Question

Decomposing inhomogeneous PDEs to facilitate the use of separation of variables Inhomogeneities may arise in the initial (ICs) or boundary (BCs) conditions, or in the PDE itself. A simple example is the falling of an elastic wire under gravity: ə?u ,02u at2 = Car2g If the ICs are: u(x,0) = f(x) and (x,0) = 0, and the BCs are: u(0,t) = 0 and u(L,t) = h(t), then there are three inhomogeneities in this equation: the gravitational attraction -g in the PDE; the shape of the wire f (x) in the first IC; and the movement of the far end of the wire, given by h(t) in the second BC. If the problem can be solved by the Fourier (FT) or Laplace (LT) transform, then inhomogeneities don't usually matter, because the transform method copes with them quite easily. However, the separation-of-variables (SoV) method can usually only cope with one inhomogeneous IC or BC (the inhomogeneity is usually in the IC, because SoV also requires a pair of homogeneous BCs in one independent variable, which is usually the spatial variable). In this coursework, we will develop some simple methods that can sometimes reduce the number of inhomogeneities in a PDE problem, and thereby allow us to solve it by SoV. Such decompositions may not always work, especially for complicated problems. But a surprising number of PDEs can be cast into standard SoV form by relatively simple decompositions, making it a useful method. We will solve two PDEs step-by-step, to show how the method works. Problem 1, Inhomogeneous BC in the heat equation. Let the temperature u(x,t) in a bar of length L and thermal diffusivity D be governed by the heat equation, du azu at – Dar2 The initial condition is that the temperature is 0 °C, and the left edge of the bar, at x = 0, is also maintained at 0 °C. But the right edge of the bar, at x = L, is maintained at a non-zero constant, T. We want to find the temperature u(x, t) at any position in the bar at any time, using Sov. (a) Show that the standard SoV method will fail, and explain why. (5 marks] (b) Decompose the problem into two parts: u(x,t) = v(x, t) + w(x), where w(x) is the steady-state solution; note that w does not depend on t. Choose BCs for w(x) so that the remaining BCs for v(x, t) are homogeneous. Write out the detailed formulation for w(x), and solve for w(x). (10 marks] (c) Having solved for w(x), write out the detailed formulation (PDE and IC and BCs) for v(x, t). Show that the PDE and BCs are homogeneous, and solve for v(x, t) by Sov. [50 marks] (d) Write out the solution u(x, t) to the original problem. (5 marks]

Explain how does w(x) been solve

Answer 1

Given poe is lau=Day Ot :: meing tempetakҷ3. ј» - → ulit) zo ant to 40,1)=0 End a=0 is maintained at . 4(36,) = o at m=0 oc . End azt is maintained at temperaturet .: Wat) at x=L - 41L,T) = T. Thus, conditions on uunity are as below: Tuco,t) zo] - 12) Tuch, t1 = 1 -13) uca,ol = 0 for applying method f separation of variables successfully, the RHS in 8) and B) both must be o. Here it is not the case, in such a case, we always but 40,t) = 12,5 tu This value Furnit) will be substituted in (1). The wa) will be chosen in such a way that vamust satisfico ( H = DBA (ii) voittoo (ii) Velitelso AL

now, from (4) au = auto un wa) is independent ft Alsen = 24 twica and more than any entusicas - putting out and any in 1) , we get, =D [ Tw"ca)] son - Donow"CA) choose Dulcano satisfice av = 2 2 2 It will enayre that y + que a la constant olux) =AHB putting in 4), 407,+) = V(ot) + AMB ~5) = y cort) =Vcati+ALO)+B I putting to oco+B 1:46oitzo Voit)=0 Alsol uch, tha Velit) TALIB putting aaL in (5) » Totalto .: UCL,t) 21 » TAFEL in wean), weget, putting values of A and B wecx) = Ixto Julex) = Ixl