Partial differential equation for temperature u(x, t) in a heat conducting insulated rod along the x-axis is given by the Heat equation:

Chapter 7 Heat Equation Partial differential equation for temperature u(x, t) in a heat conducting insulated rod along the x-axis is given by the Heat equation: u t = ku x x, x, t > (7.1) Here k is a constant and represents the conductivity coefficient of the material used to make the rod. Since we assumed k to be constant, it also means that material properties are constant and do not depend on the location in the rod. It means we are considering a homogeneous material. Note that rod is a three dimensional object and we considered x in the above model. It means that we are assuming that material properties of the rod vary only in the x-direction and are constant w.r.t. y, z directions; and thereby also expecting temperature distribution to respect these symmetries. In a more realistic model we have to consider that conductivity coefficient will vary from point to point in the rod and thus k has to be a function of the variables (x, y, z) and thus we have to look at the following model: u t = k(x, y, z) u xx + u yy + u z z, x 3, t > (7.2) In Section 7.1 we introduce fundamental solution associated to heat equation, and derive formulae for solutions of Cauchy problems for homogeneous and nonhomogeneous heat equation posed for x. We deal the case of heat equation posed on bounded x- intervals and corresponding initial-boundary value problems (IBVP) in Section 7.2. We obtain a formal solution to IBVP using method of separation of variables, and uniqueness is proved using energy method. Maximum principle for solutions to heat equation will be discussed in Section 7.3, and as an application we prove that formal solution to IBVP derived in Section 7.2 is indeed a solution (under some conditions on the data), and also prove uniqueness of solutions to IBVP. 7.1 Fundamental solution and its applications Let a >. Sicne heat equation is invariant under the change of coordinates z = ax and τ = a 2 t (see Exercise 7.1), and z2 τ = x2, we look for solutions to Laplace equation having such symmetric properties. Let us look for a solution of heat equation having the t form u(x, t) = v x t. Substituting this ansatz into heat equation yields the following 187

188 Chapter 7. Heat Equation differential equation satisfied by v d 2 v d z + z d v =. (7.3) 2 2 d z Such transformations which convert a PDE into an ODE are called similarity transformations. The equation (7.3) can be solved explicitly, and its general solution is given by z v(z) = C 1 exp s 2 d s + C 4 2, (7.4) where C 1,C 2 are arbitrary constants. Since the constant term C 2 plays no role in determining a fundamental solution to heat equation, we take C 2 =. Thus u is given by Note that u(x, t) = C 1 x t where u(x,) is interpreted as lim u(x, t). t + Differentiating u w.r.t. x using the formula (7.5) gives exp s 2 d s. (7.5) 4 π if x <, u(x,) = (7.6) π if x >, u x (x, t) = C 1 t exp x2 4t. (7.7) Choose C 1 so that u x (x, t) d x = 1, and this gives C 1 = 1 2. We define fundamental π solution, using (7.7), by K(x, t) = 1 2 πt exp x2 4t. (7.8) Remark 7.1 (Reasons behind the choice of fundamental solution). (i) Fundamental solution is expected to pick-up effects of impulses which are usually concentrated at a point. Note that for a function u(x, ) as in (7.6), the corresponding derivative function represents an impulse at x =. (ii) Of course, we need to verify that fundamental solution defined by (7.8) has the properties that are expected of a fundamental solution, which will be done in the next few results. 7.1.1 Cauchy problem for homogeneous equation Let K 1 : (, ) be defined by K 1 (x, y, t) = 1 2 πt exp x y 2 4t In the following result, we collect some properties of K 1. (7.9) Lemma 7.2. The function K 1 has the following properties: IIT Bombay

7.1. Fundamental solution and its applications 189 (i) K 1 C ( (, )). (ii) K 1 (x, y, t) > for t >. (iii) t 2 K1 (x, y, t) = for t >. x 2 (iv) K 1 (x, y, t) d y = 1 for x, t >. (v) For any δ >, lim K 1 (x, y, t) d y = (7.1) t + y x >δ uniformly for x. Proof. Step 1: proof of (i), (ii), (iii): Properties (i), (ii), and (iii) follow from the definition of K 1 as given by (7.9). Step 2: proof of (iv): Setting z = y x 2, we have t K 1 (x, y, t) d y = 1 exp z 2 d z = 1. (7.11) π Step 3: proof of (v): Introducing a change of variable as in Step 2, we get K 1 (x, y, t) d y = 1 exp z 2 d z. (7.12) π y x >δ Since exp z 2 d z = π, we have This completes the proof of (v). z > δ 2 t lim exp z 2 d z =. (7.13) t z > δ 2 t Solution to Cauchy problem for homogeneous heat equation u t u x x =, x, t >, (7.14a) u(x,) = φ(x), for x (7.14b) can be expressed using fundamental solution, and is the content of the following result. Theorem 7.3. Let φ : be a continuous and bounded function. Let u be defined by Then u(x, t) = K 1 (x, y, t)φ(y) d y = 1 exp 4πt x y 2 4t φ(y) d y. (7.15) (i) u C ( (, )). (ii) u is a solution of heat equation u t = ku x x for t >. (iii) If we extend the function to [, ) by setting u(x,) = φ(x), then u C ( [, )). October 21, 215

19 Chapter 7. Heat Equation Proof. Step 1: proof of (i), (ii): Assertions (i) and (ii) follow from Lemma 7.2. Step 2: proof of (iii): For δ > (to be chosen later), in view of K 1 (x, y, t) d y = 1, we have u(x, t) φ(ξ ) = K 1 (x, y, t)(φ(y) φ(ξ )) d y K 1 (x, y, t) φ(y) φ(ξ ) d y + K 1 (x, y, t) φ(y) φ(ξ ) d y y x <δ y x >δ In the last inequality, there are two integrals: the first integral can be made small using continuity of φ, and the second integral goes to zero as t by Lemma 7.2. We will choose δ > so that φ(y) φ(ξ ) < ε whenever y ξ < 2δ. Thus we have 2 K 1 (x, y, t) φ(y) φ(ξ ) d y K 1 (x, y, t) φ(y) φ(ξ ) d y y x <δ < ε 2 < ε 2 y ξ <2δ y ξ <2δ K 1 (x, y, t) d y K 1 (x, y, t) d y < ε 2. (7.16) By (7.1), there exists t so that for < t < t, we have K 1 (x, y, t) φ(y) φ(ξ ) d y < ε 2. (7.17) y x >δ Combining the inequalities (7.16) and (7.17), we get for (x, t) (, ) such that x y < δ and t < t, the following inequality u(x, t) φ(ξ ) < ε 2 + ε 2 = ε, thus proving that u is continuous at points of x-axis, and that u(x,) = φ(x) for each x. Remark 7.4. The Cauchy data φ need not be a bounded function for the formula (7.15) to represent a solution to Cauchy problem. If there exists M,a such that φ satisfies for every x φ(x) M e ax2, then the formula (7.15) represents a solution to Cauchy problem for < t < 1 4a, and u C ( (, 1 4a ). Remark 7.5 (infinite speed of propagation). (i) The solution having the form (7.15) has the following domain of dependence property. Solution at any point (x, t) (, ) depends on the values of initial data φ(x) at all x, and thus domain of dependence of solution at any point (x, t) for t > is the entire real line. The domain of influence of any point on x-axis is (, ). This shows that information from Cauchy data reaches all points instantly. This suggests that heat equation may not be suitable to study physical phenomenon. IIT Bombay

7.1. Fundamental solution and its applications 191 (ii) In addition to the hypotheses of Theorem 7.3 assume that the Cauchy data φ is such that φ(x) for all x. It immediately follows from the formula (7.15) that u(x, t) > for all x and t >. This is another instance illustrating infinite speed of propagation property of heat equation. Note that Theorem 7.3 asserts only the existence of a solution. In general uniqueness is not expected for Cauchy problem for heat equation, posed for x. However we can prove that Cauchy problem admits only one solution when we are looking for solutions belonging to special classes of functions, like solutions having a controlled growth of the type u(x, t) M e ax2 [2]. The following example of Tychonoff illustrates non-uniqueness of solutions. Example 7.6 (Tychonoff s example). Let g : be given by 1 exp if t >, g(t) = t 2 if t. Then the series u(x, t) := g (k) (t) k= (2k)! converges and is a solution to homogeneous heat equation, and u(x, ) =. For further details, see [2, 23]. However v(x, t) is also a solution to the same Cauchy problem. Thus a Cauchy problem may have more than one solution. x 2k 7.1.2 Duhamel s principle for nonhomogeneous equation As explained in Subsection 4.1.4, Duhamel principle gives us a way to solve nonhomogeneous problems corresponding to a linear differential operator, by superposition of solutions of a family of corresponding homogeneous problems. The Cauchy problem for nonhomogeneous heat equation is given by u t u x x = f (x, t), x, t >, (7.18a) Let S(t)ϕ denote the solution of the Cauchy problem u(x,) = φ(x), for x. (7.18b) u t u x x =, x, t >, u(x,) = φ(x) for x. Taking cue from (4.76), we expect the following formula to yield a solution of the Cauchy problem (7.18) u(x, t) = (S(t)φ)(x) + t (S(t τ) f τ )(x) dτ, (7.19) where f τ (x) := f (x,τ). We will now check that the second term on the right hand side of (7.19) solves the nonhomogeneous heat equation, as the first term corresponds to solution of the homogeneous wave equation satisfying the given Cauchy data. October 21, 215

192 Chapter 7. Heat Equation Indeed, differentiating the equation (7.19) w.r.t. t gives t t (S(t τ)f t τ )(x) dτ = S()f t + t (S(t τ) f τ )(x) dτ = f (x, t) + t t (S(t τ) f τ )(x) dτ. Since S(t τ)f τ are solutions to homogeneous heat equation, we have Thus we get t (S(t τ) f τ ) = 2 x (S(t τ)f 2 τ ). t 2 t (S(t τ) f x 2 τ )(x) dτ = f (x, t). Using the definition of the operator S(t) in the formula (7.19) for the solution of Cauchy problem for nonhomogeneous heat equation, we get the following expression for solution. u(x, t) = 1 x y 2 exp φ(y) d y 4πt 4t t 1 + 4π(t τ) x y 2 exp f (y,τ) d y dτ. (7.2) 4(t τ) Remark 7.7. Note that w(x, t;τ) := S(t τ) f τ solves the following problem and satisfies the conditions Thus the integral w t w x x =, x, t > τ (7.21a) w(x,τ;τ) = f (x,τ) for x. (7.21b) t in the Duhamel s formula has the form t (S(t τ) f τ )(x) dτ w(x, t;τ) dτ 7.2 Initial boundary value problem for Heat equation Consider the Initial Boundary Value Problem (IBVP) for heat equation given by u t = ku xx, < x < π, < t < T (7.22a) u(, t) = g 1 (t), t T (7.22b) u(π, t) = g 3 (t), t T (7.22c) u(x,) = g 2 (x), x π (7.22d) IIT Bombay

7.2. Initial boundary value problem for Heat equation 193 where k >. Definition 7.8 (solution to IBVP). Let be the the rectangle (,π) (,T ). Let C H denote the collection of all functions φ : such that the functions φ, φ x, φ xx, φ t belong to the space C (). A function v C H is said to be a solution of the IBVP (7.22) if v satisfies all the equations (7.22a) - (7.22d) Since the equation (7.22a) is linear and homogeneous, principle of superposition holds for its solutions. Thus a solution to the IBVP (7.22) may be obtained by a superposition of three solutions of IBVP, where each one of these solves IBVP with exactly one of the functions g 1, g 2, g 3 being non-zero. We illustrate how to obtain a solution of the IBVP (7.22) where g 1 = g 3 =, and other two cases are left as an exercise. Thus, we consider the Initial Boundary Value Problem for heat equation given by u t = ku xx, < x < π, < t < T, (7.23a) u(, t) =, t T, (7.23b) u(π, t) =, t T, (7.23c) u(x,) = f (x), x π, (7.23d) where k >. We show the existence of solutions to the IBVP (7.23) using the method of separation of variables. Substituting u(x, t) = X (x)t (t) in the equation (7.23a), and re-arranging the terms, we get X (x) X (x) = T (t) kt (t) (7.24) Since the LHS and RHS of the equation (7.24) are functions of the variables x and t respectively, each of them must be a constant function. Thus we have, for some constant λ, From the equation (7.25), we get the following two ODEs X (x) X (x) = T (t) kt (t) = λ (7.25) X (x) λx (x) =, T (t) λt (t) =. (7.26a) (7.26b) Since we are interested in finding a non-trivial solution, the boundary conditions (7.23b)- (7.23c) give rise to the following conditions on the function X : X () =, X (π) =, (7.27) The ODE (7.26a) has non-trivial solutions satisfying the boundary conditions (7.27) if and only if λ = n 2 for some n. Thus λ n = n 2 are eigenvalues and the corresponding eigenfunctions are given by X n (x) = sin nx. October 21, 215

194 Chapter 7. Heat Equation For each n, the solution of ODE (7.26b) with λ = λ n = n 2 (upto a constant multiple) is given by T n (t) = e n2 k t. Since for each n, the function u n (x, y) = X n (x)t n (t) is a solution (7.23a), we propose a formal solution of (7.23a) by superposition of the sequence of solutions (u n ) as follows: u(x, t) = b n e n2kt sin nx. (7.28) n=1 The coefficients b n in the formula (7.28) will be determined using the condition (7.23d). Thus we get f (x) = b n sin nx. (7.29) Thus b n are the Fourier sine coefficients of f, which are given by b n = 2 π π n=1 Thus the formal solution of (7.23) is given by u(x, t) n=1 7.2.1 Uniqueness for IBVP via energy method f (s)sin ns d s. (7.3) 2 π f (s)sin ns d s e n2kt sin nx. (7.31) π In this subsection, we establish uniqueness of solutions to most general IBVP for nonhomogeneous heat equation by energy considerations. Multiplying the equation (7.23a) with u and integrating over the interval [,π] gives π u u t d x = k π Note that the LHS of the equation (7.32) can be written as d π 1 d t 2 u2 d x = π u u xx d x. (7.32) uu t d x, and integrating by parts on RHS of the equation (7.32) yields π uu x x d x = π u x u x d x + uu x π. Thus the energy integral E(t) = 1 π 2 u2 (x, t) d x is non-increasing if kuu π x. For example if u satisfies u(, t) = and u x (π, t) =, we get d E(t). As a consequence, d t we get E(t) E() for t >. IIT Bombay

7.3. Maximum principle for Heat equation and its consequences 195 Using energy method, we can establish that solutions to an IBVP for heat equation are unique. Consider the IBVP where k >, f, g 1, g 2, g 3 are known functions. u t = ku x x + f (x, t), < x < π, < t (7.33a) u(, t) = g 1 (t), t (7.33b) u(π, t) = g 3 (t), t (7.33c) u(x,) = g 2 (x), x π (7.33d) Theorem 7.9. Let u and v be solutions to the IBVP (7.33) on the strip := (,π) (, ). Then u = v on. Proof. Define w := u v. Then w solves homogeneous heat equation with zero initialboundary conditions on the strip. Note that E(t) is a non-increasing function, and thus E(t) E(). But E() =. Thus E(t) = and hence w(x, t) = on. 7.3 Maximum principle for Heat equation and its consequences We begin this section with the notion of parabolic boundary which plays an important role in the study of maximum principle for heat equation. Definition 7.1 (Parabolic Boundary). Let be the rectangle (,π) (,T ). The boundary of the rectangle is the union of lines L i (i = 1,2,3,4) where L 1 = {(, t) : t T }, L 2 = {(x,) : x π}, L 3 = {(π, t) : t T }, L 4 = {(x,t ) : x π}. The parabolic boundary of the rectangle, which is denoted by P, is defined by P = L 1 L 2 L 3. The following maximum principle holds for solutions of heat equation. Theorem 7.11. Let u C H be a solution of the heat equation. Then the maximum value of u on is achieved on the parabolic boundary P. Proof. Since u C (), the maximum value of u is attained somewhere in. We would like to show that this maximum is also attained on the parabolic boundary P. Let M and m be defined by M := max u m := max u (7.34) P Clearly m M. The proof of the theorem will be complete if we prove that m < M is not possible. October 21, 215

196 Chapter 7. Heat Equation t (,T ) L 4 (π,t ) L 1 L 3. x (,) L 2 (π,) Figure 7.1. Rectangle for Heat equation Let L 4 denote L 4 without the end-points, i.e., L 4 = L 4 \ {(,T ), (π,t )}. Assume that m < M holds. Let (x 1, t 1 ) L 4 be such that u(x 1, t 1 ) = M. Define the function v : by For (x, t) P, we have v(x, t) = u(x, t) + M m 4π 2 (x x 1 ) 2 (7.35) v(x, t) m + M m π 2 = m + M m < M (7.36) 4π 2 4 Further, v(x 1, t 1 ) = u(x 1, t 1 ) = M. Thus the function v assumes its maximum value, namely M, on L 4. Let (x 2, t 2 ) L 4 be such that v(x 2, t 2 ) = M. Note that < x 2 < π. Note that (a) if (x 2, t 2 ), then we must have v t (x 2, t 2 ) =, and (b) if (x 2, t 2 ) L 4, then we must have v t (x 2, t 2 ). Thus we have v t (x 2, t 2 ). In view of the relations v t (x 2, t 2 ) = u t (x 2, t 2 ) = ku x x (x 2, t 2 ) = k we get v xx (x 2, t 2 ) M m 2π 2 v t (x 2, t 2 ) < kv x x (x 2, t 2 ). (7.37) However, v xx (x 2, t 2 ) since v attains a maximum at (x 2, t 2 ), which contradicts (7.37). Thus m < M is not possible, which completes the proof of the theorem., IIT Bombay

7.3. Maximum principle for Heat equation and its consequences 197 Remark 7.12. Note that the maximum principle proved here is the weak maximum principle. A strong maximum principle which is similar to that of laplace equation also holds for heat equation, and a proof may be found in [12]. Corollary 7.13. Let u C H be a solution of the heat equation. Then the minimum value of u on is achieved on the parabolic boundary P. Proof. The assertion of the corollary follows by applying Theorem 7.11 to the function v := u. Corollary 7.14 (uniqueness of solutions to IBVP). The initial boundary value problem (7.22) for the heat equation has at most one solution. Proof. Let u 1, u 2 be solutions of the initial boundary value problem (7.22). We need to show that u 1 = u 2. Consider w defined by w := u 1 u 2. Then w satisfies the following IBVP w t = kw x x, < x < π, < t < T w(, t) =, w(π, t) =, w(x,) =, t T t T x π By Theorem 7.11, and Corollary 7.13, we conclude that w attains both its maximum and minimum on the parabolic boundary, but w = there. Hence w. Theorem 7.15. Consider the IBVP (7.23), where f is such that the fourier series of f converges uniformly to f. Then the formal solution given by (7.31) is indeed a solution of the IBVP (7.23). Proof. Let us now check the validity of the formal solution given by (7.31). (i) The series in (7.31) converges uniformly for (x, t) [, π] (, T ], when f is integrable on the interval [,π]. For, b n 2 π π f (s) sin ns d s 2 π π f (s) d s c <, (7.39) b n e n2 kt sin nx ce n2 kt ce n2 kt (7.4) Since the series n=1 e n2 kt is convergent (follows from ratio test), we conclude that the series in (7.31) converges uniformly for (x, t) [,π] [t,t ], and hence u is a continuous function. Since t is arbitrary, we conclude that u is continuous on [,π] (,T ]. (ii) The derivatives u t, u x, u x x exist as continuous functions on [,π] (,T ]. This follows from the fact that the series in (7.31) can be differentiated term-by-term once w.r.t. t and twice w.r.t. x. Since proofs of assertions regarding u x, u xx are on similar lines, we present the proof for the case of u t. Note that the series ( n 2 k)b n e n2kt sin nx (7.41) n=1 October 21, 215

198 Chapter 7. Heat Equation is uniformly convergent for (x, t) [,π] [t,t ], which follows from ratio test, the convergence of the series n=1 n 2 ke n2 k t, and the inequalities b n n 2 ke n2 kt sin nx cn 2 e n2 k t ce n2 k t. (7.42) Thus u t is a continuous function on [,π] [t,t ]. Since t > is arbitrary, it follows that u t is a continuous function on [,π] (,T ]. Thus it follows that u satisfies heat equation on the domain [,π] (,T ]. It remains to show that u satisfies the initial-boundary conditions (7.23b)-(7.23d). In order to show that u is continuous on [,π] [,T ], we show that the sequence of partial sums of the series in (7.31) is uniformly Cauchy on [,π] [,T ]. Let the N th partial sum be denoted by S N (x, t) which is given by S N (x, t) = N b n e n2 k t sin nx. (7.43) For m l, let w l,m be defined by w l,m (x, t) = S m (x, t) S l (x, t). Thus w l,m (x, t) = n=1 m n=l +1 Note that w l,m is a solution of (7.23), and satisfies w l,m (x,) = b n e n2 k t sin nx. (7.44) m n=l +1 Applying maximum principle to the functions w l,m we get b n sin nx. (7.45) w l,m (x, t) w l,m (x,) (7.46) Note that m n=l +1 b n sin nx is a uniformly Cauchy sequence, since the Fourier sine series for f is assumed to converge uniformly to f on the interval [,π]. Thus the function u is continuous on [,π] [,T ], and satisfies the initial condition u(x,) = f (x). Remark 7.16. Note from the above proof that the series (7.31) was proved to be differentiable w.r.t. t by proving that series resulting from term-by-term differentiation of (7.31) is uniformly convergent, which followed from the presence of exponentially decaying term e n2 kt. By a similar argument, it follows that the function defined by the series (7.31) is infinitely differentiable w.r.t. x and t. Thus a solution of heat equation belongs to the space C (), even when u(x,) = f is not. This is described as the regularizing (smoothing) effect of heat equation. IIT Bombay

Exercises 199 General Exercises 7.1. Show that the heat equation u t u xx = under the change of coordinates z = ax and τ = a 2 t takes the form w τ w z z =. 7.2. [23] Goal of this exercise is to prove Weierstrass s approximation theorem using formula (7.15). Let f C [a, b]. (i) Extend the function to φ : by f (a) if x < a, φ(x) = f (x) if a x b, f (b) if x > b. Note that φ is continuous and bounded on. (ii) Consider Cauchy problem for homogeneous heat equation with Cauchy data φ as defined in (i) above. Write down the formula (7.15). Show that u(x, t) f (x) uniformly for a x b, as t. (iii) Approximate K 1 (x, y, t) by its truncated power series w.r.t. x y, and conclude Weierstrass approximation theorem. (iv) Further show that any f C m () and each of its derivatives upto order m can be approximated by polynomials uniformly in any bounded set containing the support of u. 7.3. [4] Solve the heat equation with constant dissipation: u t u xx + b u = for x, u(x,) = φ(x), where b > is a constant. (Hint: Try a solution having the form u(x, t) = e b t v(x, t)). 7.4. [4] If u(x, t) = f (x at) is a solution of homogeneous heat equation, then find f and show that the speed a is arbitrary. 7.5. [4] Let u be a solution of 2u t = u x x. Let v be defined by v(x, t) = 1 x 2 x exp u t 2t t, 1. t Show that v satisfies the backward heat equation 2v t = v x x for t >. IBVP 7.6. Solve the following initial boundary value problem: u t = k u x x x >, t >, u(, t) = 1 t, u(x,) = x. (Ans: u(x, t) = for x > t and u(x, t) = 1 for < x < t.) October 21, 215

2 Chapter 7. Heat Equation 7.7. [4] Consider heat equation posed for x (, l ) with Robin boundary conditions u x (, t) a u(, t) = and u x (l, t)+a l u(l, t) =, where a > and a l > are constants. Use energy method to show that the endpoints contribute to the decrease of l u2 (x, t) d x. These boundary conditions are called radiating, dissipative boundary conditions as energy is lost at the boundary. Maximum principles 7.8. [16] The initial boundary value problem for Heat equation is well-posed. Existence was proved by separation of variables method and uniqueness using maximum principle. Now prove the stability part of the well-posedness. 7.9. State and prove a maximum principle for solutions of an IBVP for u t = k u, where is the laplacian in d. 7.1. Consider the following initial boundary value problem: u t = ku x x, < x < π, < t < T u(, t) = g(t), u(π, t) = g(t), u(x,) = f (x), t T t T x π (a) Using maximum principle, show that the above initial boundary value problem has a unique solution. (b) Using maximum principle, show that the solutions to IBVP are stable in the sense that for each ε >, there exists a δ > such that for functions f 1, f 2, g 1, g 2, h 1, h 2 satisfying max f 1 (x) f 2 (x) < δ, max g 1 (t) g 2 (t) < δ, max h 1 (t) h 2 (t) < δ, x [,π] t [,T ] t [,T ] the corresponding solutions u 1, u 2 of the IBVP corresponding to the data (f, g, h) = ( f 1, g 1, h 1 ) and (f, g, h) = ( f 2, g 2, h 2 ) respectively satisfy where := (,π) (,T ). max u 1 (x, t) u 2 (x, t) ε, 7.11. [4] Let u be a solution of u t = u xx, x l, t <. (a) Let M (T ) be the maximum of the function u on the rectangle {(x, t) : x l, t T }. Is M (T ) (strictly) decreasing or increasing as a function of T? (b) Let m(t ) be the minimum of the function u on the rectangle {(x, t) : x l, t T }. Is m(t ) (strictly) decreasing or increasing as a function of T? 7.12. [4] The aim of this exercise is to show that the maximum principle does not hold for the equation u t = x u x x which has a variable coefficient. (a) Verify that the function u(x, t) = 2xt x 2 is a solution. Find its maximum on the rectangle {(x, t) : 2 x 2, t 1}. (b) Where exactly does our proof of the maximum principle fail in the case of this equation? IIT Bombay

Exercises 21 7.13. [33] For T >, let be defined by = {(x, t) : < x < π, < t < T }. Let u be a solution to the problem u t u xx = on, u(, t) = u(π, t) = for t T, u(x,) = sin 2 x for x π. Using maximum principle, show that u(x, t) e t sin x holds for (x, t). October 21, 215