Definitions, Theorems, Examples, Topics for Asmar 2nd edition Chapter 3: 3.1 and 3.2 only 3.1 PDE of Mathematical Physics ==== EXAMPLE. Classical PDEs of mathematical physics Wave equation in 1 and 2 dimensions Heat equation in 1 and 2 dimensions Laplace equation in dimension 2 Poisson equation in dimension 2 Telegraph equation Schrodinger's equation in 1 dimension DEF. Nonlinear PDE. Example: D_tD_x u(x,t) + u(x,t) D_x u(x,t) = exp(x) DEF. General linear PDE of order 2 Homogeneous equation, non-homogeneous equation Boundary conditions Boundary value problem THEOREM 1. [Superposition] If c1, c2 are constants and u1, u2 are solutions of a general linear homogeneous PDE, then u = c1 u1 + c2 u2 is also a solution. EXAMPLE. Superposition in Laplace's equation u_xx + u_yy = 0 DEF. A harmonic function is any solution of Laplace's equation. Examples of harmonic functions are: u=x+y, u=x^2-y^2, u=x/r^2, u=y/r^2, u=ln(r^2), u=exp(x)sin(y), where r^2=x^2+y^2 Any linear combination of these is also a harmonic function, by Theorem 1. EXAMPLE. Failure of superposition. The PDE u_t + u u_x = 0 is order 1, homogeneous but not linear. The superposition principle fails for this example: u=x/(t+1) is a solution but v = c u = cx/(t+1) is not a solution unless c=0 or c=1. CLASSIFICATION of 2nd Order linear PDE Use the discriminant D=B^2-AC Elliptic: D>0 [Laplace equation] Parabolic: D=0 [Heat equation] Hyperbolic: D<0 [Wave equation] 3.2 Wave Equation and Vibrating Strings ==== Model for a finite string of length L: Wave Equation 1. String has mass density rho 2. Perfectly elastic, no resistance forces 3. Horizontal component of the tension is constant. 4. Transverse vibrations have a small angle, making it possible to use the approximation sin(theta)=tan(theta) 5. The slope u_x=tan(theta) is small. DETAILS. We get u_tt = c^2 u_xx where c^2=tau/rho where tau is the magnitude of the tension in equilibrium and rho is the mass density of the string. Here's the argument: Vertical components of the tension are -tau sin(alpha) and tau sin(beta). See Figure 2 in the textbook. The vertical force by Newtons second law is ma where m=rho deltaX is the mass and a=u_tt is the acceleration. A competition is made from the tension forces to give the equation -tau sin(alpha) + tau sin(beta) = rho deltaX u_tt Replace sine by tangent, due to the small angle assumption -tau tan(alpha) + tau tan(beta) = rho deltaX u_tt Use tan(theta)=u_x for one more replacement -tau u_x(x,t) + tau u_x(x+deltaX,t) = rho deltaX u_tt Divide by deltaX and take the limit to get u_xx = (rho/tau) u_tt u_tt = c^2 u_xx [swap sides and divide] UNITS An example appears in Exercise 1: rho = 1 gram/meter, tau = 100 Newtons. Then c^2=tau/rho=100 Newton-meters per gram. If the string is 1 meter and has mass 10 grams, then rho=10 grams per meter. See Exercise 3. FORCED VIBRATIONS The force term F(x,t) deltaX is added to the tension force terms. Then the limiting PDE is u_tt = c^2 u_xx + (1/rho) F(x,t) EXAMPLE. Force due to gravity. Add to the tension terms -mg = -rho g deltaX. Then the limit gives u_tt = c^2 u_xx - g EXAMPLE. Force due to frictional resistance. In the viscous model, the force is proportional to the velocity of the string, which is u_t(x,t). Then for some damping constant we have the model for a string vibrating in a fluid u_tt + 2k u_t = c^2 u_xx ====