Week 2: First Order Semi-Linear PDEs

Transcription

1 Week 2: First Order Semi-Linear PDEs Introction We want to find a formal solution to the first order semilinear PDEs of the form a(, y)u + b(, y)u y = c(, y, u). Using a change of variables corresponding to characteristic lines, we can rece the problem to a system of 3 ODEs. The solution follows by simply solving two ODEs in the resulting system. Step 1: Formally, we want to solve the following system of PDEs a(, y) = b(, y) = c(, y, u). Step 2: We first find the characteristic curve by solving the first pair, a(, y) = b(, y) a(, y) = b(, y). We will get a characteristic line of the form C = f(, y). Step 3: We now can now find the general solution by solving either a(, y) = c(, y, u) or b(, y) = c(, y, u). We choose to solve the ODE that is easier to solve. We may need to use the characteristic curve and the implicit function theorem to write f(, y) = C as y = y(c, ) or = (C, y) to eliminate a variable to solve this ODE. When we solve the ODE, we must remember to write the constant of integration we get when we solve ODEs as a function F (C) = F (f(, y)) instead. Step 4: If we are given some initial conditions, then we can find the specific form of F (C). We will eplain why this method works in the net section. eamples using this technique. We start by formally solving several Problems Problem 1. Solve the initial value problem u t 3u = 0, u(0, ) = e 2. Solution 1. This is a linear first order PDE, so we can solve it using characteristic lines. Step 1: We want to solve u t = 3u. This gives us the system of equations 1 = 0 = 3u. 1 = 0 = 0 = C. Page 1 of 8

2 Therefore, the characteristic curves are given by C =. This is a separable ODE, which has solution 1 = 3u = 3u. log u = 3t + f(c) u = e f(c) e 3t. Since f(c) is an arbitrary function, we can redefine e f(c) =: g(c). Since C =, we have our general solution is u(t, ) = g()e 3t. Step 4: We now use the initial value to solve for g. Since u(0, ) = e 2 we have e 2 = u(0, ) = g()e 3 0 g() = e 2, so our particular solution is of the form u(t, ) = e 2 e 3t = e 3t 2. Problem 2. Solve the initial value problem u t 4u + u = 0, u(0, ) = Solution 2. This is a linear first order PDE, so we can solve it using characteristic lines. Step 1: We want to solve u t 4u = u. This gives us the system of equations 1 = 4 = u. 1 = 4 = 4 = 4t + C. Therefore, the characteristic curves are given by C = + 4t. This is a separable ODE, which has solution 1 = u = u. log u = t + f(c) u = e f(c) e t Since f(c) is an arbitrary function, we might can redefine e f(c) =: g(c). Since C = + 4t, we have our general solution is u(t, ) = g( + 4t)e t. Page 2 of 8

3 Step 4: We now use the initial value to solve for g. Since u(0, ) = we have so our particular solution is of the form = u(0, ) = g()e3 0 g() = , u(t, ) = ( + 4t) 2 e t. Problem 3. Solve the initial value problem with u(, y) = ep(/( y)) on { + y = 1}. 2yu + ( 2 + y 2 )u y = 0 Solution 3. This is a linear first order PDE, so we can solve it using characteristic lines. Step 1: We have the system of equations 2y = ( 2 + y 2 ) = 0. 2y = ( 2 + y 2 ) = 1 2 y y. This is a Homogenous ODE, which can be solved using the change of variables w = y. = dw + w, so under this change of variables we have This is a separable equation, so dw + w = 1 2 w w dw = 1 2 w w = 1 w2 2w. We have 2w 1 w 2 dw = 1 ln(1 w2 ) = ln + D e D = (1 w 2 ) = 2 y 2. If we set C = e D, then C = 2 y 2 is our characteristic curve. 2y = 0 = 0 u = f(c). Since C = 2 y 2, we have our general solution is ( 2 y 2 ) u(, y) = f. Step 4: We now use the initial value to solve for f. Since u(, y) = ep(/( y)) when + y = 1, we have ( e y = u(, y) +y=1 2 y 2 ) ( ( y)( + y) ) ( y ) = f = f = f. +y=1 +y=1 Page 3 of 8

4 If we set z = y, then the above implies e 1 z = f(z), so our particular solution is of the form u(, y) = e 2 y 2. Problem 4. Solve the initial value problem with u(, y) = 0 on {y = 2 }. u + yu y = e u Solution 4. This is a semilinear first order PDE, so we can solve it using characteristic lines. Step 1: We have the system of equations = y = e u. This is a separable ODE, which has solution Therefore, our characteristic curve is C = y. = y = y. y = C = e u = e u e u = + f(c). Since C = y, we have our general solution is ( ( y u(, y) = ln + f. )) Step 4: We now use the initial value to solve for f. Since u(, y) = 0 when y = 2, we have 0 = u(, y) ( ( y )) y= 2 = ln + f = ln( + f()) f() = 1. y=2 Therefore, our particular solution is of the form u(, y) = ln ( + 1 y ). Problem 5. Solve the initial value problem when u(1, y) = y 2. u + yu y = 2( 2 y 2 ) Page 4 of 8

5 Solution 5. This is a linear first order PDE, so we can solve it using characteristic lines. Step 1: We have the system of equations = y = 2( 2 y 2 ). This is a separable ODE, which has solution Therefore, our characteristic curve is C = y. = y = y. y = C = 2( 2 y 2 ) = 2(2 y 2 ). There is a y variable appearing in this ODE that we must eliminate it. Since y = C, we need to solve Since C = y, we have the general solution = 2(2 C 2 2 ) u = 2(1 C2 ) 3 + f(c). 3 u(, y) = 2 3 ( ( y 2 ) ( y 1 ) 3 + f. ) Step 4: We now use the initial value to solve for f. Since u(1, y) = y 2 we have y 2 = u(1, y) = 2 3 Therefore, our particular solution is of the form ( 1 y 2) + f(y) f(y) = y 2 2 ( 1 y 2) = y2. u(, y) = 2 3 ( ( y 2 ) 1 ) ( y ) 2. 3 Problem 6. Find the general solution of 2u + 4u y = e +3y 5u. Solution 6. This is a linear first order PDE, so we can solve it using characteristic lines. Step 1: We have the system of equations 2 = 4 = e +3y 5u. Page 5 of 8

6 Therefore, our characteristic curve is C = y = 4 = 2 y = 2 + C. 2 = e +3y 5u = 1 2 (e+3y 5u) 5 2 u = 1 2 e+3y. There is a y variable appearing in this ODE that we must eliminate it. Since y = C 2, we need to solve 5 2 u = 1 2 e 5+3C. This is a linear ODE, which can be solved using an integrating factor of the form φ() = e 5 2, which gives us ( u = e ) e 5+3C e 5 2 = 12 ( e 52 2e 15 2 ) + 3C + f(c) u = e 5+3C 1 2 f(c)e 5 2. Since C = y + 2, if we set g(z) = 1 2f(z) then we get the general solution We can easily verify that this solves our PDE. u(, y) = 1 15 e+3y + g(y + 2)e 5 2. Problem 7. Find the general solution of u yu y + y 2 u = y 2, where y 0 Solution 7. This is a linear first order PDE, so we can solve it using characteristic lines. Step 1: We want to solve u yu y = y 2 y 2 u. We have the system of equations = y = y 2 y 2 u. This is a separable ODE, with solution = y = y. ln y = ln + D e D = y If we set C = e D then our characteristic curve is C = y. Step 3: We now solve the second and third equation, y = y 2 y 2 u = y + yu yu = y. Page 6 of 8

7 This is a linear ODE, which can be solved using an integrating factor of the form φ(y) = e y2 /2, which gives us ( ) ( ) u = e y2 /2 ye y2 /2 = e y2 /2 e y2 /2 + f(c) = 1 + f(c)e y2 /2. Since C = y, we get the general solution We can easily verify that this solves our PDE. Proof of the Coordinate Method u(, y) = 1 + f(y)e y2 /2. We now eplain why the above method works for certain semilinear PDEs that can be reced to solvable ODEs. We essentially do a change of variables to rece the PDE into an ODE. Suppose we have a PDE of the form a(, y)u + b(, y)u y = c(, y, u). ( ) Step 1: We want to solve the equation a(, y) = b(, y) b(, y) = a(, y). Suppose we can find a family of solutions f(, y) = C, called the characteristic curves. Since this is a solution to the ODE, by implicitly differentiating, we must have 0 = d f(, y) = f (, y) + f y (, y) = f b(, y) (, y) + f y (, y) a(, y) and therefore, our the family of solutions must satisfy the condition 0 = a(, y)f (, y) + b(, y)f y (, y). (1) We will see that using this f(, y), we can rece our PDE into either an ODE with respect to (Choice 1) or an ODE with respect to y (Choice 2). Step 2 (Choice 1): We now eplain why it suffices to solve the equations a(, y) = c(, y, u) c(, y, u) = a(, y). To rece ( ) to this ODE, we do a change of variables ξ(, y) =, η(, y) = f(, y) where f(, y) = C is the function we found in Step 1. Notice that and therefore we have u = u ξ ξ + u η η = u ξ + u η f (, y) u y = u ξ ξ y + u η η y = u ηf y (, y) a(, y)u + b(, y)u y = a(, y)(u ξ f (, y)u η ) + b(, y)f y (, y)u η = a(, y)u ξ + (a(, y)f (, y) + b(, y)f y (, y))u η = a(, y)u ξ (ξ, η) Page 7 of 8

8 since f(, y) satisfies (1). Since our PDE satisfies ( ), we have shown that u ξ (ξ, η) = c((ξ, η), y(ξ, η), u) a((ξ, η), y(ξ, η)) We can write this back in our original coordinates ξ =, η = f(, y) = C. If the characteristic curve f(, y) = C can be solved implictly for y = y(, C), then we can eliminate the y variable, which gives us c(, y(, C), u) u (, C) = a(, y(, C)) This is precisely the ODE we are solving if we choose the first ODE in Step 3. This also eplains why the integration constant is of the form F (C) instead of just a constant. Step 2 (Choice 2): It might be easier to instead solve the system b(, y) = c(, y, u) c(, y, u) = b(, y). To rece ( ) to this ODE, we instead use the change of variables, ξ(, y) = y, η(, y) = f(, y), where f(, y) = C is the function we found in Step 1. Notice that u = u ξ ξ + u η η = u ηf (, y) and so we have u y = u ξ ξ y + u η η y = u ξ + u η f y (, y) a(, y)u + b(, y)u y = a(, y)(f (, y)u η ) + b(, y)(u ξ + f y (, y)u η ) = b(, y)u ξ (ξ, η) + (a(, y)f (, y) + b(, y)f y (, y))u η = b(, y)u ξ (ξ, η) since f(, y) satisfies (1). Since our PDE satisfies ( ), we have shown that u ξ (ξ, η) = c((ξ, η), y(ξ, η), u) a((ξ, η), y(ξ, η)) We can write this back in our original coordinates ξ = y, η = f(, y) = C. If the characteristic curve f(, y) = C can be solved implictly for = (y, C), then we can eliminate the variable, which gives us c((y, C), y, u) u y (C, y) = a((y, C), y) This is precisely the ODE we are solving if we choose the second ODE in Step 3. This also eplains why the integration constant is of the form F (C) instead of just a constant. Remark: Notice that the above procere works even if one of the coefficients vanish (i.e. a 0, b 0 or c 0 ). We just lose some choice as to the equations we solve, since if a 0, then we must use Choice 2 because Choice 1 does not make sense. The derivation assumed that the coefficients were sufficiently nice such that we can find solutions to the ODEs at least locally. We should verify that our formal solutions are true solutions by checking our solutions after deriving them. Page 8 of 8