Strauss PDEs 2e: Section Exercise 2 Page 1 of 8. In order to do so, we ll solve for the Green s function G(x, t) in the corresponding PDE,

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1 Strauss PDEs 2e: Section Eercise 2 Page of 8 Eercise 2 Do the same for φ() = for > and φ() = 3 for <. Solution Solution by the Similarity Method We have to solve the initial value problem, u t = ku, u(, ) = φ(). () In order to do so, we ll solve for the Green s function G(, t) in the corresponding PDE, where δ(), the Dirac delta function, is defined as { = δ() =. G t = kg, G(, ) = δ(), (2) The reason we re solving the equation with the delta function is that it has the etremely useful sifting property, f(s)δ( s) ds = f(), so the solution to the initial value problem () in terms of the Green s function is u(, t) = G( s, t)φ(s) ds. This can be verified by substituting this form for u into (). Now we will go about solving (2) for G(, t) by using the similarity method (also known as the combination of variables method). Because u is a dimensionless quantity (that is, it yields a pure number with no units) the variables, t, and k have to appear in the solution in a dimensionless combination. has units of meters, t has units of seconds, and k has units of meters 2 /second, so the combination of variables has to be 2 or any convenient multiple or power thereof. Therefore, u = u(η), where we choose η =. We choose this particular form for η so the process of getting the final answer is smoother. We re trying to solve (2) for G, though, and G is not dimensionless; as can be seen from the initial condition, it has the same dimensions as δ(). δ() has the inverse dimension of its argument, so G has dimensions of meters. Thus, G has to be of the form, G(, t) = g ( ),

2 Strauss PDEs 2e: Section Eercise 2 Page 2 of 8 where g is an arbitrary function. In order to determine g, we have to plug this form into (2) and solve the resulting ODE. To start, write the epressions for G t and G. G t = 2 g + ( ) 3 2 g 3 G = g = g Substituting these epressions into (2) gives 2 G 2 = g = k 3 t 3 g 2 g 3 2 g = 3 3 g. Cancel common terms and move everything to one side. g + g g = Use the combination variable η. g + η 2 g + 2 g = The last two terms on the left side can be written as one using the product rule. Integrate both sides of the equation. g + ( η 2 g ) = g + η 2 g = C This is an inhomogeneous first-order linear differential equation that can be solved with an integrating factor. The integrating factor is Multiply both sides by I. η I = e 2 dη = e η2 4. e η2 4 g + η 2 e η2 4 g = C e η2 4 The two terms on the left side can be written as one using the product rule. ( e η2 4 g ) = C e η2 4 Integrate both sides of the equation a second time. e η2 4 g = η C e s2 4 ds + C2 Hence, the arbitrary function g is g(η) = e η2 4 [C η e s2 4 ds + C2,

3 Strauss PDEs 2e: Section Eercise 2 Page 3 of 8 and consequently, the Green s function is G = η 2 [ g(η) = e 4 η C e s2 4 ds + C2. (3) The net order of business is to determine the constants of integration, C and C 2. We need to return to the diffusion equation and the initial condition in (2) to figure these out. G t = kg Integrate both sides of the equation with respect to over the whole line. G t d = kg d Take out the time derivative from the left side and evaluate the right side. d dt G d = kg We assume that G and G tend to as ±, so we have This implies that the quantity, d dt G d =. G d, remains constant for all time. Initially G(, ) = δ(), so G(, t) d = G(, ) d = δ() d =. (4) In order for this integral to converge, C has to be. In terms of and t, (3) becomes G(, t) = C 2 e 2. C 2 can be thought of as a normalization constant that we determine by plugging into (4). G(, t) d = Make the following substitution to solve the integral. v = dv = C 2 e 2 d = d 2 dv = d The integral becomes 2C 2 e v2 dv =,

4 Strauss PDEs 2e: Section Eercise 2 Page 4 of 8 and it evaluates to π. Solving for C 2 yields 2C 2 π = C 2 = 4π. Therefore, the Green s function is G(, t) = and the solution to the initial value problem in () is u(, t) = 4π e 2, 4π e ( s)2 φ(s) ds. (5) u(, t) can be interpreted as the convolution of the initial condition with a Gaussian filter. At every point, u(, t) is an averaged, or smoothed, version of the initial condition over an interval of width. As t increases, the range of the filter grows and u(, t) becomes increasingly smooth over. Any discontinuities or kinks that are present in the initial condition are smoothed out. In this eercise, the initial condition is { > φ() = 3 <. If we substitute this into the formula in (5), then we get u(, t) = Make the following substitutions to solve the integrals. e ( s)2 3 ds + e ( s)2 ds. 4π 4π v = s q = s dv = ds dv = ds dq = ds The integrals become u(, t) = 3 e v2 dv + π π e q2 dq. Bring the constants out in front and switch the limits of integration to eliminate the minus sign. u(, t) = [3 e v2 dv + e q2 dq π Split up the integrals now to get the appropriate limits. u(, t) = π [3 e v2 dv 3 e v2 dv + e q2 dq + e q2 dq

5 Strauss PDEs 2e: Section Eercise 2 Page 5 of 8 Make the substitution p = q and dp = dq in the third integral and combine the first and last integrals. We can do this because v and q are just dummy variables. u(, t) = [4 e v2 dv 3 e v2 dv + e p2 dp π Combine the last integrals. p and v are just dummy variables as well. u(, t) = π [4 The integral on the left evaluates to π/2. The error function, erf z, is defined as e v2 dv 2 u(, t) = π [4 π 2 2 e v2 dv e v2 dv so erf z = 2 π z u(, t) = 2 erf e v2 dv, ( ). Figure : Plot of the solution u(, t) with k = m 2 /s for si different times: t = s (red), t = s (orange), t = 2 s (yellow), t = 5 s (green), t = s (blue), and t = 2 s (purple).

6 Strauss PDEs 2e: Section Eercise 2 Page 6 of 8 Solution by Eploitation of the Invariance Properties The initial condition, φ() = { 3 < >, can be written in terms of the signum function, which is defined as { > sgn = <, as φ() = 2 sgn. Since any linear combination of solutions is a solution to the diffusion equation, we can solve () if we can find the solutions with 2 and sgn as initial conditions separately. The equations we have to solve are v t = kv, v(, ) = 2 (6) w t = kw, w(, ) = sgn. (7) Once we have v and w, the solution will be u(, t) = v(, t) w(, t). Right away we can solve (6) since 2 is a constant: v(, t) = 2. To solve (7) we will use the similarity method. Since the signum function is dimensionless, w is dimensionless as well. Hence, the variables (meters), t (seconds), and k (meters 2 /second) have to appear in the solution as a dimensionless combination. The combination variable η is thus 2 or any convenient multiple or power thereof. Choose η = to make the process of obtaining the final answer smoother. Hence, w = w(η), Write epressions for w t and w in terms of this new variable using the chain rule. w t = dw η dη t = dw ( ) dη 4 3 w = dw η dη = dw ( ) dη 2 w 2 = ( ) w = η ( ) ( ) w = d2 w η dη 2 Substituting these epressions into (7) gives dw dη ( ) 4 3 = k d2 w dη 2 ( ).

7 Strauss PDEs 2e: Section Eercise 2 Page 7 of 8 Cancel common terms and bring all terms to one side. w + w = Use the combination variable η. Make the substitution r = w = dw/dη. w + 2ηw = dr dη + 2ηr = This is a first-order differential equation that we can solve by separation of variables. Integrate both sides. Eponentiate both sides. dr r = 2η dη ln r = η 2 + D r = e η2 +D r = ±e D e η2 r = D e η2 Now that we have r, we can get w by integrating the result. w(η) = η r ds + D 2 = η D e s2 ds + D 2 The lower limit,, is arbitrary; choosing a different value leads to a different value for D 2. To evaluate the constants of integration, we look to the initial condition, w(, ) = sgn. For >, as t, η +. Conversely, for <, as t, η. Thus, For > : For < : The system of equations to solve is Solving it yields = D e s2 ds + D 2 = D e s2 ds + D 2 π = D 2 + D 2 π = D 2 + D 2. D = 2 π and D 2 =. The solution is thus w(η) = 2 π η e s2 ds = erf(η).

8 Strauss PDEs 2e: Section Eercise 2 Page 8 of 8 In terms of the original variables it is w(, t) = erf ( ). Therefore, u(, t) = 2 erf ( ), which is the same result as before.