Three times in this book a function has been called a solution of an equation it could not literally satisfy. The shocks of Chapter 2 are discontinuous, and a discontinuous function has no derivative at the jump to put into \(u_t+uu_x=0\) . The plucked string of Example 6.2 has corners, so \(u_{tt}\) and \(u_{xx}\) do not exist along certain lines, and yet d’Alembert’s formula plainly describes its motion. The fundamental solution of Chapter 4 was said to be the response to a unit point source, as if “\(-\Delta\Phi\) equals a point source” were a meaningful equation, although \(\Delta\Phi=0\) wherever \(\Delta\Phi\) is defined.

In each case there was a way round: an integral conservation law, a formula, a limiting computation on shrinking spheres. This chapter replaces the three ad hoc devices by one definition. The idea is to stop asking for the values of a function’s derivatives at points and to ask instead how the function acts when integrated against smooth functions. Every locally integrable function then has derivatives of all orders, of a generalized kind; the three examples become solutions in a precise sense; and the price is that one must keep track of when a generalized derivative is again a function. That last question leads directly to the spaces in which Chapters 9 and 10 solve boundary value problems.

8.1 Testing a function instead of evaluating it

A thermometer does not measure the temperature at a mathematical point. It reports an average of \(u\) over a small region, weighted by some sensitivity profile \(\varphi\) : a number of the form \(\int u\varphi\,dx\) . Knowing these numbers for all reasonable profiles \(\varphi\) is as good as knowing \(u\) , and they make sense for much rougher \(u\) than point values do. The profiles we allow are as nice as possible.

Test functions exist in abundance, though no analytic function is one. The standard example is the bump

It is infinitely differentiable even on the sphere \(\lvert x\rvert=1\) , because every derivative of \(e^{-1/s}\) is a polynomial in \(1/s\) times \(e^{-1/s}\) and tends to zero as \(s\to0\) from above (Exercise 1). Translating and scaling gives a test function supported in any prescribed ball. Dividing by its integral and scaling, \(\eta_\varepsilon(x)=\varepsilon^{-n}\eta(x/\varepsilon)/\!\int\eta\) , gives a nonnegative test function of total integral one supported in the ball of radius \(\varepsilon\) , called a mollifier.

Because a test function vanishes near the boundary of \(\Omega\) , integration by parts against it never produces boundary terms: for \(u\in C^1(\Omega)\) ,

This identity is the whole mechanism of the chapter. The right side makes sense when \(u\) is merely integrable.

A function is locally integrable on \(\Omega\) , written \(u\in L^1_{\mathrm{loc}}(\Omega)\) , if it is integrable over every closed bounded subset of \(\Omega\) . Such functions are determined by their action on test functions.

8.2 Distributions

Two kinds of example matter most.

Functions. Every \(f\in L^1_{\mathrm{loc}}(\Omega)\) defines a distribution \(T_f\) by \(\langle T_f,\varphi\rangle=\int_\Omega f\varphi\,dx\) . It is continuous because \(\lvert\int f(\varphi_j-\varphi)\rvert\le\max\lvert\varphi_j-\varphi\rvert\int_K\lvert f\rvert\) on the common support \(K\) . By Lemma 8.2(b), \(T_f\) determines \(f\) up to a set of measure zero, so nothing is lost by identifying \(f\) with \(T_f\) , and we write \(f\) for both. Distributions of this kind are called regular.

Point evaluations. For a point \(a\in\Omega\) the Dirac delta at \(a\) is \(\langle\delta_a,\varphi\rangle=\varphi(a)\) ; we write \(\delta\) for \(\delta_0\) . It is linear, and continuous because uniform convergence implies convergence at \(a\) . It is the idealized unit point source.

The delta is, however, a limit of functions. If \(k_\varepsilon\) is any approximate identity in the sense of Section 4.5, nonnegative with integral one and concentrating at the origin, then \(\int k_\varepsilon\varphi\,dx\to\varphi(0)\) , which says \(k_\varepsilon\to\delta\) in \(\mathcal D'\) . Theorem 4.8 showed this for the Poisson kernel as the height tends to zero, and Theorem 5.3(c) for the heat kernel as \(t\to0\) : the statement “\(S(\cdot,t)\) is the temperature after a unit point release” now has a precise form, \(S(\cdot,t)\to\delta\) .

Not every functional is a distribution. The map \(\varphi\mapsto\varphi(0)^2\) is not linear. Linearity is what will let derivatives be moved around freely, and it is also the source of the theory’s main limitation: there is no general way to multiply two distributions.

8.3 Differentiating distributions

The integration by parts identity of Section 8.1 says that for a \(C^1\) function \(u\) the distribution \(T_{u_{x_i}}\) acts on \(\varphi\) as \(-\langle T_u,\varphi_{x_i}\rangle\) . We take this as the definition in general.

The right side is linear in \(\varphi\) , and continuous because \(\varphi_j\to\varphi\) in \(\mathcal D\) implies \(\partial_i\varphi_j\to\partial_i\varphi\) in \(\mathcal D\) . So every distribution, and in particular every locally integrable function, has derivatives of all orders, and mixed derivatives may be taken in any order because that is true for test functions. For a \(C^1\) function the new derivative agrees with the old one. For rougher functions it carries more information than the pointwise derivative does.

8.4 Weak derivatives

For boundary value problems the useful question is when the distributional derivative of a function is again a function.

In other words, \(u\) has a weak derivative when its distributional derivative is regular. By Lemma 8.2(b) the weak derivative is unique up to sets of measure zero when it exists. A \(C^1\) function has its classical derivative as weak derivative. On an interval, Theorem 8.6 and Proposition 8.4 decide the piecewise smooth case completely: a piecewise \(C^1\) function has a weak derivative if and only if it has no jumps, and the weak derivative is then the pointwise one. So \(\lvert x\rvert\) has the weak derivative \(\operatorname{sgn}x\) on \((-1,1)\) , while the Heaviside function has none: its distributional derivative is \(\delta\) , which is not a function.

In several variables a function with a weak gradient need not even be bounded.

8.5 Distributional solutions of differential equations

Let \(L\) be a linear differential operator with smooth coefficients, \(Lu=\sum_\alpha a_\alpha D^\alpha u\) . Integrating by parts every derivative moves \(L\) onto the test function as its formal adjoint \(L^{\ast}\varphi=\sum_\alpha(-1)^{\lvert\alpha\rvert}D^\alpha(a_\alpha\varphi)\) . For constant coefficients, the Laplacian and the wave operator are their own adjoints, and the adjoint of \(\partial_t-k\,\partial_x^2\) is \(-\partial_t-k\,\partial_x^2\) .

A classical solution is a distributional solution, by integration by parts. Conversely a distributional solution that happens to be \(C^m\) , with \(m\) the order of \(L\) and \(f\) continuous, is classical: integrating by parts back gives \(\int(Lu-f)\varphi\,dx=0\) for all \(\varphi\) , and Lemma 8.2(a) applies. So the new notion enlarges the class of solutions without changing which smooth functions belong to it. Here are the three loose ends.

So d’Alembert’s formula with a tent-shaped \(g\) , or with any locally integrable data, gives a distributional solution of the wave equation. The corners of Example 6.2 are legitimate, and so would be jumps.

The limiting argument on shrinking spheres in Chapter 4 was a proof of this identity before the language for stating it was available. The same proof shows that Green’s function of a region satisfies \(-\Delta_yG(x,\cdot)=\delta_x\) there.

Exercises

1. Let \(h(s)=e^{-1/s}\) for \(s>0\) and \(h(s)=0\) for \(s\le0\) . Prove by induction that for \(s>0\) every derivative of \(h\) has the form \(p(1/s)e^{-1/s}\) with \(p\) a polynomial, and deduce that \(h\) is infinitely differentiable on \(\mathbb R\) and that the bump \(\eta\) is a test function.

   Hint

   \(h'(s)=s^{-2}e^{-1/s}\) ; if \(h^{(m)}(s)=p(1/s)e^{-1/s}\) then \(h^{(m+1)}(s)=\bigl(-s^{-2}p'(1/s)+s^{-2}p(1/s)\bigr)e^{-1/s}\) , again of that form. Since \(r^je^{-r}\to0\) as \(r=1/s\to\infty\) , each derivative tends to \(0\) as \(s\to0^+\) ; show by the definition of the derivative, using the same limit, that \(h^{(m)}(0)=0\) for every \(m\) . Then \(\eta(x)=h(1-\lvert x\rvert^2)\) is a composition of smooth functions.

2. Decide which of these maps are distributions on \(\mathbb R\) , giving a reason in each case: (a) \(\varphi\mapsto\int_0^1\varphi\,dx\) ; (b) \(\varphi\mapsto\varphi(0)^2\) ; (c) \(\varphi\mapsto\sum_{k=0}^{\infty}\varphi(k)\) ; (d) \(\varphi\mapsto\lvert\varphi(0)\rvert\) ; (e) \(\varphi\mapsto\varphi''(1)\) .

   Answer

   (a) Yes: it is the regular distribution of the indicator of \([0,1]\) . (b) No: not linear. (c) Yes: each test function vanishes outside a bounded set, so the sum is finite, and it is linear and continuous; it is \(\sum\delta_k\) . (d) No: not linear, since \(\lvert-\varphi(0)\rvert\ne-\lvert\varphi(0)\rvert\) . (e) Yes: it is \(\delta_1''\) .

3. Compute the first and second distributional derivatives of: (a) \(x_+=\max(x,0)\) ; (b) \(\operatorname{sgn}x\) ; (c) \(H(x)\cos x\) ; and the first derivative of (d) the integer-part function on \(\mathbb R\) .

   Answer

   (a) \(x_+'=H\) , \(x_+''=\delta\) . (b) \((\operatorname{sgn})'=2\delta\) , \((\operatorname{sgn})''=2\delta'\) . (c) \((H\cos x)'=-H\sin x+\delta\) , and \((H\cos x)''=-H\cos x+\delta'\) , since \(H\sin x\) is continuous. (d) The integer part is piecewise constant with a jump of \(1\) at each integer: its derivative is \(\sum_k\delta_k\) .

4. Show that \(x\delta'=-\delta\) and \(x^2\delta''=2\delta\) , and more generally that \(a\delta'=a(0)\delta'-a'(0)\delta\) for a smooth function \(a\) .

   Hint

   \(\langle a\delta',\varphi\rangle=\langle\delta',a\varphi\rangle=-(a\varphi)'(0)=-a'(0)\varphi(0)-a(0)\varphi'(0)\) . Take \(a=x\) , and for the second identity apply the same computation to \(\delta''\) with \(a=x^2\) , using \((x^2\varphi)''(0)=2\varphi(0)\) .

5. Prove that a distribution \(T\) on \(\mathbb R\) with \(T'=0\) is a constant function. Use the fact, which you should also prove, that a test function \(\varphi\) is the derivative of a test function if and only if \(\int\varphi\,dx=0\) .

   Solution

   The auxiliary fact. If \(\varphi=\psi'\) with \(\psi\) a test function, then \(\int\varphi\,dx=\psi(\infty)-\psi(-\infty)=0\) . Conversely let \(\varphi\) be a test function vanishing outside \([-R,R]\) with \(\int\varphi\,dx=0\) , and put \(\psi(x)=\int_{-\infty}^x\varphi(s)\,ds\) . Then \(\psi\) is infinitely differentiable with \(\psi'=\varphi\) , it vanishes for \(x<-R\) , and for \(x>R\) it equals \(\int\varphi=0\) . So \(\psi\) is a test function.

   The theorem. Fix a test function \(\rho\) with \(\int\rho\,dx=1\) and put \(c=\langle T,\rho\rangle\) . For an arbitrary test function \(\varphi\) the function \(\varphi-\bigl(\int\varphi\bigr)\rho\) has integral zero, so it equals \(\psi'\) for a test function \(\psi\) . By linearity and the definition of the derivative,

   \[ \langle T,\varphi\rangle=\Bigl(\int\varphi\,dx\Bigr)\langle T,\rho\rangle+\langle T,\psi'\rangle=c\int\varphi\,dx-\langle T',\psi\rangle=\int c\,\varphi\,dx , \]

   because \(T'=0\) . So \(T\) is the regular distribution of the constant function \(c\) . In particular a locally integrable function on \(\mathbb R\) whose weak derivative is zero equals a constant almost everywhere; the same proof works on any open interval, and Chapter 9 uses it there.

6. For \(k>0\) verify by the jump rule that \(E(x)=\dfrac{1}{2k}e^{-k\lvert x\rvert}\) satisfies \(-E''+k^2E=\delta\) on \(\mathbb R\) , and use it to write a solution of \(-u''+k^2u=f\) for bounded continuous \(f\) .

   Answer

   \(E\) is continuous; \(E'=-\tfrac12\operatorname{sgn}(x)e^{-k\lvert x\rvert}\) jumps by \(-1\) at the origin and has pointwise derivative \(\tfrac k2e^{-k\lvert x\rvert}=k^2E\) . So \(E''=k^2E-\delta\) . A solution is \(u(x)=\int_{-\infty}^{\infty}\tfrac{1}{2k}e^{-k\lvert x-y\rvert}f(y)\,dy\) , which is bounded by \(\sup\lvert f\rvert/k^2\) .

7. Show that in \(\mathcal D'(\mathbb R)\) : (a) \(\dfrac{\varepsilon}{\pi(x^2+\varepsilon^2)}\to\delta\) as \(\varepsilon\to0\) ; (b) \(\sin(nx)\to0\) as \(n\to\infty\) ; (c) \(\sin^2(nx)\to\tfrac12\) . What does the comparison of (b) and (c) say about squaring a limit of distributions?

   Solution

   (a) The functions \(P_\varepsilon(x)=\dfrac{\varepsilon}{\pi(x^2+\varepsilon^2)}\) are the Poisson kernel of the half-plane at height \(\varepsilon\) . A test function \(\varphi\) is bounded and continuous, so Theorem 4.8 gives \(\int P_\varepsilon(x)\varphi(x)\,dx=\int P_\varepsilon(0-s)\varphi(s)\,ds\to\varphi(0)\) as \(\varepsilon\to0\) , the kernel being even. That is \(P_\varepsilon\to\delta\) .

   (b) Integrating by parts, \(\int\sin(nx)\varphi(x)\,dx=\tfrac1n\int\cos(nx)\varphi'(x)\,dx\) , whose absolute value is at most \(\tfrac1n\int\lvert\varphi'\rvert\,dx\to0\) . So \(\sin(nx)\to0\) in \(\mathcal D'\) , although \(\sin(nx)\) converges at almost no point.

   (c) \(\sin^2(nx)=\tfrac12-\tfrac12\cos(2nx)\) , and \(\cos(2nx)\to0\) in \(\mathcal D'\) by the argument of (b). So \(\sin^2(nx)\to\tfrac12\) .

   The limit of the squares is \(\tfrac12\) , while the square of the limit is \(0\) . Squaring is not continuous with respect to convergence of distributions. Convergence in \(\mathcal D'\) records averages, and the average of a rapidly oscillating function says nothing about the average of its square. This is why nonlinear equations cannot be handled by distribution theory alone, and why the shock of Example 8.3 needed the function \(u^2\) to make sense separately.

8. Decide which of these functions have a weak derivative on \((-1,1)\) , and give it when it exists: (a) \(\lvert x\rvert^{1/2}\) ; (b) \(\operatorname{sgn}x\) ; (c) \(x\lvert x\rvert\) ; (d) \(\log\lvert x\rvert\) .

   Answer

   (a) Yes: \(\tfrac12\operatorname{sgn}(x)\lvert x\rvert^{-1/2}\) , which is integrable; the function is continuous, and the boundary terms at \(\pm\varepsilon\) in the integration by parts tend to zero. (b) No: the derivative is \(2\delta\) . (c) Yes: \(2\lvert x\rvert\) ; the function is \(C^1\) . (d) No: away from the origin the derivative is \(1/x\) , which is not locally integrable on \((-1,1)\) .

9. A student argues: “Since \(H^2=H\) , the product rule gives \(H'=2HH'\) , so \(H\delta=\tfrac12\delta\) . Since \(H^3=H\) , it also gives \(H'=3H^2H'=3H\delta=\tfrac32\delta\) . But \(H'=\delta\) , so \(1=\tfrac32\) .” Find the illegitimate step and explain what the paradox shows about distributions.

   Solution

   The illegitimate step is the product rule itself, applied to \(H\cdot H\) . In Section 8.3 the product \(aT\) was defined only for a smooth function \(a\) and a distribution \(T\) , by \(\langle aT,\varphi\rangle=\langle T,a\varphi\rangle\) , and the product rule was proved only in that case. Here the rule is applied with \(a=H\) , which is not even continuous, and \(T=H'=\delta\) . The expression \(H\delta\) would have to mean \(\varphi\mapsto\langle\delta,H\varphi\rangle=H(0)\varphi(0)\) , but \(H\varphi\) is not a test function and \(H\) has no value at the one point where \(\delta\) looks. The student’s first line “defines” \(H\delta=\tfrac12\delta\) , and the second line shows that no such definition can be consistent with the product rule for \(H^3\) as well.

   What the paradox shows: distributions form a vector space on which differentiation is always possible, but they do not form an algebra. There is no multiplication of distributions that extends the multiplication of functions, is associative, and obeys the product rule. Each individual identity \(H^2=H\) and \(H'=\delta\) is correct; only the manipulation connecting them is not. In practice this means that every nonlinear term in an equation must be given a meaning as a function before the equation is read distributionally.

10. In the unit disc of \(\mathbb R^2\) let \(u(x)=\log\lvert x\rvert\) . Show that \(u\) has a weak gradient, compute it, and show that \(\int\lvert\nabla u\rvert^2dx\) is infinite. Then repeat the three questions for \(w(x)=\log\log(e/\lvert x\rvert)\) , which is also unbounded.

    Answer

    \(\nabla u=x/\lvert x\rvert^2\) , with \(\lvert\nabla u\rvert=1/r\) , integrable in the plane; the boundary term on the circle of radius \(\varepsilon\) is of size \(\varepsilon\lvert\log\varepsilon\rvert\) , so this is the weak gradient. But \(\int\lvert\nabla u\rvert^2dx=2\pi\int_0^1r^{-1}dr\) diverges. For \(w\) : \(\nabla w=-\dfrac{x}{\lvert x\rvert^2\log(e/\lvert x\rvert)}\) , again a weak gradient, and with \(s=\log(e/r)\) , \(\int\lvert\nabla w\rvert^2dx=2\pi\int_0^1\dfrac{dr}{r\log^2(e/r)}=2\pi\int_1^\infty s^{-2}ds=2\pi\) . So \(w\) is unbounded with finite Dirichlet energy, even in two dimensions.

11. Let \(u\) be the indicator function of the unit disc in \(\mathbb R^2\) . Show that \(\langle\partial_iu,\varphi\rangle=-\int_{\lvert x\rvert=1}\varphi\,\nu_i\,dS\) , and explain why \(u\) has no weak derivative.

    Hint

    \(\langle\partial_iu,\varphi\rangle=-\int_{\lvert x\rvert<1}\varphi_{x_i}\,dx\) , and the divergence theorem applied to the field \(\varphi e_i\) gives the boundary integral. If a locally integrable \(v\) represented this functional, testing with \(\varphi\) supported off the circle would give \(v=0\) almost everywhere, as in Proposition 8.4.

12. Verify directly from the definition, without Theorem 8.9, that \(u(x,t)=H(x-ct)\) is a distributional solution of the transport equation \(u_t+cu_x=0\) in \(\mathbb R^2\) .

    Solution

    The formal adjoint of \(\partial_t+c\,\partial_x\) is \(-\partial_t-c\,\partial_x\) , so we must show that \(\iint H(x-ct)\bigl(\varphi_t+c\varphi_x\bigr)\,dx\,dt=0\) for every test function \(\varphi\) on \(\mathbb R^2\) .

    For each fixed \(t\) substitute \(x=\xi+ct\) in the inner integral; the substitution has Jacobian one, so

    \[ \iint H(x-ct)\bigl(\varphi_t+c\varphi_x\bigr)(x,t)\,dx\,dt=\int_{-\infty}^{\infty}H(\xi)\left(\int_{-\infty}^{\infty}\bigl(\varphi_t+c\varphi_x\bigr)(\xi+ct,t)\,dt\right)d\xi , \]

    the exchange of the order of integration being justified because the integrand is bounded and vanishes outside a bounded set. By the chain rule, \((\varphi_t+c\varphi_x)(\xi+ct,t)=\dfrac{d}{dt}\bigl[\varphi(\xi+ct,t)\bigr]\) . For fixed \(\xi\) the function \(t\mapsto\varphi(\xi+ct,t)\) is smooth and vanishes for large \(\lvert t\rvert\) , so the inner integral is zero. Hence the whole integral is zero.

    The argument used nothing about \(H\) except local integrability: every function of \(x-ct\) is a distributional solution of the transport equation. This is Theorem 2.1 without the hypothesis \(g\in C^1\) , and the proof is the same idea, differentiating along the characteristic lines, with the derivative placed on the test function.

13. Show by the method of Example 8.3 that the expansion shock \(u=0\) for \(x<\tfrac12t\) , \(u=1\) for \(x>\tfrac12t\) is a weak solution of the Burgers equation for \(t>0\) , and that the rarefaction wave of Example 2.5 is another with the same initial values. Which one is the entropy solution?

    Solution

    The expansion shock. Here \(u=1\) and \(\tfrac12u^2=\tfrac12\) on \(D'=\{x>\tfrac12t,\ t>0\}\) and \(u=0\) elsewhere, so the weak formulation requires \(\iint_{D'}(\varphi_t+\tfrac12\varphi_x)\,dx\,dt=0\) . By the divergence theorem for the field \((\tfrac12\varphi,\varphi)\) this is the flux through the line \(x=\tfrac12t\) , whose normal pointing out of \(D'\) is proportional to \((-1,\tfrac12)\) . The integrand is \(\tfrac12\varphi\cdot(-1)+\varphi\cdot\tfrac12=0\) . So the expansion shock is a weak solution, for every test function supported in \(t>0\) .

    The rarefaction. The function \(u=0\) for \(x\le0\) , \(u=x/t\) for \(0<x<t\) , \(u=1\) for \(x\ge t\) is continuous for \(t>0\) and satisfies the equation classically in each of the three regions. Apply the divergence theorem to the field \((\tfrac12u^2\varphi,\,u\varphi)\) in each region. The interior contributions are \(\iint(u_t+uu_x)\varphi=0\) . On each of the lines \(x=0\) and \(x=t\) the boundary terms from the two sides have opposite normals and equal integrands, because \(u\) and \(u^2\) are continuous across the line, so they cancel. Hence \(\iint(u\varphi_t+\tfrac12u^2\varphi_x)\,dx\,dt=0\) .

    Both functions have the initial values \(0\) for \(x<0\) and \(1\) for \(x>0\) . So the weak formulation admits at least two solutions with the same data. The entropy solution is the rarefaction: the shock has \(\phi'(u_l)=0<\tfrac12<1=\phi'(u_r)\) , the reverse of the Lax condition in Definition 2.7.

14. Let \(u(x,t)=S(x,t)\) for \(t>0\) and \(u(x,t)=0\) for \(t\le0\) , where \(S\) is the heat kernel. Prove that \(u_t-ku_{xx}=\delta\) in \(\mathcal D'(\mathbb R^2)\) , the delta being at the origin of the \((x,t)\) plane.

    Solution

    First, \(u\) is locally integrable on \(\mathbb R^2\) , since \(\int S(x,t)\,dx=1\) for each \(t>0\) . The adjoint of \(\partial_t-k\,\partial_x^2\) is \(-\partial_t-k\,\partial_x^2\) , so for a test function \(\varphi\) we must compute

    \[ I=\int_0^\infty\!\!\int_{-\infty}^{\infty}S\,\bigl(-\varphi_t-k\varphi_{xx}\bigr)\,dx\,dt=\lim_{\varepsilon\to0}I_\varepsilon,\qquad I_\varepsilon=\int_\varepsilon^\infty\!\!\int_{-\infty}^{\infty}S\,\bigl(-\varphi_t-k\varphi_{xx}\bigr)\,dx\,dt . \]

    For \(t\ge\varepsilon\) the kernel is smooth, so integrate by parts, once in \(t\) and twice in \(x\) . The \(x\) -integrations produce no boundary terms. The \(t\) -integration produces one, at \(t=\varepsilon\) :

    \[ I_\varepsilon=\int_{-\infty}^{\infty}S(x,\varepsilon)\,\varphi(x,\varepsilon)\,dx+\int_\varepsilon^\infty\!\!\int_{-\infty}^{\infty}\bigl(S_t-kS_{xx}\bigr)\varphi\,dx\,dt . \]

    The double integral vanishes by Lemma 5.2(a). In the remaining term replace \(\varphi(x,\varepsilon)\) by \(\varphi(x,0)\) ; the error is at most \(\varepsilon\max\lvert\varphi_t\rvert\int S(x,\varepsilon)\,dx=\varepsilon\max\lvert\varphi_t\rvert\) . Finally \(\int S(x,\varepsilon)\varphi(x,0)\,dx\to\varphi(0,0)\) as \(\varepsilon\to0\) , by Theorem 5.3(c) applied to the bounded continuous function \(\varphi(\cdot,0)\) at the point \(x_0=0\) . Therefore \(I=\varphi(0,0)=\langle\delta,\varphi\rangle\) .

    So the heat kernel, extended by zero to negative times, is the fundamental solution of the heat operator in the plane: the response to a unit release of heat at the point \(x=0\) at the instant \(t=0\) .

15. Let \(f\) be continuous on \(\Omega\) and let \(u\in C^2(\Omega)\) be a distributional solution of \(-\Delta u=f\) . Prove that \(u\) is a classical solution.

    Hint

    For \(\varphi\in\mathcal D(\Omega)\) , Green’s second identity on a region containing the support of \(\varphi\) gives \(\int u\,\Delta\varphi\,dx=\int\varphi\,\Delta u\,dx\) , with no boundary terms. So \(\int(-\Delta u-f)\varphi\,dx=0\) for all \(\varphi\) , and \(-\Delta u-f\) is continuous. Apply Lemma 8.2(a).

16. In \(\mathbb R^3\) compute the distributional Laplacian of \(u(x)=\lvert x\rvert\) , showing carefully that it is the locally integrable function \(2/\lvert x\rvert\) with no delta at the origin. Contrast with \(\Delta(1/\lvert x\rvert)\) .

    Solution

    Away from the origin \(u=r\) is smooth, and by the radial formula of Section 3.1 with \(n=3\) , \(\Delta u=u''+\tfrac2ru'=\tfrac2r\) . This function is locally integrable in \(\mathbb R^3\) , since \(\int_{B_1}\tfrac2r\,dx=4\pi\int_0^12r\,dr\) is finite.

    Let \(\varphi\) be a test function. Since \(u\) is continuous, \(\int u\,\Delta\varphi\,dx=\lim_{\varepsilon\to0}\int_{\lvert x\rvert>\varepsilon}u\,\Delta\varphi\,dx\) . Apply Green’s second identity on \(\{\lvert x\rvert>\varepsilon\}\) , with \(\nu\) the normal pointing toward the origin on the sphere \(\partial B_\varepsilon\) :

    \[ \int_{\lvert x\rvert>\varepsilon}u\,\Delta\varphi\,dx=\int_{\lvert x\rvert>\varepsilon}\frac{2}{\lvert x\rvert}\,\varphi\,dx+\int_{\partial B_\varepsilon}\Bigl(u\,\frac{\partial\varphi}{\partial\nu}-\varphi\,\frac{\partial u}{\partial\nu}\Bigr)dS . \]

    On the sphere \(u=\varepsilon\) and \(\partial u/\partial\nu=-1\) , and its area is \(4\pi\varepsilon^2\) . So the first boundary term is at most \(\varepsilon\max\lvert\nabla\varphi\rvert\cdot4\pi\varepsilon^2\) and the second at most \(\max\lvert\varphi\rvert\cdot4\pi\varepsilon^2\) ; both tend to zero. Hence \(\int u\,\Delta\varphi\,dx=\int\tfrac{2}{\lvert x\rvert}\varphi\,dx\) : the distributional Laplacian of \(\lvert x\rvert\) is the function \(2/\lvert x\rvert\) , with nothing concentrated at the origin.

    For \(1/\lvert x\rvert=4\pi\Phi\) the same computation has \(u=1/\varepsilon\) and \(\partial u/\partial\nu=+1/\varepsilon^2\) on the small sphere, so the second boundary term is \(-\varepsilon^{-2}\int_{\partial B_\varepsilon}\varphi\,dS\to-4\pi\varphi(0)\) : by Theorem 8.10, \(\Delta(1/\lvert x\rvert)=-4\pi\delta\) . The difference is the strength of the singularity of the gradient. A delta appears when the flux through small spheres has a nonzero limit; for \(\lvert x\rvert\) the gradient is bounded and the flux through \(\partial B_\varepsilon\) is of order \(\varepsilon^2\) .

17. Let \(G\) be the Green’s function of Example 4.2 and \(f\) integrable on \((0,1)\) . Show that \(u(x)=\int_0^1G(x,y)f(y)\,dy\) is continuous, vanishes at both ends, and satisfies \(-u''=f\) in \(\mathcal D'(0,1)\) .

    Hint

    Write \(u(x)=(1-x)\int_0^xyf(y)\,dy+x\int_x^1(1-y)f(y)\,dy\) ; each integral is a continuous function of \(x\) , and \(u(0)=u(1)=0\) . For the equation use the symmetry of \(G\) and Example 8.4: for a test function \(\varphi\) , \(-\int_0^1G(x,y)\varphi''(x)\,dx=\varphi(y)\) . Multiply by \(f(y)\) , integrate in \(y\) , and exchange the order of integration.