Dirichlet’s principle, Theorem 3.10, says that among all functions with given boundary values the harmonic one has the least energy. Riemann turned this round into an existence proof: the energy is bounded below by zero, so take a function that minimizes it, and that function is harmonic. The argument has a hole, which Weierstrass pointed out in 1870: a quantity bounded below has an infimum, but nothing guarantees a function at which the infimum is attained. One can take a sequence of smooth functions whose energies decrease to the infimum. Does the sequence converge, and to what?

In general it does not converge to anything smooth. The smooth functions \(\sqrt{x^2+1/j}\) have uniformly bounded energy on \((-1,1)\) and converge to \(\lvert x\rvert\) , which has a corner. With respect to the size that the energy measures, the space of smooth functions has holes. The repair is the same as the passage from rational to real numbers: fill the holes. The functions that fill them are exactly those with square-integrable weak derivatives, in the sense of Chapter 8. This chapter builds that space, establishes what it means for such a function to vanish on the boundary, proves the inequality that makes the energy control the function, and rewrites boundary value problems as equations in the new space.

Throughout, \(\Omega\subseteq\mathbb R^n\) is open, \(L^2(\Omega)\) is the space of square-integrable functions on \(\Omega\) , with functions equal almost everywhere identified, and

are its inner product and norm. For a vector field such as \(\nabla u\) , \(\lVert\nabla u\rVert^2\) means \(\int_\Omega\lvert\nabla u\rvert^2dx\) . The Cauchy–Schwarz inequality \(\lvert(u,v)\rvert\le\lVert u\rVert\,\lVert v\rVert\) is used constantly; it is the case \(p=2\) of Hölder’s inequality (math.measure-theory.holder-minkowski).

9.1 The space of functions with finite energy

Every function in \(C^1(\overline\Omega)\) on a bounded \(\Omega\) belongs to \(H^1(\Omega)\) , with its classical gradient. So do continuous piecewise \(C^1\) functions on an interval, by Theorem 8.6. The point of the definition is the next theorem, which is false for either of those smaller spaces.

The proof shows why weak derivatives are the right notion: the defining identity survives limits in \(L^2\) , whereas pointwise differentiability does not.

9.2 One dimension: finite energy implies continuity

On an interval the members of \(H^1\) can be described completely.

From now on we take the continuous representative and write \(u\) for it. The theorem says three things. An \(H^1\) function on an interval is continuous, indeed Hölder continuous with exponent \(\tfrac12\) . The fundamental theorem of calculus holds for it, with the weak derivative. And its values at points, in particular at the endpoints, are controlled by its \(H^1\) norm, so that convergence in \(H^1\) implies uniform convergence.

None of this survives in higher dimensions. Example 8.2 exhibited an unbounded function in the unit ball of \(\mathbb R^3\) with finite Dirichlet energy, and Exercise 8.10 another in the plane. In two or more dimensions a function in \(H^1\) need not be continuous and does not have a value at a point. Saying what it means for such a function to “vanish on the boundary” therefore takes a definition.

9.3 Functions that vanish on the boundary

A closed subspace of a Hilbert space is a Hilbert space, so \(H_0^1(\Omega)\) is one, with the inner product of \(H^1(\Omega)\) . Its members are the limits, in the energy sense, of functions that vanish near the boundary. In one dimension this is what one hopes.

In higher dimensions the corresponding statement is the trace theorem, which we quote without proof. If \(\Omega\) is bounded with \(C^1\) boundary, the restriction map \(u\mapsto u|_{\partial\Omega}\) , defined for \(u\in C^1(\overline\Omega)\) , satisfies \(\int_{\partial\Omega}u^2\,dS\le C\lVert u\rVert_{H^1}^2\) and so extends to a bounded linear map, the trace, from \(H^1(\Omega)\) to \(L^2(\partial\Omega)\) ; and \(H_0^1(\Omega)\) is exactly the set of functions with zero trace. In particular a function in \(C^1(\overline\Omega)\) that vanishes on \(\partial\Omega\) belongs to \(H_0^1(\Omega)\) . Apart from this last fact, everything below uses only Definition 9.4, and it holds for every bounded open set, however rough its boundary.

9.4 The Poincaré inequality

A function that vanishes on the boundary cannot be large without having a large gradient somewhere. Theorem 5.7 used this on an interval; here is the general statement.

We write \(C_P\) for the smallest constant with \(\lVert u\rVert\le C_P\lVert\nabla u\rVert\) on \(H_0^1(\Omega)\) ; the theorem says \(C_P\le d/\sqrt2\) . On an interval of length \(L\) , Corollary 7.9 gives the exact value \(C_P=L/\pi\) , first for continuous piecewise \(C^1\) functions and then, by the density argument just used, for all of \(H_0^1(0,L)\) . The inequality has two consequences that the rest of the theory rests on.

First, on \(H_0^1(\Omega)\) the Dirichlet energy alone is as good as the full norm:

So \(\lVert\nabla u\rVert\) is a norm on \(H_0^1(\Omega)\) equivalent to \(\lVert u\rVert_{H^1}\) . Second, the inequality is false on \(H^1(\Omega)\) : a nonzero constant has zero gradient. The boundary condition is what excludes constants, and this is the weak form of the fact, seen in Theorem 3.9, that the Dirichlet problem has a unique solution while the Neumann problem determines its solution only up to a constant.

9.5 The Dirichlet problem in weak form

Consider the problem \(-\Delta u=f\) in \(\Omega\) , \(u=0\) on \(\partial\Omega\) . Suppose for the moment that \(u\) is a classical solution, smooth up to the boundary. Multiply the equation by a test function \(v\) and use Green’s first identity; the boundary term vanishes and

Only first derivatives of \(u\) remain. The identity makes sense for \(u\in H^1\) , and the boundary condition can be expressed by membership in \(H_0^1\) .

Both sides are continuous in \(v\) with respect to the \(H^1\) norm, so it is enough to require the identity for test functions \(v\) ; then it says precisely that \(-\Delta u=f\) in the sense of distributions, Definition 8.8. A weak solution is thus a distributional solution with two further properties: finite energy, and zero boundary values in the sense of \(H_0^1\) . The same letter is used for the unknown and the test function’s space on purpose: the solution is sought in the space from which the test functions are taken.

This is the energy method of Theorem 3.9 again. There it required \(u\in C^2(\overline\Omega)\) and a boundary smooth enough for Green’s identity. Here it is one line and requires nothing, because the solution itself is an admissible test function. Applied to the difference of two solutions with different data, the estimate gives \(\lVert\nabla(u_1-u_2)\rVert\le C_P\lVert f_1-f_2\rVert\) : continuous dependence in the energy norm.

What is still missing is existence. In the classical setting that was the hard part, and Chapter 4 obtained it only for special regions. In \(H_0^1(\Omega)\) it will follow in the next chapter from a general theorem about Hilbert spaces, for every bounded open set. That is the return on the investment in completeness.

9.6 Natural boundary conditions and variable coefficients

Neumann conditions. Consider \(-\Delta u+u=f\) in \(\Omega\) with \(\partial u/\partial\nu=g\) on \(\partial\Omega\) . Multiply by a function \(v\in C^1(\overline\Omega)\) that is not required to vanish on the boundary and apply Green’s first identity:

The weak formulation is: find \(u\in H^1(\Omega)\) such that this holds for all \(v\in H^1(\Omega)\) . The boundary condition does not appear in the choice of space. It is recovered from the identity: if a weak solution happens to be in \(C^2(\overline\Omega)\) , testing with \(v\in\mathcal D(\Omega)\) gives the differential equation, and then testing with general \(v\) and integrating by parts leaves \(\int_{\partial\Omega}(\partial u/\partial\nu-g)\,v\,dS=0\) for all \(v\) , which forces \(\partial u/\partial\nu=g\) . Conditions that are consequences of the weak identity are called natural. Conditions built into the space, like the Dirichlet condition, are called essential.

Variable and discontinuous coefficients. For a symmetric matrix \(A(x)\) and a function \(c(x)\) , the equation \(-\nabla\cdot(A\nabla u)+cu=f\) with \(u=0\) on the boundary has the weak form

This makes sense when the entries of \(A\) and \(c\) are merely bounded and measurable. If \(A\) jumps across an interface, as the conductivity does between two materials, the classical equation is meaningless at the interface, but the weak form is unaffected, and it automatically enforces the correct physical condition there: continuity of the normal flux \(A\nabla u\cdot\nu\) . That is the conservation law of Section 1.2, which never required the flux to be differentiable. For an inhomogeneous Dirichlet condition \(u=g\) one writes \(u=u_g+w\) with \(u_g\in H^1(\Omega)\) any function having the boundary values \(g\) and \(w\in H_0^1(\Omega)\) , and solves for \(w\) .

Exercises

1. Decide which of the functions \(x^{2/3}\) , \(x^{1/3}\) , \(\sqrt x\) , and \(\min(x,1-x)\) belong to \(H^1(0,1)\) , and compute \(\lVert u\rVert_{H^1}^2\) for \(u=x^{2/3}\) .

   Answer

   For \(x^\alpha\) on \((0,1)\) the derivative \(\alpha x^{\alpha-1}\) is square-integrable exactly when \(\alpha>\tfrac12\) . So \(x^{2/3}\in H^1\) , while \(x^{1/3}\) and \(\sqrt x\) are not, the latter being the borderline case with \(\int_0^1\tfrac{1}{4x}dx\) divergent. The tent \(\min(x,1-x)\) is continuous and piecewise linear, so it is in \(H^1\) . For \(u=x^{2/3}\) : \(\lVert u\rVert^2=\int_0^1x^{4/3}dx=\tfrac37\) and \(\lVert u'\rVert^2=\tfrac49\int_0^1x^{-2/3}dx=\tfrac43\) , so \(\lVert u\rVert_{H^1}^2=\tfrac37+\tfrac43=\tfrac{37}{21}\) .

2. Let \(u\in H^1(a,b)\) . Use Theorem 9.3 to prove that if \(u'\ge0\) almost everywhere then \(u\) is nondecreasing, and that if \(u'=0\) almost everywhere then \(u\) is constant.

   Hint

   By Theorem 9.3, \(u(y)-u(x)=\int_x^yu'(s)\,ds\) for the continuous representative. The integral of a nonnegative function over \([x,y]\) with \(x<y\) is nonnegative; the integral of a function that vanishes almost everywhere is zero.

3. Prove that if \(u,v\in H^1(a,b)\) on a bounded interval then \(uv\in H^1(a,b)\) with \((uv)'=u'v+uv'\) , and deduce the integration by parts formula \(\int_a^bu'v\,dx=u(b)v(b)-u(a)v(a)-\int_a^buv'\,dx\) .

   Hint

   By Theorem 9.3 both \(u\) and \(v\) are continuous, hence bounded, so \(uv\) and \(u'v+uv'\) are in \(L^2\) . Approximate: choose smooth \(u_j\to u\) and \(v_j\to v\) in \(H^1\) , for instance by integrating test-function approximations of \(u'\) and \(v'\) as in the proof of Theorem 9.5; convergence is then also uniform. Pass to the limit in \(\int u_jv_j\varphi'\,dx=-\int(u_j'v_j+u_jv_j')\varphi\,dx\) . The integration by parts formula is Theorem 9.3 applied to \(uv\) .

4. Let \(u_j(x)=\sqrt{x^2+1/j}\) on \((-1,1)\) . Prove that \(u_j\to\lvert x\rvert\) in \(H^1(-1,1)\) , and explain why this shows that \(C^1[-1,1]\) is not complete in the \(H^1\) norm.

   Solution

   Each \(u_j\) is smooth on \([-1,1]\) , with \(u_j'(x)=x/\sqrt{x^2+1/j}\) .

   Convergence in mean square. Since \(\sqrt{x^2+1/j}-\lvert x\rvert=\dfrac{1/j}{\sqrt{x^2+1/j}+\lvert x\rvert}\le\dfrac{1/j}{\sqrt{1/j}}=j^{-1/2}\) , we have \(0\le u_j-\lvert x\rvert\le j^{-1/2}\) uniformly, so \(\lVert u_j-\lvert x\rvert\rVert^2\le2/j\to0\) .

   Convergence of derivatives. The weak derivative of \(\lvert x\rvert\) is \(\operatorname{sgn}x\) . For \(x\ne0\) , \(u_j'(x)\to x/\lvert x\rvert=\operatorname{sgn}x\) , and \(\lvert u_j'-\operatorname{sgn}x\rvert\le2\) . By dominated convergence \(\int_{-1}^1(u_j'-\operatorname{sgn}x)^2dx\to0\) . Explicitly, \(1-\dfrac{\lvert x\rvert}{\sqrt{x^2+1/j}}\le\min\bigl(1,\tfrac{1}{2jx^2}\bigr)\) , and splitting the integral at \(\lvert x\rvert=j^{-1/2}\) gives a bound of order \(j^{-1/2}\) .

   So \(u_j\to\lvert x\rvert\) in \(H^1(-1,1)\) . The sequence \((u_j)\) is therefore Cauchy in the \(H^1\) norm and consists of \(C^1\) functions, but its limit \(\lvert x\rvert\) is not \(C^1\) ; and it has no other limit, since limits in a normed space are unique. A Cauchy sequence in \(C^1[-1,1]\) with no limit in \(C^1[-1,1]\) shows that this space is not complete for the \(H^1\) norm. \(H^1\) is the completion: it contains exactly the limits of such sequences.

5. Prove that every \(u\in H_0^1(0,L)\) satisfies \(\max\lvert u\rvert\le\tfrac12\sqrt L\,\lVert u'\rVert\) , and find a function for which equality holds.

   Solution

   Let \(u\in H_0^1(0,L)\) , continuous by Theorem 9.3, with \(u(0)=u(L)=0\) by Theorem 9.5. For any \(x\in[0,L]\) the same theorem gives two representations,

   \[ u(x)=\int_0^xu'(s)\,ds\qquad\text{and}\qquad u(x)=-\int_x^Lu'(s)\,ds . \]

   Taking absolute values and adding, \(2\lvert u(x)\rvert\le\int_0^x\lvert u'\rvert\,ds+\int_x^L\lvert u'\rvert\,ds=\int_0^L\lvert u'\rvert\,ds\) . By the Cauchy–Schwarz inequality the last integral is at most \(\sqrt L\,\lVert u'\rVert\) . Hence \(\lvert u(x)\rvert\le\tfrac12\sqrt L\,\lVert u'\rVert\) for every \(x\) .

   Equality requires equality in Cauchy–Schwarz, so \(\lvert u'\rvert\) constant, and \(u'\) of one sign on each side of the point \(x\) where the maximum is attained; the two representations then force that point to be the midpoint. The tent \(u(x)=\min(x,L-x)\) has \(\max u=L/2\) and \(\lVert u'\rVert=\sqrt L\) , so \(\tfrac12\sqrt L\,\lVert u'\rVert=L/2\) : equality holds. The constant \(\tfrac12\) improves the constant \(1\) that Theorem 9.3 would give by using the boundary condition at both ends.

6. Decide which of \(\sin(\pi x)\) , \(x\) , \(x(1-x)\) , the constant \(1\) , and \(\min(x,1-x)\) belong to \(H_0^1(0,1)\) . Then show that the distance in the \(H^1\) norm from the constant function \(1\) to the subspace \(H_0^1(0,1)\) is at least \(\tfrac12\) .

   Answer

   By Theorem 9.5 the members are the \(H^1\) functions vanishing at both ends: \(\sin(\pi x)\) , \(x(1-x)\) , and \(\min(x,1-x)\) . The function \(x\) fails at \(x=1\) and the constant \(1\) at both ends. For \(v\in H_0^1(0,1)\) put \(w=1-v\) ; then \(w(0)=1\) , and Theorem 9.3 gives \(1\le\max\lvert w\rvert\le\lVert w\rVert+\lVert w'\rVert\le\sqrt2\,\lVert w\rVert_{H^1}\) , so the distance is at least \(1/\sqrt2>\tfrac12\) .

7. For \(u=x(1-x)\) on \((0,1)\) compute \(\lVert u\rVert^2\) and \(\lVert u'\rVert^2\) , and compare the ratio \(\lVert u\rVert^2/\lVert u'\rVert^2\) with the constant \(\tfrac12d^2\) of Theorem 9.6 and with the sharp constant \(1/\pi^2\) .

   Answer

   \(\lVert u\rVert^2=\int_0^1x^2(1-x)^2dx=\tfrac1{30}\) and \(\lVert u'\rVert^2=\int_0^1(1-2x)^2dx=\tfrac13\) , with ratio \(\tfrac1{10}\) . Theorem 9.6 guarantees at most \(\tfrac12\) ; the sharp constant is \(1/\pi^2\approx0.1013\) . The parabola is within about one percent of the extremal function \(\sin(\pi x)\) .

8. Show by example that the Poincaré inequality fails on \(H^1(0,L)\) , and prove the substitute: if \(u\in H^1(0,L)\) and \(\int_0^Lu\,dx=0\) , then \(\lVert u\rVert\le L\lVert u'\rVert\) .

   Solution

   Failure. The constant function \(u=1\) belongs to \(H^1(0,L)\) with \(\lVert u\rVert=\sqrt L>0\) and \(\lVert u'\rVert=0\) , so no inequality \(\lVert u\rVert\le C\lVert u'\rVert\) can hold on \(H^1(0,L)\) .

   The substitute. Let \(u\in H^1(0,L)\) with \(\int_0^Lu\,dx=0\) , and take the continuous representative. A continuous function with zero integral cannot be positive everywhere or negative everywhere, so by the intermediate value theorem there is \(x_0\in[0,L]\) with \(u(x_0)=0\) . By Theorem 9.3, for every \(x\) ,

   \[ \lvert u(x)\rvert=\lvert u(x)-u(x_0)\rvert\le\lvert x-x_0\rvert^{1/2}\lVert u'\rVert\le\sqrt L\,\lVert u'\rVert . \]

   Squaring and integrating over \((0,L)\) , \(\lVert u\rVert^2\le L\cdot L\,\lVert u'\rVert^2\) , that is, \(\lVert u\rVert\le L\lVert u'\rVert\) .

   The condition of zero mean removes the constants, as the boundary condition did in \(H_0^1\) . This is the inequality behind the Neumann problem: on the subspace of mean-zero functions the energy \(\lVert u'\rVert^2\) again controls the whole \(H^1\) norm. The best constant is \(L/\pi\) , attained by \(\cos(\pi x/L)\) , by the cosine analogue of Corollary 7.9.

9. Derive the weak formulation of \(-(p(x)u')'+q(x)u=f\) on \((0,1)\) with \(u(0)=0\) and \(p(1)u'(1)+\beta u(1)=\gamma\) . State the space, the bilinear form, and the functional, and say which boundary condition is essential and which is natural.

   Answer

   Multiply by \(v\) with \(v(0)=0\) and integrate by parts: \(\int_0^1(pu'v'+quv)\,dx-p(1)u'(1)v(1)=\int_0^1fv\,dx\) . Substituting \(p(1)u'(1)=\gamma-\beta u(1)\) : the space is \(V=\{v\in H^1(0,1):v(0)=0\}\) , the form is \(a(u,v)=\int_0^1(pu'v'+quv)\,dx+\beta u(1)v(1)\) , and the functional is \(\ell(v)=\int_0^1fv\,dx+\gamma v(1)\) . The condition at \(x=0\) is essential; the Robin condition at \(x=1\) is natural.

10. For \(0<a<1\) find the weak solution \(u\in H_0^1(0,1)\) of \(\int_0^1u'v'\,dx=v(a)\) for all \(v\in H_0^1(0,1)\) , compute its energy \(\lVert u'\rVert^2\) , and prove that \(\lvert v(a)\rvert\le\sqrt{a(1-a)}\,\lVert v'\rVert\) for all \(v\in H_0^1(0,1)\) , with equality for a suitable \(v\) .

    Solution

    Guided by Example 9.3, the solution should be Green’s function of Example 4.2 with source at \(a\) :

    \[ u(x)=(1-a)\,x\ \ (0\le x\le a),\qquad u(x)=a\,(1-x)\ \ (a\le x\le1). \]

    It is continuous and piecewise linear with \(u(0)=u(1)=0\) , so \(u\in H_0^1(0,1)\) , with \(u'=1-a\) on \((0,a)\) and \(u'=-a\) on \((a,1)\) . For \(v\in H_0^1(0,1)\) , using Theorem 9.3 on each piece,

    \[ \int_0^1u'v'\,dx=(1-a)\bigl(v(a)-v(0)\bigr)-a\bigl(v(1)-v(a)\bigr)=(1-a)v(a)+a\,v(a)=v(a). \]

    So \(u\) is a weak solution, and it is the only one by the argument of Theorem 9.9. Its energy is

    \[ \lVert u'\rVert^2=a(1-a)^2+(1-a)a^2=a(1-a), \]

    which also equals \(u(a)\) , as testing with \(v=u\) requires.

    For the inequality, the identity just proved and the Cauchy–Schwarz inequality give, for every \(v\in H_0^1(0,1)\) ,

    \[ \lvert v(a)\rvert=\Bigl\lvert\int_0^1u'v'\,dx\Bigr\rvert\le\lVert u'\rVert\,\lVert v'\rVert=\sqrt{a(1-a)}\,\lVert v'\rVert , \]

    with equality for \(v=u\) . The constant is largest, \(\tfrac12\) , at the midpoint, in agreement with Example 9.3, and tends to zero as \(a\) approaches either end, where functions in \(H_0^1\) are pinned.

11. A student sets up the weak form of \(-u''+u=f\) on \((0,1)\) with \(u'(0)=u'(1)=0\) in the space \(V=\{v\in H^1(0,1):v'(0)=v'(1)=0\}\) . Explain why \(V\) is not well defined as a subspace of \(H^1(0,1)\) , why its natural replacement is all of \(H^1(0,1)\) , and where the boundary conditions then come from.

    Solution

    For \(v\in H^1(0,1)\) the derivative \(v'\) is an element of \(L^2(0,1)\) : an equivalence class of functions defined up to sets of measure zero. It has no value at the single points \(0\) and \(1\) . Theorem 9.3 gives point values to \(v\) , not to \(v'\) . So the conditions \(v'(0)=v'(1)=0\) do not define a subset of \(H^1(0,1)\) at all.

    One might take instead the smooth functions satisfying the conditions and close up in the \(H^1\) norm. But that closure is all of \(H^1(0,1)\) : any smooth \(v\) can be changed near the endpoints, on intervals of length \(\varepsilon\) , into a function with zero slope there at a cost in the \(H^1\) norm that tends to zero with \(\varepsilon\) , because the \(H^1\) norm does not see the derivative at a point. The condition is not stable under \(H^1\) limits, in contrast to the Dirichlet condition, which is, by the bound in Theorem 9.3.

    So the correct space is \(H^1(0,1)\) itself, with the weak form \(\int_0^1(u'v'+uv)\,dx=\int_0^1fv\,dx\) for all \(v\in H^1(0,1)\) . The boundary conditions come out of the identity, as in Example 9.4: if the solution is \(C^2[0,1]\) , integrating by parts gives \(\int_0^1(-u''+u-f)v\,dx+u'(1)v(1)-u'(0)v(0)=0\) , and the freedom to choose \(v(0)\) and \(v(1)\) arbitrarily forces \(u'(0)=u'(1)=0\) . They are natural conditions.

12. Suppose \(u\in C^2[0,1]\) satisfies \(\int_0^1(u'v'+uv)\,dx=\int_0^1fv\,dx+\gamma v(1)\) for all \(v\in H^1(0,1)\) , where \(f\) is continuous and \(\gamma\) is a constant. Prove that \(-u''+u=f\) on \((0,1)\) , \(u'(0)=0\) , and \(u'(1)=\gamma\) .

    Solution

    Since \(u\in C^2[0,1]\) and \(v\in H^1(0,1)\) , the integration by parts of Exercise 9.3 applies to \(u'\) and \(v\) :

    \[ \int_0^1u'v'\,dx=u'(1)v(1)-u'(0)v(0)-\int_0^1u''v\,dx . \]

    The weak identity becomes

    \[ \int_0^1\bigl(-u''+u-f\bigr)v\,dx+\bigl(u'(1)-\gamma\bigr)v(1)-u'(0)\,v(0)=0\qquad\text{for all }v\in H^1(0,1). \]

    Step 1. Take \(v\) a test function on \((0,1)\) ; the boundary terms vanish, and since \(-u''+u-f\) is continuous, Lemma 8.2(a) gives \(-u''+u=f\) on \((0,1)\) .

    Step 2. With the integral now zero for every \(v\) , the identity reads \((u'(1)-\gamma)v(1)-u'(0)v(0)=0\) for all \(v\in H^1(0,1)\) . The choice \(v(x)=x\) gives \(u'(1)=\gamma\) , and the choice \(v(x)=1-x\) gives \(u'(0)=0\) .

    No boundary condition was imposed on the space \(H^1(0,1)\) . Both were forced by the identity: the term \(\gamma v(1)\) in the functional produced the inhomogeneous Neumann condition at \(x=1\) , and the absence of a corresponding term at \(x=0\) produced the homogeneous one there.

13. On \((0,1)\) let \(a(x)=1\) for \(x<\tfrac12\) and \(a(x)=3\) for \(x>\tfrac12\) . Find the function \(u\in H^1(0,1)\) with \(u(0)=0\) , \(u(1)=1\) , and \(\int_0^1au'v'\,dx=0\) for all \(v\in H_0^1(0,1)\) . Verify that the flux \(au'\) is continuous at the interface although \(u'\) is not.

    Answer

    \(u\) is piecewise linear with slopes \(s_1\) on \((0,\tfrac12)\) and \(s_2\) on \((\tfrac12,1)\) . Testing with \(v\in H_0^1(0,1)\) gives \(s_1v(\tfrac12)-3s_2v(\tfrac12)=0\) , so \(s_1=3s_2\) : the flux \(au'\) is continuous. With \(\tfrac12s_1+\tfrac12s_2=1\) this gives \(s_1=\tfrac32\) , \(s_2=\tfrac12\) , so \(u=\tfrac32x\) on the left, \(u=\tfrac12+\tfrac12x\) on the right, and \(u(\tfrac12)=\tfrac34\) . The flux is \(\tfrac32\) on both sides; the slope drops by a factor of three where the conductivity rises by that factor.

14. For \(f=1\) on \((0,1)\) minimize \(J(v)=\tfrac12\int_0^1(v')^2dx-\int_0^1v\,dx\) over the one-parameter families \(v=c\,x(1-x)\) and \(v=c\sin(\pi x)\) . Compare the two minimum values with each other and with \(J\) of the weak solution, and explain the ordering by Theorem 9.10.

    Answer

    For \(v=c\,x(1-x)\) : \(J=\tfrac16c^2-\tfrac16c\) , minimized at \(c=\tfrac12\) with \(J=-\tfrac1{24}\approx-0.04167\) . For \(v=c\sin(\pi x)\) : \(J=\tfrac{\pi^2}{4}c^2-\tfrac2\pi c\) , minimized at \(c=4/\pi^3\) with \(J=-4/\pi^4\approx-0.04106\) . The weak solution is \(u=\tfrac12x(1-x)\) , which lies in the first family, so the first minimum is the global one; the second family does not contain \(u\) , and by Theorem 9.10 its minimum must be larger.

15. Let \(u_1,u_2\in H_0^1(\Omega)\) be the weak solutions of the Dirichlet problem with data \(f_1,f_2\in L^2(\Omega)\) . Prove that \(\lVert u_1-u_2\rVert_{H^1}\le C_P\sqrt{1+C_P^2}\,\lVert f_1-f_2\rVert\) , and explain how this, with uniqueness and the existence theorem to come, makes the problem well-posed in the sense of Definition 1.4.

    Solution

    The difference \(w=u_1-u_2\) belongs to \(H_0^1(\Omega)\) and, subtracting the two weak identities, satisfies \(\int_\Omega\nabla w\cdot\nabla v\,dx=\int_\Omega(f_1-f_2)v\,dx\) for all \(v\in H_0^1(\Omega)\) . So \(w\) is the weak solution with datum \(f_1-f_2\) , and Theorem 9.9 gives \(\lVert\nabla w\rVert\le C_P\lVert f_1-f_2\rVert\) . By the Poincaré inequality \(\lVert w\rVert\le C_P\lVert\nabla w\rVert\) , hence

    \[ \lVert w\rVert_{H^1}^2=\lVert w\rVert^2+\lVert\nabla w\rVert^2\le(1+C_P^2)\,\lVert\nabla w\rVert^2\le(1+C_P^2)\,C_P^2\,\lVert f_1-f_2\rVert^2 , \]

    and taking square roots gives the claim.

    For well-posedness in the sense of Definition 1.4 one must name the classes and the measures of size: data \(f\in L^2(\Omega)\) measured in the \(L^2\) norm, solutions \(u\in H_0^1(\Omega)\) measured in the \(H^1\) norm. Requirement (ii), uniqueness, is Theorem 9.9. Requirement (iii), continuous dependence, is the inequality just proved: the map from data to solution is Lipschitz with constant \(C_P\sqrt{1+C_P^2}\) . Requirement (i), existence for every \(f\in L^2(\Omega)\) , is the theorem of Chapter 10. Compare Theorem 3.7, which gave well-posedness in the maximum norm for classical solutions but had to assume existence.

16. Use Example 8.2 to show that \(u(x)=\lvert x\rvert^{-1/4}\) belongs to \(H^1(B)\) for the unit ball \(B\subset\mathbb R^3\) and compute \(\lVert u\rVert_{H^1}^2\) . Why does this function show that Theorem 9.3 has no analogue in three dimensions?

    Answer

    By Example 8.2 with \(n=3\) and \(a=\tfrac14<\tfrac12\) , \(u\) has a weak gradient with \(\lVert\nabla u\rVert^2=\tfrac\pi2\) , and \(\lVert u\rVert^2=4\pi\int_0^1r^{-1/2}r^2dr=\tfrac{8\pi}{5}\) . So \(\lVert u\rVert_{H^1}^2=\tfrac{8\pi}{5}+\tfrac{\pi}{2}=\tfrac{21\pi}{10}\) . The function is unbounded near the origin, so it has no continuous representative and no bound of its maximum by its \(H^1\) norm can hold.

17. On the unit square compute \(\lVert u\rVert^2\) and \(\lVert\nabla u\rVert^2\) for \(u=\sin(\pi x)\sin(\pi y)\) , and compare the ratio with the bound of Theorem 9.6. Using Exercise 7.11’s eigenfunctions and Parseval’s identity in two variables, explain why \(C_P=1/(\pi\sqrt2)\) for the square.

    Hint

    \(\lVert u\rVert^2=\tfrac14\) and \(\lVert\nabla u\rVert^2=2\pi^2\cdot\tfrac14=\tfrac{\pi^2}{2}\) , with ratio \(1/(2\pi^2)\approx0.0507\) , against the bound \(\tfrac12\) . Expanding \(u\in H_0^1\) in the products \(\sin(m\pi x)\sin(n\pi y)\) with coefficients \(c_{mn}\) , Parseval gives \(\lVert u\rVert^2=\tfrac14\sum c_{mn}^2\) and \(\lVert\nabla u\rVert^2=\tfrac14\sum\pi^2(m^2+n^2)c_{mn}^2\ge2\pi^2\lVert u\rVert^2\) , with equality for \(m=n=1\) .