The Fourier transform: The heat equation with source term.
By the Fundamental Theorem of Calculus \[\begin{aligned} \dot{x}(t) & = \lambda G e^{\lambda t} + e^{\lambda(t-s)} F(s) \Big|_{s=t} + \int_0^t \lambda e^{\lambda(t-s)} F(s) \, ds \\ & = \lambda G e^{\lambda t} + F(t) + \lambda \int_0^t e^{\lambda(t-s)} F(s) \, ds \\ & = \lambda x(t) + F(t) \end{aligned}\] as claimed.
Taking the Fourier transform of \(u_t = k u_{xx} + f\) with respect to the \(x\) variable gives \[\widehat{u_t} = k \widehat{u_{xx}} + \hat{f} \quad \Longleftrightarrow \quad \hat{u}_t(\xi,t) = k (i \xi)^2 \hat{u}(\xi,t) + \hat{f}(\xi,t) = -k \xi^2 \hat{u}(\xi,t) + \hat{f}(\xi,t).\] Taking the Fourier transform of the initial condition \(u(x,0)=g(x)\) gives \[\hat{u}(\xi,0)=\hat{g}(\xi).\] We have reduced the PDE to a one-parameter family of uncoupled ODEs, indexed by \(\xi\): \[\hat{u}_t = - k \xi^2 \hat{u} + \hat{f}, \qquad \hat{u}(\xi,0)=\hat{g}(\xi).\] Applying part (i) with \(x = \hat{u}\), \(\lambda = -k \xi^2\), \(F=\hat{f}\), \(G=\hat{g}\) gives \[\hat{u}(\xi,t) = \hat{g}(\xi) e^{- k \xi^2 t} + \int_0^t e^{-k \xi^2 (t-s)} \hat{f}(\xi,s) \, ds.\] Therefore \[\begin{equation} \label{eq:uhat2} u(x,t) = \mathcal{F}^{-1}\left( \hat{g}(\xi) e^{- k \xi^2 t}\right)(x) + \int_0^t \mathcal{F}^{-1}\left( e^{-k \xi^2 (t-s)} \hat{f}(\xi,s)\right)(x) \, ds. \end{equation}\] Recall that \[\widehat{e^{-ax^2}}(\xi) = \frac{1}{\sqrt{2a}} \, e^{-\frac{\xi^2}{4a}}.\] Therefore \[e^{- k \xi^2 t} = \sqrt{2a} \, \widehat{e^{-a x^2}}(\xi) \quad \textrm{for} \quad a = \frac{1}{4 k t}.\] Since the product of Fourier transforms is the Fourier transform of a convolution, we obtain \[\begin{aligned} \hat{g}(\xi) e^{-k \xi^2 t} & = \sqrt{2a} \, \hat{g}(\xi) \, \widehat{e^{-a x^2}}(\xi) \\ & = \sqrt{2a} \frac{1}{\sqrt{2 \pi}} \, \widehat{g*e^{-a x^2}}(\xi) \\ & = \sqrt{ \frac{a}{\pi} } \; \widehat{g*e^{-a x^2}}(\xi) \\ & = \frac{1}{\sqrt{4 \pi k t}} \, \widehat{g*e^{-\frac{x^2}{4 k t}}}(\xi) \end{aligned}\] since \(a = \frac{1}{4kt}\). Therefore \[\begin{equation} \label{IFT1} \mathcal{F}^{-1}\left( \hat{g}(\xi) e^{- k \xi^2 t}\right)(x) = \frac{1}{\sqrt{4 \pi k t}} \, g*e^{-\frac{x^2}{4 k t}} = \Phi(\cdot,t) * g = \int_{-\infty}^\infty \Phi(x-y,t) g(y) \, dy \end{equation}\] where \(\Phi\) is the fundamental solution of the heat equation in \(\mathbb{R}\). Similarly \[\begin{equation} \label{IFT2} \mathcal{F}^{-1}\left( e^{-k \xi^2 (t-s)} \hat{f}(\xi,s)\right)(x) = \frac{1}{\sqrt{4 \pi k (t-s)}} \, e^{-\frac{x^2}{4 k (t-s)}}*f(\cdot,s) %= \Phi (\cdot,t-s) * f(\cdot,s) = \int_{-\infty}^\infty \Phi(x-y,t-s) f(y,s) \, dy. \end{equation}\] Combining equations \(\eqref{eq:uhat2}\), \(\eqref{IFT1}\) and \(\eqref{IFT2}\) yields \[u(x,t) = \int_{-\infty}^\infty \Phi(x - y,t) g(y) \, d y + \int_0^t \int_{-\infty}^\infty \Phi(x-y,t-s) f(y,s) \, d y \, ds\] as required.
The Fourier transform: The transport equation.
By definition \[\begin{aligned} \widehat{\tau_a v}(\xi) & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \tau_a v(x) e^{-i \xi x} \, d x \\ & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty v(x-a) e^{-i \xi x} \, d x \\ & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty v(y)e^{-i \xi (y+a)} \, d y & (y=x-a) \\ & = e^{-i \xi a} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty v(y)e^{-i \xi y} \, d y \\ & = e^{-i \xi a}\hat{v}(\xi) \end{aligned}\] as required.
Taking the Fourier transform of the transport equation \(u_t + c u_x = 0\) gives \[\widehat{u_t} + c \widehat{u_x} = 0 \quad \Longleftrightarrow \quad \hat{u}_t(\xi,t) + c i \xi \hat{u}(\xi,t) = 0\] and taking the Fourier transform of the initial condition \(u(x,0)=g(x)\) gives \[\hat{u}(\xi,0) = \hat{g}(\xi).\] We have reduced the PDE to a one-parameter family of uncoupled ODEs, indexed by \(\xi\): \[\hat{u}_t =- c i \xi \hat{u}, \qquad \hat{u}(\xi,0)=\hat{g}(\xi).\] Recall that the ODE \(\dot{x} = \lambda x\) has solution \(x(t)=x(0)e^{\lambda t}\). Applying this with \(x = \hat{u}\), \(\lambda = - c i \xi\) yields \[\begin{aligned} \hat{u}(\xi,t) & = \hat{u}(\xi,0) e^{- c i \xi t} \\ & = \hat{g}(\xi) e^{- c i \xi t} \\ & = \widehat{\tau_a g}(\xi) \end{aligned}\] where \(a = ct\), by part (i). By taking the inverse Fourier transform we obtain \[u(x,t) = \tau_a g(x) = g(x-a) = g(x-ct)\] as desired.
The Fourier transform: Schrödinger’s equation.
Taking the Fourier transform of \(i u_t = - u_{xx}\) with respect to the \(x\) variable gives \[i \widehat{u_t} = - \widehat{u_{xx}} \quad \Longleftrightarrow \quad i \hat{u}_t(\xi,t) = - (i \xi)^2 \hat{u}(\xi,t) = \xi^2 \hat{u}(\xi,t).\] By multiplying by \(-i\) we can rewrite this as \(\hat{u}_t = - i \xi^2 \hat{u}\). Taking the Fourier transform of the initial condition \(u(x,0)=g(x)\) gives \[\hat{u}(\xi,0)=\hat{g}(\xi).\] We have reduced the PDE to a one-parameter family of uncoupled ODEs, indexed by \(\xi\): \[\hat{u}_t = -i \xi^2 \hat{u}, \qquad \hat{u}(\xi,0)=\hat{g}(\xi).\] Recall that the ODE \(\dot{x} = \lambda x\) has solution \(x(t)=x(0)e^{\lambda t}\). Applying this with \(x = \hat{u}\), \(\lambda = -i \xi^2\) gives \[\begin{equation} \label{eq:uhat} \hat{u}(\xi,t) = \hat{u}(\xi,0) e^{- i \xi^2 t} = \hat{g}(\xi) e^{- i \xi^2 t}. \end{equation}\] To obtain \(u\) we need to compute the following inverse Fourier transform: \[\mathcal{F}^{-1}\left( \hat{g}(\xi) e^{- i \xi^2 t}\right)(x).\] The trick is to recognise that \(\hat{g}(\xi) e^{- i \xi^2 t}\) is the product of Fourier transforms, which follows from the fact that the Fourier transform of a Gaussian is a Gaussian. Recall that \[\widehat{e^{-ax^2}}(\xi) = \frac{1}{\sqrt{2a}} \, e^{-\frac{\xi^2}{4a}}.\] Therefore \[\begin{equation} \label{eq:a} e^{- i \xi^2 t} = \sqrt{2a} \, \widehat{e^{-a x^2}}(\xi) \quad \textrm{for} \quad a = \frac{1}{4 i t}. \end{equation}\] Since the product of Fourier transforms is the Fourier transform of a convolution, we obtain \[\begin{aligned} \hat{g}(\xi) e^{-k \xi^2 t} & = \sqrt{2a} \, \hat{g}(\xi) \, \widehat{e^{-a x^2}}(\xi) \quad & \textrm{(by equation \eqref{eq:a})} \\ & = \sqrt{2a} \frac{1}{\sqrt{2 \pi}} \, \widehat{g*e^{-a x^2}}(\xi) \\ & = \sqrt{ \frac{a}{\pi} } \; \widehat{g*e^{-a x^2}}(\xi). \end{aligned}\] Combining this with equation \(\eqref{eq:uhat}\) and taking the inverse Fourier transform gives \[\hat{u}(\xi,t) = \sqrt{ \frac{a}{\pi} } \; \widehat{g*e^{-a x^2}}(\xi) \quad \Longleftrightarrow \quad u(x,t) = \sqrt{ \frac{a}{\pi} } \, g*e^{-a x^2}.\] Since \(a = \frac{1}{4it}\) and the convolution is commutative we arrive at \[u(x,t) = \frac{1}{\sqrt{4 \pi i t}} \, g*e^{-\frac{x^2}{4it}}(x) = \frac{1}{\sqrt{4 \pi i t}} \, e^{\frac{i x^2}{4t}} * g(x) = \frac{1}{\sqrt{4 \pi i t}} \int_{-\infty}^\infty e^{\frac{i (x-y)^2}{4t}} g(y) \, dy\] as required.
In part (i) we showed that \[\hat{u}(\xi,t) = \hat{g}(\xi) e^{- i \xi^2 t}.\] Since the Fourier transform preserves the \(L^2\)–norm, \[\begin{aligned} \| u(\cdot,t) \|_{L^2(\mathbb{R})}^2 & = \| \hat{u}(\cdot,t) \|_{L^2(\mathbb{R})}^2 \\ & = \| \hat{g}(\xi) e^{- i \xi^2 t} \|_{L^2(\mathbb{R})}^2 \\ & = \int_{-\infty}^\infty \left| \hat{g}(\xi) e^{- i \xi^2 t} \right|^2 d \xi \\ & = \int_{-\infty}^\infty \left| \hat{g}(\xi) \right|^2 d \xi \\ & = \| g \|_{L^2(\mathbb{R})}^2 \end{aligned}\] as required.
The Fourier transform: The wave equation. Use the Fourier transform to derive the solution \[u(x,t)=\frac 12 [g(x-ct)+g(x+ct)]\] of the wave equation \[\begin{aligned} u_{tt} = c^2 u_{xx} \quad & \textrm{for } (x,t) \in \mathbb{R} \times (0,\infty), \\ u(x,0) = g(x) \quad & \textrm{for } x \in \mathbb{R}, \\ u_t(x,0) = 0 \quad & \textrm{for } x \in \mathbb{R}, \end{aligned}\] where the constant \(c>0\) is the wave speed. This is known as D’Alembert’s solution. Hint: Use Q2(i) and the fact that \(\cos(c \xi t) = [\exp(i c \xi t) + \exp(- i c \xi t)]/2\).
Taking the Fourier transform of \(u_{tt} = c^2 u_{xx}\) with respect to the \(x\) variable gives \[\widehat{u_{tt}} = \widehat{c^2 u_{xx}} \quad \Longleftrightarrow \quad \hat{u}_{tt}(\xi,t) = c^2 (i \xi)^2 \hat{u}(\xi,t) = - c^2 \xi^2 \hat{u}(\xi,t).\] Taking the Fourier transform of the initial condition \(u(x,0)=g(x)\) gives \[\hat{u}(\xi,0)=\hat{g}(\xi).\] Taking the Fourier transform of the initial condition \(u_t(x,0)=0\) gives \[\hat{u}_t(\xi,0)=0.\] We have reduced the PDE to a one-parameter family of uncoupled ODEs, indexed by \(\xi\): \[\hat{u}_{tt} = -c^2 \xi^2 \hat{u}, \qquad \hat{u}(\xi,0)=\hat{g}(\xi), \qquad \hat{u}_t(\xi,0) = 0.\] Recall that the ODE \(\ddot{x} = -\lambda^2 x\) has solution of the form \(x(t) = A \cos(\lambda t) + B \sin(\lambda t)\). Applying this with \(x = \hat{u}\), \(\lambda = c \xi\) gives \[\hat{u}(\xi,t) = A \cos(c \xi t) + B \sin (c \xi t).\] The initial conditions \(\hat{u}(\xi,0)=\hat{g}(\xi)\), \(\hat{u}_t(\xi,0) = 0\) imply that \(A = \hat{g}(\xi)\) and \(B=0\). Therefore \[\begin{aligned} \hat{u}(\xi,t) & = \hat{g}(\xi) \cos(c \xi t) \\ & = \hat{g}(\xi) \left[ \frac{\exp(i c \xi t) + \exp(- i c \xi t)}{2} \right] \\ & = \frac 12 \exp(i c \xi t) \hat{g}(\xi) + \frac 12 \exp(-i c \xi t) \hat{g}(\xi) \\ & = \frac 12 \, \widehat{\tau_{-a} g}(\xi) + \frac 12 \, \widehat{\tau_{a} g}(\xi) \end{aligned}\] where \(a=ct\), by Q2(i). Taking the inverse Fourier transform gives \[\begin{aligned} u(x,t)& =\frac 12 \, \tau_{-a} \, g(x) + \frac12 \, \tau_{a} \, g(x) \\ &= \frac 12 \, g(x+a) + \frac 12 \, g(x-a) \\ & = \frac 12 [g(x+ct)+g(x-ct)] \end{aligned}\] as required.
The Fourier transform of a derivative. By definition \[\begin{aligned} \widehat{u'}(\xi) & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty u'(x) e^{-i \xi x} \, dx \\ & = - \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty u(x) \frac{d}{dx} e^{-i \xi x} \, dx & \textrm{(integration by parts)} \\ & = - \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty u(x) (-i \xi) e^{-i \xi x} \, dx \\ & = i \xi \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty u(x) e^{-i \xi x} \, dx \\ & = i \xi \hat{u}(\xi) \end{aligned}\] as required.
The Fourier transform of a convolution. By definition \[\begin{aligned} \widehat{u*v}(\xi) & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty (u*v)(x) e^{-i \xi x} \, d x \\ & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \left( \int_{-\infty}^\infty u(z) v(x-z) \, dz \right) e^{-i \xi x} \, d x \\ & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \left( \int_{-\infty}^\infty u(z) v(x-z) \, dz \right) e^{- i \xi z} e^{-i \xi(x-z)} \, d x \\ & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \left( \int_{-\infty}^\infty v(x-z) e^{-i \xi(x-z)} \, d x \right) u(z) e^{- i \xi z} \, d z \\ & = \int_{-\infty}^\infty \left( \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty v(\tilde{x}) e^{-i \xi \tilde{x}} \, d \tilde{x} \right) u(z) e^{- i \xi z} \, d z & (\tilde{x} = x-z) \\ & = \int_{-\infty}^\infty \hat{v}(\xi) u(z) e^{- i \xi z} \, d z \\ & = \sqrt{2 \pi} \, \hat{v}(\xi) \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty u(z) e^{- i \xi z} \, d z \\ & = \sqrt{2 \pi} \, \hat{v}(\xi) \hat{u}(\xi) \end{aligned}\] as desired.
Proof of the Sobolev embedding using the Fourier transform.
Since the Fourier transform preserves the \(L^2\)–norm \[\begin{aligned} \| u \|_{H^1(\mathbb{R})}^2 & = \| u \|_{L^2(\mathbb{R})}^2 + \| u' \|_{L^2(\mathbb{R})}^2 \\ & = \| \hat{u} \|_{L^2(\mathbb{R})}^2 + \| \, \widehat{u'} \, \|_{L^2(\mathbb{R})}^2 \\ & = \| \hat{u} \|_{L^2(\mathbb{R})}^2 + \| i \xi \hat{u} \|_{L^2(\mathbb{R})}^2 \\ & = \int_{-\infty}^\infty |\hat{u}(\xi)|^2 \, d \xi + \int_{-\infty}^\infty | i \xi \hat{u}(\xi)|^2 \, d \xi \\ & = \int_{-\infty}^\infty (1 + |\xi|^2) |\hat{u}(\xi)|^2 \, d \xi. \end{aligned}\] as required.
Following the hint \[\begin{aligned} \| \hat{u} \|_{L^1(\mathbb{R})} & = \int_{-\infty}^{\infty} | \hat{u}(\xi) | d \xi \\ & = \int_{-\infty}^{\infty} \frac{1}{(1 + |\xi|^2)^{1/2}} \, (1 + |\xi|^2)^{1/2} \, |\hat{u}(\xi)| \, d \xi \\ & \le \left( \int_{-\infty}^{\infty} \frac{1}{(1 + |\xi|^2)} \, d \xi \right)^{1/2} \left( \int_{-\infty}^{\infty} (1 + |\xi|^2) \, |\hat{u}(\xi)|^2 \, d \xi \right)^{1/2} & \textrm{(Cauchy-Schwarz)} \\ & = \left( \int_{-\infty}^{\infty} \frac{1}{(1 + |\xi|^2)} \, d \xi \right)^{1/2} \| u \|_{H^1(\mathbb{R})} \\ & = C \| u \|_{H^1(\mathbb{R})} \end{aligned}\] where \[C = \left( \int_{-\infty}^{\infty} \frac{1}{(1 + |\xi|^2)} \, d \xi \right)^{1/2}.\] If we can show that \(C\) is finite, then we’ve completed the proof. This is a simple calculus exercise; one way is as follows: \[\begin{aligned} C^2 &= 2 \int_{0}^{\infty} \frac{1}{(1 + \xi^2)} \, d \xi \\ & = \int_{0}^{1} \frac{1}{(1 + \xi^2)} \, d \xi + \int_{1}^{\infty} \frac{1}{(1 + \xi^2)} \, d \xi \\ & \le \int_{0}^{1} \frac{1}{(1 + 0)} \, d \xi + \int_{1}^{\infty} \frac{1}{\xi^2} \, d \xi \\ & = 1 + 1 \\ & = 2 < \infty \end{aligned}\] as required.
By the Fourier Inversion Theorem \[\begin{aligned} | u(x)| & = \left| \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \hat{u}(\xi) e^{i \xi x} \, d \xi \right| \\ & \le \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} |\hat{u}(\xi)| \, |e^{i \xi x}| \, d \xi \\ & = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} |\hat{u}(\xi)| \, d \xi \\ & = \frac{1}{\sqrt{2 \pi}} \| \hat{u} \|_{L^1(\mathbb{R})} \\ & \le \frac{C}{\sqrt{2 \pi}} \, \| u \|_{H^1(\mathbb{R})} \end{aligned}\] by part (ii). Since this holds for all \(x \in \mathbb{R}\), \[\| u \|_{L^\infty(\mathbb{R})} \le \frac{C}{\sqrt{2 \pi}} \, \| u \|_{H^1(\mathbb{R})}\] as required.
Fundamental Solution of the Heat Equation. We compute \[\begin{aligned} \Phi_t(\boldsymbol{x},t) & = \frac{1}{(4 \pi k)^{\frac n2}} e^{-\frac{|\boldsymbol{x}|^2}{4kt}} \left( - \frac n2 t^{-\frac n2 - 1} + \frac{|\boldsymbol{x}|^2}{4 k t^2} \right) \\ \frac{\partial \Phi}{\partial x_i}(\boldsymbol{x},t) & = \frac{1}{(4 \pi k t)^{\frac n2}} e^{-\frac{|\boldsymbol{x}|^2}{4kt}} \left( - \frac{2 x_i}{4kt} \right), \\ \frac{\partial^2 \Phi}{\partial x_i^2}(\boldsymbol{x},t) & = \frac{1}{(4 \pi k t)^{\frac n2}} e^{-\frac{|\boldsymbol{x}|^2}{4kt}} \left( - \frac{2}{4kt} + \left( \frac{2 x_i}{4kt} \right)^2 \right) = \frac{1}{(4 \pi k t)^{\frac n2}} e^{-\frac{|\boldsymbol{x}|^2}{4kt}} \left( - \frac{1}{2kt} + \frac{ x_i^2}{4k^2t^2} \right), \\ \Delta \Phi (\boldsymbol{x},t) & = \sum_{i=1}^n \frac{\partial^2 \Phi}{\partial x_i^2}(\boldsymbol{x},t) = \frac{1}{(4 \pi k t)^{\frac n2}} e^{-\frac{|\boldsymbol{x}|^2}{4kt}} \left( - \frac{n}{2kt} + \frac{ |\boldsymbol{x}|^2}{4k^2t^2} \right). \end{aligned}\] Therefore \(\Phi\) satisfies \(\Phi_t(\boldsymbol{x},t) = k \Delta \Phi(\boldsymbol{x},t)\) for all \(\boldsymbol{x}\in \mathbb{R}^n\), \(t>0\), as required.
Finite speed of propagation for a degenerate diffusion equation.
Observe that \[\frac12 \left( \frac{3}{k \pi t} \right)^{\frac 13} - \frac{1}{6k} \frac{x^2}{t} = 0 \quad \Longleftrightarrow \quad |x| = 3^{\frac 23} k^{\frac 13} \pi^{-\frac 16} t^{\frac 13}\] and so \[\Phi(x,t) = \left\{ \begin{array}{cl} \displaystyle \frac12 \left( \frac{3}{k \pi t} \right)^{\frac 13} - \frac{1}{6k} \frac{x^2}{t} & \textrm{if } |x| < 3^{\frac 23} k^{\frac 13} \pi^{-\frac 16} t^{\frac 13}, \\ 0 & \textrm{if } |x| \ge 3^{\frac 23} k^{\frac 13} \pi^{-\frac 16} t^{\frac 13}. \end{array} \right.\] Clearly \(\Phi\) satisfies the degenerate diffusion equation for \(|x| > 3^{\frac 23} k^{\frac 13} \pi^{-\frac 16} t^{\frac 13}\). For \(|x|< 3^{\frac 23} k^{\frac 13} \pi^{-\frac 16} t^{\frac 13}\), \[\begin{aligned} \Phi_t(x,t) & = \frac{1}{6}\left( \frac{3}{k \pi t} \right)^{-\frac 23} \left( - \frac{3}{k \pi t^2} \right) + \frac{x^2}{6kt^2} = -\frac{1}{6t} \left( \frac{3}{k \pi t} \right)^{\frac 13} + \frac{x^2}{6k t^2}, \\ \Phi_x(x,t) & = - \frac{x}{3kt}, \end{aligned}\] \[\begin{aligned} (a(\Phi) \Phi_x)(x,t) & = -\frac{x}{6t} \left( \frac{3}{k \pi t} \right)^{\frac 13} + \frac{x^3}{18 k t^2}, \\ ( a(\Phi) \Phi_x)_x(x,t) & = -\frac{1}{6t} \left( \frac{3}{k \pi t} \right)^{\frac 13} + \frac{x^2}{6k t^2}. \end{aligned}\] Therefore \(\Phi\) satisfies the degenerate diffusion equation for all \(t>0\) and all \(x \in \mathbb{R}\) with \(|x| \ne 3^{\frac 23} k^{\frac 13} \pi^{-\frac 16} t^{\frac 13}\).
For fixed \(t>0\), \(\Phi(x,t)\) is the maximum of a concave quadratic function and the zero function. The support of the map \(x \mapsto \Phi(x,t)\) is the compact interval \[[- 3^{\frac 23} k^{\frac 13} \pi^{-\frac 16} t^{\frac 13}, 3^{\frac 23} k^{\frac 13} \pi^{-\frac 16} t^{\frac 13}].\]
The mathematical equation that caused the banks to crash. Let \(t(x,\tau)\) satisfy \[\tau = T-t(x,\tau).\] Differentiating this expression with respect to \(\tau\) and \(x\) gives \[\begin{gathered} t_\tau = -1, \qquad t_x = 0. \end{gathered}\] Let \(S(x,\tau)\) satisfy \[\begin{equation} \label{eq:BS} x = \ln \left( \frac{S(x,\tau)}{K} \right) + \left( r - \frac 12 \sigma^2 \right) \tau. \end{equation}\] Differentiating this expressions with respect to \(\tau\) gives \[\begin{aligned} 0 = \frac{K}{S} \frac{S_\tau}{K} + r - \frac 12 \sigma^2 \quad \Longleftrightarrow \quad S_\tau = \left( \frac12 \sigma^2 - r \right) S. \end{aligned}\] Differentiating equation \(\eqref{eq:BS}\) with respect to \(x\) gives \[1 = \frac{K}{S} \frac{S_x}{K} \quad \Longleftrightarrow \quad S_x = S.\] Therefore \[\begin{aligned} u_\tau & = C e^{r \tau} \left[ rV + V_S S_\tau + V_t t_\tau \right] = C e^{r \tau} \left[ rV + \left( \frac12 \sigma^2 - r \right) S V_s - V_t \right], \\ u_x & = Ce^{r \tau} \left[ V_S S_x + V_t t_x \right] = Ce^{r \tau} S V_S, \\ u_{xx} & = Ce^{r \tau} \left[ S_x V_S + S V_{SS} S_x \right] = Ce^{r \tau} \left[ S V_S + S^2 V_{SS} \right]. \end{aligned}\] Therefore \[\begin{aligned} u_\tau - \frac 12 \sigma^2 u_{xx} & = C e^{r \tau} \left[ rV + \left( \frac12 \sigma^2 - r \right) S V_s - V_t - \frac 12 \sigma^2 (S V_S + S^2 V_{SS} ) \right] \\ & = - C e^{r \tau} \left[ V_t + \frac12 \sigma^2 S^2 V_{SS} + rS V_S - rV \right] \\ & = 0 \end{aligned}\] since \(V\) satisfies the Black-Scholes PDE. This completes the proof.
The energy method: Uniqueness for the heat equation in a time dependent domain. Let \(u\) and \(v\) be solutions and let \(w=u-v\). Then \(w\) satisfies \[\begin{aligned} w_{t} - k w_{xx} =0 \quad & \textrm{for } (x,t) \in U, \\ w(a(t),t)=0 \quad & \textrm{for } t \in [0,T], \\ w(b(t),t)=0 \quad & \textrm{for } t \in [0,T], \\ w(x,0)=0 \quad & \textrm{for } x \in (a(0),b(0)). \end{aligned}\] Multiply the equation \(w_t = k w_{xx}\) by \(w\) and integrate over \((a(t),b(t))\) to obtain \[\begin{equation} \label{Q6:1} \int_{a(t)}^{b(t)} w w_t \, dx = k \int_{a(t)}^{b(t)} w w_{xx} \, dx. \end{equation}\] Recall the Fundamental Theorem of Calculus: \[\frac{d}{dt} \int_{a(t)}^{b(t)} f(x,t) \, dx = \int_{a(t)}^{b(t)} f_t(x,t) \, dx + \dot{b}(t) f(b(t),t) - \dot{a}(t) f(a(t),t).\] We can use this to rewrite the left-hand side of equation \(\eqref{Q6:1}\) as follows: \[\begin{align} \nonumber \int_{a(t)}^{b(t)} w w_t \, dx & = \frac{d}{dt} \frac{1}{2} \int_{a(t)}^{b(t)} w^2(x,t) \, dx - \frac12 \dot{b}(t) w^2(b(t),t) + \frac 12 \dot{a}(t) w^2(a(t),t) \\ \label{Q6:2} & = \frac{d}{dt} \frac{1}{2} \int_{a(t)}^{b(t)} w^2(x,t) \, dx \end{align}\] since \(w(a(t),t)=w(b(t),t)=0\). We rewrite the right-hand side of equation \(\eqref{Q6:1}\) using integration by parts: \[\begin{equation} \label{Q6:3} k \int_{a(t)}^{b(t)} w w_{xx} \, dx = k w w_x \Big|_{a(t)}^{b(t)} - k \int_{a(t)}^{b(t)} w_x^2 \, dx = - k \int_{a(t)}^{b(t)} w_x^2 \, dx, \end{equation}\] again using the fact that \(w(a(t),t)=w(b(t),t)=0\). Substituting \(\eqref{Q6:2}\) and \(\eqref{Q6:3}\) into \(\eqref{Q6:1}\) gives \[\frac{d}{dt} \frac{1}{2} \int_{a(t)}^{b(t)} w^2(x,t) \, dx = - k \int_{a(t)}^{b(t)} w_x^2 \, dx \le 0.\] Let \[E(t) = \frac{1}{2} \int_{a(t)}^{b(t)} w^2(x,t) \, dx.\] We have shown that \(\dot{E}(t) \le 0\). Hence \[0 \le E(t) \le E(0) = 0\] since \(w=0\) for \(t=0\). Consequently \(E(t)=0\) for all \(t\). Therefore \(w=0\) in \(U\) and so \(u=v\), as required.
The energy method: Uniqueness for a 4th-order heat equation. Let \(u\) and \(v\) be solutions and let \(w=u-v\). Then \(w\) satisfies \[\begin{align} \label{Q7:1} w_t + k w_{xxxx} = 0 \quad & \textrm{for } (x,t) \in (a,b) \times (0,T], \\ \label{Q7:2} w(a,t) = w(b,t)=0 \quad & \textrm{for } t \in [0,T], \\ \label{Q7:3} w_x(a,t) = w_x(b,t)=0 \quad & \textrm{for } t \in [0,T], \\ \label{Q7:4} w(x,0) = 0 \quad & \textrm{for } x \in (a,b). \end{align}\] Multiplying the equation \(w_t = - k w_{xxxx}\) by \(w\) and integrating over \((a,b)\) gives \[\begin{aligned} \int_a^b \underbrace{w w_t}_{\frac12 \frac{\partial}{\partial t} w^2} \, dx & = - k \int_a^b w w_{xxxx} \, dx \\ & = -k w w_{xxx} \Big|_a^b + k \int_a^b w_x w_{xxx} \, dx \\ & = k \int_a^b w_x w_{xxx} \, dx & \textrm{(by \eqref{Q7:2})} \\ & = k w_x w_{xx} \Big|_a^b - k\int_a^b w_{xx} w_{xx} \, dx \\ & = - k\int_a^b w^2_{xx} \, dx & \textrm{(by \eqref{Q7:3}).} %\\ %& = - k \int_a^b w_{xx}^2 \, dx. \end{aligned}\] Therefore \[\frac{d}{dt} \frac12 \int_a^b w^2 \, dx = - k \int_a^b w_{xx}^2 \, dx \le 0\] and so \[0 \le \int_a^b w^2(x,t) \, dx \le \int_a^b w^2(x,0) \, dx = 0\] by \(\eqref{Q7:4}\). We conclude that \(w=0\) and hence \(u=v\), as required.
We consider \(u_t + k u_{xxxx}=0\) to be the 4th-order version of the heat equation \(u_t - k u_{xx} = 0\) since it has the same energy-decay property: \[\frac{d}{dt} \| u \|^2_{L^2([a,b])} \le 0\] (provided that \(u\) and \(u_x\) vanish at \(x=a\) and \(x=b\)). The equation \(u_t - k u_{xxxx}=0\) looks more similar to the heat equation \(u_t - k u_{xx}=0\) (because of the minus sign), but its \(L^2\)–energy grows with time: \[\frac{d}{dt} \| u \|^2_{L^2([a,b])} \ge 0.\]
Asymptotic behaviour of the heat equation with time independent data. Let \(w(\boldsymbol{x},t)=u(\boldsymbol{x},t)-v(\boldsymbol{x})\). We need to prove that \(\lim_{t \to \infty} \| w \|_{L^2(\Omega)} = 0\). By subtracting the PDEs for \(u\) and \(v\) we find that \(w\) satisfies \[\begin{aligned} w_t(\boldsymbol{x},t) - k \Delta w(\boldsymbol{x},t) = 0 \quad & \textrm{for } (\boldsymbol{x},t) \in \Omega \times (0,\infty), \\ w(\boldsymbol{x},t) = 0 \quad & \textrm{for } (\boldsymbol{x},t) \in \partial \Omega \times [0,\infty), \\ w(\boldsymbol{x},0) = u_0(\boldsymbol{x}) - v(\boldsymbol{x}) \quad & \textrm{for } \boldsymbol{x}\in \Omega. \end{aligned}\] Multiplying the equation \(w_t = k \Delta w\) by \(w\) and integrating by parts over \(\Omega\) gives \[\begin{align} \nonumber \int_{\Omega} w w_t \, d \boldsymbol{x}= k \int_\Omega w \Delta w \, d \boldsymbol{x} \quad & \Longleftrightarrow \quad \frac{d}{dt} \frac12 \int_{\Omega} w^2 \, d \boldsymbol{x}= k \int_{\partial \Omega} w \, \nabla w \cdot \boldsymbol{n}\, dS - k \int_{\Omega} \nabla w \cdot \nabla w \, d \boldsymbol{x} \\ \label{Q12:1} & \Longleftrightarrow \quad \frac{d}{dt} \frac12 \int_{\Omega} w^2 \, d \boldsymbol{x}= - k \int_{\Omega} |\nabla w|^2 \, d \boldsymbol{x} \end{align}\] since \(w=0\) on \(\partial \Omega\). By the Poincaré inequality, there exists a constant \(C_p>0\) such that \[\int_\Omega |w|^2 \, d \boldsymbol{x}\le C_p \int_\Omega |\nabla w|^2 \, d \boldsymbol{x}.\] Multiplying this by \(-k/C_p\) gives \[\begin{equation} \label{Q12:2} - \frac{k}{C_p} \int_\Omega |w|^2 \, d \boldsymbol{x}\ge -k \int_\Omega |\nabla w|^2 \, d \boldsymbol{x}. \end{equation}\] Combining equations \(\eqref{Q12:1}\), \(\eqref{Q12:2}\) yields \[\frac{d}{dt} \frac12 \int_{\Omega} w^2 \, d \boldsymbol{x}\le - \frac{k}{C_p} \int_\Omega |w|^2 \, d \boldsymbol{x}.\] Define \[E(t) = \int_{\Omega} w^2(\boldsymbol{x},t) \, d \boldsymbol{x}= \| w \|^2_{L^2(\Omega)}\] and \(\lambda = \frac{2k}{C_p}\). We have shown that \[\dot{E} \le - \lambda E.\] By the Gr̈onwall inequality, \[E(t) \le e^{-\lambda t} E(0).\] Since \(\lambda > 0\), we conclude that \(E(t) \to 0\) as \(t \to \infty\). Therefore \(w \to 0\) in \(L^2(\Omega)\) as \(t \to \infty\), as required.
Asymptotic behaviour of the heat equation with time independent data in the \(L^\infty\)–norm.
By linearity, \(w\) satisfies \[\begin{aligned} w_t(x,t) - k w_{xx}(x,t) = 0 \quad & \textrm{for } (x,t) \in (a,b) \times (0,\infty), \\ w(x,0)=u_0(x)-v(x) \quad & \textrm{for } x \in (a,b), \\ w(a,t)=w(b,t)=0 \quad & \textrm{for } t \in [0,\infty). \end{aligned}\] Multiplying the PDE by \(w\) and integrating over \([a,b]\) gives \[\int_a^b w w_t \, dx = k \int_a^b w w_{xx} \, dx \quad \Longleftrightarrow \quad \frac{d}{dt} \, \frac 12 \int_a^b w^2 \, dx = \underbrace{k w w_x \Big|_a^b}_{=0} - k \int_a^b w_x^2 \, dx\] where we have used the Chain Rule and integration by parts. Note that the boundary terms vanish when we perform integration by parts since \(w(a,t)=w(b,t)=0\). Therefore \[\frac{d}{dt} \int_a^b w^2(x,t) \, dx = - 2k \int_a^b w_x^2(x,t) \, dx\] as required.
We can write the result of part (i) in terms of \(L^2\)–norms as \[\begin{equation} \label{eq:Thm630_1} \frac{d}{dt} \| w \|^2_{L^2([a,b])} = - 2k \| w_x \|^2_{L^2([a,b])}. \end{equation}\] By the Poincaré inequality, there exists a constant \(C>0\) such that \[\begin{equation} \label{eq:Thm630_2} \| w \|^2_{L^2([a,b])} \le C \| w_x \|^2_{L^2([a,b])}. \end{equation}\] Combining equations \(\eqref{eq:Thm630_1}\), \(\eqref{eq:Thm630_2}\) gives \[\begin{equation} \label{eq:Thm630_3} \frac{d}{dt} \| w \|^2_{L^2([a,b])} = - 2 k \| w_x \|^2_{L^2([a,b])} \le - \frac{2k}{C} \| w \|^2_{L^2([a,b])}. \end{equation}\] Define \[E(t) = \| w \|^2_{L^2([a,b])}.\] We can rewrite equation \(\eqref{eq:Thm630_3}\) as \[\dot{E} \le - \lambda E\] with \(\lambda = 2k/C >0\). By the Grönwall inequality, \[E(t) \le E(0) e^{- \lambda t} \to 0 \quad \textrm{as } t \to \infty.\] Therefore, by definition of \(E\), \(w \to 0\) in \(L^2([a,b])\) as \(t \to \infty\), as required.
By differentiating the PDE for \(w\) with respect to \(t\) we obtain \[\begin{aligned} w_{tt}(x,t) - k w_{txx}(x,t) = 0 \quad & \textrm{for } (x,t) \in (a,b) \times (0,\infty), \\ w_t(a,t)=w_t(b,t)=0 \quad & \textrm{for } t \in [0,\infty). \end{aligned}\] In particular, \(w_t\) satisfies the heat equation with Dirichlet boundary conditions, just like \(w\). Therefore the argument we applied in parts (i) and (ii) to \(w\) can also be applied to \(w_t\), which yields \(w_t \to 0\) in \(L^2([a,b])\) as \(t \to \infty\).
We have \[\begin{aligned} \| w_x \|^2_{L^2([a,b])} & = \int_a^b w_x^2(x,t) \, dx \\ & = - \frac{1}{2k} \frac{d}{dt} \int_a^b w^2(x,t) \, dx & \textrm{(by part (i))} \\ & = - \frac{1}{k} \int_a^b w(x,t) w_t(x,t) \, dx \\ & \le \frac 1k \left( \int_a^b w^2(x,t) \, dx \right)^{1/2} \left( \int_a^b w_t^2(x,t) \, dx \right)^{1/2} & \textrm{(Cauchy-Schwarz)} \\ & = \frac 1k \, \| w \|_{L^2([a,b])} \, \| w_t \|_{L^2([a,b])} \to 0 \quad \textrm{as } t \to \infty \end{aligned}\] by parts (ii) and (iii). Therefore \(w_x \to 0\) in \(L^2([a,b])\) as \(t \to \infty\), as required. Note that we don’t really need \(w_t \to 0\) in \(L^2([a,b])\) as \(t \to \infty\), we just need \(\| w_t \|_{L^2([a,b])}\) to be uniformly bounded in \(t\).
This final result follows from the Sobolev inequality: There exists a constant \(C>0\) such that \[\| w \|_{L^\infty([a,b])} \le C \| w \|_{H^1([a,b])} = C \left( \| w \|_{L^2([a,b])} + \| w_x \|_{L^2([a,b])} \right)^{1/2} \to 0 \quad \textrm{as } t \to \infty\] by parts (ii) and (iv).
Applications of the maximum principle: Uniqueness and bounds on solutions.
Let \(\Gamma_T=[a,b] \times \{0\} \cup \{ a , b \} \times [0,T]\) be the parabolic boundary of \(\Omega_T\). Let \(u,v \in C_1^2(\Omega_T) \cap C(\overline{\Omega}_T)\) satisfy \[\begin{aligned} u_t - u_{xx} = 1 \quad & \textrm{in } \Omega_T, \\ u = 0 \quad & \textrm{in } \Gamma_T. \end{aligned}\] Then \(w =u-v \in C_1^2(\Omega_T) \cap C(\overline{\Omega}_T)\) satisfies \[\begin{aligned} w_t - w_{xx} = 0 \quad & \textrm{in } \Omega_T, \\ w = 0 \quad & \textrm{in } \Gamma_T. \end{aligned}\] By the weak maximum principle \[\max_{\overline{\Omega}_T} w = \max_{\Gamma_T} w = 0, \qquad \min_{\overline{\Omega}_T} w = \min_{\Gamma_T} w = 0.\] Therefore \(w=0\) and \(u=v\), as required.
Since \(u_t - u_{xx} = 1 > 0\), the weak maximum principle gives \[\min_{\overline{\Omega}_T} u = \min_{\Gamma_T} u = 0.\] This is the desired lower bound on \(u\). We still need to prove the upper bound. Let \(v(x,t)=t\). Then \(v_t-v_{xx}=1\) and \(w=u-v\) satisfies \[\begin{aligned} w_t - w_{xx} = 0 \quad & \textrm{in } \Omega_T, \\ w = -t \quad & \textrm{in } \Gamma_T. \end{aligned}\] By the weak maximum principle \[\max_{\overline{\Omega}_T} w = \max_{\Gamma_T} w = \max_{\Gamma_T} (-t) = 0.\] Therefore \(w \le 0\) in \(\Omega_T\) and hence \(u \le v = t\) in \(\Omega_T\), which is the desired upper bound.
Application of the maximum principle: Comparison Principle. Define \(v = u_1 - u_2\). Then \(v\) satisfies \[\begin{aligned} \frac{\partial v}{\partial t}(\boldsymbol{x},t) - k \Delta v(\boldsymbol{x},t) = f_1(\boldsymbol{x}) - f_2(\boldsymbol{x}) \quad & \textrm{for } (\boldsymbol{x},t) \in \Omega \times (0,T], \\ v(\boldsymbol{x},t) = g_1(\boldsymbol{x})-g_2(\boldsymbol{x}) \quad & \textrm{for } (\boldsymbol{x},t) \in \partial \Omega \times [0,T], \\ v(\boldsymbol{x},0) = u^0_1(\boldsymbol{x}) - u^0_2(\boldsymbol{x}) \quad & \textrm{for } \boldsymbol{x}\in \Omega. \end{aligned}\] Since \(f_1 \le f_2\), then \[v_t - k \Delta v = f_1 - f_2 \le 0 \quad \textrm{in } \Omega_T.\] Therefore the weak maximum principle implies that \[\max_{\overline{\Omega_T}} v = \max_{\Gamma_T} v.\] For \((\boldsymbol{x},t) \in \Gamma_T\), \[v(\boldsymbol{x},t) = \left\{ \begin{array}{cl} g_1(\boldsymbol{x}) - g_2(\boldsymbol{x}) & \textrm{if } (\boldsymbol{x},t) \in \partial \Omega \times [0,T], \\ u^0_1(\boldsymbol{x}) - u^0_2(\boldsymbol{x}) & \textrm{if } t=0, \, \boldsymbol{x}\in \Omega. \end{array} \right.\] But \[g_1 - g_2 \le 0, \qquad u^0_1 - u^0_2 \le 0.\] Therefore \(v \le 0\) on \(\Gamma_T\) and hence \[\max_{\overline{\Omega_T}} v = \max_{\Gamma_T} v \le 0.\] Hence \(v \le 0\) in \(\Omega_T\) and so \(u_1 \le u_2\) in \(\Omega_T\), as required.
Eigenfunctions of the Laplacian and an application to the heat equation. Formally (not worrying about interchanging limits and infinite sums), \[\begin{aligned} 0 & = v_t - k \Delta v \\ & = \sum_{n=1}^\infty \dot{c}_n(t) u_n(\boldsymbol{x}) - k \sum_{n=1}^\infty c_n(t) \Delta u_n(\boldsymbol{x}) \\ & = \sum_{n=1}^\infty \dot{c}_n(t) u_n(\boldsymbol{x}) + k \sum_{n=1}^\infty c_n(t) \lambda_n u_n(\boldsymbol{x}) \\ & = \sum_{n=1}^\infty (\dot{c}_n(t) + k \lambda_n c_n(t) ) \, u_n(\boldsymbol{x}). \end{aligned}\] Since \(\{u_n\}_{n \in \mathbb{N}}\) forms an orthogonal basis, it follows that \[\dot{c}_n(t) + k \lambda_n c_n(t) = 0\] for all \(n\). We also have \[v(\boldsymbol{x},0) = g(\boldsymbol{x}) \quad \Longleftrightarrow \quad \sum_{n=1}^\infty c_n(0) u_n(\boldsymbol{x}) = \sum_{n=1}^\infty g_n u_n(\boldsymbol{x}) \quad \Longleftrightarrow \quad \sum_{n=1}^\infty (c_n(0)-g_n) u_n(\boldsymbol{x}) = 0.\] Again, since \(\{u_n\}_{n \in \mathbb{N}}\) forms an orthogonal basis, it follows that \[c_n(0)=g_n\] for all \(n\). We have reduced the PDE for \(v\) to a one-parameter family of uncoupled ODEs, indexed by \(n\): \[\dot{c}_n(t) = - k \lambda_n c_n(t), \qquad c_n(0)=g_n.\] These ODEs have solutions \[c_n(t) = g_n e^{- k \lambda_n t}.\] Therefore \[v(\boldsymbol{x},t) = \sum_{n=1}^\infty g_n e^{- k \lambda_n t} u_n(\boldsymbol{x})\] as required.