Partial Dierential Equations Linear P.D.Es. contains no owers roducts o the deendent variables / an o its derivatives can occasionall be solved. Consider eamle ( ) a (sometimes written as a ) we can integrate to get a We need a boundar condition to comlete the solution e.g. i (o course i ( ) a ) ( ) ( ) at ( ) ( ) a ( ) What i Let Similarl t α β s γ δ α γ β δ α. ( β ) ( γ δ )
Take δ γ but the Jacobian must not vanish ( t s) ( ) α β αδ γβ γ δ Within the constraint we are ree to take α β γ δ Then ( t s) ( ) t ( s) ( γ ) ( ) δ As another eamle consider Let where Assume c c c c c ( α β) g( γ δ) Recall rom the Jacobian that αδ βγ. The derivatives are: c α γg c α γ g c αβ γδg Then c β δg c β δ g. α γ g αβ γδg β δ g ( αβ β ) ( γ γδ δ ) g α
3 Both coeicients are ero i We can take We can take ( β )( α β ) ( γ δ )( γ δ ) g α. α β δ γ αδ βγ βγ βγ 3βγ c to be an solution o β γ α δ ( ) g( ) F eamle Also ( ) Since which checks. c is the general solution o the equation. ( ) g( )
4 Linear First Order P.D.E. A general st der linear P.D.E. can be written P ( ) Q( ) R ( ) R ( ) R( ) A solution can be written as ( ) c u where cconstant the solution is reerred to as an integral surace. Then u u & u u u / u & u / u u u P Q R u u Pu Qu Ru Let since r r r r V Pi Qj Rk r r r r u u i u j u k V r r u Recall that r u is erendicular to u ( ) c V r is erendicular to r u tangent to u ( ) c. An curve traced out b a article moving such that its direction is along V r on the surace is a characteristic curves. Let ( s) r be a characteristic curve where s is the arc length r d dr d r d r i j k ds ds ds ds which must oint along V r r d r µ V ds d d d ds d d d P µ Q µ R µ ds ds ds µ P Q R There are three equations the last two can be integrated to ield u ( ) C u ( ) C an C C ields a oint on the surace F ( C C ) reresents the surace.
5 Eamle : Consider Since we must have as one equation P Q R. d P d Q d R d d d d ( ) C u the second making Finall becomes Equivalentl Note that in general has the solution d d d d ( ) C u F F ( C C ) ( ) ( ). a b F ( b a ) ( b a)
6 Eamle : Consider P Q R d d d ln ln C ln ln ln C ln C u C F u
7 Second Order Equations The linear equations can be written as: a b c d e g We will be concerned with a b c d e constant (not g). The characteristics are determined b a b& c. Consider Let I d e g a b c ( m) m m m am bm c b ± b 4ac a There are three cases: () b 4ac > is herbolic () b 4ac is arabolic (3) b 4ac < is ellitic ( m ) g( m ). I a b c Note that ( ) ( ) b ( b c ) b c b ( ) Since has a solution Then b c ( m) g
8 mg g ( c) g mb m c / b c g b The comlete solution is I I b 4ac B looking at m m m m Eing in a Tal Series c g b a b c ( ) ( ) ( ) g( ) m m m b a ( m) g [ ( m ) ] g [ ( m ) ] g ( m) g ( m) ( m) g ( m) g ( m) 4444 44443 4 44 3 ( m) ( m) g( m) g 4 ( m)
9 Eamle: Consider Then m ± 4 4 ( ) g( ) Note: Since b 4ac 4 4 the equation is arabolic. Lalace s Equation Eamle: making Then Note that a b c b 4ac < ± 4 m ± ±i. ( i) g( i) i( i) i ( i) ( i) i i
Wave Equation The wave equation is Hence a b c b 4ac > m ± 4 ± ( ) g( ) Usuall the wave equation is written as is subject to initial conditions ϕ c ϕ ϕ ( ) F( ) ϕ ( ) G( ) t Letting ct ϕ ( t) ( ct) g( ct) F the initial conditions this ields ϕ t ϕ ( ) F( ) ( ) g( ) ( t) c ( ct) cg ( ct) Hence ϕ ( ) G( ) c[ ( ) g ( ) ] t
Integrating d C c ( ) g( ) G( ) But ( ) g( ) F( ) Hence c ( ) F( ) G( ) d C g ( ) F ( ) G( ) d C c Then ct ct ϕ ( t) F( ct) G( ) d C ( ) ( ) F ct G d C c c ϕ c ct ( t) [ F( ct) F( ct) ] G( ) ct d
Constant Coeicient Case F the case o all the derivatives resent (but the coeicients constant) a b c d e Assume now Solving β α β Ae aα bαβ cβ dα eβ ( bα e) β aα dα cβ α α ϕ ( α ) ( ) β ϕ α ϕ ( α ) Ae α Be ϕ ( α ) α α ϕ ( α ) α ϕ ( α ) [ A( α ) e B( α ) e ] dα Note this is not a general solution! Eamle: Consider α β α β β α α α ( ) ( α e ) A α 4443 4 ( ) ( α β )( α β ) α ( ) ( α ) e B α 44 443 e g ( ) ( ) e g( ) This does require some imagination!
3. Find the general solution o. Find the general solution o 3. Find the solution o that satisies all t satisies < < l. ϕ Homewk No. ϕ c ϕ ( t) ϕ ( l t) ϕ ( ) F ( ) ϕ ( ) G( ) t a. Show the boundar conditions at l result in ( u) g( u) (where u ct ) ( u l) ( u) (where u ct l ) show the initial conditions result in ( u) g( u) F( u) c '( u) g '( u) G( u) < u < l. b. Show that i let H '( u) G( u) that the initial conditions become ( u) F( u) H ( u) c g( u) F( u) H ( u) c < u < l. Also show that all u ( u ) is eriodic with eriod l that ( u) g '( u). c. Show that all the conditions are satisied i we set ( u) ( u) ( u) F c H ( ) g u ( u ) ( u ) F H c u G u are odd eriodic unctions with eriod l. all u where F ( ) ( ) Note that F ( u) G ( u) are equal to F ( u ) ( ) that H '( u) G ( u). The general solution becomes G u in < u < l ct ϕ ( t) ( ct) ( ct ) ( ξ ) dξ F F c G. ct