MATH 220 solution to homework 1
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1 MATH solution to homework Problem. Define z(s = u( + s, y + s, then z (s = u u ( + s, y + s + y ( + s, y + s = e y, z( y = u( y, = f( y, u(, y = z( = z( y + y If we prescribe the data u(, = f(, then z (s ds = f( y + ye y.. f( must satisfy that f ( = e =, otherwise there will be no solutions.. If f ( =, then there are infinitely many solutions. Problem. Define z(s = φ(t + s, + us, then { z (s + cz(s( z(s =, z( t = φ(, ut =. Solving the equation we have φ(t, = z( = + e ct (. From the formula we see that, if g(, then. The numerator.. The denominator is at t = and remains positive at the points where the solution eists. the positivity is preserved. (Note that φ(t, might not eist for all (t,. For eample, when > and c >, then at t = ( c log the denominator becomes, and the solution only eists for t < t. But at the points where the solution eists, the positivity is preserved. The maimum principle doesn t hold. For eample, if c =, g( /, then φ(, > /,.
2 If g(, then by φ(t, = + e ct we see that φ(t,. ( Problem 3. (i Define E(t = φ(t, d, then E (t = φ(t, φ t (t, d = R n φ(t, (u( φ(t, d = u( (φ (t, d = φ (t, ( u( d =. (Here we assume that, for fied t, φ(t, is compactly supported in. This assumption holds if, for eample, u( is bounded. φ(t, d = E(t = E( = f ( d. (ii Note that if we take f( as the indicator function of the set S, then φ(t, is the indicator function of S(t for fied t, assuming that f( is smooth. by (i we immediately get (ii. The problem is, the indicator function is not smooth. So we need to make some modification. ɛ >, we can find set S and S + such that S S S + and µ(s \ S < ɛ, µ(s + \ S < ɛ. And there eists smooth functions f (, f + ( such that:. f ( is on S and on S c. And f (.. f + ( is on S and on S c +. and f + (. Suppose we solve the equation in (i with initial data f (, f + ( and we get φ (t,, φ + (t, respectively, then we observe that φ (t, χ S(t ( φ + (t,, (t, [,, where χ S(t ( is the indicator function of the set S(t. Then by (i we have vol(s ɛ f ( d = φ (t, d χ S(t ( d R n = vol(s(t φ +(t, d = f+( d vol(s + ɛ. Now let ɛ, we have vol(s(t = vol(s. (iii Similar as in (i, define E(t = φ(t, d, then E (t = φ(t, φ t (t, d = R n φ(t, (u( φ(t, d = u( (φ (t, d = φ (t, ( u( d E(t.
3 Problem 4. e t (E (t E(t d(e t E(t e t E(t E( E(t e t E( φ (t, d e t f( d. u( = ( k u ( = k = ( u ( = e k(y f(y dy + ( f( k e k( y f(y dy, e k(y f(y dy f( + k e k(y f(y dy + e k( y f(y dy, ( f( + k = f( + k u(. u ( + k u( = f(. e k(y f(y dy f( + k Note that for the corresponding homogeneous equation v ( + k v( =, the solution has the form v( = Ae k + Be k. the solution to the equation u ( + k u( = f( has the form u( = Ae k + Be k + k e k y f(y dy. e k( y f(y dy e k( y f(y dy 3
4 where A, B are arbitrary numbers. So there are infinitely many solutions. To ensure uniqueness, we need to add constraint to guarantee that A, B are. function k e k y f(y dy Note that the is bounded because f( is compactly supported, and v( = Ae k + Be k is unbounded unless both A and B are. So a reasonable constraint is, u( is a bounded function. In other words, u( = k e k y f(y dy is the unique solution to the problem u ( + k u( = f(, sup u( <. R Problem 5. For this problem, we use forward difference in time and backward difference in space. The matlab code: h =.; % eperiment with different tau tau =.; = : h : ; t = : tau : ; u = zeros(numel(t, numel(; u(, : = sin(; u(:, = t; for j t = : numel(t u(j t, : end = ( (tau / h * u(j t, : end + (tau / h * u(j t,... : end ; end 4
5 The numerical results τ =., t = τ =., t = τ =., t = Figure : The numerical results for different t where τ =.. In this case we get the eact solution, since the time-space step matches the transportation speed. τ =.5, t = τ =.5, t = τ =.5, t = Figure : The numerical results for different t where τ =.5. In this case we get a first order accurate solution. τ =.44, t = τ =.44, t = τ =.44, t = Figure 3: The numerical results for different t where τ =.. In this case we get a wrong solution (note the magnitude of the y-ais, since the CFL condition is violated. 5
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