Mathematics
Need explanations in solving partial differential equations

Question Description

I have 2 study guides with questions an full solutions to Chapters 5, 6, and 12 of my Partial Differential Equations textbook but I am having trouble understanding how certain steps are derived.

I need detailed, dumbed down explanations for all my questions on the study guides which are labeled in red font with a " Q. " before it.

Chapter 5 & 6 Study Guide.docx 

Chapter 12.docx 

I would like this to be completed by Wednesday - 8:00 am EST. Thank you.

P.S. I'm very slow when it comes to PDE's so please explain in really simple terms that a total math illiterate would understand.

Unformatted Attachment Preview

Chapter 5 & 6 – Study Guide 5.3.3 Q. Why do we have to divide p’ by p? 5.3.5 Q. Where did cosine come from? Is it because B=0 in part a? Q. Where did the 2L/(n-1) come from? How would I know that is has that period? Q. Does that mean there are no zero eigenvalues, can ∅ = constant? Why or why not? 5.3.8 5.3.9 Q. What does “equidimensional equation” mean? Q. Why is the ∅ = xp used, how do we know this? Q. Is it trivial if c and d both equal 0, so we reject it? 5.4.1  u ( x, t ) =  cn e − ntn ( x ) n =1 Q. Where did this formula come from? where 1  2  ... are the eigenvalues of the Sturm-Liouville eigenvalue problem ( K 0  ) +  +  c = 0,  ( 0) =  ( L ) = 0, 1 , 2 ,... are the corresponding eigenfunctions, and L  f ( x )  ( x ) c ( x )  ( x ) dx n cn = 0 L   ( x ) c ( x )  ( x ) dx 2 n 0 Q. Is this a formula I should know? What do cn, c(x), and p(x) mean? If 1  0 then lim u ( x, t ) = 0 t →+ A) We are seeking for the solution as a superposition of solutions u ( x, t ) =  ( x ) G ( t ) with separated variables of the differential equation that satisfy the boundary conditions. Substituting u ( x, t ) =  ( x ) G ( t ) into the given PDE gives: ( K ( x )  ( x )) G (t ) +  ( x ) G (t ) = c ( x )  ( x ) ( x ) G (t ) 0 ( K ( x )  ( x ) ) +  ( x ) = G ( t ) 0 c ( x )  ( x ) ( x ) G (t ) Since the left hand side is function of x only and the right hand side is function of y only, we can deduce that: ( K ( x )  ( x ) ) +  ( x ) = G ( t ) = − 0 c ( x )  ( x ) ( x ) G (t ) where  is constant. Therefore, ( K0 ) +  +  c = 0, G = −G Conversely, if G and  are solutions to the above pair of ODEs, then u ( x, t ) = G ( t )  ( x ) is solution of given PDE. Substituting u ( x, t ) = G ( t )  ( x ) into the boundary conditions get:  ( L ) G (t ) = 0,  ( 0) G ( t ) = 0, Assume G ( t )  0 , then boundary conditions become  ( 0) =  ( L ) = 0 Consider the regular Sturm-Liouville eigenvalue problem ( K 0  ) +  +  c = 0, This problem has infinitely many eigenvalues 1  2  ...  ( 0) =  ( L ) = 0 Denote n to be the eigenfunction corresponding to n . According to Rayleigh quotient L 2 2  − K 0nn +   K 0 n −  n  dx 0  0 L n = L  c  2 n dx 0 Because n ( 0) = n ( L ) = 0 , the non-integral term L − K 0nn = − K 0 ( L ) n ( L ) n ( L ) + K 0 ( 0 ) n ( 0 ) n ( 0 ) = 0 0 vanishes. Therefore, if   0 ,  = −  then it follows that 2   +  n 2  dx K  0 n  0   L n = L  c  n 2 0 dx 0 B) The function G can be determined from G = −G , the general solution of this equation is G ( t ) = c0e−t , where c0 is a constant. Thus we obtain the following solutions: un ( x, t ) = e−ntn ( x ) , n = 1, 2,...   n =1 n =1 u ( x, t ) =  un ( x, t ) = cn e − ntn ( x ) with constants c1 , c2 ,..., cn ,... . Substituting the series into initial condition u(x,0) = f(x), we have:  f ( x ) =  cnn ( x ) n =1 where L  f ( x )  ( x ) c ( x )  ( x ) dx n cn = 0 L   ( x ) c ( x )  ( x ) dx 2 n 0 C) If all eigenvalues are positive then the solution u(x,t) vanishes as t → + because each term of the corresponding series contains decaying factor e − n . 5.4.3 5.5.1 bc 5.5.2 Q. How did we get L and what is a differential operator? Q. What was the point of simplifying the right side of the equation? 5.5.9 Q. Why and how did we expand ∅(d4∅/dx4)? 5.5.15 Define the differential operator L as following: L= d  d  r  dr  dr  Q. Why is L defined like the one above? So the differential equation can be written as: L ( ) +  r = 0 Let the eigenfunctions n ( r ) and m ( r ) correspond to two different eigenvalues n and m . Therefore, n L (m ) − m L (n ) = n d  dm  d  dn  r  − m  r  dr  dr  dr  dr  = d   dm   dn  dm  d   dv   dm  dn  n  r m  r  − r −  + r  dr   dr   dr  dr  dr   dr   dr  dr  = d   dm   d   dv   n  r m  r  −  dr   dr   dr   dr   = d   d   dm r  n − m n    dr   dr dr   Integrating over the interval defined by the boundary conditions 1 d   d 0 n L (m ) − m L (n ) dr = r  n drm − m drn  0 1 And substituting in L ( ) = − r gives 1 d   d 0  −nm rm + mn rn dr = r  n drm − m drn  0 1 1   ( n − m )  nm rdr = 1 n (1) 0 dm d d d (1) − m (1) n (1)  − 0  n ( 0 ) m ( 0 ) − m ( 0 ) n ( 0 )  = 0 dr dr dr dr    Since n  m , n − m  0 , we have 1  r  n m dr = 0 0 This relation tells that the eigenfunctions corresponding to different eigenvalues are orthogonal with weighting function r. 5.6.1 ac Q, How can I find a good trial function, it seems difficult? Q. How can I find a trial solution that satisfies the boundary conditions? 5.6.4 (a) The Rayleigh quotient relates an eigenfunction  to the corresponding eigenvalues  1 −r  0 +  r   ( r ) dr 2 1 = 0 1  r  (r ) 2 dr 0 The boundary condition implies that: −r  0 = −1 (1)   (1) + 0 ( 0 )   ( 0 ) = 0 1 It follows that   0 . If  = 0 , then 1  r (r ) 0 2 dr = 0    ( r ) = 0   ( r ) is constant function over 0 < r < 1, contradiction. Q. Why is it a contradiction, can’t ∅(r) be a constant? Therefore,   0 (b) The problem d d  d  (1) = 0 has infinitely many eigenvalues r  +  r = 0 ,  ( 0 )  , dr dr  dr  1  2   n  Assume that the eigenfunctions are known, according to the Rayleigh quotient: 1 1 n = −rnn +  r n ( r ) dr 0 2 0 1  r  (r ) n 2 dr 0 The non-integral term −rnn 0 = −1n (1) n (1) + 0n ( 0 ) n ( 0 ) = 0 vanishes 1 Then the solution is:   ( r ) =  cnn ( r ) n =1 where L   ( r )  ( r ) rdr n cn = 0 L   (r ) n 0 using orthogonality. 5.7.1 2 rdr Q. Where did 1 < c2 < 1+a2 come from, wasn’t c2 = 1+4a2(x- ½)2? 5.7.3 (a) Assume that the solution is u ( x, t ) =  ( x ) G ( t ) The given PDE becomes:  2u  2u 2 = c x ( ) 2 t 2 x  G ( t )  ( x ) = c 2 ( x ) G ( t )   ( x )  c2 ( x )   ( x ) G ( t ) = = −  ( x ) G (t ) 2 c ( x )   ( x ) +  ( x ) = 0,   G ( t ) + G ( t ) = 0 This problem has infinitely many eigenvalues 1  2  ...  n  The differential operator L is defined as: L = c2 ( x )  d2 dx 2 Therefore, L ( ) +  = 0 Let the eigenfunctions n ( x ) and m ( x ) correspond to two different eigenvalues n and m . Therefore, d 2m d 2n 2 n L (m ) − m L (n ) = n  c ( x )  2 − m  c ( x )  2 dx dx 2  = c 2 ( x ) n   = c2 ( x ) d 2m d 2n  −   m  dx 2 dx 2  d  d  d m n  − m  n   dx  dx dx  Integrating over the interval defined by the boundary conditions b  n L (m ) − m L (n ) c2 ( x ) a b d   d dx =  n m − m n  dx  a  dx And substituting in L ( ) = − gives  −n mm + mnn dx =   dm −  dn  m  n  a c2 ( x ) dx  a  dx b b 1 c ( x) 2 ( n − m )  nm dx = b  n ( b ) b a  dm d d d ( b ) − m ( a ) n ( a )  − a  n ( a ) m ( a ) − m ( a ) n ( a )  = 0 dx dx dx dx    Since n  m , n − m  0 , we have b   ( x )  ( x ) dx = 0 n m a This relation tells that the eigenfunctions corresponding to different eigenvalues are orthogonal with weighting function = 1. (b) The Reyleigh quotient for the eigenfunctions states that: d d −  +  c2 ( x ) dx dx a a dx b = 2 b b  u ( x) 2 dx a According to the boundary conditions of the given one-dimensional wave equation, we always have that: b du du du −u  = −u ( b )  ( b ) + u ( a )  ( a ) = 0 dx a dx dx Hence, b = 2  c ( x) a d dx dx b 2   ( x) 2 0 dx a If  = 0 , then b  c ( x) ( x) 2 2 dx = 0    ( x ) = 0   ( r ) is constant function over (a, b), a contradiction. Therefore,   0 (c) Since   0 , the equation G ( t ) + G ( t ) = 0 has the following solution: G ( t ) = c1 cos ( t ) + c sin ( t ) 2 Thus we obtain the following solutions: ( un ( x, t ) = c1 cos ( t ) + c sin ( t )) ( x ) , n = 1, 2,...   n =1 n =1 2 ( u ( x, t ) =  un ( x, t ) = c1n cos with constants c1n , c2 n ,..., . n ( t ) ( x ) + c n 2n sin ( t ) ( x )) n 5.8.3 Q. What does “transcendental equation” mean? Q. How do I know what the horizontal axis equal and the bounds too? 5.8.8 Q. What is Green’s Formula? Q. How did we get the ∅n? Q. What formula is this, the An? 5.8.11 Q. Isn’t it usually = 0, < 0, > 0. Why are there 5’s? 6.2.5 Q. How is this sum obtained, where do the 8’s come from? Q. How did we get the magnitude of error? Chapter 12 – Final Exam EX. 12.2.1 EX. 12.2.2 Q. Where did the w, s, and the x+3t all come from? Q. What was the point of integrating dx/dt and dt/ds when ultimately we integrate dw/ds? Q. I thought t = 3s +x, why does t = x+3t? Q. 12.3.1 Q. 12.3.3 Q. Where’s the xx after the u? Q. Why is this a solution? Q. Where did that formula come from and how did we get this? Q. 12.3.4 ...
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