MATH 220: MIDTERM OCTOBER 29, 2015
|
|
- Anne Stanley
- 6 years ago
- Views:
Transcription
1 MATH 22: MIDTERM OCTOBER 29, 25 This is a closed book, closed notes, no electronic devices exam. There are 5 problems. Solve Problems -3 and one of Problems 4 and 5. Write your solutions to problems and 2 in blue book #, and your solutions to problems 3, 4 and 5 in blue book #2. Within each book, you may solve the problems in any order. Total score: points. Problem. (2 points) Solve the PDE xu x + u y = u, u(x, ) = e x, on as large of a subset of R 2 containing the x-axis as you can. Do all characteristics intersect the x-axis, and do so in a unique point, non-tangentially? What does this mean for the solutions of the initial value problem of the PDE? Sketch the characteristics. Solution. The PDE is semilinear. The associated vector field in R 2 is V (x, y) = (x, ), and the surface on which we prescribe the function values is Γ = {(r, ) : r R}. It can be readily seen that the vector field is non-characteristic, i.e., at no point (x, y) Γ the vector V (x, y) is tangent to Γ at (x, y). We write Γ(r) = (Γ x (r), Γ y (r)) = (r, ). The equations for the projected characteristics are and ẋ r (s) = x r (s), x r () = r, ẏ r (s) =, y r () =, v r (s) = v r (s), v r () = e r, where v r (s) = u(x r (s), y r (s)). It follows that x r (s) = re s, y r (s) = s, v r (s) = e r e s. To check whether we can perform the change of variables (r, s) (x, y) note that [ ] [ ] r x det r (s) r y r (s) e s = det s x r (s) s y r (s) re s = e s, for any (r, s), which implies the inverse map exists locally. We now address the question whether the characteristics reach every point in the plane. This is equivalent to finding the inverse map explicitly. Fix (x, y). We need to find (r, s) such
2 2 MATH 22: MIDTERM OCTOBER 29, y y x x Figure. Vector field and characteristics for the ODE in Problem. that x = re s, y = s, from where we deduce s = y and it follows that x = re y, r = xe y, and s = y. Since this is valid for any (x, y) R 2 we conclude for every point there is a characteristic curve that goes through it (see Fig. for a sketch of the characteristics). We conclude u(x, y) = e xe y e y. Problem 2. (25 points) Consider the (real-valued) variable coefficient wave equation on [, l] x [, ) t with Dirichlet boundary conditions: u tt = (c(x, t) 2 u x ) x, u(, t) = = u(l, t). where c > is allowed to depend on x and t, and there are constants c, c 2 > such that c c(x) c 2 for all x. Assume throughout that u is C 2. Let E(t) = 2 (u t (x, t) 2 + c(x, t) 2 u x (x, t) 2 ) dx. (i) Show that if c is independent of t then E is constant. (ii) Show that even if c depends on t, but t c C 3, E(t) e αt E() where α = 2C 3 /c. (Hint: consider d dt (e αt E(t)).) (iii) Show the solution of the variable coefficient wave equation (under the conditions mentioned above) with given initial condition is unique. Solution: We break down the solution by parts. (i) First note E is well-defined for t > as the integral is over a bounded domain, and the integrand is C 2. Since the boundary condition implies
3 MATH 22: MIDTERM OCTOBER 29, 25 3 E (t) = d 2 dt u t (, t) = u t (l, t) = for all t >, exchanging the derivative with respect to t and the integral shows that = 2 = = =. (u t (x, t) 2 + c(x) 2 u x (x, t) 2 ) dx d dt (u t(x, t) 2 + c(x) 2 u x (x, t) 2 ) dx (u t (x, t)u tt (x, t) + c(x) 2 u x (x, t)u xt (x, t)) dx (u tt (x, t) (c(x) 2 u x (x, t)) x ) }{{} = as u solves the PDE u t (x, t) dx + c(x) 2 u x (x, t)u t (x, t) x=l x= }{{} = by the boundary conditions (ii) Since c is bounded above, the same arguments as in (i) show E is welldefined. Since the boundary condition implies u t (, t) = u t (l, t) = for all t >, exchanging the derivative with respect to t and the integral shows that E (t) = d 2 dt = 2 (u t (x, t) 2 + c(x, t) 2 u x (x, t) 2 ) dx d dt (u t(x, t) 2 + c(x, t) 2 u x (x, t) 2 ) dx = (u t (x, t)u tt (x, t) + c(x, t) 2 u x (x, t)u xt (x, t)) dx + }{{} = = by the same computation performed in (i) c t (x, t)c(x, t)u x (x, t) 2 dx. Following the hint, define Ẽ(t) = e αt E(t). Then Ẽ (t) = αe αt E(t) + e αt E (t) = αe αt 2 = αe αt 2 c t (x, t)c(x, t)u x (x, t) 2 dx (u t (x, t) 2 + c(x, t) 2 u x (x, t) 2 ) dx + e αt c t (x, t)c(x, t)u x (x, t) 2 dx (u t (x, t) 2 + ( 2α c t (x, t) + c(x, t))c(x, t)u x (x, t) 2 ) dx. Since we want to show Ẽ is decreasing, we need to show the right-hand side is non-positive, i.e., the integral on the right-hand side is non-negative. Using the value of α = 2C 3 /c we can write 2 α c t(x, t) + c(x, t) c(x, t) 2 C 3 α = c(x, t) c, from where we conclude Ẽ(t). In particular this implies Ẽ(t) Ẽ() = E(t) eαt E().
4 4 MATH 22: MIDTERM OCTOBER 29, 25 There is an alternative way to prove the desired result. Both ways are valid, but we include both for completeness. Note that E (t) = C 3 c c t (x, t)c(x, t)u x (x, t) 2 dx 2C 3 c 2 αe(t), c c(x, t)u x (x, t) 2 dx c(x, t) 2 u x (x, t) 2 dx from where it follows that d dt log (E(t)) α = E(t) eαt E(), as E(t). This proves the desired inequality. (iii) If we fix the initial conditions, e.g., u(x, ) = φ(x) and u t (x, ) = ψ(x) for x (, l), then if u and u 2 are two solutions to the PDE, then w = u u 2 is a solution to the homogeneous problem w tt = (c(x, t) 2 w x ) x, w(, t) = w(l, t) =, w(x, ) = w t (x, ) =. In this case E() = 2 (w t (x, ) 2 + c(x, t) 2 w x (x, ) 2 ) dx =, and by (ii) we conclude E(t). Since E(t) it follows that E(t) for t and we have = E(t) = 2 2 (w t (x, t) 2 + c(x, t) 2 w x (x, t) 2 ) dx w t (x, t) 2 dx + 2 c2 w x (x, t) 2 dx. which implies w t (x, t) = w x (x, t) = for (x, t) [, l] x [, ) t as c >. Therefore w is a constant which has to be equal to zero by either the initial or boundary conditions. We conclude u = u 2 and the solution to the PDE with given initial conditions is unique. Continue to the next page!
5 MATH 22: MIDTERM OCTOBER 29, 25 5 Problem 3. (i) (8 points) Find the general C 2 solution of the PDE u xx + 5u xt + 6u tt =. (ii) (8 points) State what it means that a distribution u solves this PDE, and give an example of a discontinuous solution. Justify your answer. (Hint: if u j are distributions solving the PDE, and u j u in distributions, what can you say about u?) (iii) (8 points) Solve the initial value problem with initial condition u(x, x) = φ(x), u t (x, x) = ψ(x), with φ, ψ given C 2 functions. (iv) (6 points) If φ and ψ are C except at, where can you say for sure that u is C? Explain the geometric meaning. Solution: We break down the solution by parts. (i) We first determine the type of second-order operator. Note that [ ] 5/2 det = /2 6 4 = 4 <, and the PDE is hyperbolic. We can factor it as the product of two firstorder operators as (a x + b t )(c x + d t ) = ac xx + (ad + bc) xt + bd tt, from where it follows that ac =, ad + bc = 5, bd = 6. Choosing a = we see that c =, d + b = 5, and bd = 6. We conclude b = 2 and d = 3 and thus xx + 5 xt + 6 tt = ( x + 2 t )( x + 3 t ). As we have seen throughout the class, the general form of the C 2 solution is u(x, t) = f(2x t) + g(3x t), for some C 2 functions f and g. (ii) If u D (R 2 ) solves the PDE then φ C c (R 2 ) : ( xx u + 5 xt u + 6 tt u)(φ) =. Using the fact that D (R 2 ) is a linear space, the definition of the distributional derivative, and the fact that distributions are linear maps, we have that ( xx u + 5 xt u + 6 tt u)(φ) = xx u(φ) + 5 xt u(φ) + 6 tt u(φ) = u( xx φ) + 5u( xt φ) + 6u( tt φ) = u( xx φ + 5 xt φ + 6 tt φ). We conclude u D (R 2 ) solves the PDE if φ C c (R 2 ) : u( xx φ + 5 xt φ + 6 tt φ) =,
6 6 MATH 22: MIDTERM OCTOBER 29, 25 where the second-order operator acting on the argument of u is the same as in the original PDE. We claim h(x, t) = H(2x t) is a discontinuous solution to the PDE in the sense of distributions. To justify this is the case, consider a sequence {f k } in C 2 (R) with f k H pointwise, and satisfying f k, f k (x) = for x <, and f k (x) = for x > /k, for all k. Define h k (x, t) = f k (2x t). Clearly ι hk ι h in the sense of distributions. In fact, for π Cc (R 2 ) we have ι hk (φ) ι h (φ) f k (2x t) H(2x t) φ(x, t) dxdt = f k (y) H(y) φ(x, 2x y) dxdy = 2, /k /k ( f k (y) H(y) φ(x, 2x y) dx ) φ(x, 2x y) dx dy as k. Furthermore, by definition of the distributional derivative (in particular the fact it is consistent) we have ( xx + 5 xt + 6 tt )ι hk = ( x + 3 t )( x + 2 t )ι hk = ( x + 3 t )ι ( x+2 t)h k = ( x + 3 t )ι =, and ι hk solves the PDE in the sense of distributions for any k. Therefore for any fixed φ Cc (R 2 ) we have = lim ( xx + 5 xt + 6 tt )ι hk (φ) k = lim ι h k (( xx + 5 xt + 6 tt )φ) k = ι h (( xx + 5 xt + 6 tt )φ) = ( xx + 5 xt + 6 tt )ι h (φ), which justifies why h(x, t) = H(2x t) is a solution in the sense of distributions. (iii) Since the general solution to the PDE is u(x, t) = f(2x t) + g(3x t), we only need to find f, g C 2 (R) such that u satisfies the desired conditions. Since u t (x, t) = f (2x t) g (3x t), we obtain the system of equations φ(x) = f(x) + g(2x), ψ(x) = f (x) g (2x),
7 MATH 22: MIDTERM OCTOBER 29, 25 7 and by differentiating the first equation we obtain the system φ (x) = f (x) + 2g (2x), ψ(x) = f (x) g (2x). Adding both equations leads to φ (x) + ψ(x) = g (2x) from where it follows that x ( ) x ( ) ( ) g(x) g() = φ 2 y dy+ ψ 2 y dy = 2φ 2 x 2 x 2φ()+2 ψ(y) dy. Adding two times the second to the first leads to φ (x)+2ψ(x) = f (x) and f(x) f() = x Consequently φ (y) dy 2 x u(x, t) = f()+g() φ() φ(2x t)+2φ ψ(y) dy = φ(x) + φ() 2 ( 3 2 x ) 3 2 t +2 Note that g() + f() = φ() and thus ( 3 u(x, t) = φ(2x t) + 2φ 2 x ) 2 t x 2 t 2 x 2 t 2x t x ψ(y) dy 2 ψ(y) dy. ψ(y) dy. 2x t (iv) If φ, ψ are C except at the origin, then we can say for sure that the solution is C except at points (x, t) for which either 2x t = or 3x t =. Geometrically, this means that singularities propagate along the characteristics of the first-order operators. Solve one of Problems 4 and 5. Problem 4. Consider the equation u t + u 2 u x =, u(x, ) = φ(x), t. (i) (6 points) State the definition of u being a weak solution of this PDE. (ii) ( points) Suppose c R, and u(x, t) = if x < ct and u(x, t) = if x > ct. For what value of c does this give a weak solution of the PDE? Derive this directly from the definition. (iii) (9 points) Suppose the initial condition is φ(x) =, x <, x, < x <,, x >. Will the solution develop a shock in t >? Why/why not? Solution: We break down the solutions by parts. (i) First note the PDE corresponds to a conservation law with f(u) = u 3 /3. We say u is a weak solution of the PDE if (uψ t + f(u)ψ x ) dxdt + φ(x)ψ(x, ) dx =, R x [, ) t R for all ψ C (R 2 ). ψ(y) dy.
8 8 MATH 22: MIDTERM OCTOBER 29, 25 (ii) We proceed in the same way as in Problem 4, Homework 2. Define ξ(t) = ct and Ω = {(x, t) : x < ξ(t)}, Ω + = {(x, t) : x > ξ(t)}, Γ = {(x, t) : x = ξ(t)}. By construction the restriction to u to Ω ± is constant. In particular { f(u) = 3 if (x, t) Ω, if (x, t) Ω +. Let φ C (R x (, ) t ). If u is a weak solution then = (uφ t + f(u)φ x ) dxdt ( = φ t + ) Ω 3 φ x dxdt = φν t ds + φν x ds Γ 3 Γ ( = ν t + ) 3 ν x φ ds, Γ where ν = (ν x, ν t ) is the outward-pointing unit normal vector to Ω. We can parameterize Γ using the function γ(s) = (cs, s). From this it follows that γ(s) = (c, ) and ν = (, c). Therefore ( = c + ) φ(cs, s) + c 3 2 ds ( = c + 3) + c 2 φ(cs, s) ds, for any such φ. This is only possible if c = /3. Note this coincides with the Rankine-Hugoniot condition. (iii) This is a quasilinear equation for which we impose initial conditions on the surface Γ = {(r, ) : r R}, and we parameterize it as Γ(r) = (r, ). We obtain the system of ODEs ṫ r (s) =, t r () =, ẋ r (s) = z r (s) 2, x r () = r, ż r (s) =, z r () = φ(r), from where it readily follows that t r (s) = s, z r (s) = φ(r), and thus x r (s) = sφ(r) 2 + r. Consequently x r (t) = tφ(r) 2 + r. We know the solution will develop a shock if the characteristics intersect. Let r, r 2 R and r r 2. Without loss we can assume r < r 2 and x r2 (t) x r (t) = r 2 r + t(φ(r 2 ) 2 φ(r ) 2 ) = (r 2 r ) ( + t φ(r 2) 2 φ(r ) 2 ). r 2 r
9 MATH 22: MIDTERM OCTOBER 29, t -5 5 x Figure 2. Characteristics for the ODE in Problem 4. However, note that φ(r 2 ) 2 φ(r ) 2 r 2 r = 2 r 2 r r2 r φ (r)φ(r) dr, as the integrand is non-negative. Therefore the characteristics do not intersect for any t > (see also Fig. 2 for a sketch of the characteristics). Alternatively, using x r (s) = sφ(r) 2 + r and t r (s) = s we see that [ ] [ ] r t det r (s) r x r (s) + 2sφ = det (r)φ(r) s t r (s) s x r (s) φ(r) 2 = + 2sφ (r)φ(r), Problem 5. if s > as φ (r)φ(r). (i) (5 points) State the definition of a function f being Schwartz: f S(R n ). (ii) (7 points) Show that for f Schwartz, the Fourier transform of D j f = i jf is ξ j (Ff)(ξ). (iii) (7 points) On R, use the Fourier transform (by taking the F.T. of both sides) to solve the ODE d dx u + u = 2 e x. You may leave your answer as the inverse Fourier transform of a function. You may use that the Fourier transform of f(x) = e ax2, a >, is (π/a) /2 e ξ2 /(4a). (iv) (6 points) Is the solution of the ODE obtained by the Fourier transform unique? How do you reconcile this with the fact that the ODE existence and uniqueness theorem guarantees at least local solvability and uniqueness to the initial value problem, where one also imposes u() = c, with c C arbitrary? (Hint: what are the solutions of the homogeneous ODE?) Solution: We break down the solutions by parts.
10 MATH 22: MIDTERM OCTOBER 29, 25 (i) A function f C (R n ) is in S(R n ) if sup x R n x α β f(x) <, for any multiindices α, β. (ii) Fix f S(R n ) and define g(x) = f(x)e ix ξ. Then D j g(x) = D j f(x)e ix ξ ξ j f(x)e ix ξ, is absolutely integrable. Then = D j g(x) dx = D j f(x)e ix ξ dx ξ j f(x)e ix ξ dx, from where we conclude (FD j f)(ξ) = ξ j (Ff)(ξ). (iii) By taking Fourier transform on both sides we have iξ(fu)(ξ) + (Fu)(ξ) = π /2 e ξ2 /4 (Fu)(ξ) = 2 + iξ e ξ /4, which shows Fu is in fact in S(R). Therefore we can take inverse Fourier transform to obtain u(x) = e ξ 2 /4 2π + iξ eixξ dξ, and conclude that u S(R). (iv) Using the Fourier transform we obtain a unique solution which is also Schwartz. Note this leaves no degrees of freedom to find a solution, not even locally, for a given initial condition. However, note that the solution of the ODE for given initial conditions is the sum of a solution to the homogeneous problem plus a particular solution to the inhomogeneous problem. Note the solutions to the homogeneous problem are du dx + u = u(x) = Ae x, for some A R. It can be seen that u is not Schwartz, as it does not decay as x. Therefore taking Fourier transform allows us only to find a particular solution to the inhomogeneous problem when the inhomogeneous term is a Schwartz function. To understand why this is the case, if we were to solve the homogeneous ODE we would have (iξ + )(Fu)(ξ) =, implying that (Fu)(ξ) = for all ξ.
MATH 173: PRACTICE MIDTERM SOLUTIONS
MATH 73: PACTICE MIDTEM SOLUTIONS This is a closed book, closed notes, no electronic devices exam. There are 5 problems. Solve all of them. Write your solutions to problems and in blue book #, and your
More informationMATH 220: Problem Set 3 Solutions
MATH 220: Problem Set 3 Solutions Problem 1. Let ψ C() be given by: 0, x < 1, 1 + x, 1 < x < 0, ψ(x) = 1 x, 0 < x < 1, 0, x > 1, so that it verifies ψ 0, ψ(x) = 0 if x 1 and ψ(x)dx = 1. Consider (ψ j )
More informationMATH 220: Problem Set 3 Solutions
MATH 220: Problem Set 3 Solutions Problem 1. Let ψ C() be given by: 0, < 1, 1 +, 1 < < 0, ψ() = 1, 0 < < 1, 0, > 1, so that it verifies ψ 0, ψ() = 0 if 1 and ψ()d = 1. Consider (ψ j ) j 1 constructed as
More informationMath Partial Differential Equations 1
Math 9 - Partial Differential Equations Homework 5 and Answers. The one-dimensional shallow water equations are h t + (hv) x, v t + ( v + h) x, or equivalently for classical solutions, h t + (hv) x, (hv)
More informationMath 342 Partial Differential Equations «Viktor Grigoryan
Math 342 Partial Differential Equations «Viktor Grigoryan 15 Heat with a source So far we considered homogeneous wave and heat equations and the associated initial value problems on the whole line, as
More informationMATH 425, FINAL EXAM SOLUTIONS
MATH 425, FINAL EXAM SOLUTIONS Each exercise is worth 50 points. Exercise. a The operator L is defined on smooth functions of (x, y by: Is the operator L linear? Prove your answer. L (u := arctan(xy u
More informationPropagation of discontinuities in solutions of First Order Partial Differential Equations
Propagation of discontinuities in solutions of First Order Partial Differential Equations Phoolan Prasad Department of Mathematics Indian Institute of Science, Bangalore 560 012 E-mail: prasad@math.iisc.ernet.in
More informationMATH 220: THE FOURIER TRANSFORM TEMPERED DISTRIBUTIONS. Beforehand we constructed distributions by taking the set Cc
MATH 220: THE FOURIER TRANSFORM TEMPERED DISTRIBUTIONS Beforehand we constructed distributions by taking the set Cc ( ) as the set of very nice functions, and defined distributions as continuous linear
More informationMATH 126 FINAL EXAM. Name:
MATH 126 FINAL EXAM Name: Exam policies: Closed book, closed notes, no external resources, individual work. Please write your name on the exam and on each page you detach. Unless stated otherwise, you
More informationApplied Math Qualifying Exam 11 October Instructions: Work 2 out of 3 problems in each of the 3 parts for a total of 6 problems.
Printed Name: Signature: Applied Math Qualifying Exam 11 October 2014 Instructions: Work 2 out of 3 problems in each of the 3 parts for a total of 6 problems. 2 Part 1 (1) Let Ω be an open subset of R
More informationPartial Differential Equations, Winter 2015
Partial Differential Equations, Winter 215 Homework #2 Due: Thursday, February 12th, 215 1. (Chapter 2.1) Solve u xx + u xt 2u tt =, u(x, ) = φ(x), u t (x, ) = ψ(x). Hint: Factor the operator as we did
More informationMath 5587 Midterm II Solutions
Math 5587 Midterm II Solutions Prof. Jeff Calder November 3, 2016 Name: Instructions: 1. I recommend looking over the problems first and starting with those you feel most comfortable with. 2. Unless otherwise
More informationMTH 847: PDE I (Fall 2017) Exam 2,
MTH 847: PDE I (Fall 2017) Exam 2, 2017.12.8 Name: Standard exam rules apply: You are not allowed to give or receive help from other students. All electronic devices must be turned off for the duration
More informationPartial Differential Equations
Part II Partial Differential Equations Year 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2015 Paper 4, Section II 29E Partial Differential Equations 72 (a) Show that the Cauchy problem for u(x,
More informationMATH 425, HOMEWORK 5, SOLUTIONS
MATH 425, HOMEWORK 5, SOLUTIONS Exercise (Uniqueness for the heat equation on R) Suppose that the functions u, u 2 : R x R t R solve: t u k 2 xu = 0, x R, t > 0 u (x, 0) = φ(x), x R and t u 2 k 2 xu 2
More informationUNIVERSITY OF MANITOBA
Question Points Score INSTRUCTIONS TO STUDENTS: This is a 6 hour examination. No extra time will be given. No texts, notes, or other aids are permitted. There are no calculators, cellphones or electronic
More informationMATH 6337: Homework 8 Solutions
6.1. MATH 6337: Homework 8 Solutions (a) Let be a measurable subset of 2 such that for almost every x, {y : (x, y) } has -measure zero. Show that has measure zero and that for almost every y, {x : (x,
More informationMATH 819 FALL We considered solutions of this equation on the domain Ū, where
MATH 89 FALL. The D linear wave equation weak solutions We have considered the initial value problem for the wave equation in one space dimension: (a) (b) (c) u tt u xx = f(x, t) u(x, ) = g(x), u t (x,
More informationThe first order quasi-linear PDEs
Chapter 2 The first order quasi-linear PDEs The first order quasi-linear PDEs have the following general form: F (x, u, Du) = 0, (2.1) where x = (x 1, x 2,, x 3 ) R n, u = u(x), Du is the gradient of u.
More informationMath 172 Problem Set 8 Solutions
Math 72 Problem Set 8 Solutions Problem. (i We have (Fχ [ a,a] (ξ = χ [ a,a] e ixξ dx = a a e ixξ dx = iξ (e iax e iax = 2 sin aξ. ξ (ii We have (Fχ [, e ax (ξ = e ax e ixξ dx = e x(a+iξ dx = a + iξ where
More informationOscillatory integrals
Chapter Oscillatory integrals. Fourier transform on S The Fourier transform is a fundamental tool in microlocal analysis and its application to the theory of PDEs and inverse problems. In this first section
More informationModeling Rarefaction and Shock waves
April 30, 2013 Inroduction Inroduction Rarefaction and shock waves are combinations of two wave fronts created from the initial disturbance of the medium. Inroduction Rarefaction and shock waves are combinations
More informationThere are five problems. Solve four of the five problems. Each problem is worth 25 points. A sheet of convenient formulae is provided.
Preliminary Examination (Solutions): Partial Differential Equations, 1 AM - 1 PM, Jan. 18, 16, oom Discovery Learning Center (DLC) Bechtel Collaboratory. Student ID: There are five problems. Solve four
More informationDifferential equations, comprehensive exam topics and sample questions
Differential equations, comprehensive exam topics and sample questions Topics covered ODE s: Chapters -5, 7, from Elementary Differential Equations by Edwards and Penney, 6th edition.. Exact solutions
More informationMATH FALL 2014
MATH 126 - FALL 2014 JASON MURPHY Abstract. These notes are meant to supplement the lectures for Math 126 (Introduction to PDE) in the Fall of 2014 at the University of California, Berkeley. Contents 1.
More information6 Non-homogeneous Heat Problems
6 Non-homogeneous Heat Problems Up to this point all the problems we have considered for the heat or wave equation we what we call homogeneous problems. This means that for an interval < x < l the problems
More informationMath 220a - Fall 2002 Homework 6 Solutions
Math a - Fall Homework 6 Solutions. Use the method of reflection to solve the initial-boundary value problem on the interval < x < l, u tt c u xx = < x < l u(x, = < x < l u t (x, = x < x < l u(, t = =
More informationMATH-UA 263 Partial Differential Equations Recitation Summary
MATH-UA 263 Partial Differential Equations Recitation Summary Yuanxun (Bill) Bao Office Hour: Wednesday 2-4pm, WWH 1003 Email: yxb201@nyu.edu 1 February 2, 2018 Topics: verifying solution to a PDE, dispersion
More informationFinal: Solutions Math 118A, Fall 2013
Final: Solutions Math 118A, Fall 2013 1. [20 pts] For each of the following PDEs for u(x, y), give their order and say if they are nonlinear or linear. If they are linear, say if they are homogeneous or
More informationMATH 205C: STATIONARY PHASE LEMMA
MATH 205C: STATIONARY PHASE LEMMA For ω, consider an integral of the form I(ω) = e iωf(x) u(x) dx, where u Cc (R n ) complex valued, with support in a compact set K, and f C (R n ) real valued. Thus, I(ω)
More informationz x = f x (x, y, a, b), z y = f y (x, y, a, b). F(x, y, z, z x, z y ) = 0. This is a PDE for the unknown function of two independent variables.
Chapter 2 First order PDE 2.1 How and Why First order PDE appear? 2.1.1 Physical origins Conservation laws form one of the two fundamental parts of any mathematical model of Continuum Mechanics. These
More informationMath 5588 Final Exam Solutions
Math 5588 Final Exam Solutions Prof. Jeff Calder May 9, 2017 1. Find the function u : [0, 1] R that minimizes I(u) = subject to u(0) = 0 and u(1) = 1. 1 0 e u(x) u (x) + u (x) 2 dx, Solution. Since the
More informationMath 220A - Fall 2002 Homework 5 Solutions
Math 0A - Fall 00 Homework 5 Solutions. Consider the initial-value problem for the hyperbolic equation u tt + u xt 0u xx 0 < x 0 u t (x, 0) ψ(x). Use energy methods to show that the domain of dependence
More informationMathematical Methods - Lecture 9
Mathematical Methods - Lecture 9 Yuliya Tarabalka Inria Sophia-Antipolis Méditerranée, Titane team, http://www-sop.inria.fr/members/yuliya.tarabalka/ Tel.: +33 (0)4 92 38 77 09 email: yuliya.tarabalka@inria.fr
More informationMATH 220: PROBLEM SET 1, SOLUTIONS DUE FRIDAY, OCTOBER 2, 2015
MATH 220: PROBLEM SET 1, SOLUTIONS DUE FRIDAY, OCTOBER 2, 2015 Problem 1 Classify the following PDEs by degree of non-linearity (linear, semilinear, quasilinear, fully nonlinear: (1 (cos x u x + u y =
More informationMATH 131P: PRACTICE FINAL SOLUTIONS DECEMBER 12, 2012
MATH 3P: PRACTICE FINAL SOLUTIONS DECEMBER, This is a closed ook, closed notes, no calculators/computers exam. There are 6 prolems. Write your solutions to Prolems -3 in lue ook #, and your solutions to
More informationUNIVERSITY OF MANITOBA
DATE: May 8, 2015 Question Points Score INSTRUCTIONS TO STUDENTS: This is a 6 hour examination. No extra time will be given. No texts, notes, or other aids are permitted. There are no calculators, cellphones
More informationDiffusion on the half-line. The Dirichlet problem
Diffusion on the half-line The Dirichlet problem Consider the initial boundary value problem (IBVP) on the half line (, ): v t kv xx = v(x, ) = φ(x) v(, t) =. The solution will be obtained by the reflection
More informationChapter 3 Second Order Linear Equations
Partial Differential Equations (Math 3303) A Ë@ Õæ Aë áöß @. X. @ 2015-2014 ú GA JË@ É Ë@ Chapter 3 Second Order Linear Equations Second-order partial differential equations for an known function u(x,
More informationPARTIAL DIFFERENTIAL EQUATIONS. Lecturer: D.M.A. Stuart MT 2007
PARTIAL DIFFERENTIAL EQUATIONS Lecturer: D.M.A. Stuart MT 2007 In addition to the sets of lecture notes written by previous lecturers ([1, 2]) the books [4, 7] are very good for the PDE topics in the course.
More informationIntroduction to Partial Differential Equations, Math 463/513, Spring 2015
Introduction to Partial Differential Equations, Math 463/513, Spring 215 Jens Lorenz April 1, 215 Contents Department of Mathematics and Statistics, UNM, Albuquerque, NM 87131 1 First Order Scalar PDEs
More informationSection 12.6: Non-homogeneous Problems
Section 12.6: Non-homogeneous Problems 1 Introduction Up to this point all the problems we have considered are we what we call homogeneous problems. This means that for an interval < x < l the problems
More informationMath 124A October 11, 2011
Math 14A October 11, 11 Viktor Grigoryan 6 Wave equation: solution In this lecture we will solve the wave equation on the entire real line x R. This corresponds to a string of infinite length. Although
More informationMATH 425, HOMEWORK 3 SOLUTIONS
MATH 425, HOMEWORK 3 SOLUTIONS Exercise. (The differentiation property of the heat equation In this exercise, we will use the fact that the derivative of a solution to the heat equation again solves the
More informationHyperbolic PDEs. Chapter 6
Chapter 6 Hyperbolic PDEs In this chapter we will prove existence, uniqueness, and continuous dependence of solutions to hyperbolic PDEs in a variety of domains. To get a feel for what we might expect,
More information0.3.4 Burgers Equation and Nonlinear Wave
16 CONTENTS Solution to step (discontinuity) initial condition u(x, 0) = ul if X < 0 u r if X > 0, (80) u(x, t) = u L + (u L u R ) ( 1 1 π X 4νt e Y 2 dy ) (81) 0.3.4 Burgers Equation and Nonlinear Wave
More informationSobolev Spaces. Chapter 10
Chapter 1 Sobolev Spaces We now define spaces H 1,p (R n ), known as Sobolev spaces. For u to belong to H 1,p (R n ), we require that u L p (R n ) and that u have weak derivatives of first order in L p
More information[2] (a) Develop and describe the piecewise linear Galerkin finite element approximation of,
269 C, Vese Practice problems [1] Write the differential equation u + u = f(x, y), (x, y) Ω u = 1 (x, y) Ω 1 n + u = x (x, y) Ω 2, Ω = {(x, y) x 2 + y 2 < 1}, Ω 1 = {(x, y) x 2 + y 2 = 1, x 0}, Ω 2 = {(x,
More informationLECTURE 5: THE METHOD OF STATIONARY PHASE
LECTURE 5: THE METHOD OF STATIONARY PHASE Some notions.. A crash course on Fourier transform For j =,, n, j = x j. D j = i j. For any multi-index α = (α,, α n ) N n. α = α + + α n. α! = α! α n!. x α =
More informationThe second-order 1D wave equation
C The second-order D wave equation C. Homogeneous wave equation with constant speed The simplest form of the second-order wave equation is given by: x 2 = Like the first-order wave equation, it responds
More informationNonlinear stabilization via a linear observability
via a linear observability Kaïs Ammari Department of Mathematics University of Monastir Joint work with Fathia Alabau-Boussouira Collocated feedback stabilization Outline 1 Introduction and main result
More informationEXPOSITORY NOTES ON DISTRIBUTION THEORY, FALL 2018
EXPOSITORY NOTES ON DISTRIBUTION THEORY, FALL 2018 While these notes are under construction, I expect there will be many typos. The main reference for this is volume 1 of Hörmander, The analysis of liner
More informationA proof for the full Fourier series on [ π, π] is given here.
niform convergence of Fourier series A smooth function on an interval [a, b] may be represented by a full, sine, or cosine Fourier series, and pointwise convergence can be achieved, except possibly at
More informationMath 4263 Homework Set 1
Homework Set 1 1. Solve the following PDE/BVP 2. Solve the following PDE/BVP 2u t + 3u x = 0 u (x, 0) = sin (x) u x + e x u y = 0 u (0, y) = y 2 3. (a) Find the curves γ : t (x (t), y (t)) such that that
More informationGreen s Functions and Distributions
CHAPTER 9 Green s Functions and Distributions 9.1. Boundary Value Problems We would like to study, and solve if possible, boundary value problems such as the following: (1.1) u = f in U u = g on U, where
More informationMy signature below certifies that I have complied with the University of Pennsylvania s Code of Academic Integrity in completing this exam.
My signature below certifies that I have complied with the University of Pennsylvania s Code of Academic Integrity in completing this exam. Signature Printed Name Math 241 Exam 1 Jerry Kazdan Feb. 17,
More informationChap. 1. Some Differential Geometric Tools
Chap. 1. Some Differential Geometric Tools 1. Manifold, Diffeomorphism 1.1. The Implicit Function Theorem ϕ : U R n R n p (0 p < n), of class C k (k 1) x 0 U such that ϕ(x 0 ) = 0 rank Dϕ(x) = n p x U
More informationFOURIER METHODS AND DISTRIBUTIONS: SOLUTIONS
Centre for Mathematical Sciences Mathematics, Faculty of Science FOURIER METHODS AND DISTRIBUTIONS: SOLUTIONS. We make the Ansatz u(x, y) = ϕ(x)ψ(y) and look for a solution which satisfies the boundary
More informationMath 241 Final Exam, Spring 2013
Math 241 Final Exam, Spring 2013 Name: Section number: Instructor: Read all of the following information before starting the exam. Question Points Score 1 5 2 5 3 12 4 10 5 17 6 15 7 6 8 12 9 12 10 14
More information1 Assignment 1: Nonlinear dynamics (due September
Assignment : Nonlinear dynamics (due September 4, 28). Consider the ordinary differential equation du/dt = cos(u). Sketch the equilibria and indicate by arrows the increase or decrease of the solutions.
More informationHere we used the multiindex notation:
Mathematics Department Stanford University Math 51H Distributions Distributions first arose in solving partial differential equations by duality arguments; a later related benefit was that one could always
More informationMath 311, Partial Differential Equations, Winter 2015, Midterm
Score: Name: Math 3, Partial Differential Equations, Winter 205, Midterm Instructions. Write all solutions in the space provided, and use the back pages if you have to. 2. The test is out of 60. There
More informationClass Meeting # 1: Introduction to PDEs
MATH 18.152 COURSE NOTES - CLASS MEETING # 1 18.152 Introduction to PDEs, Spring 2017 Professor: Jared Speck Class Meeting # 1: Introduction to PDEs 1. What is a PDE? We will be studying functions u =
More informationPDEs, Homework #3 Solutions. 1. Use Hölder s inequality to show that the solution of the heat equation
PDEs, Homework #3 Solutions. Use Hölder s inequality to show that the solution of the heat equation u t = ku xx, u(x, = φ(x (HE goes to zero as t, if φ is continuous and bounded with φ L p for some p.
More informationNAME: MATH 172: Lebesgue Integration and Fourier Analysis (winter 2012) Final exam. Wednesday, March 21, time: 2.5h
NAME: SOLUTION problem # 1 2 3 4 5 6 7 8 9 points max 15 20 10 15 10 10 10 10 10 110 MATH 172: Lebesgue Integration and Fourier Analysis (winter 2012 Final exam Wednesday, March 21, 2012 time: 2.5h Please
More informationTraffic flow problems. u t + [uv(u)] x = 0. u 0 x > 1
The flow of cars is modelled by the PDE Traffic flow problems u t + [uvu)] x = 1. If vu) = 1 u and x < a) ux, ) = u x 2 x 1 u x > 1 b) ux, ) = u e x, where < u < 1, determine when and where a shock first
More informationCLASSIFICATION AND PRINCIPLE OF SUPERPOSITION FOR SECOND ORDER LINEAR PDE
CLASSIFICATION AND PRINCIPLE OF SUPERPOSITION FOR SECOND ORDER LINEAR PDE 1. Linear Partial Differential Equations A partial differential equation (PDE) is an equation, for an unknown function u, that
More informationSolutions to Homework # 1 Math 381, Rice University, Fall (x y) y 2 = 0. Part (b). We make a convenient change of variables:
Hildebrand, Ch. 8, # : Part (a). We compute Subtracting, we eliminate f... Solutions to Homework # Math 38, Rice University, Fall 2003 x = f(x + y) + (x y)f (x + y) y = f(x + y) + (x y)f (x + y). x = 2f(x
More informationAnalysis III (BAUG) Assignment 3 Prof. Dr. Alessandro Sisto Due 13th October 2017
Analysis III (BAUG Assignment 3 Prof. Dr. Alessandro Sisto Due 13th October 2017 Question 1 et a 0,..., a n be constants. Consider the function. Show that a 0 = 1 0 φ(xdx. φ(x = a 0 + Since the integral
More informationStability of Feedback Solutions for Infinite Horizon Noncooperative Differential Games
Stability of Feedback Solutions for Infinite Horizon Noncooperative Differential Games Alberto Bressan ) and Khai T. Nguyen ) *) Department of Mathematics, Penn State University **) Department of Mathematics,
More informationPARTIAL DIFFERENTIAL EQUATIONS MIDTERM
PARTIAL DIFFERENTIAL EQUATIONS MIDTERM ERIN PEARSE. For b =,,..., ), find the explicit fundamental solution to the heat equation u + b u u t = 0 in R n 0, ). ) Letting G be what you find, show u 0 x) =
More informationReminder Notes for the Course on Distribution Theory
Reminder Notes for the Course on Distribution Theory T. C. Dorlas Dublin Institute for Advanced Studies School of Theoretical Physics 10 Burlington Road, Dublin 4, Ireland. Email: dorlas@stp.dias.ie March
More informationLecture Notes Math 632, PDE Spring Semester Sigmund Selberg Visiting Assistant Professor Johns Hopkins University
Lecture Notes Math 63, PDE Spring Semester 1 Sigmund Selberg Visiting Assistant Professor Johns Hopkins University CHAPTER 1 The basics We consider the equation 1.1. The wave equation on R 1+n u =, where
More informationMath 5440 Problem Set 6 Solutions
Math 544 Math 544 Problem Set 6 Solutions Aaron Fogelson Fall, 5 : (Logan,.6 # ) Solve the following problem using Laplace transforms. u tt c u xx g, x >, t >, u(, t), t >, u(x, ), u t (x, ), x >. The
More informationMathematics for Engineers II. lectures. Differential Equations
Differential Equations Examples for differential equations Newton s second law for a point mass Consider a particle of mass m subject to net force a F. Newton s second law states that the vector acceleration
More informationOptimal Transportation. Nonlinear Partial Differential Equations
Optimal Transportation and Nonlinear Partial Differential Equations Neil S. Trudinger Centre of Mathematics and its Applications Australian National University 26th Brazilian Mathematical Colloquium 2007
More informationLecture No 1 Introduction to Diffusion equations The heat equat
Lecture No 1 Introduction to Diffusion equations The heat equation Columbia University IAS summer program June, 2009 Outline of the lectures We will discuss some basic models of diffusion equations and
More informationSome Aspects of Solutions of Partial Differential Equations
Some Aspects of Solutions of Partial Differential Equations K. Sakthivel Department of Mathematics Indian Institute of Space Science & Technology(IIST) Trivandrum - 695 547, Kerala Sakthivel@iist.ac.in
More informationOn some weighted fractional porous media equations
On some weighted fractional porous media equations Gabriele Grillo Politecnico di Milano September 16 th, 2015 Anacapri Joint works with M. Muratori and F. Punzo Gabriele Grillo Weighted Fractional PME
More informationMidterm Exam, Thursday, October 27
MATH 18.152 - MIDTERM EXAM 18.152 Introduction to PDEs, Fall 2011 Professor: Jared Speck Midterm Exam, Thursday, October 27 Answer questions I - V below. Each question is worth 20 points, for a total of
More informationsech(ax) = 1 ch(x) = 2 e x + e x (10) cos(a) cos(b) = 2 sin( 1 2 (a + b)) sin( 1 2
Math 60 43 6 Fourier transform Here are some formulae that you may want to use: F(f)(ω) def = f(x)e iωx dx, F (f)(x) = f(ω)e iωx dω, (3) F(f g) = F(f)F(g), (4) F(e α x ) = α π ω + α, F( α x + α )(ω) =
More informationFinal Exam May 4, 2016
1 Math 425 / AMCS 525 Dr. DeTurck Final Exam May 4, 2016 You may use your book and notes on this exam. Show your work in the exam book. Work only the problems that correspond to the section that you prepared.
More information1/12/05: sec 3.1 and my article: How good is the Lebesgue measure?, Math. Intelligencer 11(2) (1989),
Real Analysis 2, Math 651, Spring 2005 April 26, 2005 1 Real Analysis 2, Math 651, Spring 2005 Krzysztof Chris Ciesielski 1/12/05: sec 3.1 and my article: How good is the Lebesgue measure?, Math. Intelligencer
More informationClass Meeting # 2: The Diffusion (aka Heat) Equation
MATH 8.52 COURSE NOTES - CLASS MEETING # 2 8.52 Introduction to PDEs, Fall 20 Professor: Jared Speck Class Meeting # 2: The Diffusion (aka Heat) Equation The heat equation for a function u(, x (.0.). Introduction
More informationModule 2: First-Order Partial Differential Equations
Module 2: First-Order Partial Differential Equations The mathematical formulations of many problems in science and engineering reduce to study of first-order PDEs. For instance, the study of first-order
More informationStrauss PDEs 2e: Section Exercise 2 Page 1 of 6. Solve the completely inhomogeneous diffusion problem on the half-line
Strauss PDEs 2e: Section 3.3 - Exercise 2 Page of 6 Exercise 2 Solve the completely inhomogeneous diffusion problem on the half-line v t kv xx = f(x, t) for < x
More information2) Let X be a compact space. Prove that the space C(X) of continuous real-valued functions is a complete metric space.
University of Bergen General Functional Analysis Problems with solutions 6 ) Prove that is unique in any normed space. Solution of ) Let us suppose that there are 2 zeros and 2. Then = + 2 = 2 + = 2. 2)
More informationNon-linear Scalar Equations
Non-linear Scalar Equations Professor Dr. E F Toro Laboratory of Applied Mathematics University of Trento, Italy eleuterio.toro@unitn.it http://www.ing.unitn.it/toro August 24, 2014 1 / 44 Overview Here
More information7. Let X be a (general, abstract) metric space which is sequentially compact. Prove X must be complete.
Math 411 problems The following are some practice problems for Math 411. Many are meant to challenge rather that be solved right away. Some could be discussed in class, and some are similar to hard exam
More informationPH.D. PRELIMINARY EXAMINATION MATHEMATICS
UNIVERSITY OF CALIFORNIA, BERKELEY Dept. of Civil and Environmental Engineering FALL SEMESTER 2014 Structural Engineering, Mechanics and Materials NAME PH.D. PRELIMINARY EXAMINATION MATHEMATICS Problem
More informationPartial Differential Equations
M3M3 Partial Differential Equations Solutions to problem sheet 3/4 1* (i) Show that the second order linear differential operators L and M, defined in some domain Ω R n, and given by Mφ = Lφ = j=1 j=1
More informationIn this chapter we study elliptical PDEs. That is, PDEs of the form. 2 u = lots,
Chapter 8 Elliptic PDEs In this chapter we study elliptical PDEs. That is, PDEs of the form 2 u = lots, where lots means lower-order terms (u x, u y,..., u, f). Here are some ways to think about the physical
More information1. Differential Equations (ODE and PDE)
1. Differential Equations (ODE and PDE) 1.1. Ordinary Differential Equations (ODE) So far we have dealt with Ordinary Differential Equations (ODE): involve derivatives with respect to only one variable
More informationHomework 3 solutions Math 136 Gyu Eun Lee 2016 April 15. R = b a
Homework 3 solutions Math 136 Gyu Eun Lee 2016 April 15 A problem may have more than one valid method of solution. Here we present just one. Arbitrary functions are assumed to have whatever regularity
More informationODE Homework 1. Due Wed. 19 August 2009; At the beginning of the class
ODE Homework Due Wed. 9 August 2009; At the beginning of the class. (a) Solve Lẏ + Ry = E sin(ωt) with y(0) = k () L, R, E, ω are positive constants. (b) What is the limit of the solution as ω 0? (c) Is
More informationWeek 4 Lectures, Math 6451, Tanveer
1 Diffusion in n ecall that for scalar x, Week 4 Lectures, Math 6451, Tanveer S(x,t) = 1 exp [ x2 4πκt is a special solution to 1-D heat equation with properties S(x,t)dx = 1 for t >, and yet lim t +S(x,t)
More informationMidterm 1 Solutions Thursday, February 26
Math 59 Dr. DeTurck Midterm 1 Solutions Thursday, February 26 1. First note that since f() = f( + ) = f()f(), we have either f() = (in which case f(x) = f(x + ) = f(x)f() = for all x, so c = ) or else
More informationFourier Transform & Sobolev Spaces
Fourier Transform & Sobolev Spaces Michael Reiter, Arthur Schuster Summer Term 2008 Abstract We introduce the concept of weak derivative that allows us to define new interesting Hilbert spaces the Sobolev
More informationName: ID.NO: Fall 97. PLEASE, BE NEAT AND SHOW ALL YOUR WORK; CIRCLE YOUR ANSWER. NO NOTES, BOOKS, CALCULATORS, TAPE PLAYERS, or COMPUTERS.
MATH 303-2/6/97 FINAL EXAM - Alternate WILKERSON SECTION Fall 97 Name: ID.NO: PLEASE, BE NEAT AND SHOW ALL YOUR WORK; CIRCLE YOUR ANSWER. NO NOTES, BOOKS, CALCULATORS, TAPE PLAYERS, or COMPUTERS. Problem
More informationDynamics of Propagation and Interaction of Delta-Shock Waves in Conservation Law Systems
Dynamics of Propagation and Interaction of Delta-Shock Waves in Conservation Law Systems V. G. Danilov and V. M. Shelkovich Abstract. We introduce a new definition of a δ-shock wave type solution for a
More information