:limits} \lim_{s\downarrow\pi/2} h(s) = \lim_{s\uparrow 3\pi/2} h(s) = +\infty. \end{equation}\] Since \(h(\pi) = 1\), we certainly have that \(t_c\le 1\). In fact, as the maximum of \(|\cos(s)|\) is \(1\), \(t_c=1\).
In this problem we consider harmonic function on the unit ball in \(\mathbb{R}^3\), \(B_1(0)\).
Using the fact that the Laplacian in spherical coordinates is given by \[\Delta \psi = \frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2 \frac{\partial \psi}{\partial r}\right)+\frac{1}{r^2\sin\theta}\left( \sin\theta \frac{\partial \psi}{\partial \theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2 \psi}{\partial \phi^2}\] show that if \(u\in C^2\left( \overline{B_1\left( 0\right)}\right)\) is radial (i.e. only depends on \(r\) in spherical coordinates) and harmonic in \(B_1\left( 0\right)\) then it must be constant.
Show that there exists no radial solution in \(C^2\left( \overline{B_1(0)}\right)\) to the equation \[\nonumber \left\{ \begin{array}{ll} -\Delta u(\vec{x})=0, &\vec{x}\in B_1\left( 0\right),\\ u(\vec{x})=f(\vec{x}), & \vec{x}\in \partial B_1\left( 0\right), \end{array} \right.\] for \(f(\vec{x})=x_1^2\).
Solution:
Given a harmonic function \(u(r)\) that solves the equation \(\Delta u (x)=0\) we find that \[\frac{1}{r^2}\frac{d}{d r}\left( r^2 u^\prime(r)\right)=0\] which implies that \[r^2 u^\prime(r) = C\] or \[u^\prime(r) = \frac{C}{r^2}.\] Integration gives us that the general radial solution for Laplace’s equation is \[u(r)=-\frac{C}{r}+B.\] As \(u\) needs to be continuous we find that the only harmonic radial function in \(B_1(0)\) is a constant function.
Since the boundary condition is not a constant function there can be no radial harmonic function which solves the equation.
Consider the heat-like equation \[\begin{equation} \label{eq:heat_like} \left\{ \begin{array}{ll} u_t-u_{xx} +cu =0, & \left( x,t\right)\in\mathbb{R}\times\left( 0,+\infty\right),\\ u(x,0)=g(x), & x\in\mathbb{R}, \end{array} \right. \end{equation}\] where \(c\in\mathbb{R}\) is a fixed constant and \(g\in C_c\left( \mathbb{R}\right)\).
Define \(v(x,t)=e^{ct}u(x,t)\). Show that \(v(x,t)\) solves the heat equation \[\begin{equation} \nonumber %\label{eq:heat} \left\{ \begin{array}{ll} v_t-v_{xx} =0, & \left( x,t\right)\in\mathbb{R}\times\left( 0,+\infty\right),\\ v(x,0)=g(x), & x\in\mathbb{R}. \end{array} \right. \end{equation}\]
Show that there exists a solution to \(\eqref{eq:heat_like}\) that satisfies \[\sup_{x\in\mathbb{R}}\left\lvert u(x,t)\right\rvert \leq e^{-ct}\left\lVert g\right\rVert_{L^\infty\left( \mathbb{R}\right)}.\] You may use the following inequality without proof: For any \(f\in L^1\left( \mathbb{R}\right)\) and \(g\in L^\infty\left( \mathbb{R}\right)\) we have that \[\left\lvert\int_{\mathbb{R}}f\left( x-y\right)g(y)\right\rvert \leq \left\lVert f\right\rVert_{L^1\left( \mathbb{R}\right)}\left\lVert g\right\rVert_{L^\infty\left( \mathbb{R}\right)},\qquad \forall x\in\mathbb{R}.\]
Solution:
We have that \[v_t-v_xx = e^{ct}u_t +ce^{ct}u - e^{ct}u_{xx}=e^{ct}\left( u_t-u_{xx}+cu\right)=0.\] Additionally \[v(x,0)=e^0 u(x,0)=g(x).\]
From class we know that \(v(x,t)=\left( \Phi(\cdot,t)\ast g \right)(x)\) is a solution to the equation.
Moreover, \(\Phi\) is non-negative and \(\int_{\mathbb{R}}\Phi(x,t)dx=1\). Using the given inequality we find that \[\left\lvert v(t,x)\right\rvert =\left\lvert\Phi\ast g(x,t)\right\rvert\leq \left\lVert\Phi\right\rVert_{L^1\left( \mathbb{R}\right)}\left\lVert g\right\rVert_{L^\infty\left( \mathbb{R}\right)}=\left\lVert g\right\rVert_{L^\infty\left( \mathbb{R}\right)}\] from which we conclude that for any \(x\in\mathbb{R}\) the solution \(u\) that is associated to the above \(v\) satisfies \[\left\lvert u(x,t)\right\rvert=e^{-ct}\left\lvert v(x,t)\right\rvert\leq e^{-ct}\left\lVert g\right\rVert_{L^\infty\left( \mathbb{R}\right)},\] which shows the desired result.
Section B
We consider the following conservation law \[\begin{equation} \label{eq:q5} \left\{ \begin{array}{ll} \partial_{t}u(x,t)-u(x,t)\partial_x u(x,t) = 0, & (x,t)\in\mathbb{R}\times (0,+\infty),\\ u(x,0) = u_0(x), & x\in\mathbb{R}. \end{array} \right. \end{equation}\] [Notice that this is not Burgers’ equation.]
Suppose that \(u_0\) is bounded, differentiable with bounded derivative. Give a formula of the critical time \(t_c\), for which we know that \(\eqref{eq:q5}\) has a classical solution on \(\mathbb{R}\times(0,t_c)\).
Let \(u_0(x)=-\arctan(x)\). Show that in this case \(\eqref{eq:q5}\) has a global in time classical solution.
Let \(u_0\) now be given by \[u_0(x)=\left\{ \begin{array}{ll} 0, & x<0,\\ 1, & x\ge 0. \end{array} \right.\] By drawing the characteristics, show that there is instantaneous crossing of characteristics. Find a shock that satisfies the Rankine–Hugoniot condition. Give the expression of the weak solution in this case.
Solution:
We notice here that the flux function is \(f(s)=-\frac12 s^2\) and the wave speed is \(c(s)=-s.\) From the lectures we know that for \(s\in\mathbb{R}\), the characteristics in the \((x,t)\)-half-plane are given by \[x = s + c(u_0(s))t,\] and the critical time is defined as \[t_c:=\inf\left\{-\frac{1}{\partial_s(c(u_0(s)))} = \frac{1}{u_0'(s)} : s\in I \right\},\] where \(I=\{s\in\mathbb{R}: \partial_s(c(u_0(s)))<0\}=\{s\in\mathbb{R}: u_0'(s)>0\}.\)
As \(u_0'(s)=-\frac{1}{1+s^2}<0\), we have that \(I=\emptyset\), and by definition \(t_c=+\infty.\)
We have that the characteristics are given by \[x=s,\ \ {\rm{if}}\ s<0;\ \ \ {\rm{and}}\ \ \ x=s-t,\ {\rm{if}}\ s\ge 0.\]
To introduce a shock that satisfies the Rankine–Hugoniot condition, this must satisfy \(\sigma(0)=0\) and \[(u_\ell-u_r)\dot\sigma = f(u_\ell) - f(u_r)=-\frac{1}{2}(u_\ell-u_r)(u_\ell+u_r),\] from where \(\dot\sigma = -\frac12\) and so, \(\sigma(t)=-\frac{t}{2}\). The figure below includes this shock.
Therefore, the weak integral solution in this case writes as \[u(x,t)=\left\{ \begin{array}{ll} 0, & x<-t/2,\\ 1, & x>t/2. \end{array} \right.\]
We aim to solve the following problem by the method of characteristics \[\begin{equation} \label{eq:q6} \left\{ \begin{array}{ll} \partial^2_{xx}u - 3\partial^2_{xy}u+2\partial^2_{yy}u=0, & (x,y)\in\mathbb{R}^2,\\ u(1,y) = g(y), & y\in\mathbb{R},\\ \partial_xu(1,y) = h(y), & y\in\mathbb{R}, \end{array} \right. \end{equation}\] where \(g,h:\mathbb{R}\to\mathbb{R}\) are given smooth functions.
Identify the Cauchy data and the Cauchy curve in the above problem.
Rewrite the PDE in \(\eqref{eq:q6}\) as a system of two linear first order PDEs. [Hint: think about the algebraic relation \((a-b)(a-2b)=a^2-3ab+2b^2,\ (a,b\in\mathbb{R})\).]
By solving the two first order PDEs arising from [qAin6.2] using the method of characteristics, find the solution to \(\eqref{eq:q6}\).
Solution:
The Cauchy data are the functions \(g\) and \(h\), while the Cauchy curve is given by \(\{\gamma(s)=(1,s):s\in\mathbb{R}\}\subset\mathbb{R}^2\).
Using the hint, we can write \[\partial^2_{xx}u - 3\partial^2_{xy}u+2\partial^2_{yy}u = (\partial_x-\partial_y)(\partial_x-2\partial_y)u=0.\] Therefore, introduce \(v\) such that \[\left\{ \begin{array}{l} \partial_x u-2\partial_yu=v,\\ \partial_xv-\partial_yv = 0. \end{array} \right.\] The corresponding boundary conditions read as \[\left\{ \begin{array}{ll} v(1,y)=h(y) - 2g'(y), & y\in\mathbb{R},\\ u(1,y) = g(y), & y\in\mathbb{R}. \end{array} \right.\]
We first solve the equation for \(v\). Using the method of characteristics, we can write the ODE system \[\left\{ \begin{array}{l} \partial_\tau X(\tau,s) = 1,\\ \partial_\tau Y(\tau,s) = -1,\\ \partial_\tau z(\tau,s) = 0. \end{array} \right. \ \ \left\{ \begin{array}{l} X(0,s) = 1,\\ Y(0,s) = s,\\ z(0,s) = h(s) - 2g'(s). \end{array} \right.\] Solving this system, we obtain \[\left\{ \begin{array}{l} X(\tau,s) = \tau+1,\\ Y(\tau,s) =-\tau+ s,\\ z(\tau,s) = h(s) - 2g'(s). \end{array} \right.\] For a given arbitrary point \((x,y)\in\mathbb{R}^2\), by inverting the flow map, we find \[\left\{ \begin{array}{l} \tau+1=x,\\ -\tau+ s = y, \end{array} \right. \ \ {\rm{and\ so\ }}\ \left\{ \begin{array}{l} \tau=x-1,\\ s = y+ x -1, \end{array} \right.\] from where the solution reads as \[v(x,y) = h(y+ x -1) - 2g'(y+ x -1).\]
Now we focus on solving the PDE for \(u\), for this we write the ODE system \[\left\{ \begin{array}{l} \partial_\tau X(\tau,s) = 1,\\ \partial_\tau Y(\tau,s) = -2,\\ \partial_\tau z(\tau,s) = v(X(\tau,s),Y(\tau,s)), \end{array} \right. \ \ \left\{ \begin{array}{l} X(0,s) = 1,\\ Y(0,s) = s,\\ z(0,s) = g(s). \end{array} \right.\]
Solving this system, we obtain \[\left\{ \begin{array}{l} X(\tau,s) = \tau+1,\\ Y(\tau,s) =-2\tau+ s,\\ z(\tau,s) = g(s)+\int_0^\tau v(X(t,s),Y(t,s))dt = \int_0^\tau \left(h(s-t) - 2g'(s-t)\right)dt. \end{array} \right.\] For a given arbitrary point \((x,y)\in\mathbb{R}^2\), by inverting the flow map, we find \[\left\{ \begin{array}{l} \tau+1=x,\\ -2\tau+ s = y, \end{array} \right. \ \ {\rm{and\ so\ }}\ \left\{ \begin{array}{l} \tau=x-1,\\ s = y+ 2x -2, \end{array} \right.\] therefore, the general solution for \(u\), and hence of the original problem reads as \[\begin{aligned} u(x,y) &= g(y+2x-2)+ \int_0^{x-1} \left(h(y+2x-2-t) - 2g'(y+2x-2-t)\right)dt\\ & = \int_0^{x-1} h(y+2x-2-t)dt + 2 g(y+x-1) - g(y+2x-2). \end{aligned}\] From this expression we see \[\begin{aligned} u(1,y) &= g(y),\\ \partial_x u (1,y) &= h(y), \end{aligned}\] and by direct computation, we can also check that this \(u\) satisfies also the PDE.
Let \(\Omega\) be an open bounded set with smooth boundary in \(\mathbb{R}^n\) and let \(u_1\) and \(u_2\) be \(C^2\left( \Omega\right)\cap C\left( \overline{\Omega}\right)\) solutions to Poisson equation \[\nonumber \left\{ \begin{array}{ll} -\Delta u_i(\vec{x})=f\left( \vec{x}\right), & \vec{x}\in \Omega,\\ u_i(\vec{x})=g_i(\vec{x}), & \vec{x}\in \partial \Omega, \end{array} \right.\] \(i=1,2\), where \(f\in C^1\left( \overline{\Omega}\right)\) and \(g_1,g_2\in C\left( \partial \Omega\right)\).
Show that for any \(\vec{x}\in\overline{\Omega}\) \[u_2(\vec{x})-u_1(\vec{x}) \leq \max_{\vec{x}\in \partial \Omega}\left( g_2(\vec{x})-g_1(\vec{x})\right).\]
Show that \[\max_{\vec{x}\in\overline{\Omega}}\left\lvert u_2(\vec{x})-u_1(\vec{x})\right\rvert \leq \max_{\vec{x}\in \partial \Omega}\left\lvert g_2(\vec{x})-g_1(\vec{x})\right\rvert.\]
For \(n\in\mathbb{N}\) let \(u_n\in C^2\left( \Omega\right)\cap C\left(
\overline{\Omega}\right)\) solve the system \[\nonumber
\left\{
\begin{array}{ll}
-\Delta u_n(\vec{x})=f\left( \vec{x}\right), &
\vec{x}\in \Omega,\\
u_n(\vec{x})=g_n(\vec{x}), & \vec{x}\in \partial
\Omega,
\end{array}
\right.\] and let \(u\in
C^2\left( \Omega\right)\cap C\left( \overline{\Omega}\right)\)
solve the system \[\nonumber
\left\{
\begin{array}{ll}
-\Delta u(\vec{x})=f\left( \vec{x}\right), & \vec{x}\in
\Omega,\\
u(\vec{x})=g(\vec{x}), & \vec{x}\in \partial \Omega.
\end{array}
\right.\] Show that if \(\left\lbrace
g_n\right\rbrace_{n\in\mathbb{N}}\) converges uniformly to \(g\) on \(\partial
\Omega\) then \(\left\lbrace
u_n\right\rbrace_{n\in\mathbb{N}}\) converges uniformly to \(u\) on \(\overline{\Omega}\).
Recall that we say that a sequence of functions \(\left\lbrace
f_n\right\rbrace_{n\in\mathbb{N}}\) in \(C\left( K\right)\) converges uniformly to
\(f\in C\left( K\right)\) if \[\sup_{x\in K}\left\lvert
f_n(x)-f(x)\right\rvert\underset{n\to\infty}{\longrightarrow}0.\]
Solution:
Since \(w=u_2-u_1\) solves Laplace’s equation \[\nonumber \begin{aligned} -\Delta w(\vec{x})=0 &\qquad \vec{x}\in \Omega\\ w(\vec{x})=g_2(\vec{x})-g_1(\vec{x}) &\qquad \vec{x}\in \partial \Omega \end{aligned}\] we can use the weak maximum principle to conclude that \[\max_{\overline{\Omega}}w(\vec{x}) \leq \max_{\vec{x}\in \partial \Omega}\left( g_2(\vec{x})-g_1(\vec{x})\right),\] from which we conclude that for any \(\vec{x}\in\overline{\Omega}\) \[u_2(\vec{x})-u_1(\vec{x}) \leq \max_{\vec{x}\in \partial \Omega}\left( g_2(\vec{x})-g_1(\vec{x})\right).\]
Interchanging \(u_1\) with \(u_2\) we find that for any \(\vec{x}\in\overline{\Omega}\) \[u_1(\vec{x})-u_2(\vec{x}) \leq \max_{\vec{x}\in \partial \Omega}\left( g_1(\vec{x})-g_2(\vec{x})\right).\] Consequently \[\max_{\vec{x}\in \overline{\Omega}}\left\lvert u_2(\vec{x})-u_1(\vec{x})\right\rvert =\max_{\vec{x}\in \overline{\Omega}}\max\left( u_2(\vec{x})-u_1(\vec{x}),u_1(\vec{x})-u_2(\vec{x})\right)\] \[\leq \max_{\vec{x}\in\Omega}\max\left( \max_{\vec{x}\in \partial \Omega}\left( g_2(\vec{x})-g_1(\vec{x})\right),\max_{\vec{x}\in \partial \Omega}\left( g_1(\vec{x})-g_2(\vec{x})\right)\right)\] \[\leq \max_{x\in\partial \Omega}\left\lvert g_1(\vec{x})-g_2\left( \vec{x}\right)\right\rvert.\]
Using our previous estimation we find that \[\max_{\vec{x}\in \overline{\Omega}}\left\lvert u\left( \vec{x}\right)-u_n\left( \vec{x}\right)\right\rvert \leq \max_{\vec{x}\in \partial \Omega}\left\lvert g\left( \vec{x}\right)-g_n\left( \vec{x}\right)\right\rvert\underset{n\to\infty}{\longrightarrow}0.\] This concludes the proof.
Let \(u\in L^1\left( \mathbb{R}\right)\cap L^2\left( \mathbb{R}\right)\) be a classical solution to the equation \[\nonumber \left\{ \begin{array}{ll} u_t + ku_{xxxx}=0,& \left( x,t\right)\in \mathbb{R}\times\left( 0,+\infty\right),\\ u(x,0)=f(x),& x\in\mathbb{R}, \end{array} \right.\] where \(k>0\) is a fixed constant and \(f\) is a smooth function on \(\mathbb{R}\) that belongs to \(L^1\left( \mathbb{R}\right)\cap L^2\left( \mathbb{R}\right)\).
Show that \(\widehat{u}\), the Fourier transform of \(u\) in the \(x-\)variable, satisfies \[\widehat{u}(\xi,t)=\widehat{f}(\xi)e^{-k\xi^4 t}.\]
Using the fact that the Fourier transform preserves the \(L^2\) norm (Plancherel’s identity) as well as the fact that it is in \(L^\infty\) to show that \[\left\lVert u(\cdot,t)\right\rVert^2_{L^2\left( \mathbb{R}\right)} \leq \frac{\left( \int_{\mathbb{R}}e^{-x^4}dx\right)^{\frac{1}{2}}}{\displaystyle\sqrt[8]{4kt}}\left\lVert f\right\rVert_{L^1\left( \mathbb{R}\right)}\left\lVert f\right\rVert_{L^2\left( \mathbb{R}\right)}.\]
Solution:
Using the linearity of the Fourier transform together with the identity \[\widehat{u^\prime}(\xi)=i\xi \widehat{u}(\xi)\] we find that using the Fourier transform on the spatial variable in our equation yields the equation \[\widehat{u}_t(\xi,t)+k\left( i\xi\right)^4\widehat{u}(\xi,t)=0.\] This is a separable equation whose solution is \[\widehat{u}(\xi,t)=e^{-k\xi^4 t}\widehat{u}(\xi,0)=e^{-k\xi^4 t}\widehat{f}(\xi)\] since \(u(x,0)=f(x)\).
Using the fact that for any \(f\in L^1\left( \mathbb{R}\right)\) \[\left\lVert\widehat{f}\right\rVert_{L^\infty\left( \mathbb{R}\right)} \leq \left\lVert f\right\rVert_{L^1\left( \mathbb{R}\right)}\] together with Plancherel identity and Cauchy-Schwarz we find that \[\left\lVert u(\cdot,t)\right\rVert_{L^2\left( \mathbb{R}\right)}^2=\left\lVert\widehat{u}(\cdot,t)\right\rVert_{L^2\left( \mathbb{R}\right)}^2=\int_{\mathbb{R}}\left\lvert e^{-k\xi^4 t}\widehat{f}(\xi)\right\rvert^2d\xi\] \[\leq \left\lVert\widehat{f}\right\rVert_{L^\infty\left( \mathbb{R}\right)}\left( \int_{\mathbb{R}}e^{-2k\xi^4 t}\left\lvert\widehat{f}\left( \xi\right)\right\rvert d\xi\right) \leq \left\lVert f\right\rVert_{L^1\left( \mathbb{R}\right)}\left\lVert\widehat{f}\right\rVert_{L^2\left( \mathbb{R}\right)}\left( \int_{\mathbb{R}}e^{-4k\xi^4 t}d\xi\right)^{\frac{1}{2}}\] \[\underset{x=\sqrt[4]{4kt}\xi}{=}\frac{\left( \int_{\mathbb{R}}e^{-x^4}dx\right)^{\frac{1}{2}}}{\sqrt[8]{4kt}}\left\lVert f\right\rVert_{L^1\left( \mathbb{R}\right)}\left\lVert f\right\rVert_{L^2\left( \mathbb{R}\right)}.\]