A metal rod is hot in the middle and cold at the ends. A plucked string is displaced in the middle and held at the ends. A drop of dye sits in a pipe of flowing water. Each situation is described by a function of position and time, and each function obeys an equation relating its partial derivatives. The equations for the rod and the string look almost the same, namely \(u_t=ku_{xx}\) and \(u_{tt}=c^2u_{xx}\) , yet the rod’s temperature profile flattens smoothly and never returns, while the string’s shape splits into two copies that travel away undamaged. Where do these equations come from, and what in their form predicts such different behaviour?

This chapter answers both questions at the level of first principles. The equations come from a single bookkeeping identity, the conservation law, combined with one physical assumption about how the conserved quantity flows. The behaviour is predicted by the highest-order derivatives alone, through a classification into elliptic, parabolic, and hyperbolic type that survives every change of coordinates.

1.1 Equations, order, and linearity

When one variable plays the role of time we write it separately as \(t\) , and “order” counts derivatives in all variables together. The heat equation \(u_t=ku_{xx}\) is of second order even though it contains only one time derivative.

Each class contains the previous one. The inviscid Burgers equation \(u_t+uu_x=0\) is quasilinear and not semilinear, because the coefficient of the first derivative \(u_x\) is \(u\) itself. The equation \(u_{tt}-u_{xx}+\sin u=0\) is semilinear. The eikonal equation \(u_x^2+u_y^2=1\) is fully nonlinear.

Linearity matters because of superposition. If \(L\) is a linear operator, meaning \(L(\alpha u+\beta v)=\alpha Lu+\beta Lv\) for all numbers \(\alpha,\beta\) and all sufficiently smooth \(u,v\) , then any linear combination of solutions of \(Lu=0\) is again a solution, and the difference of two solutions of \(Lu=f\) solves \(Lu=0\) . Almost every uniqueness proof in this book begins with the second remark: to show that a linear problem has at most one solution, subtract two solutions and show that the difference, which solves the problem with zero data, must vanish. For a nonlinear equation no such reduction is available, and Chapter 2 shows what can go wrong.

1.2 From a conservation law to an equation

Let \(u(x,t)\) be the density of some quantity: heat energy per unit volume, mass of dye per unit volume, cars per unit length of road. Let the vector field \(\mathbf q(x,t)\) be its flux: the quantity crosses a small surface element with unit normal \(\nu\) at the rate \(\mathbf q\cdot\nu\) per unit area per unit time. Let \(f(x,t)\) be the rate, per unit volume, at which the quantity is created inside the medium. The basic principle is that the amount inside any fixed region \(V\) changes only by flow through its boundary or by creation inside:

The minus sign is there because \(\nu\) points outward, so \(\mathbf q\cdot\nu>0\) means loss. This is the integral form of the conservation law. It makes sense for rough \(u\) , and Chapter 2 and Chapter 8 will come back to that.

To reach a differential equation, assume \(u\) and \(\mathbf q\) are continuously differentiable and \(f\) is continuous. The time derivative may then be taken inside the integral. The divergence theorem, valid for a \(C^1\) vector field on a bounded region with piecewise smooth boundary, converts the boundary integral into \(\int_V\nabla\cdot\mathbf q\,dx\) (math.multivariable-calculus.generalized-stokes-vector-calculus). Hence

An integrand whose integral over every ball vanishes must itself vanish, provided it is continuous.

Lemma 1.3 turns the integral identity into the differential form of the conservation law,

This is one equation for two unknowns, \(u\) and \(\mathbf q\) . It becomes a closed equation only after a constitutive law, a statement drawn from physics about how the flux depends on the density. Different laws produce the different model equations of this book.

Transport. If the quantity is simply carried along by a medium moving with a known velocity field \(\mathbf c\) , then \(\mathbf q=\mathbf cu\) and the law reads \(u_t+\nabla\cdot(\mathbf cu)=f\) . For a constant velocity and no sources this is the transport equation \(u_t+\mathbf c\cdot\nabla u=0\) . In one space dimension, if the velocity is not prescribed but depends on the density, as it does for cars that slow down in heavy traffic, the flux is a nonlinear function \(q=\phi(u)\) and the law is the scalar conservation law \(u_t+\phi(u)_x=0\) . These are the first-order equations of Chapter 2.

Diffusion. Heat flows from hot to cold, at a rate proportional to the temperature gradient. This is Fourier’s law, \(\mathbf q=-\kappa\nabla u\) for the heat flux in terms of the temperature \(u\) , with thermal conductivity \(\kappa>0\) . The conserved density is the heat energy \(\rho c_pu\) , where \(\rho\) is the mass density and \(c_p\) the specific heat. The conservation law gives \(\rho c_pu_t=\nabla\cdot(\kappa\nabla u)+f\) , and for a homogeneous medium

The constant \(k\) is the diffusivity; its units are length squared per time. The same equation, under the name Fick’s law, governs the concentration of a dilute substance spreading by molecular motion. This is the heat equation of Chapter 5.

Equilibrium. If the temperature has stopped changing, then \(u_t=0\) and the heat equation reduces to the Poisson equation \(-k\,\Delta u=f/(\rho c_p)\) , or to the Laplace equation \(\Delta u=0\) when there are no sources. These are the subjects of Chapters 3 and 4.

Waves. A taut string lies along the \(x\) -axis and moves a small distance \(u(x,t)\) transversally. Let \(\rho\) be its mass per unit length and \(T\) its tension, both constant. The conserved quantity is transverse momentum, with density \(\rho u_t\) . The piece of string to the right of the point \(x\) pulls the piece to the left with a force of magnitude \(T\) directed along the string, whose transverse component is \(T\sin\theta\approx Tu_x\) for small slopes. So momentum enters the segment \([a,b]\) at the rate \(Tu_x(b,t)-Tu_x(a,t)\) , which says that the momentum flux is \(q=-Tu_x\) . The conservation law is \(\rho u_{tt}-Tu_{xx}=0\) , that is,

the wave equation of Chapter 6. The constant \(c\) has the units of a speed, and Chapter 6 shows that it is one.

So the heat equation and the wave equation are the same conservation law with different conserved quantities: energy with flux proportional to \(-u_x\) in one case, momentum with flux proportional to \(-u_x\) in the other. The difference that matters is that momentum density is itself a time derivative, which puts two time derivatives into the equation.

1.3 Side conditions and well-posed problems

A differential equation alone has far too many solutions; even \(u_x=0\) in the plane is solved by every function of \(y\) . A problem is determined by adding side conditions, and the physics suggests which ones.

For an evolution equation on a spatial region \(\Omega\) one prescribes the state at the initial time. The heat equation has one time derivative and needs one initial condition, \(u(x,0)=g(x)\) . The wave equation has two and needs two, \(u(x,0)=g(x)\) and \(u_t(x,0)=h(x)\) : position and velocity, as for Newton’s law. On the boundary of \(\Omega\) one prescribes how the medium interacts with the outside. The three standard choices are the Dirichlet condition \(u=g\) on \(\partial\Omega\) , in which the value is held; the Neumann condition \(\partial u/\partial\nu=g\) on \(\partial\Omega\) , in which the flux through the boundary is prescribed, with \(g=0\) meaning insulation; and the Robin condition \(\partial u/\partial\nu+\alpha u=g\) , which models a flux proportional to the difference between the boundary value and an outside value. An equilibrium equation such as \(\Delta u=0\) has no time variable and takes boundary conditions only.

The third requirement is what makes a model usable, because measured data always carry errors. It is not automatic, and it depends on matching the kind of side condition to the kind of equation.

1.4 The type of a second-order equation

Consider a linear second-order equation in two variables,

with coefficients that are functions of \((x,y)\) . The terms with second derivatives form the principal part. The factor \(2\) is a convention that makes the coefficients the entries of a symmetric matrix,

so that the principal part is the sum of \(a_{ij}u_{x_ix_j}\) over \(i,j\in\{1,2\}\) with \((x_1,x_2)=(x,y)\) .

For the wave equation \(u_{tt}-c^2u_{xx}=0\) in the variables \((x,t)\) we have \(a_{11}=-c^2\) , \(a_{12}=0\) , \(a_{22}=1\) , and \(D=c^2>0\) : hyperbolic. For the heat equation \(u_t-ku_{xx}=0\) we have \(a_{11}=-k\) and \(a_{12}=a_{22}=0\) , so \(D=0\) : parabolic. For the Laplace equation \(a_{11}=a_{22}=1\) , \(a_{12}=0\) , and \(D=-1<0\) : elliptic. The names come from the conic sections. The level curves of the quadratic form \(a_{11}\xi^2+2a_{12}\xi\eta+a_{22}\eta^2\) are hyperbolas when \(D>0\) and ellipses when \(D<0\) . When \(D=0\) the form is plus or minus a perfect square \((\alpha\xi+\beta\eta)^2\) and its level curves degenerate into pairs of parallel lines; this is the borderline case in which a conic with linear terms added, such as \(\xi^2-\eta=0\) , is a parabola.

A classification is useful only if it does not depend on the coordinates in which the equation happens to be written.

The lower-order terms change under a change of variables and play no part in the type. That is the first lesson of the classification: only the principal part matters.

The second lesson is that each type has a simplest representative. For constant coefficients the reduction is explicit. Suppose \(a_{11}\ne 0\) and look for a new variable of the form \(\xi=y-\lambda x\) . Its gradient is \((-\lambda,1)\) , and the new coefficient of \(w_{\xi\xi}\) is

This vanishes when \(\lambda\) is a root of the characteristic equation \(a_{11}\lambda^2-2a_{12}\lambda+a_{22}=0\) , whose roots are \(\lambda_\pm=(a_{12}\pm\sqrt D)/a_{11}\) . The number of real roots is decided by the sign of \(D\) .

In the hyperbolic case the lines \(y-\lambda_\pm x=\text{constant}\) are called the characteristic lines of the equation. Along them the equation loses one of its second derivatives, and Chapter 6 shows that they are the paths along which signals travel. An elliptic equation has no real characteristics, and a parabolic one has a single family.

1.5 What the three types do to a single wave

The classification would be a curiosity if it did not predict behaviour. A quick experiment shows that it does. Take a single oscillation in \(x\) with wave number \(\xi>0\) , namely \(\sin(\xi x)\) , and ask how each model equation evolves it. Try \(u(x,t)=T(t)\sin(\xi x)\) .

For the heat equation \(u_t=ku_{xx}\) we need \(T'=-k\xi^2T\) , so \(T(t)=e^{-k\xi^2t}\) : the oscillation decays, and the faster it oscillates the faster it decays. For the wave equation \(u_{tt}=c^2u_{xx}\) we need \(T''=-c^2\xi^2T\) , so \(T(t)=\cos(c\xi t)\) or \(\sin(c\xi t)\) : the oscillation persists for ever with unchanged amplitude. For the Laplace equation \(u_{xx}+u_{yy}=0\) with \(y\) in place of \(t\) we need \(T''=\xi^2T\) , so \(T(y)=\cosh(\xi y)\) or \(\sinh(\xi y)\) : the oscillation grows exponentially in \(y\) , and the faster it oscillates the faster it grows.

Three consequences follow, each of which a later chapter turns into a theorem. A parabolic equation damps fine detail, so its solutions are smooth for \(t>0\) whatever the initial state; it also cannot be run backwards, because reversing time turns violent damping into violent growth. A hyperbolic equation transports detail without damping it, so corners and jumps in the data persist. An elliptic equation treated as an evolution amplifies fine detail without bound, which is the mechanism of Example 1.2, and so it must be given boundary conditions all round instead of initial conditions.

Exercises

1. For each equation give the order and say whether it is linear, semilinear, quasilinear, or fully nonlinear, naming the term that decides: (a) \(u_t+uu_x=u_{xx}\) ; (b) \(xu_x+yu_y=u\) ; (c) \(u_t=(u^2)_{xx}\) ; (d) \(u_{xx}u_{yy}-u_{xy}^2=1\) ; (e) \(u_{tt}-u_{xx}+u^3=0\) .

   Answer

   (a) Order \(2\) , semilinear: the top-order term \(u_{xx}\) has a constant coefficient and the nonlinearity \(uu_x\) is of lower order. (b) Order \(1\) , linear. (c) Order \(2\) , quasilinear: expanding, \((u^2)_{xx}=2uu_{xx}+2u_x^2\) , so the coefficient of \(u_{xx}\) depends on \(u\) . (d) Order \(2\) , fully nonlinear: the second derivatives are multiplied together. (e) Order \(2\) , semilinear: the nonlinear term \(u^3\) contains no derivative.

2. A substance with concentration \(u(x,t)\) diffuses according to Fick’s law with diffusivity \(k\) and decays at the rate \(\gamma u\) per unit volume, where \(\gamma>0\) is constant. Derive the equation \(u_t=k\Delta u-\gamma u\) from the integral conservation law, stating where the divergence theorem and Lemma 1.3 are used, and show that \(v=e^{\gamma t}u\) solves the heat equation.

   Solution

   Let \(V\) be a ball inside the medium. The amount of substance in \(V\) is \(\int_Vu\,dx\) . It leaves through \(\partial V\) at the rate \(\int_{\partial V}\mathbf q\cdot\nu\,dS\) and is destroyed inside at the rate \(\int_V\gamma u\,dx\) , so

   \[ \frac{d}{dt}\int_Vu\,dx=-\int_{\partial V}\mathbf q\cdot\nu\,dS-\int_V\gamma u\,dx . \]

   Fick’s law is \(\mathbf q=-k\nabla u\) . Assuming \(u\in C^2\) in space and \(C^1\) in time, differentiate under the integral on the left and apply the divergence theorem to the \(C^1\) field \(\mathbf q\) on the right: \(\int_{\partial V}\mathbf q\cdot\nu\,dS=\int_V\nabla\cdot\mathbf q\,dx=-\int_Vk\Delta u\,dx\) . Therefore \(\int_V(u_t-k\Delta u+\gamma u)\,dx=0\) for every ball \(V\) . The integrand is continuous, so Lemma 1.3 gives \(u_t=k\Delta u-\gamma u\) at every point.

   For \(v=e^{\gamma t}u\) we have \(v_t=e^{\gamma t}(u_t+\gamma u)=e^{\gamma t}k\Delta u=k\Delta v\) , since the factor \(e^{\gamma t}\) does not depend on \(x\) . So \(v\) solves the heat equation: decay at a constant rate can be removed by an exponential change of the unknown.

3. Dye in a channel is carried with a velocity \(c(x)\) that depends on position. Derive \(u_t+(c(x)u)_x=0\) from the conservation law, explain why this is not the same as \(u_t+c(x)u_x=0\) unless \(c\) is constant, and for \(c(x)=x\) show that \(u(x,t)=e^{-t}g(xe^{-t})\) is a solution that conserves the total amount of dye on the line.

   Solution

   In one dimension the region is an interval \([a,b]\) and the flux is \(q=c(x)u\) , the amount crossing the point \(x\) per unit time in the positive direction. The conservation law reads \(\frac{d}{dt}\int_a^bu\,dx=q(a,t)-q(b,t)=-\int_a^b(cu)_x\,dx\) , so \(\int_a^b\bigl(u_t+(cu)_x\bigr)dx=0\) for every interval and, by Lemma 1.3, \(u_t+(c(x)u)_x=0\) . Expanding, \(u_t+c(x)u_x+c'(x)u=0\) . The extra term \(c'(x)u\) is absent only when \(c'=0\) . It records that where the flow accelerates, \(c'>0\) , the fluid is stretched and the concentration falls even for an observer moving with the flow.

   For \(c(x)=x\) the equation is \(u_t+xu_x+u=0\) . With \(u=e^{-t}g(xe^{-t})\) and \(s=xe^{-t}\) ,

   \[ u_t=-e^{-t}g(s)-xe^{-2t}g'(s),\qquad u_x=e^{-2t}g'(s), \]

   so \(u_t+xu_x+u=-e^{-t}g-xe^{-2t}g'+xe^{-2t}g'+e^{-t}g=0\) . The total amount is \(\int_{-\infty}^{\infty}e^{-t}g(xe^{-t})\,dx=\int_{-\infty}^{\infty}g(s)\,ds\) by the substitution \(s=xe^{-t}\) , independent of \(t\) . Along a path \(x=x_0e^{t}\) moving with the flow the value is \(e^{-t}g(x_0)\) , which decays: the quantity conserved is the integral, not the pointwise value.

4. Verify that \(u(x,t)=e^{-kt}\cos x\) solves \(u_t=ku_{xx}\) , compute the heat content of the segment \(0\le x\le\pi/2\) in units where \(\rho c_p=1\) , and check directly that its rate of change equals the net inflow through the two endpoints.

   Answer

   \(u_t=-ke^{-kt}\cos x=ku_{xx}\) . The heat content of \([0,\pi/2]\) is \(\int_0^{\pi/2}e^{-kt}\cos x\,dx=e^{-kt}\) , with rate of change \(-ke^{-kt}\) . The flux is \(q=-ku_x=ke^{-kt}\sin x\) . At \(x=0\) it is zero, so nothing crosses the left end; at \(x=\pi/2\) it equals \(ke^{-kt}\) in the outward direction \(+x\) . Net inflow is \(-ke^{-kt}\) , as required.

5. Find the discriminant and the type of each equation, and for (d) describe the regions of each type: (a) \(u_{xx}+2u_{xy}+2u_{yy}=0\) ; (b) \(u_{xx}+6u_{xy}+u_{yy}+u_x=0\) ; (c) \(4u_{xx}-4u_{xy}+u_{yy}=0\) ; (d) \(u_{xx}+(x^2+y^2-1)u_{yy}=0\) .

   Answer

   (a) \(a_{12}=1\) , \(D=1-2=-1\) : elliptic. (b) \(a_{12}=3\) , \(D=9-1=8\) : hyperbolic; the term \(u_x\) is irrelevant. (c) \(a_{12}=-2\) , \(D=4-4=0\) : parabolic. (d) \(a_{12}=0\) , \(D=1-x^2-y^2\) : hyperbolic inside the unit circle, parabolic on it, elliptic outside.

6. Reduce \(u_{xx}+u_{xy}-2u_{yy}=0\) to canonical form, write its general solution, and then find the solution satisfying \(u(x,0)=\sin x\) and \(u_y(x,0)=0\) .

   Solution

   Here \(a_{11}=1\) , \(a_{12}=\tfrac12\) , \(a_{22}=-2\) , so \(D=\tfrac14+2=\tfrac94>0\) and the equation is hyperbolic. The characteristic equation \(\lambda^2-\lambda-2=0\) has roots \(\lambda_+=2\) and \(\lambda_-=-1\) , giving \(\xi=y-2x\) and \(\eta=y+x\) . By Theorem 1.7(a) the equation becomes \(-9w_{\xi\eta}=0\) , so \(w=F(\xi)+G(\eta)\) . Renaming the arbitrary functions, the general solution is

   \[ u(x,y)=F(x+y)+G(2x-y). \]

   The conditions on \(y=0\) are \(F(x)+G(2x)=\sin x\) and \(F'(x)-G'(2x)=0\) . The second gives \(F(x)=\tfrac12G(2x)+C\) for a constant \(C\) ; substituting in the first, \(\tfrac32G(2x)+C=\sin x\) , so \(G(s)=\tfrac23\sin(s/2)-\tfrac23C\) and \(F(x)=\tfrac13\sin x+\tfrac23C\) . The constants cancel in the sum and

   \[ u(x,y)=\tfrac13\sin(x+y)+\tfrac23\sin\!\left(x-\tfrac{y}{2}\right). \]

   Check: at \(y=0\) this is \(\sin x\) , and \(u_y=\tfrac13\cos(x+y)-\tfrac13\cos(x-\tfrac y2)\) vanishes at \(y=0\) . For the equation, the first term contributes \(\tfrac13(-1-1+2)\sin(x+y)=0\) and the second contributes \(\tfrac23(-1+\tfrac12+\tfrac12)\sin(x-\tfrac y2)=0\) .

7. Prove that if \(L\) is a linear differential operator then every linear combination of solutions of \(Lu=0\) is a solution, and that the difference of two solutions of \(Lu=f\) solves \(Lu=0\) . Then show that superposition fails for \(u_t+uu_x=0\) by checking that \(u=1\) and \(u=x/(1+t)\) are solutions for \(t>-1\) whose sum is not.

   Hint

   Linearity of \(L\) is the identity \(L(\alpha u+\beta v)=\alpha Lu+\beta Lv\) ; both statements are one line from it. For the Burgers equation compute \(u_t+uu_x\) for \(u=x/(1+t)\) , obtaining \(-x/(1+t)^2+x/(1+t)^2=0\) , and then for the sum \(w=1+x/(1+t)\) , obtaining \(w_t+ww_x=1/(1+t)\ne 0\) . The cross term \(1\cdot u_x\) is what superposition cannot absorb.

8. A student classifies \(u_{xx}+2u_{xy}+2u_{yy}=0\) by writing “\(b=2\) , so \(b^2-ac=4-2>0\) : hyperbolic, with two families of real characteristics.” Find the error, give the correct type, and say what the characteristic equation’s roots are.

   Answer

   The coefficient of \(u_{xy}\) is \(2a_{12}\) , so \(a_{12}=1\) , not \(2\) . Then \(D=1-2=-1<0\) and the equation is elliptic. The characteristic equation \(\lambda^2-2\lambda+2=0\) has the complex roots \(1\pm i\) , so there are no real characteristics.

9. For the backward heat equation \(u_t=-u_{xx}\) consider \(u_n(x,t)=\tfrac1n e^{n^2t}\sin(nx)\) . Show that each \(u_n\) is a solution, that the initial data tend to zero uniformly as \(n\to\infty\) , and that for every fixed \(t>0\) the maximum of \(\lvert u_n(\cdot,t)\rvert\) tends to infinity. Which requirement of Definition 1.4 fails?

   Solution

   Differentiating, \(\partial_tu_n=n\,e^{n^2t}\sin(nx)\) and \(\partial_{xx}u_n=-n\,e^{n^2t}\sin(nx)\) , so \(\partial_tu_n=-\partial_{xx}u_n\) and each \(u_n\) solves the backward heat equation. The initial datum is \(u_n(x,0)=\tfrac1n\sin(nx)\) , with maximum \(1/n\to 0\) , so the data converge uniformly to the zero datum, whose solution is \(u=0\) . At a fixed time \(t>0\) the maximum over \(x\) of \(\lvert u_n(x,t)\rvert\) is \(e^{n^2t}/n\) , which tends to infinity because the exponential outgrows \(n\) ; for instance at \(t=0.1\) and \(n=10\) it is \(e^{10}/10\approx 2203\) .

   So data arbitrarily close to zero give solutions arbitrarily far from the zero solution at every positive time. Requirement (iii), continuous dependence on the data, fails. The mechanism is the one in Section 1.5: reversing time in the heat equation turns the damping factor \(e^{-k\xi^2t}\) into the amplification factor \(e^{k\xi^2t}\) , strongest for the finest oscillations.

10. Using the eigenvalues of the coefficient matrix, classify in three variables: (a) \(2u_{xx}+2u_{yy}+u_{zz}+2u_{xy}=0\) ; (b) \(u_{xx}+u_{yy}-u_{zz}+4u_{xy}=0\) .

    Answer

    (a) The matrix has rows \((2,1,0)\) , \((1,2,0)\) , \((0,0,1)\) and eigenvalues \(3,1,1\) , all positive: elliptic. (b) The matrix has rows \((1,2,0)\) , \((2,1,0)\) , \((0,0,-1)\) and eigenvalues \(3,-1,-1\) : all nonzero with exactly one of opposite sign to the others, so hyperbolic.

11. A string of constant linear density \(\rho\) and tension \(T\) hangs under gravity, which exerts a transverse force \(-\rho g\) per unit length. Add this source to the momentum balance to derive \(\rho u_{tt}=Tu_{xx}-\rho g\) , and find the time-independent shape of a string with \(u(0)=u(L)=0\) .

    Hint

    The transverse momentum of the segment \([a,b]\) changes through the tension at its ends and through the body force: \(\frac{d}{dt}\int_a^b\rho u_t\,dx=Tu_x(b,t)-Tu_x(a,t)-\int_a^b\rho g\,dx\) . Write the tension terms as \(\int_a^bTu_{xx}\,dx\) and localize. For the static shape solve \(Tu''=\rho g\) with \(u(0)=u(L)=0\) ; you should find the parabola \(u(x)=-\dfrac{\rho g}{2T}\,x(L-x)\) .

12. Show that \(u_{xx}+2u_{xy}+u_{yy}=0\) is parabolic, reduce it to canonical form with Theorem 1.7(b), and deduce that its general solution is \(u=F(x-y)+x\,G(x-y)\) . Verify the answer by substitution.

    Solution

    The coefficients are \(a_{11}=1\) , \(a_{12}=1\) , \(a_{22}=1\) , so \(D=1-1=0\) : parabolic. The double root of \(\lambda^2-2\lambda+1=0\) is \(\lambda=1\) , and Theorem 1.7(b) prescribes \(\xi=y-x\) , \(\eta=x\) , which turns the equation into \(w_{\eta\eta}=0\) . Integrating twice in \(\eta\) with \(\xi\) held fixed, \(w=F_0(\xi)+\eta\,G_0(\xi)\) for arbitrary functions \(F_0,G_0\) . Returning to the original variables and writing \(F(s)=F_0(-s)\) , \(G(s)=G_0(-s)\) ,

    \[ u(x,y)=F(x-y)+x\,G(x-y). \]

    Verification. For \(u=F(x-y)\) : \(u_{xx}=F''\) , \(u_{xy}=-F''\) , \(u_{yy}=F''\) , and \(F''-2F''+F''=0\) . For \(u=xG(x-y)\) : \(u_x=G+xG'\) , so \(u_{xx}=2G'+xG''\) , \(u_{xy}=-G'-xG''\) , \(u_{yy}=xG''\) , and

    \[ u_{xx}+2u_{xy}+u_{yy}=2G'+xG''-2G'-2xG''+xG''=0 . \]

    There is a single family of characteristic lines, \(x-y=\text{constant}\) , in agreement with the parabolic type.

13. Show that if \(u(x,t)\) solves \(u_t=ku_{xx}\) then so does \(v(x,t)=u(\lambda x,\lambda^2t)\) for every \(\lambda>0\) . Deduce that the time for a rod of length \(L\) to equilibrate scales like \(L^2/k\) , and for the illustrative value \(k=10^{-4}\ \mathrm{m^2/s}\) compare the time scales for \(L=1\ \mathrm{cm}\) and \(L=1\ \mathrm{m}\) .

    Answer

    \(v_t=\lambda^2u_t\) and \(v_{xx}=\lambda^2u_{xx}\) , evaluated at \((\lambda x,\lambda^2t)\) , so \(v_t=kv_{xx}\) . Lengths scale by \(\lambda\) when times scale by \(\lambda^2\) , so the only time scale that can be built from \(L\) and \(k\) is \(L^2/k\) . For \(L=1\ \mathrm{cm}\) it is \(10^{-4}/10^{-4}=1\ \mathrm{s}\) ; for \(L=1\ \mathrm{m}\) it is \(10^{4}\ \mathrm{s}\) , about \(2.8\) hours.

14. Let \(u\) be a classical solution of \(u_t=ku_{xx}\) on \(0\le x\le L\) with \(u_x(0,t)=u_x(L,t)=0\) . Prove that the heat content \(\int_0^Lu\,dx\) is constant in time. If instead \(u(0,t)=u(L,t)=0\) and \(u\ge 0\) everywhere, prove that the heat content cannot increase, and explain the sign of each boundary flux.

    Solution

    Since \(u\) is a classical solution on the closed interval we may differentiate under the integral sign and use the equation:

    \[ \frac{d}{dt}\int_0^Lu\,dx=\int_0^Lku_{xx}\,dx=k\,u_x(L,t)-k\,u_x(0,t). \]

    With insulated ends both terms vanish and the heat content is constant.

    Now suppose \(u(0,t)=u(L,t)=0\) and \(u\ge 0\) . At \(x=0\) the function \(x\mapsto u(x,t)\) has a minimum over \([0,L]\) , attained at the left endpoint, so its one-sided derivative satisfies \(u_x(0,t)\ge 0\) . At \(x=L\) it again has a minimum, attained at the right endpoint, so \(u_x(L,t)\le 0\) . Hence \(k\,u_x(L,t)-k\,u_x(0,t)\le 0\) and the heat content is nonincreasing. In terms of flux \(q=-ku_x\) : at the right end \(q\ge 0\) points in the outward direction \(+x\) , and at the left end \(q\le 0\) points in the outward direction \(-x\) . Heat can only leave a nonnegative temperature distribution whose ends are held at zero.

15. Substitute \(u=e^{i(\xi x-\omega t)}\) with real \(\xi\) into each equation, solve for \(\omega\) , and say whether the mode travels, decays, or grows: (a) \(u_t+cu_x=0\) ; (b) \(u_t=ku_{xx}\) ; (c) \(u_{tt}=c^2u_{xx}\) ; (d) \(u_t=-ku_{xx}\) ; (e) \(iu_t+u_{xx}=0\) .

    Answer

    (a) \(\omega=c\xi\) : the mode travels with speed \(c\) . (b) \(\omega=-ik\xi^2\) , so \(e^{-i\omega t}=e^{-k\xi^2t}\) : decay. (c) \(\omega=\pm c\xi\) : travels in either direction with speed \(c\) . (d) \(\omega=ik\xi^2\) , so \(e^{-i\omega t}=e^{k\xi^2t}\) : growth. (e) \(\omega=\xi^2\) : the mode travels with constant amplitude at the speed \(\omega/\xi=\xi\) , which depends on the wave number.

16. For the wave equation \(u_{tt}-c^2u_{xx}=0\) in the variables \((x,t)\) , write the matrix \(A\) , take \(\xi=x-ct\) and \(\eta=x+ct\) , compute \(JAJ^{\top}\) , and confirm both the formula \(\tilde D=(\det J)^2D\) and the form \(-4c^2w_{\xi\eta}\) of the transformed principal part.

    Solution

    With \((x_1,x_2)=(x,t)\) the principal part \(-c^2u_{xx}+u_{tt}\) has matrix \(A=\operatorname{diag}(-c^2,1)\) and \(D=0-(-c^2)(1)=c^2\) . For \(\xi=x-ct\) and \(\eta=x+ct\) the gradients are \(\nabla\xi=(1,-c)\) and \(\nabla\eta=(1,c)\) , so

    \[ J=\begin{pmatrix}1&-c\\1&c\end{pmatrix},\qquad \det J=2c . \]

    The entries of \(JAJ^{\top}\) are \((\nabla\xi)^{\top}A\nabla\xi=-c^2+c^2=0\) , \((\nabla\eta)^{\top}A\nabla\eta=-c^2+c^2=0\) , and \((\nabla\xi)^{\top}A\nabla\eta=-c^2-c^2=-2c^2\) . Hence

    \[ JAJ^{\top}=\begin{pmatrix}0&-2c^2\\-2c^2&0\end{pmatrix}, \]

    and the principal part in the new variables is \(2\cdot(-2c^2)\,w_{\xi\eta}=-4c^2w_{\xi\eta}\) . The new discriminant is \(\tilde D=(-2c^2)^2-0=4c^4\) , and indeed \((\det J)^2D=4c^2\cdot c^2=4c^4\) . The same result comes from Theorem 1.7(a) after matching conventions: the characteristic lines are \(x\pm ct=\text{constant}\) .

17. One end \(x=L\) of a rod loses heat to surrounding air at temperature \(u_0\) according to Newton’s law of cooling: the outward heat flux equals \(\beta(u-u_0)\) with \(\beta>0\) . Write the resulting boundary condition at \(x=L\) and the corresponding one at \(x=0\) , identify the type of condition, and explain the difference in sign between the two ends.

    Answer

    The outward heat flux at \(x=L\) is \(-\kappa u_x(L,t)\) , since the outward direction is \(+x\) ; setting it equal to \(\beta(u-u_0)\) gives \(\kappa u_x(L,t)+\beta u(L,t)=\beta u_0\) . At \(x=0\) the outward direction is \(-x\) , so the outward flux is \(+\kappa u_x(0,t)\) and the condition is \(-\kappa u_x(0,t)+\beta u(0,t)=\beta u_0\) . Both are Robin conditions, and both read \(\kappa\,\partial u/\partial\nu+\beta u=\beta u_0\) in terms of the outward normal derivative; the sign in front of \(u_x\) differs only because the outward normal does.

18. Let \(\Omega\) be a bounded region with smooth boundary and suppose \(u\in C^2(\overline\Omega)\) solves \(\Delta u=0\) in \(\Omega\) with \(\partial u/\partial\nu=g\) on \(\partial\Omega\) . Show that necessarily \(\int_{\partial\Omega}g\,dS=0\) , show that solutions are never unique, interpret both facts in terms of steady heat flow, and say which requirements of Definition 1.4 fail.

    Solution

    Apply the divergence theorem to the \(C^1\) vector field \(\nabla u\) on \(\Omega\) :

    \[ \int_{\partial\Omega}g\,dS=\int_{\partial\Omega}\nabla u\cdot\nu\,dS=\int_\Omega\Delta u\,dx=0 . \]

    So a solution can exist only when the boundary datum has total integral zero. If \(u\) is a solution, so is \(u+C\) for every constant \(C\) , since constants have zero Laplacian and zero normal derivative; hence solutions are never unique.

    In terms of steady heat flow, \(-\kappa g\) is the flux pushed out through the boundary. A temperature that does not change in time is possible only if the net heat entering through the boundary is zero; otherwise the body would warm or cool. And prescribing only fluxes fixes temperature differences, not the absolute level. Requirement (i) fails for data with nonzero integral and requirement (ii) fails always. The repair is to change the classes in Definition 1.4: admit only data with \(\int_{\partial\Omega}g\,dS=0\) and normalize solutions, for example by \(\int_\Omega u\,dx=0\) .