Through the Heat Equation
The Heat Equation
Derivation of the heat equation
Consider an infinite metal plate which we model as the plane \(\mathbb{R}^{2}\), and suppose we are given an initial heat distribution at time \(t=0\). Let the temperature at the point \((x, y)\) at time \(t\) be denoted by \(u(x,y,t)\).
Consider a small square centered at \((x_0,y_0)\) with sides parallel to the axis and of side length \(h\). The amount of heat energy in \(S\) at time \(t\) is given by \[ H(t)= \sigma \iint_{S} u(x,y,t) \mathrm{d}x \mathrm{d}y, \] where \(\sigma>0\) is a constant called the specific heat of the material. Therefore, the heat flow into \(S\) is \[ \frac{\partial H}{\partial t}= \sigma \iint_{S} \frac{\partial u}{\partial t} \mathrm{d}x\mathrm{d}y, \] which is approximately equal to \[ \sigma h^{2} \frac{\partial u}{\partial t}(x_0,y_0,t), \] since the area of \(S\) is \(h^{2}\). Now we apply Newton's law of cooling, which states that heat flows from the higher to lower temperature at a rate proportional to the difference, that is, the gradient.
The heat flow through the vertical side on the right is therefore \[ - \kappa h \frac{\partial u}{\partial x} (x_0+\frac{h}{2},y_0,t), \] where \(\kappa>0\) is the conductivity of the material. A similar argument for the other sides shows that the total heat flow through the square \(S\) is given by \[ \kappa h \left[ \frac{\partial u}{\partial x}(x_0+\frac{h}{2},y_0,t)- \frac{\partial u}{\partial x}(x_0-\frac{h}{2},y_0,t)+\frac{\partial u}{\partial y}(x_0,y_0+\frac{h}{2},t)-\frac{\partial u}{\partial y}(x_0,y_0-\frac{h}{2},t)\right]. \] Applying the mean value theorem and letting \(h\) tend to zero, we find that \[ \frac{\sigma}{\kappa} \frac{\partial u}{\partial t}= \frac{\partial ^{2}u}{\partial x^{2}}+ \frac{\partial ^{2}u}{\partial y^{2}}; \] this is called the time-dependent heat equation, often abbreviated to the heat equation.
Steady-state heat equation in the disc
After a long period of time, there is no more heat exchange, so that the system reaches thermal equilibrium and \(\displaystyle \frac{\partial u}{\partial t}=0\). In this case, the time-dependent heat equation reduces to the steady-state heat equation \[ \frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}=0. \tag{1} \] The operator \(\displaystyle \frac{\partial ^{2}}{\partial x^{2}}+\frac{\partial ^{2}}{\partial y^{2}}\) is of such importance in mathematics and physics that it is often abbreviated as \(\Delta\) and given a name: the Laplace operator or Laplacian. So the steady-state heat equation is written as \[ \Delta u=0, \] and solutions to this equation are called harmonic functions.
Consider the unit disc in the plane \[ D=\{(x, y) \in \mathbb{R}^{2}\colon x^{2}+y^{2}<1\}, \] whose boundary is the unit circle \(C\). In polar coordinates \((r,\theta)\), with \(0\leqslant r\) and \(0\leqslant \theta<2\pi\), we have \[ D=\{(r,\theta) \colon 0\leqslant r<1\} \quad\text{and} \quad C=\{(r,\theta)\colon r=1\}. \] The problem. often called the Dirichlet problem (for the Laplacian on the unit disc), is to solve the steady-state heat equation in the unit disc subject to the boundary condition \(u=f\) on \(C\). This corresponds to fixing a predetermined temperature distribution on the circle, waiting a long time, and then looking at the temperature distribution inside the disc.
While the method of separation of variables will turn out to be useful for equation (1), a difficulty comes from the fact that the boundary condition is not easily expressed in terms of rectangular coordinates. Since this boundary condition is best described by the coordinates \((r,\theta)\), namely \(u(1,\theta)=f(\theta)\), we rewrite the Laplacian in polar coordinates. An application of the chain rule gives: \[ \Delta u= \frac{\partial ^{2}u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^{2}} \frac{\partial ^{2}u}{\partial \theta^{2}}. \] We now multiply both sides by \(r^{2}\), and since \(\Delta u=0\), we get \[ r^{2}\frac{\partial ^{2}u}{\partial r^{2}}+r\frac{\partial u}{\partial r}=-\frac{\partial ^{2}u}{\partial \theta^{2}}. \] Separating these variables, and looking for a solution of the form \(u(r,\theta)=F(r)G(\theta)\), we find \[ \frac{r^{2}F''(r)+rF'(r)}{F(r)}=- \frac{G''(\theta)}{G(\theta)}. \] Since the two sides depend on different variables, they must both be constant, say equal to \(\lambda\). We therefore get the following equations: \[ \begin{cases} G''(\theta)+\lambda G(\theta)=0, \\ r^{2} F''(r)+rF'(r)-\lambda F(r)=0. \end{cases} \] Since \(G\) must be periodic of period \(2\pi\), this implies that \(\lambda\geqslant 0\) and (as we have seen before) that \(\lambda=m^{2}\) where \(m\) is an integer; hence \[ G(\theta)= \tilde{A} \mathrm{e}^{im\theta} +\tilde{B} \mathrm{e}^{-im\theta} . \] An application of Euler's identity, \(\mathrm{e}^{ix} =\cos x+ i \sin x\), allows one to rewrite \(G\) in terms of complex exponentials, \[ G(\theta)= A \mathrm{e}^{im\theta} +B \mathrm{e}^{-im \theta} . \]
With \(\lambda=m^{2}\) and \(m\neq 0\), two simple solution of the equation in \(F\) are \(F(r)=r^{m}\) and \(F(r)=r^{-m}\). If \(m=0\), then \(F(r)=1\) and \(F(r)=\log r\) are two solutions. If \(m>0\), we note that \(r^{-m}\) grows unboundedly large as \(r\) tends to zero, so \(F(r)G(\theta)\) is unbounded at the originl; the same occurs when \(m=0\) and \(F(r)= \log r\). We reject these solutions as countrary to our intuition. Therefore, we are left with the following special functions: \[ u_{m}(r,\theta)= r^{\lvert m \rvert }\mathrm{e}^{im\theta} , \quad m \in \mathbb{Z}. \] We now make the important observation that (1) is linear, and so as in the case of the vibrating string, we may superpose the above special solutions to obtain the presumed general solution: \[ u(r,\theta)= \sum_{m=-\infty}^{\infty} a_m r^{\lvert m \rvert }\mathrm{e}^{im \theta} . \] If this expression gave all the solutions to the steady-state heat equation, then for a reasonable \(f\) we should have \[ u(1,\theta)= \sum_{m=-\infty}^{\infty} a_m \mathrm{e}^{im \theta} = f(\theta). \] We therefore ask again in this context: given any reasonable function \(f\) on \([0,2\pi]\) with \(f(0)= f(2\pi)\), can we find coefficients \(a_m\) so that \[ f(\theta)= \sum_{m=-\infty}^{\infty} a_m \mathrm{e}^{im \theta} \]
solution to Euler equation: \[ r^{2}F''(r)+rF'(r)-n^{2}F(r)=0, \] which are twice differentiable when \(r>0\), are given by linear combinations of \(r^{n}\) ad \(r^{-n}\) when \(n\neq 0\), and \(1\) and \(\log r\) when \(n=0\). Write \(F(r)=g(r)r^{n}\), then \(rg'(r)+2ng(r)=c\) where \(c\) is a constant.
Dirichlet problem in the rectangle \(R=\{(x, y)\colon 0\leqslant x\leqslant \pi,0\leqslant y\leqslant 1\}\) and \[ u(x,0)=f_0(x) \quad u(x,1)=f_1(x) \quad u(0,y)=u(\pi,y)=0, \] If \(f_0\) and \(f_1\) have Fourier expansions \[ f_0(x)=\sum_{k=1}^{\infty} A_k \sin kx \quad f_1(x)= \sum_{k=1}^{\infty} B_k \sin kx, \] then \[ u(x, y)= \sum_{k=1}^{\infty} \left( \frac{\sinh k(1-y)}{\sinh k}A_k+ \frac{\sinh ky}{\sinh k}B_k\right) \sin kx \]
The Poisson kernel and Dirichlet's problem in the unit disc
To adapt Abel summability to the context of Fourier series, we define the Abel means of the function $f() _{n=-}^{} a_n ^{in } $ by \[ A_r(f)(\theta)= \sum_{n=-\infty}^{\infty} r^{\lvert n \rvert }a_n \mathrm{e}^{in \theta} . \] It is natural to write \(c_0=a_0\), and \(c_n= a_n \mathrm{e}^{in \theta} +a_{-n}\mathrm{e}^{-in \theta}\) for \(n>0\).
Since \(f\) is integrable, $a_n $ is uniformly bounded in \(n\), so that \(A_r(f)\) converges absolutely and uniformly for each \(0\leqslant r<1\). \[ A_r(f)(\theta)=(f * P_r)(\theta), \] where \(P_{r}(\theta)\) is the Poisson kernel given by \[ P_{r}(\theta)= \sum_{n=-\infty}^{\infty} r^{\lvert n \rvert }\mathrm{e}^{in \theta} \] Actually, if \(0\leqslant r<1\), then \[ P_{r}(\theta)= \frac{1-r^{2}}{1-2r \cos \theta+r^{2}}. \] The Poisson kernel is a good kernel, as \(r\) tends to \(1\) from below.
Theorem The Fourier series of an integrable function on the circle is Abel summable to \(f\) at every point of continuity. Moreover, if \(f\) is continuous on the circle, then the Fourier series of \(f\) is uniformly Abel summable to \(f\).
We now return to Dirichlet problem in the unit disc with boundary condition \(u=f\) on the circle. We expected that a solution was given by \[ u(r,\theta)= \sum_{m=-\infty}^{\infty} a_mr^{\lvert m \rvert }\mathrm{e}^{im \theta} , \] where \(a_m\) was the \(m^{th}\) Fourier coefficient of \(f\). In other words, we were led to take \[ u(r,\theta)= A_{r}(f)(\theta)= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(\varphi)P_{r}(\theta-\varphi) \mathrm{d}\varphi. \]
Theorem Let \(f\) be an integrable function defined on the unit circle. Then the function \(u\) defined in the unit disc by the Poisson integral \[ u(r,\theta)=(f*P_{r})(\theta) \] has the following properties: - u has two continuous derivatives in the unit disc and satisfies \(\Delta u=0\). - If \(\theta\) is any point of continuity of \(f\), then \[ \lim_{r \to 1}u(r,\theta)=f(\theta). \] If \(f\) is continuous everywhere, then this limit is uniform. - If \(f\) is continuous, then \(u(r,\theta)\) is the unique solution to the steady-state heat equation in the disc which satisfies the above two conditions.
We only prove the last property. Suppose \(v\) solves the steady-state heat equation in the disc and converges to \(f\) uniformly as \(r\) tends to \(1\) from below. For each fixed \(r\) with \(0<r<1\), the function \(v(r,\theta)\) has a Fourier series \[ \sum_{n=-\infty}^{\infty} a_n(r)\mathrm{e}^{in \theta} \quad\text{where}\quad a_n(r)=\frac{1}{2\pi} \int_{-\pi}^{\pi} v(r,\theta)\mathrm{e}^{-in \theta} \mathrm{d}\theta. \] Taking into account that \(v(r,\theta)\) solves the equation \[ \frac{\partial ^{2}v}{\partial r^{2}}+\frac{1}{r}\frac{\partial v}{\partial r}+\frac{1}{r^{2}}\frac{\partial ^{2}v}{\partial \theta^{2}}=0, \tag{7} \] we find that \[ a''_n(r)+\frac{1}{r}a_n'(r)- \frac{n^{2}}{r^{2}}a_n(r)=0. \tag{8} \] Indeed, we may first multiply (7) by $^{-in } $ and integrate in \(\theta\). Then, since \(v\) is periodic, two integrations by parts give \[ \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{\partial ^{2}v}{\partial \theta^{2}}(r,\theta)\mathrm{e}^{-in \theta} \mathrm{d}\theta = -n^{2} a_n(r). \] Finally, we may interchange the order of differentiation and integration, which is permissible since \(v\) has two continuous derivatives; this yields (8).
Therefore, we must have \(a_n(r)= A_n r^{n}+B_n r^{-n}\) for some constants \(A_n\) and \(B_n\), when \(n \neq 0\). To evaluate the constants, we first observe that each term \(a_n(r)\) is bounded because \(v\) is bounded, therefore \(B_n=0\). Tho find \(A_n\) we let \(r \to 1\). Since \(v\) converges uniformly to \(f\) we find that \[ A_n= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(\theta)\mathrm{e}^{-in \theta} \mathrm{d}\theta. \] By a similar argument, this formula also holds when \(n=0\). Our conclusion is that for each \(0<r<1\), the Fourier series of \(v\) is given by the series of \(u(r,\theta)\), so by the uniqueness of Fourier series for continuous functions, we must have \(u=v\).
Remark. By part (iii) of the theorem, we may conclude that if \(u\) solves \(\Delta u=0\) in the disc, and converges to \(0\) uniformly as \(r \to 1\), then \(u\) must be identically \(0\). However, if uniform convergence is replaced by pointwise convergence, this conclusion may fail.
Actually, If \(P_{r}(\theta)\) denotes the Poisson kernel, then the function \[ u(r,\theta)= \frac{\partial P_r}{\partial \theta}=\frac{2(r^{2}-1)r \sin \theta}{(1-2r\cos \theta+r^{2})^{2}} \] defined for \(0\leqslant r<1\) and \(\theta \in \mathbb{R}\), satisfies: - \(\Delta u=0\) in the disc. - \(\lim_{r \to 1}u(r,\theta)=0\) for each \(\theta\).
However, \(u\) is not identically zero.
Solve Laplace's equation \(\Delta u=0\) in the semi infinite strip \[ S=\{(x, y)\colon 0<x<1,0<y\}, \] subject to the following boundary conditions \[ \begin{cases} u(0,y)=0 \quad y\geqslant 0, \\ u(1,y)=0 \quad y\geqslant 0, u(x,0)=f(x) \quad 0\leqslant x\leqslant 1 \end{cases} \] where \(f\) is a given function, with of course \(f(0)=f(1)=0\). Write \[ f(x)= \sum_{n=1}^{\infty} a_n \sin (n \pi x) \] and expand the general solution in terms of the special soluions given by \[ u_n(x,y)= \mathrm{e}^{-n\pi y} \sin (n\pi x). \] Express \(u\) as an integral involving \(f\), analogous to the Poisson integral formula.
Answer:
By considering the odd extension of \(f\) and following the derivation of Poisson's kernel with $^{-y} $ and $^{i t} $ replacing \(r\) and $^{it} $, respectively, we obtain \[ u(x,y)= \frac{1}{2} \int_{-1}^{1} f(t)Q_y(x-t) \mathrm{d}t \] where \[ Q_y(t)= \frac{1-\mathrm{e}^{-2\pi y} }{1-2\mathrm{e}^{-\pi y} \cos \pi t+ \mathrm{e}^{-2\pi y} }. \] or, using the fact that \(f\) is odd, we have the alternate form \[ u(x,y)=\frac{1}{2} \int_{0}^{1} f(t)Q(x,t) \mathrm{d}t \] where \[ Q(x,t)=\frac{1-\mathrm{e}^{-2\pi y} }{1-2\mathrm{e}^{-\pi y} \cos \pi (x-t)+ \mathrm{e}^{-2\pi y} }-\frac{1-\mathrm{e}^{-2\pi y} }{1-2\mathrm{e}^{-\pi y} \cos \pi (x+t)+ \mathrm{e}^{-2\pi y} }. \]
Consider the Dirichlet problem in the annulus defined by \(\{(r,\theta)\colon \rho<r<1\}\), where \(0<\rho<1\) is the inner radius. The problem is to solve \[ \frac{\partial ^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}} \frac{\partial ^{2}u}{\partial \theta^{2}}=0 \] subject to the boundary conditions \[ \begin{cases} u(1,\theta)=f(\theta), \\ u(\rho,\theta)=g(\theta), \end{cases} \] where \(f\) and \(g\) are given continuous functions.
Arguing as we have previously for the Dirichlet problem in the disc, we can hope to write \[ u(r,\theta)= \sum_{}^{} c_n(r) \mathrm{e}^{in \theta} \] with \(c_n(r)=A_nr^{n}+B_nr^{-n},n\neq 0\). Set \[ f(\theta) \sim \sum_{}^{} a_n \mathrm{e}^{in \theta} \quad \text{and}\quad g(\theta) \sim \sum_{}^{} b_n \mathrm{e}^{in \theta} . \] We want \(c_n(1)=a_n\) and \(c_n(\rho)=b_n\). This leads to the solution \[ u(r,\theta)= \sum_{n\neq 0}^{} \left( \frac{1}{\rho^{n}-\rho^{-n}}\right)[((\rho/r)^{n}-(r/\rho)^{n})a_n+(r^{n}-r^{-n})b_n] \mathrm{e}^{in \theta}+a_0+(b_0-a_0) \frac{\log r}{\log \rho}. \]
Show that as a result we have \[ u(r,\theta)-(P_{r}*f)(\theta)\to 0 \quad \text{as}\ r\to 1\ \text{uniformly in}\ \theta \] and \[ u(r,\theta)-(P_{\rho/r}*g)(\theta)\to 0 \quad \text{as}\ r\to \rho\ \text{uniformly in}\ \theta \]
The heat equation on the circle
As a final illustration, we return to the original problem of heat diffusion considered by Fourier.
Suppose we are given an initial temperature distribution at \(t=0\) on a ring and that we are asked to describe the temperature at points on the ring at times \(t>0\).
The ring is modeled by the unit circle. A point on this circle is described by its angle \(\theta=2\pi x\), where the variable \(x\) lies between \(0\) and \(1\). If \(u(x,t)\) denotes the temperature at time \(t\) of a point described by the angle \(\theta\), then consideration similar to the ones given in Chapter 1 show that \(u\) satisfies the differential equation \[ \frac{\partial u}{\partial t}= c\frac{\partial ^{2}u}{\partial x^{2}}. \tag{9} \]
The constant \(c\) is a positive physical constant which depends on the material of which thhe ring is made. After rescaling the time variable, we may assume that \(c=1\). If \(f\) is our initial data, we impose the condition \[ u(x,0)=f(x). \] To solve the problem, we separate variables and look for special solutions of the form \[ u(x,t)=A(x)B(t). \] Then inserting this expression for \(u\) into the heat equation we get \[ \frac{B'(t)}{B(t)}= \frac{A''(x)}{A(x)}. \] Both sides are therefore constant, say equal to \(\lambda\). Since \(A\) must be periodic of period \(1\), we see that the only possibility is \(\lambda=-4\pi^{2}n^{2}\), where \(n \in \mathbb{Z}\). Then \(A\) is a linear combination of the exponentials $^{2inx} $ and $^{-2inx} $, and \(B(t)\) is a multiple of ${-4{2}n^{2}t} $. By superposing these solutions, we are led to \[ u(x,t)=\sum_{n=-\infty}^{\infty} a_n \mathrm{e}^{-4\pi^{2}n^{2}t} \mathrm{e}^{2\pi inx}, \tag{10} \] where, setting \(t=0\), we see that \(\{a_n\}\) are the Fourier coefficients of \(f\).
Note that when \(f\) is Riemann integrable, the coefficients \(a_n\) are bounded, and since the factor \(\mathrm{e}^{-4\pi^{2}n^{2}t}\) tends to zero extremely fast, the series defining \(u\) converges. In fact, in this case, \(u\) is twice differentiable and solves equation (9).
For a better understanding of the properties of our solution (10), we write it as \[ u(x,t)=(f*H_t)(x), \] where \(H_t\) is the heat kernel for the circle, given by \[ H_{t}(x)=\sum_{n=-\infty}^{\infty} \mathrm{e}^{-4\pi^{2}n^{2}t} \mathrm{e}^{2\pi inx} , \tag{11} \] and where the convolution for functions with period \(1\) is defined by \[ (f*g)(x)= \int_{0}^{1} f(x-y)g(y) \mathrm{d}y. \]
The natural question with regard to the boundary condition is the following: do we have $u(x,t) f(x) $ as \(t\) tends to \(0\), and in what sense? A simple application of the Parsevel identity shows that this limit holds in the mean square sense, namely \[ \int_{0}^{1} \lvert u(x,t)-f(x) \rvert ^{2} \mathrm{d}x \to 0 \quad \text{as}\ t\to 0. \]
An analogy between the heat kernel and the Poisson kernel:
\[ u(\theta,\tau)= \sum_{}^{} a_n \mathrm{e}^{-n^{2}\tau} \mathrm{e}^{in \theta} =(f* h_{\tau})(\theta) \] of the equation \[ \frac{\partial u}{\partial \tau}=\frac{\partial ^{2}u}{\partial \theta^{2}} \quad \text{with}\ 0\leqslant \theta\leqslant 2\pi \ \text{and} \ \tau>0, \] with boundary condition $u(,0)= f() {}^{} a_n ^{in } $. Here \(h_{\tau}(\theta)= \sum_{n=-\infty}^{\infty} \mathrm{e}^{-n^{2}\tau} \mathrm{e}^{in \theta}\). This version of the heat kernel on \([0,2\pi]\) is the analogue of the Poisson kernel, which can be written as $P{r}()=_{n=-}^{} e^{-n }^{in } $ with $r= ^{-} $(and so \(0<r<1\) corresponds to \(\tau>0\)).
Unlike in the case of the Poisson kernel, there is no elementary formula for the heat kernel. Nevertheless, it turns out that it is a good kernel. The proof is not obvious and requires the use of the celebrated Poisson summation formula. As a corollary, we will also find that \(H_t\) is everywhere positive, a fact that is also not obvious from its defining expression (11). We can, however, give the following heuristic argument for the positivity of \(H_t\). Suppose that we begin with an initial temperature distribution \(f\) which is everywhere \(\leqslant 0\). Then it is physically reasonable to expect \(u(x,t)\leqslant 0\) for all \(t\) since heat travels from hot to cold. Now \[ u(x,t)= \int_{0}^{1} f(x-y)H_t(y) \mathrm{d}y. \] If \(H_{t}\) is negative for some \(x_0\), then we may choose \(f\leqslant 0\) supported near \(0\), and this would imply \(u(x_0,t)>0\), which is a contradiction.
The time-dependent heat equation on the real line
Here we study the analogous problem on the real line.
Consider an infinite rod, which we model by the real line, and suppose that we are given an initial temperature distribution \(f(x)\) on the rod at time \(t=0\). We wish now to determine the temperature \(u(x,t)\) at a point \(x\) at time \(t>0\). Considerations similar to the ones given in Chapter 1 show that when \(u\) is appropriately normalized, it solves the following PDE: \[ \frac{\partial u}{\partial t}= \frac{\partial ^{2}u}{\partial x^{2}}, \tag{12} \] called the heat equation. The initial condition we impose is \(u(x,0)=f(x)\).
Just as in the case of the circle, the solution is given in terms of a convolution. Indeed, define the heat kernel of the line by \[ \mathcal{H}_{t}(x)= K_{\delta}(x), \quad \text{with} \ \delta=4\pi t, \] so that \[ \mathcal{H}_{t}(x)=\frac{1}{(4\pi t)^{1/2}} \mathrm{e}^{-x^{2}/4t} \quad \text{and} \quad \hat{\mathcal{H}}_{t}(\xi)= \mathrm{e}^{-4\pi^{2} t \xi ^{2}} . \]
Taking the Fourier transform of equation (12) in the \(x\) variable leads to \[ \frac{\partial \hat{u}}{\partial t}(\xi, t)= -4\pi^{2} \xi^{2} \hat{u}(\xi,t). \] Fixing \(\xi\), this is an ODE in the variable \(t\), so there exists a constant \(A(\xi)\) so that \[ \hat{u}(\xi,t)= A(\xi) \mathrm{e}^{-4\pi^{2}\xi^{2}t} . \] We may also take the Fourier transform of the initial condition and obtain \(\hat{u}(\xi,0)= \hat{f}(\xi)\), hence \(A(\xi)=\hat{f}(\xi)\). This leads to the following theorem.
Theorem 2.1 Given \(f\in \mathcal{S}(\mathbb{R})\), let \[ u(x,t)=(f*\mathcal{H}_{t})(x) \quad t>0 \] where \(\mathcal{H}_{t}\) is the heat kernel. Then: 1. The function \(u\) is \(C^{2}\) when \(x \in \mathbb{R}\) and \(t>0\), and \(u\) solves the heat equation. 2. \(u(x,t) \to f(x)\) uniformly in \(x\) as \(t \to 0\). Hence if we set \(u(x,0)=f(x)\), then \(u\) is continuous on the closure of the upper half-plane \(\overline{\mathbb{R}_{+}^{2}}=\{(x,t)\colon x\in \mathbb{R}, t\geqslant 0\}\). 3. \(\int_{-\infty}^{\infty} \lvert u(x,t)-f(x) \rvert ^{2} \mathrm{d}x \to 0\) as \(t \to 0\).
proof Because \(u= f* \mathcal{H}_{t}\), taking the Fourier transform in the \(x-variable\) gives \(\hat{u}=\hat{f} \hat{\mathcal{H}}_{t}\), and so $(,t)=() {-4{2}^{2}t} $. The Fourier inversion formula gives \[ u(x,t)= \int_{-\infty}^{\infty} \hat{f}(\xi)\mathrm{e}^{-4\pi^{2}t \xi^{2}} \mathrm{e}^{2\pi i \xi x} \mathrm{d}\xi . \] By differentiating under the integral sign, one verifies 1. In fact, one observes that \(u\) is indefinitely differentiable. Note that 2 is an immediate consequence of Corollary 1.7. Finally, by Plancherel's formula, we have \[ \begin{aligned} \int_{-\infty}^{\infty} \lvert u(x,t)-f(x) \rvert ^{2} \mathrm{d}x &= \int_{-\infty}^{\infty} \lvert \hat{u}(\xi,t)-\hat{f}(\xi) \rvert ^{2} \mathrm{d}\xi \\ &= \int_{-\infty}^{\infty} \lvert \hat{f}(\xi) \rvert ^{2} \lvert \mathrm{e}^{-4\pi^{2}t \xi^{2}} -1 \rvert ^{2} \mathrm{d}\xi. \end{aligned} \]
To see that this last integral goes to \(0\) as \(t \to 0\), we argue as follows: since \(\lvert \mathrm{e}^{-4\pi^{2}t \xi ^{2}} -1 \rvert \leqslant 2\) and \(f\in \mathcal{S}(\mathbb{R})\), we can find \(N\) so that \[ \int_{\lvert \xi \rvert \geqslant N}^{} \lvert \hat{f}(\xi) \rvert ^{2}\lvert \mathrm{e}^{-4\pi^{2}t \xi^{2}} -1 \rvert \mathrm{d}\xi <\varepsilon, \] and for all small \(t\) we have \(\sup_{\lvert \xi \rvert \leqslant N}\lvert \hat{f}(\xi) \rvert ^{2}\lvert \mathrm{e}^{-4\pi^{2}t \xi^{2}} -1 \rvert ^{2}<\varepsilon/2N\) since \(\hat{f}\) is bounded. Thus \[ \int_{\lvert \xi \rvert \leqslant N}^{} \lvert \hat{f}(\xi) \rvert ^{2}\lvert \mathrm{e}^{-4\pi^{2}t \xi^{2}} -1 \rvert ^{2} \mathrm{d}\xi < \varepsilon \quad \text{for all small}\ t. \] This completes the proof of the theorem.
The steady-state heat equation in the upper half-plane
The equation we are now concerned with is \[ \Delta u= \frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}=0 \tag{13} \] in the upper half-plane \(\mathbb{R}_{+}^{2}=\{(x, y)\colon x\in \mathbb{R},y>0\}\). The boundary condition we require is \(u(x,0)=f(x)\). The kernel that solves this problem is called the Poisson kernel for the upper half-plane, and is given by \[ \mathcal{P}_{y}(x)=\frac{1}{\pi}\frac{y}{x^{2}+y^{2}} \quad\text{where} \ x\in \mathbb{R}\ \text{and} y>0. \] This is the analogue of the Poisson kernel for the disc.
Note that for each fixed \(y\) the kernel \(\mathcal{P}_{y}\) is only of moderate decrease as a function of \(x\), so we will use the theory of the Fourier transform appropriate for these types of functions.
We proceed as in the case of the time-dependent heat equation, by taking the Fourier transform of equation (13) in the \(x\) variable, thereby obtaining \[ -4\pi^{2}\xi^{2}\hat{u}(\xi,y)+\frac{\partial ^{2}\hat{u}}{\partial y^{2}}(\xi,y)=0 \] with the boundary condition \(\hat{u}(\xi,0)=\hat{f}(\xi)\). The general solution of this ODE in \(y\) (with \(\xi\) fixed) takes the form \[ \hat{u}(\xi,y)= A(\xi) \mathrm{e}^{-2\pi \lvert \xi \rvert y} +B(\xi) \mathrm{e}^{2\pi\lvert \xi \rvert y} . \] If we disregard the second term because of its rapid exponential increase we find, after setting \(y=0\), that \[ \hat{u}(\xi,y)=\hat{f}(\xi) \mathrm{e}^{-2\pi \lvert \xi \rvert y} . \] Therefore \(u\) is given in terms of the convolution of \(f\) with a kernel whose Fourier transform is $^{-2y} $. This is precisely the Poisson kernel given above, as we prove next.
Lemma 2.4 The following two identities hold: \[ \int_{-\infty}^{\infty} \mathrm{e}^{-2\pi\lvert \xi \rvert y} \mathrm{e}^{2\pi i \xi x} \mathrm{d}\xi = \mathcal{P}_y(x), \] \[ \int_{-\infty}^{\infty} \mathcal{P}_y(x)\mathrm{e}^{-2\pi ix \xi} \mathrm{d}x= \mathrm{e}^{-2\pi\lvert \xi \rvert y} . \]
Lemma 2.5 The Poisson kernel is a good kernel on \(\mathbb{R}\) as \(y \to 0\).
The following theorem establishes the existence of a solution to our problem.
Theorem 2.6 Given \(f \in \mathcal{S}(\mathbb{R})\),let \(u(x, y)=(f*\mathcal{P}_{y})(x)\). Then: 1. \(u (x, y)\) is \(C^{2}\) in \(\mathbb{R}^{2}_{+}\) and \(\Delta u=0\). 2. \(u(x, y) \to f(x)\) uniformly as \(y \to 0\). 3. \(\int_{-\infty}^{\infty} \lvert u(x, y)-f(x) \rvert ^{2} \mathrm{d}x \to 0\) as \(y \to 0\). 4. If \(u(x,0)=f(x)\), then \(u\) is continuous on the closure \(\overline{\mathbb{R}^{2}_{+}}\) of the upper half-plane, and vanishes at infinity in the sense that \[ u(x, y) \to 0 \quad \text{as} \ \lvert x \rvert +y \to \infty. \]
Heat kernel
Another application related to the Poisson summation formula and the theta function is the time-dependent heat equation on the circle. A solution to the equation \[ \frac{\partial u}{\partial t}=\frac{\partial ^{2}u}{\partial x^{2}} \] subject to \(u(x,0)=f(x)\), where \(f\) is periodic of period \(1\), was given in the previous chapter by \[ u(x,t)=(f*H_t)(x) \] where \(H_t(x)\) is the heat kernel on the circle, that is, \[ H_t(x)= \sum_{n=-\infty}^{\infty} \mathrm{e}^{-4\pi^{2}n^{2}t} \mathrm{e}^{2\pi inx} . \] Note in particular that with our definition of the generalized theta function in the previous section, we have \(\Theta(x|4\pi it)=H_t(x)\). Also, recall that the heat equation on \(\mathbb{R}\) gave rise to the heat kernel. \[ \mathcal{H}_{t}(x)=\frac{1}{(4\pi t)^{1/2}}\mathrm{e}^{-x^{2}/4t} \] where \(\hat{\mathcal{H}}_{t}(\xi)=\mathrm{e}^{-4\pi^{2}\xi^{2}t}\). The fundamental relation between these two objects is an immediate consequence of the Poisson summation formula:
Theorem 3.3 The heat kernel on the circle is the periodization of the heat kernel on the real line: \[ H_t(x)= \sum_{n=-\infty}^{\infty} \mathcal{H}_{t}(x+n). \] Although the proof that \(\mathcal{H}_t\) is a good kernel on \(\mathbb{R}\) was fairly straightforward, we left open the harder problem that \(H_t\) is a good kernel on the circle. The above results allow us to resolve this matter.