AMS 212A Applied Mathematical Methods I Lecture 14 Copyright by Hongyun Wang, UCSC. with initial condition

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1 Lecture 14 Copyright by Hongyun Wang, UCSC Recap of Lecture 13 Semi-linear PDE a( x, t) u t + b( x, t)u = c ( x, t, u ) x Characteristics: dt d = a x, t dx d = b x, t with initial condition t ( 0)= t 0 x( 0)= x 0 *) Observation #1: (x(), t()) can be solved separately from u() for a semi-linear PDE. *) Observation #: characteristics do not intersect in (x, t) space. Along a characteristic: du d u( 0)= u 0 = c ( x, t, u ) Additional condition in parametric form: x 0 ( s) t 0 ( s) u( x 0 ( s), t 0 ( s) )= f s Set t 0 = t 0 ( s), x 0 = x 0 s Solution parameterized in (s,) x s, t( s, ) u s, and u 0 = f s ( ) Inverse mapping of x( s, ), t s, s x, t x, t, we obtain: - 1 -

2 Solution in (x, t): ( ) u( x, t)= u s( x, t), x, t Example: u u x t x = u u( x,0)= f( x) Characteristics: dt d = 1 dx d = x Along a characteristic: ==> t = + t 0 x = x 0 exp du d = u u ==> u( )= 0 1+ u u( 0)= u 0 0 (See Exercise #14 for derivation) Additional condition: x 0 ( s)= s t 0 ( s)= 0 u( x 0 ( s), t 0 ( s) )= f( s) Set t 0 = t 0 ( s), x 0 = x 0 ( s) and u 0 = f( s), we obtain Solution in parametric form: x( s, )= s exp( ) t( s, )= f( s) u( s, )= 1+ f( s) ( ) Inverse mapping of x( s, ), t s, s( x, t)= x exp t ( x, t)= t () - -

3 Solution in (x, t): u( x, t)= ( ()) f xexp t 1+ f( xexp() t )t Method of characteristics for quasi-linear PDE a( x, t, u) u + b ( x, t, u t )u = c ( x, t, u ) x Characteristics: Assuming u(x, t) is known, it becomes a semi-linear PDE and we can write out the equations for x and t, and write out a separate equation for u along a characteristic. dt = a ( x, t, u ) d t ( 0)= t dx = b ( x, t, u ) 0 with initial condition x( 0)= x 0 d du = c ( x, t, u ) u( 0)= u 0 d Of course, u is unknown. So we have to solve for (x, t, u) simultaneously. Additional condition in parametric form: x 0 ( s) t 0 ( s) u( x 0 ( s), t 0 ( s) )= f( s) Set t 0 = t 0 ( s), x 0 = x 0 s Solution parameterized in (s,) x s, t( s, ) u s, and u 0 = f s ( ) Inverse mapping of x( s, ), t s, s x, t x, t Solution in (x, t):, we obtain ( ) u( x, t)= u s( x, t), x, t - 3 -

4 We start with an example for the special case of c( x, t, u) 0 Example: u u u t x = 0 u( x,0)= x Characteristics: ==> dt d = 1 dx d = u du d = 0 t = + t 0 x = u 0 + x 0 u = u 0 with initial condition t ( 0)= t 0 x( 0)= x 0 u( 0)= u 0 Note: a characteristic is completely determined/specified by (x 0, t 0, u 0 ). Additional condition in parametric form: x 0 ( s)= s t 0 ( s)= 0 u( x 0 ( s), t 0 ( s) )= s Set t 0 = t 0 ( s)= 0, x 0 = x 0 ( s)= s and u 0 = f( s)= s, we obtain Solution parameterized in (s,) t( s, )= x( s, )= s + s u( s, )= s Before we look at the inverse mapping, we examine the characteristics. They do not intersect in the (x, t, u) space! Question: Do they intersect in the (x, t) space? We write the characteristics as x = s t ( + 1) This is a family of curves of the form - 4 -

5 x = s ht () where ht ()= t + 1 For s 1 s, x = s 1 ht () and x = s ht () intersect <==> s 1 ht ( c )= s ht ( c ) at some t c <==> ( s 1 s ) ht ( c )= 0 <==> ht ( c )= 0 Conclusion: the characteristics {x = x 0 h(t)} intersect at and only at h(t c ) = 0. ht ( c )= 0 <==> t c + 1 = 0 <==> t c = 1 For > 0, the characteristics intersect at t c = 1 > 0 where the classical solution ceases to exist. For < 0, the characteristics do not intersect for t > 0. (Draw characteristics {x = s( t + 1)} for > 0 and for < 0) Inverse mapping of x( s, ), t( s, ) t( s, )= x( s, )= s + s Solution in (x, t): u( x, t)= x t + 1 ==> = t x s = t + 1 For > 0, the classical solution is defined only up to t = 1 > 0. Note: in this example, characteristics intersect only for > 0. Next, we look at a slightly more sophisticated example. Example: u u u t x = x u( x,0)= x Characteristics: - 5 -

6 dt d = 1 dx d = u du d = x with initial condition Approach #1 for solving this ODE system d x du = d d = x Approach # for solving this ODE system d( x+ iu) = u + ix= i( x+ iu) d d( x iu) = u ix= i( x iu) d t ( 0)= t 0 x( 0)= x 0 u( 0)= u 0 The solution is (See Appendix A for derivation of the solution) t = + t 0 x = x 0 cos( ) u 0 sin u = x 0 sin( )+ u 0 cos Additional condition in parametric form: x 0 ( s)= s t 0 ( s)= 0 u( x 0 ( s), t 0 ( s) )= s Set t 0 = t 0 ( s)= 0, x 0 = x 0 ( s)= s and u 0 = f( s)= s, we obtain Solution parameterized in (s,) t = x = s cos( ) ssin u = ssin( )+ s cos Before we look at the inverse mapping, we examine the characteristics. We write the characteristics as x = s ht () where ht ()= cos() t sin() t This is a family of curves obtained by stretching x = h(t) in the x-direction: They intersect at and only at h(t c ) = 0.

7 ht ( c )= 0 <==> cos( t c ) sin( t c )= 0 <==> + 1 cos( t c + 0 )= 0 where 0 satisfies cos( 0 )= 1 + 1, sin ( 0 )= + 1 No matter what value takes, sooner or later we will have cos( t c + 0 )= 0. At that point, the characteristics intersect. For = 1, 0 = 4 Characteristics are x = s cos t + 4 Starting at t = 0, characteristics intersect at t c + 4 = ==> t c = /4. The classical solution is defined only up to t c = /4. (Draw the curve x = x 0 cos t + 4 to illustrate the family of characteristics) For = 1, 0 = 4 Characteristics are x = s cos t 4 Starting at t = 0, characteristics intersect at t c 4 = ==> t c = 3/4. The classical solution is defined only up to t c = 3/4. (Draw the curve x = x 0 cos t 4 to illustrate the family of characteristics) Note: In this example, characteristics intersect for any value of. ( ) Inverse mapping of x( s, ), t s, - 7 -

8 = t x s = cos() t sin() t = x + 1 cos t + 0 Of course, the inverse mapping is well defined only for cos( t + 0 ) 0. Solution in (x, t): We arrive at u( x, t)= s sin()+ t cos() t = x sin ()+ t cos t cos() t sin t = x tan ( t + 0 ) = x sin t + 0 cos t + 0 u( x, t)= x tan( t + 0 ) where 0 satisfies cos( 0 )= Note: As t 0, we have u( x, t). () () 1 + 1, sin ( 0 )= + 1 Let us change the PDE just a little bit (on the right hand side, change x to x). Example: u u u t x = x u( x,0)= x Characteristics: dt d = 1 dx d = u du d = x Approach #1 d x du = d d = x Approach # with initial condition t ( 0)= t 0 x( 0)= x 0 u( 0)= u 0-8 -

9 d( x+ u) = u x = ( x + u) d d d x u = u + x = ( x u) The solution is (See Appendix B for derivation of the solution) t = + t 0 x = x + u 0 0 exp( )+ x u 0 0 exp( ) u = x + u 0 0 exp( ) x u 0 0 exp( ) Additional condition in parametric form: x 0 ( s)= s t 0 ( s)= 0 u( x 0 ( s), t 0 ( s) )= s Set t 0 = t 0 ( s)= 0, x 0 = x 0 ( s)= s and u 0 = f( s)= s, we obtain Solution parameterized in (s,) t = x = s exp( )+ exp ( ) u = s exp( ) exp ( ) Before we look at the inverse mapping, we examine the characteristics. We write the characteristics as x = s ht () 1+ 1 where ht ()= exp( t)+ exp() t This is a family of curves obtained by stretching x = ht () in the x-direction: They intersect at and only at h(t c ) = 0. ht ( c )= 0 <==> 1 + exp t 1 ( c )+ exp ( t c )= 0 <==> ( 1+ )+ ( 1 )exp( t c )= 0 (Eq. 1) - 9 -

10 For = -1, We have 1 + = 0, 1 =. (Eq. 1) becomes exp( t c )= 0 ==> h(t c ) = 0 has no solution. Characteristics do not intersect. The classical solution is defined for all t. For -1 < 1, We have 1 + > 0, 1 0. (Eq. 1) becomes ( positive)+ ( non-negative)exp( t c )= 0 ==> h(t c ) = 0 has no solution. Characteristics do not intersect. The classical solution is defined for all t. For < -1, We have 1 + < 0, 1 > 0, 0< ( 1 + )< 1. (Eq. 1) becomes 1 + exp( t c )= 1 < 1 ==> h(t c ) = 0 has a solution t c = log 1 < 0 Starting at t = 0, characteristics do not intersect for t > 0. The classical solution is defined for all t > 0. For > 1, We have 1+ > 0, 1 < 0, 0< ( 1 )< 1+ (Eq. 1) becomes exp( t c )= > 1 ==> h(t c ) = 0 has a solution t c = 1 log > 0 Starting at t = 0, characteristics intersect at t c = 1 log 1 + ( 1 ) solution is defined only up to that time (Draw function ht ()= exp( t)+ exp () t and draw characteristics {x = x 0 h(t)} for various values of ) > 0. The classical

11 = -: = -1/: ()= 1 exp ( t )+ 3 exp() t ht Starting at h(0) = 1, h(t) increases monotonically to infinity. ht ()= 1 exp ( t 4 )+ 3 4 exp() t Starting at h(0) = 1, h(t) increases monotonically to infinity. = 0: ht ()= 1 exp ( t )+ 1 exp() t Starting at h(0) = 1, h(t) increases monotonically to infinity (with h (0) = 0). = +1/: ht ()= 3 exp ( t 4 )+ 1 4 exp() t = +1: = +: h(t) is always positive for all t. Starting at h(0) = 1, h(t) decreases initially. Then it increases to infinity. ht ()= exp( t) h(t) is always positive for all t. ht ()= 3 exp ( t ) 1 exp() t Starting at h(0) = 1, h(t) decreases monotonically and exponentially to negative infinity. h(t) hits zero at t c = log 3 > 0. ( ) Inverse mapping of x( s, ), t s, = t x s = 1+ 1 exp( t)+ exp t Solution in (x, t): u( x, t)= x () 1+ 1 exp( t) exp t 1+ 1 exp( t)+ exp t exp( t) = x exp( t)+ Note: in this example, characteristics intersect only for > 1. () ()

12 Let us see another example of characteristics intersecting at finite time. Example: u t + u u x = 0 u( x,0)= u 0 x Initial condition in parametric form: x 0 ( s)= s t 0 ( s)= 0 u( x 0 ( s), t 0 ( s) )= u 0 s Characteristics: dt d = 1 dx d = u du d = 0 Solution in parametric form: t( s, )= x( s, )= u 0 ( s) + s u( s, )= u 0 ( s) 1, s < 1, u 0 ( s)= s 1 < s < 1 1 s > 1 with initial condition We examine the characteristics in the (x, t) space. x = s + t, s < 1 x = s st, 1 < s < 1 x = s t, s > 1 The characteristics intersect at t = 1. (Draw characteristics up to t = 1). Inverse mapping of x( s, ), t( s, ) for t < 1: x t, x < 1+ t x s( x, t)=, 1 + t < x < 1 t 1 t x + t x > 1 t ( x, t)= t x( 0)= s t ( 0)= 0 u( 0)= u 0 s - 1 -

13 Solution in (x, t) for t < 1: u( x, t)= 1, x < 1 + t x, 1 + t < x < 1 t 1 t 1 x > 1 t (Illustrate the solution at t = 0, t = 0.5 and t = 1-) Appendix A: dt d = 1 dx d = u du d = x solution of with initial condition The first differential equation gives us t = + t 0. t ( 0)= t 0 x( 0)= x 0 u( 0)= u 0 Differentiating the second equation and using the third equation, we get d x du = d d = x ==> x() is a linear combination of cos() and sin(). x = c 1 cos( )+ c sin( ) u = dx d = c 1 sin c cos We enforce the initial conditions for x and u. x( 0)= x 0 ==> c 1 = x 0 u( 0)= u 0 ==> c = u 0 Therefore, we obtain t = + t 0 x = x 0 cos( ) u 0 sin u = x 0 sin( )+ u 0 cos

14 Appendix B: dt d = 1 dx d = u du d = x solution of with initial condition The first differential equation gives us t = + t 0. t ( 0)= t 0 x( 0)= x 0 u( 0)= u 0 Differentiating the second equation and using the third equation, we get d x du = d d = x ==> x() is a linear combination of exp(-) and exp(). x = c 1 exp( )+ c exp( ) u = dx d = c 1 exp c exp We enforce the initial conditions for x and u. x( 0)= x 0 u( 0)= u 0 The solution is t = + t 0 x = x 0 + u 0 u = x 0 + u 0 ==> c 1 + c = x 0 c 1 c = u 0 exp( )+ x 0 u 0 exp( ) exp( ) x 0 u 0 exp( ) ==> c 1 = x + u 0 0 c = x u