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Classify \(u_{xx}+4u_{xy}+3u_{yy}=0\) , find its general solution, and determine the solution with \(u(x,0)=0\) and \(u_y(x,0)=\cos x\) .
Solution
The coefficients are \(a_{11}=1\) , \(a_{12}=2\) , \(a_{22}=3\) , so \(D=4-3=1>0\) : hyperbolic. The characteristic equation \(\lambda^2-4\lambda+3=0\) has roots \(1\) and \(3\) , and the general solution is
\[ u(x,y)=F(y-x)+G(y-3x). \]The conditions on \(y=0\) are \(F(-x)+G(-3x)=0\) and \(F'(-x)+G'(-3x)=\cos x\) . With \(s=-x\) : \(F(s)+G(3s)=0\) and \(F'(s)+G'(3s)=\cos s\) . Differentiating the first, \(F'(s)+3G'(3s)=0\) ; subtracting, \(-2G'(3s)=\cos s\) , so \(G'(r)=-\tfrac12\cos(r/3)\) and \(G(r)=-\tfrac32\sin(r/3)+C\) , \(F(s)=-G(3s)=\tfrac32\sin s-C\) . Therefore
\[ u(x,y)=\tfrac32\sin(y-x)-\tfrac32\sin\!\bigl(\tfrac y3-x\bigr). \]Check: at \(y=0\) the two terms cancel, and \(u_y=\tfrac32\cos(y-x)-\tfrac12\cos(\tfrac y3-x)\) equals \(\cos x\) there.
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In the traffic model with flux \(\phi(\rho)=v_m\rho(1-\rho/\rho_m)\) the initial density is \(\tfrac14\rho_m\) for \(x<0\) and \(\tfrac34\rho_m\) for \(x>0\) . Decide whether a shock or a rarefaction forms, justify the choice, and describe the solution for \(t>0\) . Then answer the same question with the two states exchanged.
Solution
The flux is concave, so by Definition 2.7 a shock is admissible when the density jumps upward from left to right. First case, \(\rho_l=\tfrac14\rho_m<\rho_r=\tfrac34\rho_m\) : a shock. Both states have the flux \(\tfrac3{16}v_m\rho_m\) , so the Rankine–Hugoniot speed is zero: the jump stays at \(x=0\) for all time, with \(\phi'(\rho_l)=\tfrac12v_m>0>-\tfrac12v_m=\phi'(\rho_r)\) . Cars arrive at the back of the dense region exactly as fast as they are released into it from the front. Second case, \(\rho_l=\tfrac34\rho_m>\rho_r=\tfrac14\rho_m\) : not admissible as a shock, so a rarefaction: \(\rho=\tfrac34\rho_m\) for \(x\le-\tfrac12v_mt\) , \(\rho=\tfrac12\rho_m\bigl(1-x/(v_mt)\bigr)\) for \(\lvert x\rvert<\tfrac12v_mt\) , and \(\rho=\tfrac14\rho_m\) for \(x\ge\tfrac12v_mt\) .
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A function \(u\) is harmonic in the unit disc and continuous on its closure, with boundary values \(g(\theta)=\theta(2\pi-\theta)\) for \(0\le\theta\le2\pi\) . Find \(u\) at the centre, and give the sharpest bounds on \(u\) in the interior that follow without computing it.
Answer
By the mean value property \(u(0)=\tfrac{1}{2\pi}\int_0^{2\pi}\theta(2\pi-\theta)\,d\theta=\tfrac{2\pi^2}{3}\) . The boundary values range over \([0,\pi^2]\) and are not constant, so by the strong maximum principle \(0<u<\pi^2\) in the open disc.
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A rod of length \(\pi\) with diffusivity \(1\) has both ends held at zero and initial temperature \(\sin x+\sin3x\) . Find the temperature for \(t>0\) , the time after which the second term is less than one percent of the first at \(x=\pi/2\) , and the limit of \(e^{t}u(x,t)\) as \(t\to\infty\) .
Answer
\(u=e^{-t}\sin x+e^{-9t}\sin3x\) . At \(x=\pi/2\) the ratio of the terms is \(e^{-8t}\) , below \(0.01\) for \(t>\tfrac18\log100\approx0.576\) . And \(e^{t}u\to\sin x\) uniformly.
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An infinite string with wave speed \(c=2\) is released from rest with displacement \(g(x)=\max(0,1-\lvert x\rvert)\) . Find the first time at which the point \(x=7\) moves, the displacement of the string at \(x=1\) when \(t=\tfrac12\) , and the total energy \(\tfrac12\int(u_t^2+c^2u_x^2)\,dx\) at all times.
Answer
The support \([-1,1]\) is at distance \(6\) from \(x=7\) , so the point first moves at \(t=6/c=3\) . At \(x=1\) , \(t=\tfrac12\) : \(ct=1\) and \(u=\tfrac12[g(2)+g(0)]=\tfrac12\) . The energy is that of the initial state, \(\tfrac12c^2\int(g')^2dx=\tfrac12\cdot4\cdot2=4\) , for all \(t\) .
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Say whether each statement is true or false and give a one-sentence reason citing a theorem or an example: (a) a bounded solution of the heat equation on the line with continuous initial data is infinitely differentiable for \(t>0\) ; (b) a solution of the wave equation with smooth, compactly supported data is bounded by the maximum of its initial displacement; (c) a harmonic function on a bounded region is determined by its boundary values; (d) the Fourier series of a continuous periodic function converges uniformly; (e) every function with a weak derivative on an interval is continuous.
Answer
(a) True: by Theorem 5.6 the solution is the one of Theorem 5.3, which is smooth for \(t>0\) . (b) False: take \(g=0\) and \(h\) a smooth nonnegative bump with bounded support, not identically zero; by Theorem 6.2, \(u(x,t)=\tfrac1{2c}\int_{x-ct}^{x+ct}h\,ds\) is positive wherever the interval meets the bump, although the initial displacement is zero, exactly as for the struck string of Example 6.2. (c) True, for functions continuous up to the boundary: Theorem 3.7. (d) False as stated: Theorem 7.7 needs the function to be piecewise \(C^1\) as well, and continuity alone does not even give pointwise convergence, a fact not proved in this book. (e) True, in the sense of having a continuous representative: the argument of Theorem 9.3 uses only the integrability of the weak derivative.
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Let \(f(x)=x^2\) for \(x<1\) and \(f(x)=2-x\) for \(x>1\) on the interval \((0,2)\) . Compute the first and second distributional derivatives of \(f\) . Does \(f\) belong to \(H^1(0,2)\) ? Does \(f'\) ?
Solution
On \((0,1)\) , \(f=x^2\) with \(f'=2x\) ; on \((1,2)\) , \(f=2-x\) with \(f'=-1\) . At \(x=1\) both formulas give \(1\) , so \(f\) is continuous and by Theorem 8.6 its distributional derivative is the pointwise one: \(f'=2x\) on \((0,1)\) and \(-1\) on \((1,2)\) , a bounded function. Hence \(f\in H^1(0,2)\) .
The function \(f'\) jumps at \(x=1\) from \(2\) to \(-1\) , a jump of \(-3\) , and its pointwise derivative is \(2\) on \((0,1)\) and \(0\) on \((1,2)\) . By Theorem 8.6 again,
\[ f''=2\cdot\mathbf 1_{(0,1)}-3\,\delta_1 . \]This contains a delta, so \(f'\) has no weak derivative and \(f'\notin H^1(0,2)\) . In the language of Chapter 9, \(f\) is in \(H^1\) and not in the space of functions with two square-integrable weak derivatives; a corner costs exactly one derivative.
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For \(-u''+u=1\) on \((0,1)\) with \(u'(0)=0\) and \(u(1)=0\) , write the weak formulation, naming the space and saying which condition is essential and which is natural, and compute the Galerkin approximation in the one-dimensional subspace spanned by \(1-x^2\) . Compare its value at \(x=0\) with the exact one.
Solution
Weak form. Multiply by \(v\) with \(v(1)=0\) and integrate by parts: \(\int_0^1(u'v'+uv)\,dx-[u'v]_0^1=\int_0^1v\,dx\) . The boundary term at \(x=1\) vanishes because \(v(1)=0\) and at \(x=0\) because \(u'(0)=0\) . So: find \(u\in V=\{v\in H^1(0,1):v(1)=0\}\) with \(\int_0^1(u'v'+uv)\,dx=\int_0^1v\,dx\) for all \(v\in V\) . The condition \(u(1)=0\) is essential, built into \(V\) ; the condition \(u'(0)=0\) is natural.
Galerkin. With \(\phi=1-x^2\in V\) : \(a(\phi,\phi)=\int_0^14x^2\,dx+\int_0^1(1-x^2)^2dx=\tfrac43+\tfrac8{15}=\tfrac{28}{15}\) and \(\ell(\phi)=\int_0^1(1-x^2)\,dx=\tfrac23\) . So \(c=\tfrac23\cdot\tfrac{15}{28}=\tfrac5{14}\) and \(u_h(0)=\tfrac5{14}\approx0.3571\) .
Exact. \(u=1-\cosh x/\cosh1\) satisfies \(-u''+u=1\) , \(u'(0)=0\) , \(u(1)=0\) , and \(u(0)=1-1/\cosh1\approx0.3519\) . One basis function that respects both boundary conditions gives the value at the origin to within two percent.
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Use Poisson’s formula for the half-plane to find the bounded harmonic function in \(y>0\) with boundary values \(1\) for \(\lvert x\rvert<1\) and \(0\) for \(\lvert x\rvert>1\) at the point \((0,\sqrt3)\) , and interpret the answer as an angle.
Answer
By Example 4.3, \(u\) is \(1/\pi\) times the angle subtended by \([-1,1]\) . From \((0,\sqrt3)\) each half of the segment subtends \(\arctan(1/\sqrt3)=\pi/6\) , so the angle is \(\pi/3\) and \(u=\tfrac13\) .
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Let \(\Omega\) be a bounded region with smooth boundary, \(k>0\) , and \(u\in C^2(\overline\Omega\times[0,\infty))\) a solution of \(u_t=k\Delta u\) with \(u=0\) on \(\partial\Omega\) . Prove that \(\int_\Omega u^2\,dx\) is nonincreasing and deduce uniqueness for the initial-boundary value problem. Which inequality from the later chapters would turn the argument into an exponential decay estimate?
Hint
\(\tfrac{d}{dt}\tfrac12\int_\Omega u^2dx=k\int_\Omega u\,\Delta u\,dx=-k\int_\Omega\lvert\nabla u\rvert^2dx\le0\) by Green’s first identity and the boundary condition. Apply this to the difference of two solutions. The Poincaré inequality of Theorem 9.6 gives \(\int\lvert\nabla u\rvert^2\ge C_P^{-2}\int u^2\) and hence decay like \(e^{-2kt/C_P^2}\) .
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Find the Fourier sine series of \(g(x)=x(\pi-x)\) on \((0,\pi)\) and use Parseval’s identity to evaluate \(\sum_{j\ge0}(2j+1)^{-6}\) .
Solution
Two integrations by parts, as in Example 7.1, give \(b_n=\tfrac2\pi\int_0^\pi x(\pi-x)\sin nx\,dx=\dfrac{4\bigl(1-(-1)^n\bigr)}{\pi n^3}\) , that is, \(\dfrac{8}{\pi n^3}\) for odd \(n\) and \(0\) for even \(n\) :
\[ x(\pi-x)=\frac8\pi\sum_{j\ge0}\frac{\sin\bigl((2j+1)x\bigr)}{(2j+1)^3}\qquad(0\le x\le\pi), \]with uniform convergence by Theorem 7.7. Parseval’s identity with \(L=\pi\) reads \(\int_0^\pi x^2(\pi-x)^2dx=\tfrac\pi2\sum b_n^2\) . The left side is \(\pi^5\int_0^1s^2(1-s)^2ds=\tfrac{\pi^5}{30}\) . The right side is \(\tfrac\pi2\cdot\tfrac{64}{\pi^2}\sum_{j\ge0}(2j+1)^{-6}\) . Hence
\[ \sum_{j\ge0}\frac{1}{(2j+1)^6}=\frac{\pi^5}{30}\cdot\frac{\pi}{32}=\frac{\pi^6}{960}\approx1.00145 . \] -
Solve \(u_t-u_{xx}=\sin x\) on \(0<x<\pi\) with zero boundary values and zero initial values, and find the limit of the solution as \(t\to\infty\) . Then solve \(u_{tt}-u_{xx}=\sin x\) with the same side conditions and zero initial velocity, and contrast the long-time behaviour.
Answer
Heat: with \(u=T(t)\sin x\) , \(T'+T=1\) , \(T(0)=0\) , so \(u=(1-e^{-t})\sin x\to\sin x\) , the steady state. Wave: \(T''+T=1\) , \(T(0)=T'(0)=0\) , so \(u=(1-\cos t)\sin x\) , which oscillates for ever between \(0\) and \(2\sin x\) about the same steady state and never settles. Diffusion dissipates the initial mismatch; the wave equation conserves its energy.
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Decide whether the bilinear form \(a(u,v)=\int_0^1(u'v'+u'v)\,dx\) is coercive on \(H_0^1(0,1)\) , and whether \(a(u,v)=\int_0^1u'v'\,dx-20\int_0^1uv\,dx\) is. In each case say what follows about the corresponding boundary value problem.
Solution
First form. By Exercise 10.5, \(\int_0^1v'v\,dx=0\) on \(H_0^1(0,1)\) , so \(a(v,v)=\lVert v'\rVert^2\ge(1+\pi^{-2})^{-1}\lVert v\rVert_{H^1}^2\) : coercive. The form is bounded with \(M\le2\) . By the Lax–Milgram theorem the problem \(-u''+u'=f\) , \(u(0)=u(1)=0\) , has exactly one weak solution for every \(f\in L^2(0,1)\) , although the form is not symmetric and there is no energy that the solution minimizes.
Second form. Here \(\lambda=20\) exceeds the first eigenvalue \(\pi^2\approx9.87\) . For \(v=\sin(\pi x)\) , \(a(v,v)=\tfrac12\pi^2-20\cdot\tfrac12=\tfrac12(\pi^2-20)<0\) , so the form is not coercive and Theorem 10.2 does not apply. This does not mean the problem is unsolvable: \(20\) is not one of the eigenvalues \(n^2\pi^2\) , the next being \(4\pi^2\approx39.5\) , and expanding in sines gives the unique solution \(u=\sum f_n\sin(n\pi x)/(n^2\pi^2-20)\) for every \(f\in L^2(0,1)\) . What is lost is the mechanism: there is no minimum principle, the first coefficient has the opposite sign from the load, and stability is governed by the distance from \(20\) to the nearest eigenvalue. Lax–Milgram gives a sufficient condition for well-posedness, not a necessary one.
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A thin plate occupies a region with a re-entrant corner and is made of two materials with different conductivities. It is heated by a source that is switched on at time zero and held steady. Describe which methods of this book you would use to establish that a steady state exists, to bound it, to compute it, and to estimate how fast the plate approaches it, and say where each method’s hypotheses must be checked.
Hint
Existence of the steady state: weak formulation with a discontinuous conductivity and Theorem 10.4; check uniform ellipticity and that the source is in \(L^2\) . No classical solution should be expected at the corner or the interface. Bounds: the weak maximum principle and barriers of Chapter 3 on each smooth piece, or the energy estimate. Computation: Galerkin with hat functions on a mesh aligned with the interface, with Céa’s lemma for the error. Rate of approach: the difference between the evolving and the steady temperature solves the homogeneous heat equation, and the energy argument of Exercise R.10 with the Poincaré constant, or the first eigenvalue, gives exponential decay.