1 Partial derivatives and Chain rule

Transcription

1 Math 62 Partial derivatives and Chain rule Question : Let u be a solution to the PDE t u(x, t) + 2 xu 2 (x, t) ν xx u(x, t) =, x (, + ), t >. (a) Let ψ(x, t) = x tu(ξ, t)dξ + 2 u2 (x, t) ν x u(x, t). Compute x ψ(x, t). The definition of ψ implies that x ψ(x, t) = x ( x t u(ξ, t)dx + 2 u2 (x, t) ν x u(x, t)) = t u(x, t)x + 2 xu 2 (x, t) xx u(x, t) = i.e., x ψ(x, t) =. This means that ψ depends on t only. (b) Let φ(x, t) := e x 2ν u(ξ,t)dξ. Compute t φ, x φ, and xx φ. The definition of φ, together with the chain rule, implies that ( t φ(x, t) = t x ) u(ξ, t)dξ 2ν ( = x ) t u(ξ, t)dξ 2ν e x 2ν u(ξ,t)dξ e x 2ν u(ξ,t)dξ and and xx φ(x, t) = ( x φ(x, t) = x x 2ν ( = u(x, t) 2ν ) ) u(ξ, t)dξ e x 2ν u(ξ,t)dξ e x 2ν u(ξ,t)dξ ( ) 2ν xu(x, t) e x 2ν u(ξ,t)dξ + ( ) 2 u(x, t) e x 2ν u(ξ,t)dξ 2ν (c) Compute t φ ν xx φ, assuming ψ(x, t) =. The above computations give In conclusion ν xx φ(x, t) = 2ν t φ ν xx φ = 2ν This means t φ ν xx φ =. ( x = 2ν ψ(x, t)e 2ν ( ν x u(x, t) + 2 u2 (x, t) t u(ξ, t)dξ + 2 u2 (x, t) ν x u(x, t) x u(ξ,t)dξ. ) e x 2ν u(ξ,t)dξ ) e x 2ν u(ξ,t)dξ Question 2: Let d be positive integer and let u : R d R + R be the solution to the nonlinear PDE t u(x, t) a u(x, t) + b( u(x, t)) ( u(x, t)) =, with u(x, ) = f(x), where a > and b are real numbers, and f(x) is a smooth function. Let v(x, t) = e b a u(x,t). (i) Compute v(x, t) and ( v(x, t)) (Hint: compute xi v(x, t) for i =,..., d, then d i= x ix i v(x, t). It is the same question as question in mdt; it is worded differently.) The chain rule gives xi v(x, t) = b a e b a u(x,t) xi u(x, t),

2 Math 62 2 and d xix i v(x, t) = b a e b a u(x,t) i= This implies that d i= xix i u(x, t) + b2 a 2 e b a u(x,t) v(x, t) = b a e b a u(x,t) u(x, t), d xi u(x, t) xi u(x, t). i= and v(x, t) = b a e b a u(x,t) u(x, t) + b2 a 2 e b a u(x,t) ( u(x, t)) ( u(x, t)). (ii) Compute t v(x, t), t v(x, t) a v(x, t), and v(x, ). Using again the chain rule we obtain Combining the above results gives t v(x, t) = b a e b a u(x,t) t u(x, t). t v(x, t) a v(x, t) = b a e b a u(x,t) ( t u(x, t) a u(x, t) + b( u(x, t)) ( u(x, t))) = i.e., v solves the heat equation t v(x, t) a v(x, t) = with initial condition v(x, ) = e b a f(x). (iii) Recall that the solution to the heat equation t φ(x, t) a φ(x, t) = with initial condition φ(x, ) = ψ (x) is given by ψ (x)(4πat) d/2 e x y 2 R d 4at dy. Give the integral representation of u(x, t) in terms of a, b, and f. Since t v(x, t) a v(x, t) = with initial condition v(x, ) = e b a f(x), the integral representation of the function u is then given by u(x, t) = a b log(v(x, t)), with, v(x, t) = e b a f(x) (4πat) d/2 4at dy e x y 2 R d Question 3: Let u be a vector field in R d, (d is the space dimension). Let p be a scalar field in R d. (a) Using the product rule, express div(pu) in terms of u and p. (You may use divu or u to denote the divergence of u and p or gradp to denote the gradient of p. The space dimension is d.) Using the product rule we have (pu) = In conclusion (pu) = u p + p u. d i (pu i ) = i= = u p + p u d u i i p + p i u i (b) Let u be a smooth vector field in R d with zero divergence and zero normal component at the boundary of. Let p be a smooth scalar field in. Compute p(x) u(x)dx and u(x) p(x)dx. (i) Using (a) and the fact that u is divergence free, we have u pdx = ( (pu) p u)dx = (pu)dx. The divergence theorem (fundamental theorem of calculus in d space dimension) implies that u pdx = pu nds, i=

3 Math 62 3 where denotes the boundary of. u pdx =. (ii) Note finally that p udx = since u =. Since u n = at the boundary, we finally infer that Question 4: Let φ be a smooth scalar field in R d, (d is the space dimension). (a) Prove that φ i ( φ) = φ i φ. Give all the details. Using the product rule, we have φ i ( φ) = d j φ i ( j φ) = i ( 2 j= (b) Let ɛ >. Show that ( φ 2 + ɛe i ) = the direction i. Using the chain rule, we have d ( j φ) 2 ) = i ( 2 φ 2 ) = φ i φ. j= φ +ɛ φ i( φ) where e i is the unit vector in 2 ( φ 2 + ɛe i ) = i ( φ 2 + ɛ) = 2 φ 2 + ɛ i( φ 2 + ɛ) = φ φ 2 + ɛ i φ. Then using (a) we infer that ( φ 2 + ɛe i ) = φ φ 2 + ɛ i( φ). (c) Assume that φ(x) = for all x R where R is a positive real number. φ R +ɛ d φ i( φ)dx, for all i =,..., d. Give all the details. 2 Compute Using (b) we have φ φ 2 + ɛ i( φ)dx = ( φ 2 + ɛe i )dx, R d R d where e i is the unit vector in the direction i. The divergence theorem together with the assumption that φ(x) = for all x R implies that φ φ 2 + ɛ i( φ)dx =. R d Question 5: Let y(x, t) = x cos(2t + log( x )). Compute tt y + x 2 xx y x x y + 6y. This exercise is meant to check whether you understand the notion of partial derivatives and the chain rule tt y(x, t) = 4x cos(2t + log( x )) = 4y(x, t), x y(x, t) = cos(2t + log( x )) x sin(2t + log( x )) = cos(2t + log( x )) sin(2t + log( x )), x xx y(x, t) = x sin(2t + log( x )) cos(2t + log( x )) x In conclusion tt y + x 2 xx y x x y + 6y = 4x cos(2t + log( x )) x sin(2t + log( x )) x cos(2t + log( x )) x cos(2t + log( x )) + x sin(2t + log( x )) + 6x cos(2t + log( x )) =, that is to say, y(x, t) solve the PDE tt y + x 2 y x x y + 6y =.

4 Math 62 4 Question 6: Let be the curl operator acting on vector fields: i.e., let A = (A, A 2, A 3 ) : R 3 R 3 be a three-dimensional vector field over R 3, then A = ( 2 A 3 3 A 2, 3 A A 3, A 2 2 A ). Accept as a fact that (A B) = B A A B for all smooth vector fields A and B. Let be a subset of R 3 with a smooth boundary. Find an integration by parts formula for B Adx. Using the divergence Theorem we infer that (B A A B)dx = (A B) = which implies that B Adx = (A B) nds. A Bdx + (A B) nds. Question 7: Let u, f : R R be two functions of class C. (a) Compute x f(u(x)). Using the chain rule we obtain where f denotes the derive of f. x f(u(x)) = f (u(x)) x u. (b) Let ψ : R R be functions of class C. Let F : R R be defined by F (v) = v f (t)ψ (t)dt. Use (a) to compute x (F (u(x)) x (f(u(x)))ψ (u(x)). Using the chain rule we obtain x (F (u(x)) = F (u(x)) x u(x) = f (u(x))ψ (u(x)) x u(x) = x (f(u(x)))ψ (u(x)). This means that x (F (u(x)) = x (f(u(x)))ψ (u(x)). (c) Using the notation of (a) and (b), assume that u(± ) = and compute x(f(u(x)))ψ (u(x))dx. Using (b) and u(± ) = we have x (f(u(x)))ψ (u(x))dx = x (F (u(x)))dx = F (u(x)) + = F () F () =. Question 8: Let φ(x, y) = cos(x + sin(x y)). Compute x φ(x, y) and y φ(x, y). (Do not try to simplify the results). We apply the chain rule repeatedly x φ(x, y) = sin(x + sin(x y))( + cos(x y)). y φ(x, y) = sin(x + sin(x y))( cos(x y)). Question 9: Let φ = x 4 y 4 (a) Compute φ(x, y). (b) Consider the square = [, ] [, ] and let Γ be the boundary of. Compute Γ nφdγ. (a) The definition φ = xx φ + yy φ implies that φ = xx φ + yy φ = 2x 2 2y 2 = 2(x 2 y 2 ). (b)the definition φ = div( φ) and the fundamental theorem of calculus (also known as the divergence theorem) imply that n φdγ = n φdγ = div( φ)d = φd = 2(x 2 y 2 )dxdy =. Γ Γ

5 Math 62 5 Question : Let φ = sin(x) sin(y) (a) Compute φ(x, y). (b) Consider the square = [, ] [, ] and let Γ be the boundary of. Compute Γ nφdγ. (a) The definition φ = xx φ + yy φ implies that φ = xx φ + yy φ = sin(x) + sin(y). (b)the definition φ = div( φ) and the fundamental theorem of calculus (also known as the divergence theorem) imply that n φdγ = n φdγ = div( φ)d = φd = sin(x)dxdy + sin(x)dxdy =. Γ Γ Question : Let φ(x, y) = log(2 + sin(x y)). Compute x φ(x, y) and y φ(x, y). (Do not try to simplify the results). We apply the chain rule repeatedly x φ(x, y) = cos(x y) 2 + sin(x y) y φ(x, y) = cos(x y). 2 + sin(x y) Question 2: Let φ(x, y) = sin(x) log(+y 2 +z 2 )dz. Given α, β, γ R, compute x φ(α, α+β) and y φ(α β, γ). (Do not try to simplify the results). Recalling the fundamental theorem of calculus ( t ) t f(z)dz = f(t), we apply the chain rule repeatedly This means that Recalling that we apply the chain rule repeatedly This means that x φ(x, y) = log( + y 2 + sin(x) 2 ) cos(x) x φ(α, α + β) = log( + (α + β) 2 + sin(α) 2 ) cos(α) v t f(s, t)ds = u y φ(x, y) = y φ(α β, γ) = sin(x) v u sin(α β) t f(s, t)ds, 2y + y 2 + z 2 dz 2γ + γ 2 + z 2 dz. Question 3: Let φ = x 2 + 2y 2 (a) Compute φ(x, y). (b) Consider the disk of radius centered at (, ) and let Γ be the boundary of. Compute Γ nφdγ. (a) The definition φ = xx φ + yy φ implies that φ = xx φ + yy φ = = 6. (b)the definition φ = div( φ) and the fundamental theorem of calculus (also known as the divergence theorem) imply that n φdγ = n φdγ = div( φ)d = φd = 6 d = 6π, Γ Γ because the surface of, d, is equal to π.

6 Math Heat equation Question 4: Let u be the solution of t u = xx u + 2, x (, L), with x u(, t) = α, x u(l, t) = 3, u(x, ) = f(x). (a) Let α =. Compute u(x, t)dx as a function of t. d t u(x, t)dx = xu L L + 2L. That is d t u(x, t)dx = 2 + 2L. This implies L u(x, t)dx = (2 + 2L)t + f(x)dx. (b) Let α be an arbitrary number. For which value of α, u(x, t)dx does not depend on t? The above computation yields u(x, t)dx = (3 α + 2L)t + f(x)dx. This is independent of t if (3 α + 2L) =, meaning α = 3 + 2L. Question 5: Let u solve t u xx u = 5, x (, L), with x u(, t) = α, x u(l, t) = 3, u(x, ) = f(x). (a) Let α =. Compute u(x, t)dx as a function of t. Integrate the equation over (, L): That is d t d t u(x, t)dx = t u(x, t)dx = xx udx + 5L = x u L + 5L. u(x, t)dx = 4 + 5L. This implies u(x, t)dx = ( 4 + 5L)t + f(x)dx. (b) Let α be an arbitrary number. For which value of α, u(x, t)dx does not depend on t? The above computation yields u(x, t)dx = ( 3 α + 5L)t + f(x)dx. This is independent of t if ( 3 α + 5L) =, meaning α = 3 + 5L. Question 6: Let u solve t u x ((2x+) x u) = 3, x (, L), with x u(, t) = α, x u(l, t) =, u(x, ) = f(x). (a) Let α =. Compute u(x, t)dx as a function of t. Integrate the equation over the domain (, L): That is d t d t u(x, t)dx = t u(x, t)dx = x ((2x + ) x u)dx + 3L = (2L + ) x u(l, t) x u(, t) + 3L = (2L + ) α + 3L = L α u(x, t)dx = L 2. This implies u(x, t)dx = (L 2)t + f(x)dx. (b) Let α be an arbitrary number. For which value of α, u(x, t)dx does not depend on t? The above computation yields u(x, t)dx = (L α)t + f(x)dx. This is independent of t if L α =, meaning α = L. Question 7: Let u solve t u x ((3x+) x u) = 3, x (, L), with x u(, t) =, x u(l, t) = α, u(x, ) = f(x). (a) Compute u(x, t)dx as a function of t. Integrate the equation over the domain (, L): d t u(x, t)dx = t u(x, t)dx = x ((3x + ) x u)dx 3L = (3L + ) x u(l, t) x u(, t) 3L = (3L + )α 3L = (3L + )(α ) That is d t u(x, t)dx = (3L+)(α ). This implies u(x, t)dx = (3L+)(α )t+ f(x)dx.

7 Math 62 7 (b) For which value of α the quantity u(x, t)dx does not depend on t? The above computation yields u(x, t)dx = (3L + )(α )t + f(x)dx. This is independent of t if (3L + )(α ) =, meaning α =. Question 8: Consider the differential equation d2 φ dt = λφ, t (, π), supplemented with the 2 boundary conditions φ() =, 3φ(π) = φ (π). (a) Prove that it is necessary that λ be positive for a non-zero solution to exist. (i) Let φ be a non-zero solution to the problem. Multiply the equation by φ and integrate over the domain. π Using the BCs, we infer (φ (t)) 2 dt φ (π)φ(π) + φ ()φ() = λ π (φ (t)) 2 dt + 3φ(π) 2 = λ which means that λ is non-negative since φ is non-zero. π π φ 2 (t)dt, φ 2 (t)dt. (ii) If λ =, then π (φ (t)) 2 dt = and φ(π) 2 =, which implies that φ (t) = and φ(π) =. The fundamental theorem of calculus implies φ(t) = φ(π) + t π φ (τ)dτ =. Hence, φ is zero if λ =. Since we want a nonzero solution, this implies that λ cannot be zero. (iii) In conclusion, it is necessary that λ be positive for a nonzero solution to exist. (b) Find the equation that λ must solve for the above problem to have a nonzero solution. Since λ is positive, φ is of the following form φ(t) = c cos( λt) + c 2 sin( λt). The boundary condition φ() = implies c =. The other boundary condition φ (π) = 3φ(π) implies λc 2 cos( λπ) = 3c 2 sin( λπ). The constant c 2 cannot be zero since we want φ to be nonzero; as a result, λ must solve the following equation λ cos( λπ) + 3 sin( λπ) =, for a nonzero solution φ to exist. Question 9: Consider the Laplace equation u = in the rectangle {(x, y); x [, L], y [, H]} with the boundary conditions u(, y) =, u(l, y) = 3 cos( 5 2 π y H ), yu(x, ) =, u(x, H) =. Solve the equation using the method of separation of variables. (Give all the details.) Let u(x) = φ(x)ψ(y). Then, provided ψ and φ are non zero functions, this implies φ (x) φ(x) = ψ (y) ψ(y) = λ. Observe that ψ () = and ψ(h) =. The energy technique applied to ψ (y) = λψ(y) gives H ψ (y)ψ(y)dy = H H ψ (y) 2 dy ψ (H)ψ(H) + ψ ()ψ() = λ ψ(y) 2 y, which implies H ψ (y) 2 dy = λ H ψ(y)2 y since ψ () = and ψ(h) =. This in turn implies that λ is nonnegative. Actually λ cannot be zero since it would mean that ψ =, which would contradict the fact that the solution u is nonzero (λ = ψ (y) = ψ(y) = ψ(h) = for all y [, H]). As a result λ is positive and ψ(y) = a cos( λy) + b sin( λy). The Neumann condition at y = gives b =. The Dirichlet condition at H implies cos( λh) =, which implies λh = (n + 2 )π, where n is any integer. This means that ψ(y) = a cos((n + 2 )π y H ). The fact that λ is positive implies φ(x) = c cosh( λx) + d sinh( λx). The boundary condition at x = implies c =. Then u(x, y) = A cos((n + 2 )π y H ) sinh((n + 2 )π x H ).

8 Math 62 8 The boundary condition at x = L gives which implies n = 2 and A = sinh ( 5 3 cos( 5 2 π y H ) = A cos((n + 2 )π y H ) sinh((n + 2 )π L H ), 2 πl H ), i.e., sinh u(x, y) = 3 sinh ( 5 2 πx H ( 5 2 πl H ) ) cos( 5 2 π y H ). Question 2: Let k : [, +] R be such that k(x) =, if x [, ] and k(x) = 2 if x (, ]. Solve the boundary value problem x (k(x) x T (x)) = with T ( ) = and x T () =. (i) What should be the interface conditions at x = for this problem to make sense? The function T and the flux k(x) x T (x) must be continuous at x =. Let T denote the solution on [, ] and T + the solution on [, +]. One should have T () = T + () and k () x T () = k + () x T + (), where k () = and k + () = 2. (ii) Solve the problem, i.e., find T (x), x [, +]. On [, ] we have k (x) =, which implies xx T (x) =. This in turn implies T (x) = a + bx. The Dirichlet boundary condition at x = implies T ( ) = = a b. This gives a = b and T (x) = a( + x). We proceed similarly on [, +] and we obtain T + (x) = c + dx. The Neumann boundary condition at x = + gives x T + (+) = = d and T + (x) = c + x. The interface conditions T () = T + () and k () x T () = k + () x T + () give a = c, and a = 2. In conclusion T (x) = { 2( + x) if x [, ], 2 + x if x [, +]. Question 2: Let u solve t u x ((sin(x) + 2) x u) = g(x)e t, x (, L), with x u(, t) = sin(l) + 2, x u(l, t) = 2, u(x, ) = f(x), where f and g are two smooth functions. (a) Compute d dt u(x, t)dx as a function of t. Integrate the equation over the domain (, L) and apply the fundamental Theorem of calculus: That is d dt u(x, t)dx = t u(x, t)dx = x ((sin(x) + 2) x u)dx + e t g(x)dx = (sin(l) + 2) x u(l) (sin() + 2) x u() + e t g(x)dx = (sin(l) + 2)2 2(sin(L) + 2) + e t g(x)dx = e t g(x)dx. d dt u(x, t)dx = e t g(x)dx. (b) Use (a) to compute u(x, t)dx as a function of t.

9 Math 62 9 Applying the fundamental Theorem of calculus again gives u(x, T )dx = = T u(x, )dx + d dt f(x)dx + ( e T ) u(x, t)dxdt g(x)dx. (c) What is the limit of u(x, t)dx as t +? The above formula gives lim T + u(x, T )dx = f(x)dx + g(x)dx. ( Question 22: Let u solve t u + x v(x, t)u µ(x, t) x u ) = g(x)e t, x (, L), t >, with µ(, t) x u(, t) =, µ(l, t) x u(l, t) = + 2e t, u(x, ) = f(x), where v, µ >, f and g are four smooth functions and v() = v(l) =. (a) Compute d dt u(x, t)dx as a function of t. Integrate the equation over the domain (, L) and apply the Fundamental Theorem of calculus: d dt That is u(x, t)dx = t u(x, t)dx = x ( v(x, t)u + µ(x, t) x u)dx + e t g(x)dx = µ(l, t) x u(l) µ(, t) x u() + e t g(x)dx = + 2e t + e t g(x)dx. d dt u(x, t)dx = e t ( g(x)dx + 2). (b) Use (a) to compute u(x, t)dx as a function of t. Applying the Fundamental Theorem of calculus again (with respect to time this time) gives u(x, T )dx = = T u(x, )dx + e t dt( f(x)dx + ( e T )( g(x)dx + 2). g(x)dx + 2). (c) What is the limit of u(x, t)dx as t +? The above formula gives lim T + u(x, T )dx = f(x)dx + g(x)dx + 2. Question 23: Let u solve t u x (µ(x, t) x u) = g(x)e t, x (, L), t >, with µ(, t) x u(, t) =, µ(l, t) x u(l, t) = + 2e t, u(x, ) = f(x), where µ >, f and g are three smooth functions. (a) Compute d dt u(x, t)dx as a function of t.

10 Math 62 Integrate the equation over the domain (, L) and apply the fundamental Theorem of calculus: That is d dt u(x, t)dx = t u(x, t)dx = x (µ(x, t) x u)dx + e t g(x)dx = µ(l, t) x u(l) µ(, t) x u() + e t g(x)dx = + 2e t + e t g(x)dx. d dt u(x, t)dx = e t ( g(x)dx + 2). (b) Use (a) to compute u(x, t)dx as a function of t. Applying the fundamental Theorem of calculus again gives u(x, T )dx = = u(x, )dx + T e t dt( f(x)dx + ( e T )( g(x)dx + 2). g(x)dx + 2). (c) What is the limit of u(x, t)dx as t +? The above formula gives lim T + u(x, T )dx = f(x)dx + g(x)dx + 2. Question 24: Consider the heat equation t T k xx T = f(x), x [a, b], t >, with f(x) = kx, where k >. Compute the steady state solution (i.e., t T = ) assuming the boundary conditions: k n T (a) =, T (b) = ( n is the normal derivative) At steady state, T does not depend on t and we have xx T (x) = x, which implies x T (x) = α 2 x2, and T (x) = β + αx 6 x3, where α, β R. The two constants α and β are determined by the boundary conditions. = n T (a) = x T (a) = α 2 a2 and = T (b) = β + αb 6 b3. We conclude that α = 2 a2 and β = αb + 6 b3 = 2 a2 b + 6 b3. In conclusion T (x) = 2 a2 b + 6 b3 + 2 a2 x 6 x3 = 2 a2 (b x) + 6 (b3 x 3 ). Question 25: Consider the equation t c(x, t) xx c(x, t) = x, where x [, L], t >, with c(x, ) = f(x), n c(, t) = 6, n c(l, t) = 5, ( n is the normal derivative). Compute E(t) := c(ξ, t)dξ. We integrate the equation with respect to x over [, L] t c(ξ, t)dξ ξξ c(ξ, t)dξ = Using that T tc(ξ, t)dξ = d t c(ξ, t)dξ together with the fundamental theorem of calculus, we infer that d t E(t) x c(l, t) + x c(, t) = 2 L2. The boundary conditions x c(, t) = n c(, t) = 6, x c(l, t) = n c(l, t) = 5 give d t E(t) = 2 L2. ξdξ.

11 Math 62 We now apply the fundamental theorem of calculus with respect to t In conclusion E(t) E() = E(t) = t τ E(τ)dτ = ( 2 L2 )t. f(ξ)dξ + ( 2 L2 )t. Question 26: Consider the heat equation t u(x, t) 3 xx u(x, t) =, x u(, t) =, x u(, t) =, u(x, ) = u (x), t >, x (, ). The general solution is u(x, t) = n= A n cos(nπx)e 3n2 π 2t. Compute the solution corresponding to the initial data u (x) = 5 cos(4πx). The solution contains one term only, corresponding to n = 4, u(x, t) = 5 cos(4πx)e 48π2t. Question 27: Consider the heat equation t u(x, t) 2 xx u(x, t) =, u(, t) =, u(, t) =, u(x, ) = u (x), t >, x (, ). The general solution is u(x, t) = n= A n sin(nπx)e 2n2 π 2t. Compute the solution corresponding to the initial data u (x) = 3 sin(4πx) 5 sin(πx). The solution contains two terms only, corresponding to n = and n = 4, u(x, t) = 5 sin(πx)e 2π2t + 3 sin(4πx)e 32π2t. Question 28: Consider the equation t c(x, t) xx c(x, t) = 6x/L 2, where x [, L], t >, with c(x, ) = f(x), n c(, t) =, n c(l, t) = 2, ( n is the normal derivative). Compute E(t) := c(ξ, t)dξ. We integrate the equation with respect to x over [, L] t c(ξ, t)dξ ξξ c(ξ, t)dξ = 6 L 2 ξdξ. Using that T tc(ξ, t)dξ = d t c(ξ, t)dξ together with the fundamental theorem of calculus, we infer that d t E(t) x c(l, t) + x c(, t) = 3. The boundary conditions x c(, t) = n c(, t) =, x c(l, t) = n c(l, t) = 2 give d t E(t) = 3. We now apply the fundamental theorem of calculus with respect to t In conclusion E(t) E() = E(t) = t τ E(τ)dτ =. f(ξ)dξ, t. ( Question 29: Consider the equation t c(x, t) x (+x 2 ) x c(x, t) ) = 6x/L 2, where x [, L], t >, with c(x, ) = f(x), n c(, t) =, n c(l, t) = 2 +L, ( 2 n is the normal derivative). Compute E(t) := c(ξ, t)dξ. We integrate the equation with respect to x over [, L] t c(ξ, t)dξ ( ξ ( + ξ 2 ) ξ c(ξ, t) ) dξ = 6 L 2 ξdξ.

12 Math 62 2 Using that T tc(ξ, t)dξ = d t c(ξ, t)dξ together with the fundamental theorem of calculus, we infer that d t E(t) ( + L 2 ) x c(l, t) + x c(, t) = 3. The boundary conditions x c(, t) = n c(, t) =, x c(l, t) = n c(l, t) = 2 +L 2 give d t E(t) = 3. We now apply the fundamental theorem of calculus with respect to t In conclusion E(t) E() = E(t) = t τ E(τ)dτ =. f(ξ)dξ, t. Question 3: Consider the heat equation t u(x, t) 2 xx u(x, t) =, u(, t) =, u(, t) =, u(x, ) = u (x), t >, x (, ). The general solution is u(x, t) = n= A n sin(nπx)e 2n2 π 2t. Compute the solution corresponding to the initial data u (x) = 3 sin(4πx). The solution contains one term only, corresponding to n = 4, u(x, t) = 3 sin(4πx)e 32π2t. Question 3: Consider the heat equation t T k xx T = f(x), x [a, b], t >, with f(x) =, where k >. Compute the steady state solution (i.e., t T = ) assuming the boundary conditions: k n T (a) =, T (b) = ( n is the normal derivative). At steady state, T does not depend on t and we have xx T (x) =, which implies x T (x) = α, and T (x) = β + αx, where α, β R. The two constants α and β are determined by the boundary conditions. = k n T (a) = k x T (a) = kα and = T (b) = β + αb. We conclude that α = k and β = αb = b k. In conclusion T (x) = x b k.

13 Math Laplace equation Question 32: (a) Find a function U(x, y) = a + bx + cy + dxy, such that U(x, ) = x, U(, y) = + y, U(x, ) = 3x, and U(, y) = y. solves the problem. U(x, y) = x y + 2xy (b) Use (a) to solve the PDE u xx + u yy =, (x, y) (, ) (, ), with the boundary conditions u(x, ) = 3 sin(πx) + x, u(, y) = + y, u(x, ) = 3x,and u(, y) = sin(2πy) y. By setting φ = u U, we observe that φ xx + φ yy = and at the boundary of the domain we have It is clear that φ(x, ) = 3 sin(πx), φ(, y) =, φ(x, ) =, and φ(, y) = sin(2πy). sinh(π( y)) sinh(2π( x)) φ(x, y) = 3 sin(πx) + sin(2πy) sinh(π) sinh(2π) Then, u(x, y) = φ(x, y) + U(x, y) Question 33: Consider the Laplace equation u = in the rectangle x [, L], y [, H] with the boundary conditions u(, y) =, u(l, y) =, u(x, ) =, u(x, H) = f(x). (a) Is there any compatibility condition that f must satisfy for a smooth solution to exist? f must be such that f() = and f(l) =, otherwise u would not be continuous at the two upper corners of the domain. (b) Solve the Equation. Use the separation of variable technique. Let u(x) = φ(x)ψ(y). Then, provided ψ and φ are non zero functions, this implies φ φ = ψ ψ = λ. Observe that φ() = φ(l) =. The usual energy technique implies that λ is negative. That is to say φ(x) = a cos( λx) + b sin( λx). The boundary conditions imply a = and λl = nπ, i.e., φ(x) = b sin(nπx/l). The fact that λ is negative implies ψ(y) = c cosh( λy) + d sinh( λy). The boundary condition at y = implies c =. Then u(xy) = n= A n sin( nπx L ) sinh(nπy L ) Question 34: Solve the PDE (note that the width and the height of the rectangle are not equal) xx u + yy u =, < x <, < y < 2, u(x, ) = 8 sin(9πx), u(x, 2) =, < x <, u(, y) = sin(2πy), u(, y) =, < y < 2. The method of separation of variables tells us that the solution is a sum of terms like sin(nπx) sinh(nπ(y 2)) and sin(mπy/2) sinh(mπ(x )/2). By looking at the boundary conditions we infer that there are two nonzero terms in the expansion: one corresponding to n = 9 and one corresponding to m = 4. This gives sinh(9π(2 y)) sinh(2π( x)) u(x, y) = 8 sin(9πx) + sin(2πy) sinh(8π)) sinh(2π))

14 Math 62 4 Question 35: Consider the Laplace equation u = in the rectangle x [, L], y [, H] with the boundary conditions u(, y) =, u(l, y) =, u(x, ) =, u(x, H) = f(x). (a) Is there any compatibility condition that f must satisfy for a smooth solution to exist? f must be such that f() = and f(l) =, otherwise u would not be continuous at the two upper corners of the domain. (b) Solve the Equation. Use the separation of variable technique. Let u(x) = φ(x)ψ(y). Then, provided ψ and φ are non zero functions, this implies φ φ = ψ ψ = λ. Observe that φ() = φ(l) =. The usual technique implies that λ is negative. That is to say φ(x) = a cos( λx)+b sin( λx). The boundary conditions imply a = and λl = nπ, i.e., φ(x) = b sin(nπx/l). The fact that λ is negative implies ψ(y) = c cosh( λy) + d sinh( λy). The boundary condition at y = implies c =. Then the ansatz is and the usual computation gives u(x, y) = A n = n= 2 L sinh( nπh L ) A n sin( nπx L ) sinh(nπy L ), f(ξ) sin( nπξ L )dξ. Question 36: Consider the Laplace equation u = in the rectangle x [, L], y [, H] with the boundary conditions u(, y) =, x u(l, y) =, u(x, ) =, u(x, H) = sin( 3 2 πx/l). Solve the Equation using the method of separation of variables. (Give all the details.) Let u(x) = φ(x)ψ(y). Then, provided ψ and φ are non zero functions, this implies φ φ = ψ ψ = λ. Observe that φ() = and φ (L) =. The usual energy technique implies that λ is negative. That is to say φ(x) = a cos( λx) + b sin( λx). The Dirichlet condition at x = implies a =. The Neumann condition at L implies cos( λl) =, which implies λl = (n+ 2 )π, where n is an integer. This means that φ(x) = b sin((n + 2 )πx/l). The fact that λ is negative implies ψ(y) = c cosh( λy) + d sinh( λy). The boundary condition at y = implies c =. Then The boundary condition at y = H gives which implies n = and A = sinh ( 3 u(x, y) = A sin((n + 2 )π x L ) sinh((n + 2 )π y L ). sin( 3 2 π x L ) = A sin((n + 2 )π x L ) sinh((n + 2 )π H L ), 2 πh L sinh u(x, y) = sinh ), i.e., ( 3 2 πy L ( 3 2 πh L ) ) sin( 3 2 πx/l) Question 37: Consider the equation u = in the rectangle {(x, y); x [, L], y [, H]} with the boundary conditions u(, y) =, u(l, y) = 5 cos( 3 2 π y H ), yu(x, ) =, u(x, H) =. Solve the equation using the method of separation of variables. (Give all the details.) Let u(x) = φ(x)ψ(y). Then, provided ψ and φ are non zero functions, this implies φ (x) φ(x) = ψ (y) ψ(y) = λ. Observe that ψ () = and ψ(h) =. The energy technique applied to ψ (y) = λψ(y) gives H H H ψ (y)ψ(y)dy = ψ (y) 2 dy ψ (H)ψ(H) + ψ ()ψ() = λ ψ(y) 2 y, which implies H ψ (y) 2 dy = λ H ψ(y)2 dy since ψ () = and ψ(h) =. This in turn implies that λ is nonnegative. Actually λ cannot be zero since it would mean that ψ =, which would

15 Math 62 5 contradict the fact that the solution u is nonzero (λ = ψ (y) = ψ(y) = ψ(h) = for all y [, H]). As a result λ is positive and ψ(y) = a cos( λy) + b sin( λy). The Neumann condition at y = gives b =. The Dirichlet condition at H implies cos( λh) =, which implies λh = (n+ 2 )π, where n is any integer. This means that ψ(y) = a cos((n+ 2 )π y H ). The fact that λ is positive implies φ(x) = c cosh( λx) + d sinh( λx). The boundary condition at x = implies c =. Then The boundary condition at x = L gives u(x, y) = A cos((n + 2 )π y H ) sinh((n + 2 )π x H ). 5 cos( 3 2 π y H ) = A cos((n + 2 )π y H ) sinh((n + 2 )π L H ), which, by identification, implies = 2 and A = 5 sinh ( 3 sinh u(x, y) = 5 sinh ( 3 2 πx H ( 3 2 πl H ) 2 πl H ), i.e., ) cos( 3 2 π y H ). Question 38: Does any of the following expressions solve the Laplace equation inside the rectangle x L, y H, with the following boundary conditions x u(, y) = 5π 5πy 5πL H sin( H ) cosh( H ), xu(l, y) =, u(x, ) =, u(x, H) =? (justify clearly your answer): u (x, y) = 3 cos( 5πy H u 3 (x, y) = 3 cos( 5πy H ) cosh(5π(x L) H ) sinh(5π(x L) H ), u 2 (x, y) = 3 sin( 5πy H ), u 4 (x, y) = 3 sin( 5πy H L) ) cosh(5π(x ), H L) ) sinh(5π(x ). H From class, we know that all the above expressions solve the Laplace equation, hence we just need to verify the boundary conditions. We observe that u and u 3 do not satisfy the Dirichlet boundary conditions u(x, ) =, u(x, H) = ; therefore u and u 3 must be discarded. Both u 2 and u 4 satify that Dirichlet conditions: u 2 (x, ) =, u 2 (x, H) =, and u 4 (x, ) =, u 4 (x, H) =. Now we need to check the Neumann conditions. Note that u 4 is such that x u 4 (L, y) = 3 5π H well. sin( 5πy H ) cosh() ; a result u 4 must be discarded as Finally u 2 is such that x u 2 (L, y) = 3 5π 5πy H sin( H ) sinh() =, but xu 2 (, y) = 3 5π H which shows that u 2 is not the solution to our problem either. In conclusion, none of the proposed solutions solve the problem. The correct solution is 5πL cosh( H u(x, y) = 3 ) ) sin(5πy H sinh( 5πL H L) ) cosh(5π(x ). H 5πy 5πL sin( H ) sinh( H ), Question 39: Does any of the following expressions solve the Laplace equation inside the rectangle x L, y H, with the following boundary conditions u(, y) =, u(l, y) = 3 sinh( 5πL H ) sin( 5πy H ), u(x, ) =, u(x, H) =? (justify clearly your answer): u (x, y) = 3 cos( 5πy H ) cosh(5π(x L) H u 3 (x, y) = 3 cos( 5πy H ) sinh(5πx H ), ), u 2 (x, y) = 3 sin( 5πy H u 4(x, y) = 3 sin( 5πy H ) sinh(5πx H ). L) ) cosh(5π(x ), H

16 Math 62 6 From class, we know that all the above expressions solve the Laplace equation, hence we just need to verify that the boundary conditions are met. We observe that u and u 3 do not satisfy the Dirichlet boundary conditions u(x, ) =, u(x, H) = ; therefore u and u 3 must be discarded. Both u 2 and u 4 satisfy that Dirichlet conditions: u 2 (x, ) =, u 2 (x, H) =, and u 4 (x, ) =, u 4 (x, H) =. Now we need to check the Neumann conditions. Note that u 2 is such that u 2 (, y) = 3 sin( 5πy 5π( L) H ) cosh( H ), which shows that u 2 is not the solution to our problem either. Finally u 4 (, y) = 3 sin( 5πy H ) sinh() = and u 4(L, y) = 3 sin( 5πy 5πL H ) sinh( H ); a result u 4 is the solution. 3. Cylindrical coordinates Question 4: Using cylindrical coordinates and the method of separation of variables, solve the Laplace equation, r r(r r u) + r 2 θθ u =, inside the domain D = {θ [, π 2 ], r [, ]}, subject to the boundary conditions θ u(r, ) =, u(r, π 2 ) =, u(, θ) = cos(3θ). We set u(r, θ) = φ(θ)g(r). This means φ = λφ, with φ () = and φ( π 2 ) =, and rd r (rd r g(r)) = λg(r). Then using integration by parts plus the boundary conditions we prove that λ is non-negative. Then φ(θ) = c cos( λθ) + c 2 sin( λθ). The boundary condition φ () = implies c 2 =. The boundary condition φ( π 2 ) = implies λ π 2 = (2n + ) π 2 with n N. This means λ = (2n + ). From class we know that g(r) is of the form r α, α. The equality rd r (rd r r α ) = λr α gives α 2 = λ. The condition α implies 2n + = α. The boundary condition at r = gives cos(3θ) = 2n+ cos((2n + )θ). This implies n =. The solution to the problem is u(r, θ) = r 3 cos(3θ). Question 4: Using cylindrical coordinates and the method of separation of variables, solve the Laplace equation, r r(r r u) + r 2 θθ u =, inside the domain D = {θ [, π 2 ], r [, ]}, subject to the boundary conditions u(r, ) =, u(r, π 2 ) =, u(, θ) = sin(2θ). (Give all the details.) We set u(r, θ) = φ(θ)g(r). This means φ = λφ, with φ() = and φ( π 2 ) =, and rd r (rd r g(r)) = λg(r). The usual energy argument applied to the two-point boundary value problem φ = λφ, φ() =, φ( π 2 ) =, implies that λ is non-negative. If λ =, then φ(θ) = c + c 2 θ and the boundary conditions imply c = c 2 =, i.e., φ =, which in turns gives u = and this solution is incompatible with the boundary condition u(, θ) = sin(2θ). Hence λ > and φ(θ) = c cos( λθ) + c 2 sin( λθ). The boundary condition φ() = implies c =. The boundary condition φ( π 2 ) = implies λ π 2 = nπ with n N. This means λ = 2n. From class we know that g(r) is of the form r α, α. The equality rd r (rd r r α ) = λr α gives α 2 = λ. The condition α implies 2n = α. The boundary condition at r = gives sin(2θ) = 2n sin(2nθ) for all θ [, π 2 ]. This implies n =. The solution to the problem is u(r, θ) = r 2 sin(2θ). Question 42: Using cylindrical coordinates and the method of separation of variables, solve the equation, r r(r r u) + r 2 θθ u =, inside the domain D = {θ [, π], r [, ]}, subject to the boundary conditions u(r, ) =, u(r, π) =, u(, θ) = 2 sin(5θ). (Give all the details.)

17 Math 62 7 () We set u(r, θ) = φ(θ)g(r). This means φ = λφ, with φ() = and φ(π) =, and r d dr (r d dr g(r)) = λg(r). (2) The usual energy argument applied to the two-point boundary value problem φ = λφ, φ() =, φ(π) =, implies that λ is non-negative. If λ =, then φ(θ) = c + c 2 θ and the boundary conditions imply c = c 2 =, i.e., φ =, which in turns gives u = and this solution is incompatible with the boundary condition u(, θ) = 2 sin(5θ). Hence λ > and φ(θ) = c cos( λθ) + c 2 sin( λθ). (3) The boundary condition φ() = implies c =. The boundary condition φ(π) = implies λπ = nπ with n N \ {}. This means λ = n, n =, 2,.... (4) From class we know that g(r) is of the form r α, α. The equality r d dr (r d dr rα ) = λr α gives α 2 = λ. The condition α implies n = α. The boundary condition at r = gives 2 sin(5θ) = c 2 n sin(nθ) for all θ [, π]. This implies n = 5 and c 2 = 2. (5) Finally, the solution to the problem is u(r, θ) = 2r 5 sin(5θ). Question 43: Using cylindrical coordinates and the method of separation of variables, solve the equation, r r(r r u) + r 2 θθ u =, inside the domain D = {θ [, 3 2π], r [, 3]}, subject to the boundary conditions u(r, ) =, u(r, 3 2π) =, u(3, θ) = 8 sin(2θ). (Give all the details.) () We set u(r, θ) = φ(θ)g(r). This means φ = λφ, with φ() = and φ( 3 2π) =, and r d dr (r d dr g(r)) = λg(r). (2) The usual energy argument applied to the two-point boundary value problem φ = λφ, φ() =, φ( 3 π) =, 2 implies that λ is non-negative. If λ =, then φ(θ) = c + c 2 θ and the boundary conditions imply c = c 2 =, i.e., φ =, which in turns gives u = and this solution is incompatible with the boundary condition u(3, θ) = 8 sin(2θ). Hence λ > and φ(θ) = c cos( λθ) + c 2 sin( λθ). (3) The boundary condition φ() = implies c =. The boundary condition φ( 3 2π) = implies λ 3 2 π = nπ with n N \ {}. This means λ = 2 3n, n =, 2,.... (4) From class we know that g(r) is of the form r α, α. The equality r d dr (r d dr rα ) = λr α gives α 2 = λ. The condition α implies 2 3 n = α = λ. The boundary condition at r = 3 gives 8 sin(2θ) = c n sin( 2 3 nθ) for all θ [, 3 2 π]. This implies n = 3 and c 2 = 2. (5) Finally, the solution to the problem is u(r, θ) = 2r 2 sin(2θ). Question 44: Using cylindrical coordinates and the method of separation of variables, solve the equation, r r(r r u) + r 2 θθ u =, inside the domain D = {θ [, 3 2π], r [, 3]}, subject to the boundary conditions θ u(r, ) =, u(r, 3 2π) =, u(3, θ) = 9 cos(θ). (Give all the details of all the steps.) () We set u(r, θ) = φ(θ)g(r). This means φ = λφ, with φ () = and φ( 3 2π) =, and r d dr (r d dr g(r)) = λg(r).

18 Math 62 8 (2) The usual energy argument applied to the two-point boundary value problem φ = λφ, φ () =, φ( 3 π) =, 2 implies that λ is non-negative. If λ =, then φ(θ) = c + c 2 θ and the boundary conditions imply c = c 2 =, i.e., φ =, which in turns gives u = and this solution is incompatible with the boundary condition u(3, θ) = 9 sin(2θ). Hence λ > and φ(θ) = c cos( λθ) + c 2 sin( λθ). (3) The boundary condition φ () = implies c 2 =. The boundary condition φ( 3 2π) = implies that cos( λ 3 2 π) =, i.e., λ 3 2 π = (2n + ) π 2 with n N. This means λ = 3 (2n + ), n =,, 2,.... (4) From class we know that g(r) is of the form r α, α. The equality r d dr (r d dr rα ) = λr α gives α 2 = λ. The condition α implies 3 (2n + ) = α = λ. The boundary condition at r = 3 gives 9 cos(θ) = c 3 3 (2n+) cos( 3 (2n + )θ) for all θ [, 3 2 π]. This implies n = and c = 3. (5) Finally, the solution to the problem is u(r, θ) = 3r cos(θ). Question 45: The solution of the equation, r r(r r u) + r 2 θθ u =, inside the domain D = {θ [, π], r [, 2]}, subject to the boundary conditions u(r, ) =, u(r, π) =, u(2, θ) = g(θ) is u(r, θ) = n= b nr n sin(nθ). What is the solution corresponding to g(θ) = 5 sin(2θ) + 2 sin(5θ)? (Give all the details.) The only non-zero terms in the expansion are a 2 r 2 sin(2θ) + b 5 r 5 sin(5θ). The boundary condition u(2, θ) = 5 sin(2θ) + 2 sin(5θ) = a sin(2θ) + a sin(5θ) is satisfied if a 2 = 5/(2 2 ) and b 5 = 2/(2 5 ), i.e., u(r, θ) = 5 r2 r5 sin(2θ) sin(5θ). Question 46: The solution of the equation, r r(r r u) + r 2 θθ u =, inside the domain D = {θ [, π 2 ], r [, 2]}, subject to the boundary conditions θu(r, ) =, u(r, π 2 ) =, u(2, θ) = g(θ) is u(r, θ) = n= a 2n+r (2n+) cos((2n+)θ). What is the solution corresponding to g(θ) = 5 cos(3θ) + 2 cos(5θ)? (Give all the details.) The only non-zero terms in the expansion are a 3 r 3 cos(3θ)+a 5 r 5 cos(5θ). The boundary condition u(2, θ) = 5 cos(3θ) + 2 cos(5θ) = a cos(3θ) + a cos(5θ) is satisfied if a 3 = 5/(2 3 ) and a 5 = 2/(2 5 ), i.e., u(r, θ) = 5 r3 r5 cos(3θ) cos(5θ). Question 47: The solution of the equation, r r(r r u) + r 2 θθ u =, inside the domain D = {θ [, π 2 ], r [, 3]}, subject to the boundary conditions θu(r, ) =, u(r, π 2 ) =, u(3, θ) = g(θ) is u(r, θ) = n= a 2n+r (2n+) cos((2n+)θ). What is the solution corresponding to g(θ) = 5 cos(θ) + 2 cos(3θ)? (Give all the details.) The only non-zero terms in the expansion are a r cos(θ) + a 3 r 3 cos(3θ). The boundary condition u(3, θ) = 5 cos(θ) + 2 cos(3θ) = a 3 cos(3θ) + a cos(5θ) is satisfied if a = 5/3 and a 3 = 2/(3 3 ), i.e., u(r, θ) = 5 r 3 cos(θ) + 2 r3 3 3 cos(3θ).

19 Math Variable coefficients Question 48: Let k : [, +] R be such that k(x) =, if x [, ] and k(x) = 2 if x (, ]. Solve the boundary value problem x (k(x) x T (x)) = with x T ( ) = T ( ) and x T () =. (i) What should be the interface conditions at x = for this problem to make sense? The function T and the flux k(x) x T (x) must be continuous at x =. Let T denote the solution on [, ] and T + the solution on [, +]. One should have T () = T + () and k () x T () = k + () x T + (), where k () = and k + () = 2. (ii) Solve the problem, i.e., find T (x), x [, +]. On [, ] we have k (x) =, which implies xx T (x) =. This in turn implies T (x) = a + bx. The Robin boundary condition at x = implies x T ( ) T ( ) = = 2b a. This gives a = 2b and T (x) = b(2 + x). We proceed similarly on [, +] and we obtain T + (x) = c + dx. The Neumann boundary condition at x = + gives x T + (+) = = d and T + (x) = c + x. The interface conditions T () = T + () and k () x T () = k + () x T + () give 2b = c, and b = 2. In conclusion T (x) = { 2(2 + x) if x [, ], 4 + x if x [, +]. Question 49: Let k : [, +] R be such that k(x) = 6, if x [, ] and k(x) = 3 if x (, ]. Solve the boundary value problem x (k(x) x T (x)) = with 6 x T ( ) = T ( ) + 3 and T () = 5. (i) What should be the interface conditions at x = for this problem to make sense? The function T and the flux k(x) x T (x) must be continuous at x =. Let T denote the solution on [, ] and T + the solution on [, +]. One should have T () = T + () and k () x T () = k + () x T + (), where k () = 6 and k + () = 3. (ii) Solve the problem, i.e., find T (x), x [, +]. On [, ] we have k (x) =, which implies xx T (x) =. This in turn implies T (x) = a + bx. The Robin boundary condition at x = implies 6 x T ( ) T ( ) = 3 = 6b (a b). This gives a = 7b 3 and T (x) = 7b 3 + bx. We proceed similarly on [, +] and we obtain T + (x) = c + dx. The Dirichlet boundary condition at x = + gives T + () = 5 = d + c. This implies c = 5 d and T + (x) = 5 d + dx. The interface conditions T () = T + () and k () x T () = k + () x T + () give 7b 3 = 5 d, and 6b = 3d. This implies d = 4 and b = 2. In conclusion { 2x + if x [, ], T (x) = 4x + if x [, +]. Question 5: Let k : [, +] R be such that k(x) = 2, if x [, ] and k(x) = 3 if x (, ]. Solve the boundary value problem x (k(x) x T (x)) = with x T ( ) = T ( ) + 3 and x T () = T () 7. (i) What should be the interface conditions at x = for this problem to make sense? The function T and the flux k(x) x T (x) must be continuous at x =. Let T denote the solution on [, ] and T + the solution on [, +]. One should have T () = T + () and k () x T () = k + () x T + (), where k () = 2 and k + () = 3.

20 Math 62 2 (ii) Solve the problem, i.e., find T (x), x [, +]. On [, ] we have k (x) =, which implies xx T (x) =. This in turn implies T (x) = a + bx. The Robin boundary condition at x = implies x T ( ) T ( ) = 3 = 2b a. This gives a = 2b 3 and T (x) = 2b 3 + bx. We proceed similarly on [, +] and we obtain T + (x) = c + dx. The Robin boundary condition at x = + gives x T + (+) T + () = 7 = 2d c. This implies c = 2d + 7 and T + (x) = 2d dx. The interface conditions T () = T + () and k () x T () = k + () x T + () give 2b 3 = 2d + 7, and 2b = 3d. This implies d = 2 and b = 3. In conclusion { 3x + 3 if x [, ], T (x) = 2x + 3 if x [, +]. Question 5: Let k, f : [, +] R be such that k(x) = 2, f(x) = if x [, ] and k(x) =, f(x) = 2 if x (, ]. Consider the boundary value problem x (k(x) x T (x)) = f(x) with T ( ) = and T () = 3. (a) What should be the interface conditions at x = for this problem to make sense? The function T and the flux k(x) x T (x) must be continuous at x =. Let T denote the solution on [, ] and T + the solution on [, +]. One should have T () = T + () and k () x T () = k + () x T + (), where k () = 2 and k + () =. Question 52: Let k, f : [, +] R be such that k(x) = 2 + x, f(x) = if x [, ] and k(x) = + 2x, f(x) = 2 if x (, ]. Consider the boundary value problem x (k(x) x T (x)) = f(x) with T ( ) = 5 and T () =. (a) What should be the interface conditions at x = for this problem to make sense? The function T and the flux k(x) x T (x) must be continuous at x =. Let T denote the solution on [, ] and T + the solution on [, +]. One should have T () = T + () and k () x T () = k + () x T + (), where k () = 2 and k + () =, i.e., 2 x T () = x T + (). Question 53: Let k, f : [, +] R be such that k(x) = 2, f(x) = if x [, ] and k(x) =, f(x) = 2 if x (, ]. Consider the boundary value problem x (k(x) x T (x)) = f(x) with T ( ) = 2 and T () = 2. (a) What should be the interface conditions at x = for this problem to make sense? The function T and the flux k(x) x T (x) must be continuous at x =. Let T denote the solution on [, ] and T + the solution on [, +]. One should have T () = T + () and k () x T () = k + () x T + (), where k () = 2 and k + () =. (b) Solve the problem, i.e., find T (x), x [, +]. On [, ] we have k (x) = 2 and f (x) = which implies xx T (x) =. This in turn implies T (x) = ax + b. The Dirichlet condition at x = implies that T ( ) = 2 = a + b. This gives a = b + 2 and T (x) = (b + 2)x + b. We proceed similarly on [, +] and we obtain xx T (x) = 2, which implies that T + (x) = x 2 + cx + d. The Dirichlet condition at x = implies T + () = 2 = + c + d. This gives c = 3 d and T (x) = x 2 + (3 d)x + d. The interface conditions T () = T + () and k () x T () = k + () x T + () give b = d and 2(b + 2) = 3 d, respectively. In conclusion b = 3, d = 3 and { 5 T (x) = 3 x 3 if x [, ], x x 3 if x [, ]. Question 54: Let k, f : [, +] R be such that k(x) = 2, f(x) = if x [, ] and k(x) =, f(x) = 2 if x (, ]. Consider the boundary value problem x (k(x) x T (x)) = f(x)

21 Math 62 2 with T ( ) = and T () = 2. (a) What should be the interface conditions at x = for this problem to make sense? The function T and the flux k(x) x T (x) must be continuous at x =. Let T denote the solution on [, ] and T + the solution on [, +]. One should have T () = T + () and k () x T () = k + () x T + (), where k () = 2 and k + () =. (b) Solve the problem, i.e., find T (x), x [, +]. Give all the details. On [, ] we have k (x) = 2 and f (x) = which implies xx T (x) =. This in turn implies T (x) = ax+b. The Dirichlet condition at x = implies that T ( ) = = a+b. This gives a = b and T (x) = bx + b. We proceed similarly on [, +] and we obtain xx T (x) = 2, which implies that T + (x) = x 2 + cx + d. The Dirichlet condition at x = implies T + () = 2 = + c + d. This gives c = 3 d and T (x) = x 2 + (3 d)x + d. The interface conditions T () = T + () and k () x T () = k + () x T + () give b = d and 2b = 3 d, respectively. In conclusion b =, d = and { x + if x [, ], T (x) = x 2 + 2x + if x [, ]. Question 55: Let k, f : [, +] R be such that k(x) = 3, f(x) = 6 if x [, ] and k(x) =, f(x) = 2 if x (, ]. Consider the boundary value problem x (k(x) x T (x)) = f(x) with T ( ) = and x T () =. (a) What should be the interface conditions at x = for this problem to make sense? The function T and the flux k(x) x T (x) must be continuous at x =. Let T denote the solution on [, ] and T + the solution on [, +]. One should have T () = T + () and k () x T () = k + () x T + (), where k () = 3 and k + () =. (b) Solve the problem, i.e., find T (x), x [, +]. Give all the details. On [, ] we have k (x) = 3 and f (x) = 6 which implies 3 xx T (x) = 6. This in turn implies T (x) = x 2 + ax + b. The Dirichlet condition at x = implies that T ( ) = = a + b. This gives a = b and T (x) = x 2 + bx + b. We proceed similarly on [, +] and we obtain xx T (x) = 2, which implies that T + (x) = x 2 + cx + d. The Neumann condition at x = implies T + () = = 2 + c. This gives c = 3 and T (x) = x 2 + 3x + d. The interface conditions T () = T + () and k () x T () = k + () x T + () give b = d and 3b = 3, respectively. In conclusion b =, d = and { x 2 + x + if x [, ], T (x) = x 2 + 3x + if x [, ]. 3.3 Maximum principle Question 56: Consider the square D = (, +) (, +). Let f(x, y) = x 2 y 2 3. Let u C 2 (D) C (D) solve u = in D and u D = f. Compute min (x,y) D u(x, y) and max (x,y) D u(x, y). We use the maximum principle (u is harmonic and has the required regularity). Then min u(x, y) = (x,y) D min f(x, y), and max u(x, y) = (x,y) D (x,y) D max f(x, y). (x,y) D A point (x, y) is at the boundary of D if and only if x 2 = and y (, ) or y 2 = and x (, ). In the first case, x 2 = and y (, ), we have f(x, y) = y 2 3, y (, ).

22 Math The maximum is 2 and the minimum is 3. In the second case, y 2 = and x (, ), we have f(x, y) = x 2 3, x (, ). The maximum is 3 and the minimum is 4. We finally can conclude min f(x, y) = min (x,y) D x x2 4, = 4, max f(x, y) = max 2 (x,y) D y y2 = 2. In conclusion min u(x, y) = 4, (x,y) D max u(x, y) = 2 (x,y) D Question 57: (i) Let be an open connected set in R 2. Let u be a real-valued nonconstant function continuous on and harmonic on. Assume that there exists x in such u(x ) =. Do we have a minimum, a maximum, or a saddle point at x? (explain) The Maximum principle implies that u cannot be either minimum or maximum at x. This point is a saddle point. (ii) Let = (, ) (note that = [, ]), and let u : R be such that u(x) = for all x, u() =, and u() =. Is u harmonic on? Find a point in where u reaches its maximum? Does this example contradict the Maximum Principle? (explain) Yes u is harmonic on since u (x) = for all x in. Note however that u is not continuous on ; as a consequence, the hypotheses for the Maximum principle are not satisfied. In other words, the above example does not contradict the Maximum principle. Question 58: Let be an open connected set in R 2. Let u be a real-valued nonconstant function continuous on and harmonic on and of class C 2 in. Assume that there exists x in such u(x ) =. Is the point x a minimum, a maximum, or a saddle point? (explain) The keyword here is that is open, meaning that the points at the boundary of are not in (it is not possible to punch a hole around the boundary points). The point x is in, that is x is not at the boundary. Since u is continuous on, harmonic on and of class C 2 in, the maximum principle can be applied. The Maximum principle implies that u cannot have either a minimum or a maximum at x. This means that x is a saddle point. Question 59: Let u be a continuous function on D where D is some open, connected set in R 2. Explain why, if u is harmonic, it is generally a waste of time to locate a point where u achieves its maximum by solving x u = and y u = simultaneously. From the Maximum Principle, we know that if u is not constant, the maximum of u is achieved at the boundary of D. The zero gradient condition does not apply for maximums at the boundary. Question 6: Let D be the open disk of radius 2 centered at (, 2). Let u C 2 (D) C (D) be a harmonic function in the disk D. Assume that u(x, y) = (x + y) 2 on the boundary of disk. Compute min (x,y) D u(x, y) and max (x,y) D u(x, y). You can give a geometric answer. Since u is in C 2 (D) C (D) and is harmonic, we can apply the maximum principle (Theorem). This theorem says that the maximum and minimum of u are attained at the boundary of the disk. The problem then amounts to finding the maximum and minimum of (x + y) 2 over the circle of radius 2 centered at (, 2). The iso-values of the function (x + y) 2 are parallel lines of slope. These iso-line are perpendicular to the gradient of (x + y) 2 which is the vector (2(x + y), 2(x + y)) = 2(x + y)(, ). We must find the two tangent lines to the circle that are perpendicular to the vector (, ). One easily verify that (, ) and (2, 3) are the tangent points (verify that the segment connecting these two points is parallel to the vector (, ) and passes through the center of the circle). As a result min u(x, y) = and max u(x, y) = 25. (x,y) D (x,y) D

23 Math The dashed lines are iso-lines of (x + y) 2. Question 6: Consider the disk D centered at (, ) of radius. Let f(x, y) = x 2 y 2 +4y 3. Let u C 2 (D) C (D) solve u = in D and u D = f. Compute min (x,y) D u(x, y) and max (x,y) D u(x, y). We use the maximum principle (u is harmonic and has the required regularity). Then min u(x, y) = (x,y) D min f(x, y), and max u(x, y) = (x,y) D (x,y) D max f(x, y). (x,y) D A point (x, y) is at the boundary of D if and only if x 2 + y 2 = ; as a result, the following holds for all (x, y) D: We obtain f(x, y) = x 2 y 2 + 4y 3 = y 2 y 2 + 4y 3 = 2( y) 2. min f(x, y) = min 2( (x,y) D y y)2 = 8, max f(x, y) = max 2( (x,y) D y y)2 =. In conclusion min u(x, y) = 8, (x,y) D max u(x, y) = (x,y) D Question 62: Consider the square D = (, +) (, +). Let f(x, y) = x 2 y 2 3. Let u C 2 (D) C (D) solve u = in D and u D = f. Compute min (x,y) D u(x, y) and max (x,y) D u(x, y). We use the maximum principle (u is harmonic and has the required regularity). Then min u(x, y) = (x,y) D min f(x, y), and max u(x, y) = (x,y) D (x,y) D max f(x, y). (x,y) D A point (x, y) is at the boundary of D if and only if x 2 = and y (, ) or y 2 = and x (, ). In the first case, x 2 = and y (, ), we have f(x, y) = y 2 3, y (, ).