> <\body> |>> <\table-of-contents|toc> |.>>>>|> 1-D Wave Equation> |.>>>>|> Boundary conditions |.>>>>|> > Method of Spherical Means |.>>>>|> > Wave equation in > |.>>>>|> > Method of spherical means |.>>>>|> > Hadamard's Method of Descent |.>>>>|> > Hadamard's Solution for all odd 3> |.>>>>|> > Distributions> |.>>>>|> The Schwartz Class |.>>>>|> > Fourier Transform |.>>>>|> > Basic Estimates |.>>>>|> > Symmetries and the Fourier Transform |.>>>>|> > Inversion Formula |.>>>>|> > More about the Wave Equation> |.>>>>|> Duhamel's Principle |.>>>>|> > Hyperbolicity and the Standard Problem |.>>>>|> > \; Send corrections to . <\equation> u=cu=0 for \> an d 0> with , (x,0)=g(x)>. D'Alembert's formula: <\equation*> u(x,t)=f(x+c*t)+f(x-c*t)+g(y) d y. Geometric identity: <\equation> u(A)+u(C)=u(B)+u(D). |gr-frame|>|gr-geometry||gr-grid||gr-edit-grid-aspect|||>|gr-edit-grid||gr-line-arrows|none||||>>>||>>||||>>>||>>||>||>||>||>||>||>||>||>|>||>>>|Sketch for the geometric identity.> We have: > solution of ()>() for every characteristic parallogram. Good and bad boundary conditions: <\equation*> 0=u+c*u, supposing 0>. |gr-frame|>|gr-geometry||gr-edit-grid-aspect|||>|gr-line-arrows|||>>>|gr-color|default|gr-line-width|default||>||||>>>||>>||||>>>||>>||>>|||>>||||>>>||>>||>||||>>>||>>||>|||>>||||>>>||>>||||>>>||>>||||>>>||>>||||>>>||>>|>|>||||>>>||>>>>|Good and bad boundary conditions for the transport equation.> Example: <\equation*> u-cu=0,x\(0,\),t\0 , (x,0)=g(x)> for \>. for 0> with the assumption that . |gr-frame|>|gr-geometry||gr-edit-grid-aspect|||>|gr-line-arrows|none||||>>>||>>||||>>>||>>||>||>|||>|||>|||>||||>>|Domain of dependence.> The dependency on ICs outside of the domain is solved by the . Extend to all of >, say >. <\equation*> (x,t)=(x+c*t)+(x-c*t)+(y) d y. \; <\equation*> (0,t)=(c*t)+(c*t)+(y) d y. Choose odd extension: <\equation*> (x,t)=|0,>>||0.>>>>> Similarly for >, >. Then (0,t)=0=u(0,t)>. (x,t)> for 0>. |gr-frame|>|gr-geometry||gr-edit-grid-aspect|||>|gr-line-arrows|none||||>>>||>>||||>>>||>>||>||>||>|||>|0>|>|0>|>>>|Different cases arising for the determination of the domain of dependence.> Case 1: D'Alembert as before. Case 2: <\equation*> u(x,t)=f(x+c*t)+>>+g(y) d y. If 0>, this corresponds to reflection as follows: |gr-frame|>|gr-geometry||gr-as-visual-grid|on|gr-edit-grid-aspect|||>|gr-line-arrows|||>>>||||>>>||>>||||>>>||>>||||>>>||>>|||||||||>||||>>>||>>||||>>>||>>|||||||>||||>>>||>>||||>>>||>>||||||>||||>>>||>>>>|Series of snapshots of solutions with .> Initial boundary value problem: |gr-frame|>|gr-geometry||gr-edit-grid-aspect|||>|gr-line-arrows|none|gr-line-width|default|gr-color|default||||>>>||>>||||>>>||>>||||>>|||>|||>||||>||>||>||>|(t)>|>|(t)>|>>>|Initial boundary value problem. We can satisfy the parallelogram identity using geometry.> For arbitrary ,\> the equation need not have a continuous solution: |gr-frame|>|gr-geometry||gr-edit-grid-aspect|||>|gr-line-arrows|none||||>>>||>>||||>>>||>>||||>||||>||>||>||>||>|>|>|>|>|>|>|>|>|>>>|Discontinuous solutions in corners.> Assume C((0,L]\(0,\))>. <\eqnarray*> ||(B),>>|||>>> . D\u(A)\u(D)>, lim0>\(t)=lim0>f(x)>. Similarly, if we want C>, this requires (0)=g(0)>, etc. <\equation*> \u-c\u=0 for all \> and 0> with <\eqnarray*> ||>|(x,0)>||>>> If \\>, let <\eqnarray*> (x,r)>||r>h(y)d S>>|||>\|=1>h(x+r\)d S>.>>>> Assume that is continuous. Then <\enumerate> 0>M(x,r)=h(x)> for every \>. (x,r)> is a continuous and even function. If C(\)>, then <\equation*> \M(x,r)=|\r>M+M|\>. If you view > as a function :\\\\\> which is spherically symmetric, then the above equation states that the Laplacian in the first variables equals the Laplacian in the second . Spherical means of <\equation*> \u-c\u=0. Then <\equation*> \M-c\M=0 and <\equation*> \M-|\M>+M|\r>=0. >> <\equation-lab> \u\u-c\u=0 >> for \\(0,\)> with and =g> for \> and . Now do Fourier analysis: If hL(\)>, consider <\equation*> (\)\>e>h(x)d x. If we take the FT of )>, we get <\equation*> +c\|\\|=0 for \\> and 0>, (\,0)=>, (\,0)=>. (\,t)=A*cos(c\|\\|t)+B*sin(c\|\\|t)>. Use ICs to find <\eqnarray*> (\,t)>||(\)cos(c\|\\|t)+(\)*\|t)|c\|\\|>.>>>> <\quote-env> Analogous caclulation for heat equation: <\equation*> u-u=0\+\|\\|=0,(\,0)= yields (\,t)=e\|t>(\)>. Then observe that multiplication becomes convolution. Observe that <\equation*> cos(c\|\\|t)=\\|t)|c\|\\|>. If we could find a such that <\equation*> \|t)|c\|\\|>=)>>e\x>k(x,t)d x, this would lead to a solution formula <\equation*> u(x,t)=>k(x-y,t)g(y)d y+\>k(x-y,t)f(y)d y. Suppose , we know that our solution formula must coincide with D'Alembert's formula <\equation*> u(x,t)=f(x+c*t)+f(x-c*t)+g(y) d y. Here <\eqnarray*> ||\c*t}>,>>|k(x,t)>||\+\.>>>> Solution formula for : <\theorem> C>(\\\)> is a solution to the wave equation with >> initial data , if and only if <\equation*> u(x,t)=| ->t*g(y)+f(y)+D f(y)(y-x)d S. Here, <\equation*> k(x,t)=ct>\d S=t\>. \; <\big-figure> |gr-frame|>|gr-geometry||gr-edit-grid-aspect|||>|gr-line-arrows|none|||>>>||>>||>||>|||>>>||>>||>||>|>||>||>||>>> <\definition> Suppose \\> is continuous. Define :\\\\\> by <\equation*> M(x,r)=| ->h(y)d S=>\1>h(x+r\)\d\. Notice that <\equation*> lim0>M(x,r)=h(x) if is continuous.\ Suppose C(\)>. Then <\equation*> \M(x)=|\r>M+\M|r>. <\proof> Similar to the mean value property for Laplace's equation. <\eqnarray*> \M(x,\)\\d\>||\>\|=1>h(x+\\)\d\*\d\>>|||\h(x+y)\d y=>h|\n>(x+y)d y>>|,d y=rd\)>||>D h(x+y)\n d y>>||||\>\|=1>(h(x+r\)\d\=rM|\r>.>>>> Then <\equation*> \M(x,\)\\d\=rM. Differentiate <\eqnarray*> Mr>||r\|d r>>>|||\|d r>+(n-1)r|d r>.>>>> Altogether <\equation*> \M=M|\r>+M|\r>. Look at spherical means of )>: <\equation*> u-c\u=0 Assume C(\\(0,\))>. Take spherical means: <\equation*> M>=(M), which means <\eqnarray*> | ->u(y,t)d S>||| ->\u(y,t)d y,>>|)>||>.>>>> And <\equation*> M(\u)>M|\r>+(n-1)M|\r>. Therefore, we have <\equation*> (M)=cM|\r>+(n-1)M|\r>. If , we can solve by D'Alembert. For : <\eqnarray*> |\r>(r*M)>|||\r>rM|\r>+M=rM|\r>+2\M|\r>.>>>> So if , we have <\equation*> (r*M)=c|\r>(r*M) This is a 1D wave equation (in !). Solve for > by D'Alembert. <\eqnarray*> (x,r,t)>||(r+c*t)M(x,r+c*t)+>>M(x,r-c*t)+rM(x,r)d r|\>>>>>> Pass to limit 0> in b) <\eqnarray*> rM(x,r)d r>||rM(x,r)d r>>>> > is even, > is odd. So <\equation*> lim0>=\c*t*M(x,c*t)=t*M(x,c*t). <\equation*> t*M(x,c*t)=t| ->g(y)d S. Similarly, a): (> even in ) <\eqnarray*> ||M(x,r+c*t)+M(x,c*t-r)+c*t*M(x,c*t+r)-M(x,c*t-r)>>|0>\>||(x,c*t)+c*t*\M(x,c*t)=\(t*M(x,c*t)).>>>> For any \C>(\)> define <\equation*> (K\\)(x)\t\(y)d S. Then if C>>, our solution to u=0> is <\equation*> ||||||||\g)(x)+\(K\f)(x).>>>>> Aside: Check that <\equation*> e\y>d S=\|)|c\|\\|>. <\remark> Huygens' principle: |Huygens' principle.> We consider data with compact support. Let <\equation*> \(t)=supp(u(x,t))\\, where obviously <\equation*> \(0)=supp(f)\supp(g). Then Huygens' principle is stated as <\equation*> \(t)\{x:dist(x,\(0))=c*t}. \; <\example> Consider radial data and 0>. <\equation*> u(x,t)=t| ->g(y)d S. <\equation*> u(x,t)\0\S(x,c*t)\B(0,\)\\. |gr-frame|>|gr-geometry||gr-edit-grid-aspect|||>|gr-line-arrows|none|gr-line-width|2ln|gr-color|default|gr-dash-style||gr-text-at-halign|center||||>>>||>>||||>>>||>>||>>|||>>|||>>|||>>|||>>|>|>>|>|>>||>>||>>|>|>>|>|>>>>|How radial data spreads in time.> Assume , radial. <\eqnarray*> ||(t*M(x,c*t))=M(x,c*t)+t*\M(x,c*t)>>|M(x,c*t)>||| ->f(y)d S>>|||| ->\|=1>f(x+c*t*\)d \>>|||| ->\|=1>D f(x+c*t\)\(c*\)d\>>|||| ->\|=1>f|\n>>(x+c*t\)d\.>>>> <\big-figure> a) b) . b) The sharp dropoff in .> <\equation*> u(x,t)=| ->f(y)*d S+c*t| ->f|\n>d S. Thus <\equation*> \<\|\|\>u(x,t)\<\|\|\>>\C\<\|\|\>u(x,0)\<\|\|\>>. More precisely, there exists a sequence >\C>(\)> and >> such that <\equation*> lim\0>\|u>(x,t>)\||sup\|u>(x)>=+\. Contrast with solution in : <\equation*> \<\|\|\>S(t)u\<\|\|\>>\\<\|\|\>u\<\|\|\>>,1\p\\, where is the shift operator. ``'' solution to the wave equation. (t)=>solution operator for wave equation in >. <\equation*> supL(\)>S(t)u\<\|\|\>>|\<\|\|\>u\<\|\|\>>>=+\. Treat as 3-dimensional wave equation. \>, =(x,x)\\>. If \\>, define :\\\> by ()=((x,x))=h(x)>. Suppose solves u-c\u=0> for \> and 0> with and (x,t)=g(x)>. Then <\eqnarray*> -c\>>||>|(,0)>||(x)>>|(,0)>||(x)>>>> for \\>, 0>. <\equation*> (,)=\(\)+\, where <\equation*> = <\eqnarray*> \>||| ->-\|=c*t>(y)d S>>|||| ->|~>\|=1>(x+c*t*|~>)d |~>.>>>> with |~>\\=(\,\)> for \\>. Then <\equation*> (+c*t*|~>)=h(x+c*t*\). <\equation*> | ->|~>\|=1>h(x+c*t*\*d |~>. |~>=(\,\)>. On |~>\|=1>, we have <\equation*> \=\\|>=\+\)>. Then <\equation*> \|\\>=|\|>> for . Thus the Jacobian is <\equation*> \|\\>+\|\\>>=\|>>. Thus <\equation*> t| ->|~>\|=1>h(x+c*t*\)d|~>=>\|\1>)|\|>>d\d\. |gr-frame|>|gr-geometry||gr-edit-grid-aspect|||>|gr-line-arrows|none|gr-line-width|2ln|gr-fill-color|blue|gr-color|default||||>>>||>>||||>>>||>>||||>>>||>>||||>>>||>>|>|||>>|||>>|>>>|Domains of dependence, conceptually, for 2D and 3D.> 3>> [cf. Evans, 2.4] , 1>. , . The general formula is <\equation*> u(x,t)=\(K\f)+K\g where for any C>> we have <\equation*> (K\h)(x)=|\2>\|\t>t| ->h(y)d S. Check: If , =4\>, so we get our usual formula. Now, Consider 0> in -\u=0>, \>, 0>, , (x,0)=0>. Extend to 0> by (which is OK because u=0> at ) Consider for 0> <\eqnarray*> |>|t)>>e/4 t>u(x,s)d s>>|||>k(s,t)u(x,s)*d s(\)>>>> Find solution for the heat equation in 1D. Use that k=\k>. <\eqnarray*> v>||>\k*u(x,s)d s>>|||>k(s,t)\u(x,s)d s>>|||>k(s,t)\u(x,s)d s=\>k(s,t)u(x,s)d s.>>>> v=\v>, \>, 0>. Also, as 0>, f(x)>. Therefore, <\eqnarray*> ||t)>>e/4t>f(x-y)d y>>|||t)>>e/4t>r| ->f(x-r\)\d\*d r>>||||(4\t)>>e/4t>rM(x,r)d r(#)>>>> Change variables using =1/4t> and equate )> and : <\equation*> >er>u(x,r)d r=|2>\>>er>\rM(x,r)d r. Then, use the Laplace transform for L(\)>: <\equation*> h(\)=>e\>h(\)d\. Basic fact: > is invertible. Observe that <\equation*> (er>)=-\er>. In particular, <\equation*> -\er>=\er>. Therefore <\eqnarray*> >\er>rM(x,r)d r>|||2>>\er>(rM(x,r))d r>>|||>>er>r\\(rM(x,r))d r.>>>> Now have Laplace transforms on both sides, use uniqueness of the Laplace transform to find <\eqnarray*> |||\2>t\|\t>tM(x,t)>>||||\2>t\|\t>tM(x,t)>>>> Let \> be open. <\definition> The set of is the set of >(U)> (>> with compact support). The topology on this set is given by \\> in iff <\enumerate-alpha> there is a fixed compact set U> such that \F> for every \|\>\-\>\\|\0> for every multi-index >. <\definition> A is a continuous linear functional on . We write D(U)> and ||>>. <\definition> [Convergence on >] A sequence |D>L> iff |\||>\||>> for every test function >. <\example> (U)\{f:U\\:f , >\|f\|d x\\ \U\\U}.> An example of this is > and >.> We associate to every L(U)> a distribution > (here: p\\>). <\equation*> |\||>\f(x)\(x)d x. Suppose |D>\>. Need to check <\equation*> |\||>\|\||>. Since \F\\U>, we have <\eqnarray*> |\||>-|\||>>||f(x)(\-\(x))d x>>||>|\|f(x)\|d x|\>>\|\-\\||\>0>.>>>> If p>, <\eqnarray*> \|f(x)\|d x>|>|1 d x\|f(x)\|.>>>> Thus, (U)\L(U)> for every q>. ( This is true for (U)>.) <\example> If > is a on , then we can define <\equation*> >|\||>=\(x)\(d x). <\example> If =\>, <\equation*> >|\||>=\(y). <\definition> If is a distribution, we define >L> for every multi-index > by <\equation*> >L|\||>\(-1)\|>>\||>. This definition is motivated through integration by parts, noting that the boundary terms do not matter since we are on a bounded domain. <\example> If is generated by >, <\equation*> >L|\||>=(-1)>\>\(0). <\theorem> >:D\D> is continuous. That is, if |D>L>, then >L|D>\>L>. <\proof> Fix \D(U)>. Consider <\eqnarray*> >L|\||>>|>|>L|\||>>>|>||\<\|\|\>>>|>|\>\||>>|>|>>\||>.>>>> <\definition> Suppose is a partial differential operator of order , that is <\equation*> P=\|\N>c>(x)\> with >\C>(U)>. <\example> > is an operator of order 2. -\>. -c\>. Fundamental solution for >: <\equation*> \K(x-y)=\ D. All this means is for every \D> <\equation*> \K(x-y)\(x)d x=\(x)\(d x)=\(y). <\definition> We say that solves in > iff <\equation*> >\||>=0 for every test function >. Here, >> is the adjoint operator obtained through integration by parts: If >(x)=c>> independent of , then <\equation*> P>=\|\N>(-1)\|>c>\>. <\example> -D>>>=-\-\>. <\example> More nontrivial examples of distributions: <\enumerate> >:> <\equation*> \lim\0>\> \ \(x)d x. <\equation*> ||>=>|d x>\, which is well-defined because > has compact support. > <\definition> (\)> Set \C>(\)> with rapid decay: <\equation*> \<\|\|\>\\<\|\|\>,\>\sup\|x>\>\(x)\|\\ for all multiindices ,\>. Topology on this class: \\> on (\)> iff \-\\<\|\|\>,\>\0> for all ,\>. <\example> If \D(\)> then \\(\)>. If \\> in )>>\\> in (\)>. <\example> (x)=e>> is in (\)>, but not in )>. <\equation*> \>\(x)=>(x)|\>>e>, so x>\>\(x)\<\|\|\>>(\)>\\>. <\example> )>>\\(\)> for every \0>. <\example> <\equation*> )>\C>, but not in (\)> for any . For example, <\equation*> sup>|(1+\|x\|)>=\ if =(3N,0,\,0)>. We can define a on (\)>: <\equation*> \(\,\)=>>\|+\|\\|=k>\-\\<\|\|\>,\>|1+\<\|\|\>\\\\<\|\|\>,\>>. Claim: \\> in (\)>>(\,\)\0>. <\theorem> (\)> is a complete metric space. <\proof> Arzelà-Ascoli. For the wave equation, we find formally that <\equation*> \K=\|t|c\|\\|>. <\definition> The Fourier transform on (\)> is given by <\equation*> (\\)(\)=)>>e\>\(x)d x. For brevity, also let |^>(\)=(\\)(\)>. <\theorem> > is an isomorphism of (\)>, and \>=Id>, where <\equation*> (\>\)(\)=(\\)(-\). <\eqnarray*> |^>(\)\|>|>|)>\|\(x)\|\x>>|||)>>(1+\|x\|)(x)\||(1+\|x\|)>\x>>||>|(1+\|x\|)\(x)\<\|\|\>>\\.>>>> Also, <\eqnarray*> >>|^>(\)>||)>>\>>e\>\(x)\x>>|||)>>(-i*x)>e\>\(x)\x>>|\<\|\|\>\>>|^>(\)\<\|\|\>>>>|>|(1+\|x\|)x>\\<\|\|\>>>.>>>> Thus show |^>\C>(\)>: <\eqnarray*> )>|^>(\)>||)>>(-i\)>e\>\(x)\x>>|||)>>\>(e\>)\(x)\x>>|||\|>|(2\)>>e\>\>\(x)\x>>|\<\|\|\>\>|^>(\)\<\|\|\>>>>|>|(1+\|x\|)\>\\<\|\|\>>>.>>>> Combine both estimates to find <\eqnarray*> |^>\<\|\|\>,\>=\<\|\|\>\>\>>|^>\<\|\|\>>>>|>|(1+\|x\|)x>\>\\<\|\|\>>>.>>>> <\example> If (x)=e/2>>. Then |^>(\)=e\|/2>>. \=\>. <\enumerate> >\)(x)=\(x/\)>. <\eqnarray*> (\(x/\))(\)>|||\>>e\>\(x/\)\(x/\)=\(\\)(\\).>>>> Thus >\|^>=\\>|^>>. \(x)=\(x-h)> for \>. (\\)(\)=e\>|^>(\)>. For every \\(\)> <\equation*> \(x)=)>>e\>|^>(\)\\. (x)=\>|^>=(\\)|^>>, where \)(x)=\(-x)>. <\proof> (of Schwartz's Theorem) Show >\e/2>=e/2>>. Extend to dilations and translations. Thus find >\=Id> on >, because it is so on a dense subset. > is 1-1, >> is onto>but >=\\>, so the claim is proven. <\theorem> > defines a continuous linear operator from (\)\L>(\)>, with <\equation*> \<\|\|\>\<\|\|\>>>\)>\<\|\|\>f\<\|\|\>>. <\theorem> > defines an isometry of (\)>. <\theorem> > defines a continuous linear operator from (\)\L>(\)> with p\2> and <\equation*> +>=1. Ideas: <\itemize> Show (\)> dense in (\)> with p\\>. Extend > from > to >. <\proposition> >(\)> is dense in (\)>. <\proof> Take a function <\equation*> \(x)\|N-1,>>||N+1.>>>>> Given \\(\)>, consider \\\>. <\eqnarray*> >\>||>(\\)=\|\\|\\|>\>\\-\>\.>>>> So x>\>\\<\|\|\>>>\\>. <\theorem> >(\)> is dense in (\)> for p\\>. <\proof> . Choose \C>(\)> with )\B(0,1)> and <\equation*> >\(x)\x=1. For any , define (x)=N\(N*x)>. Then <\equation*> >\(x)\x=1. To show: <\equation*> f\\|L>f for any L(\)>. Suppose (x)> for a rectangle . In this case, we know \f=f> at any with Q)\1/N>. Therefore, \f\f> a.e. as \>. <\equation*> >\|\\f(x)-f(x)\|\x\0 by Dominated Convergence. (>> in (\)>>. (?>) Given \\(\)>, consider \\\>. We have \-\\<\|\|\>,\>\0> for every ,\>. In particular, we have <\equation*> \<\|\|\>(\|x\|+1)(\-\)\<\|\|\>>>\0. <\eqnarray*> >\|\-\\|\x>||>|(1+x)>\|\-\\|\x\>>\x\?>>>> End aside.) (\)>> for p\\>. , i.e. <\equation*> \<\|\|\>f\\\<\|\|\>>\C\<\|\|\>f\<\|\|\>>, which we obtain by Young's inequality. <\eqnarray*> \|L|>>|>|\<\|\|\>\\<\|\|\>>\<\|\|\>f\<\|\|\>>>>|||\<\|\|\>\\<\|\|\>>\<\|\|\>f\<\|\|\>>,>>>> where the constant depends on >, but not on . Suppose L(\)>. Pick to be a step function such that f-g\<\|\|\>>\\> for p\\>. Then <\eqnarray*> f\\-f\<\|\|\>>>|>|f\\-g\\\<\|\|\>>+\<\|\|\>g\\-g\<\|\|\>>+\<\|\|\>f-g\<\|\|\>>>>||>|\<\|\|\>\\<\|\|\>>+1)\<\|\|\>f-g\<\|\|\>>+\<\|\|\>g\\-g\<\|\|\>>.>>>> Onwards to prove the > isometry, we define <\equation*> (\)|>\>f(x)>\x. <\proposition> Suppose \(\)>. Then <\equation*> f|\g|L(\)|>=(\)|>. <\proof> \; <\eqnarray*> f|\g|L(\)|>>|>>|>(\)(\)|\>\\>>|||>>(x))>>e\>>(\)\\\x>>|||>f(x)>(x)d x.>>>> <\definition> :L(\)\>(\)> is the extension of :\(\)\\(\)> to (\)>, where <\equation*> >(\)\{h:\\\0> as \>>}. <\proposition> This extension is well-defined. <\proof> Suppose <\eqnarray*> >||L>>|>|>||L>>|>>> Then \\-\\\<\|\|\>\0>: <\eqnarray*> |^>-|^>)(\)\|>||)>>e\>(\-\)>>||>|)>\<\|\|\>\-\\<\|\|\>>>>||>|)>\<\|\|\>\-f\<\|\|\>>+\<\|\|\>f-\\<\|\|\>>\0.>>>> There is something to be proved for (\)> because <\equation*> )>e\>f(x)\x is defined when L(\)>. However f> in the sense of >-lim \> where \\(\)\f> in >. We had proven <\eqnarray*> |L>|>>|>|)>|>,>>||L|>>|||>.>>>> <\definition> A linear operator (\)\\(\)> is of > if <\equation*> |L|>\C(r,s)|L|>. <\example> > is of type )> and . <\theorem> Suppose is of type ,s)> for . Then is of type where <\eqnarray*> >|||r>+|r>,>>|>|||s>+|s>>>>> for \\1>. Moreover, <\equation*> C(r,s)\C>C>. <\proof> Yosida/Hadamard's 3-circle theorem (maximum principle). <\corollary> :\\\> has a unique extension :L(\)\L>(\)> where p\2> and +1/p=1>. <\itemize> :\\\> isomorphism :L\>> (either by extension or directly) not an isomorphism :L\L> (by extension) isomorphism :L\L>> (by interpolation) <\definition> (\)> is the space of continuous linear functionals on (\)>, called the space of . Its topology is given by \L> in > iff <\equation*> |\||>\||> for all \\>. Altogether, we have \\\\D>. <\example> <\enumerate> Suppose L>. Define a tempered distribution <\equation*> ||>\>f\, which is obviously continuous. (A -example) If >>, then L>, so it defines a distribution, but not a distribution. >\\(\)>, but <\equation*> >f\=\. If is such that <\equation*> )f|L|>\\ for some , then \>. <\proof> <\equation*> \|||>\|=f\\)f|L|>)\|L>|>. \; <\proposition> Suppose \>. Then there exists 0>, \> such that <\equation> \|||>\|\C|N|> for every \\(\)>, where <\equation*> |N|>=\|,\|\\|\N>>\>\|L>|>. <\corollary> A distribution D> defines a tempered distribution>there exist , such that () holds for \\(\)>. <\proof> Suppose () is not true. Then there exist >, > such that <\equation*> \|||>\|\k|N|>. Let <\equation*> \\||N|>>\. Then <\equation*> |N|>=\0. But ||>\|\1>. But \0> in (\)>> not continuous. <\definition> If \\> is linear, continuous, then the of is the linear operator such that for every \> <\equation*> ||>=L|\||>. <\theorem> <\enumerate-alpha> (\)> is dense in (\)>. )> is dense in (\)>. <\proof> Mollification, but first verify some properties. Fix \D(\)>, <\equation*> >\=1. Let (x)=m\(m*x)>. We want to say \L> is a >> function for a distribution . <\definition> D(\)>, \D(\)>, \L> is the distribution defined by <\equation*> \L|\||>=)\\||>, where (x)=\(-x)>. If were a function , <\eqnarray*> \L|\||>>||>(\\f)(x)\(x)\x>>|||>>\(x-y)f(y)\y\(x)\x>>|||>>\(x-y)\(x)\xf(y)>>|||>(R\*\\)(y)f(y)\y.>>>> <\theorem> )> is dense in (\)>. That is, if is a distribution, then there exists a sequence of \D> such that \L> in >. <\proof> By 1) Mollification and 2) Truncation. <\proposition> \> is a >> function. More precisely, \> is equivalent to the distribution defined by the >> function <\equation*> \(x)=(R\)||>, where f(y)=f(y-x)>. <\proof> 1) :\\\> is clear. 2) > is continuous: If \x>, then (x)\\(x)>. Check <\equation*> \(x)=>(R\)||>. And >(R\)\\(R\)> >. <\itemize> We can choose s.t. >(R\))\F> for all . (y-x)\R\(y-x)>, >(R\)(y-x)\\>R\(y-x)>, where the last two properties hold uniformly on . 3) \C>: \ Use finite differences. Consider <\equation*> (x+h*e)-\(x)|h>=>(R\)-\(R\)|h>||>. Observe that\ <\equation*> \>(R\)-\(R\)\\(\>R\) in . 4) \C>:> Induction. 5) Show that \>\>. That is <\equation*> \|\||>\\||>>\(x)\(x)\x. <\eqnarray*> >\(x)\(x)\x>||0>hh\>\(y)\(y)>>|||0>hh\>(R\)||>\(y)>>|||0>h\>\(R\)\(y)||>.>>>> Show that the Riemann sum <\equation*> hh\>\(R\)\(y)\R\\\ in . <\with|par-first|0> >:> <\enumerate> \L\L\\>. >(L\\)>\>L\\>L\\>\>. Fix D(\)>. Fix \D(\)> with >\(x)\x=1>. Let (x)=m\(m*x)>. Then <\equation*> >\(x) \x=1. We know from our proposition from that \L> is >>. Consider the cutoff function <\equation*> \(x)\|m,>>||m.>>>>> Consider =\(\\L)>. \D(\)>. <\equation*> \>(\\)=|\>\-\>\\-\>\ Claim: \L> in >. <\eqnarray*> |\||>>||(\\L)|\||>=\L|\\||>>>||>>|)\(\\)||>.>>>> Finally, show <\eqnarray*> )\\ large>>(R\)\\\>|>| D.>>>> <\definition> Suppose \\> is linear. We define :\\\> as the linear operator <\equation*> L|\||>\||>. <\proposition> Suppose \\> is linear and continuous. Suppose that \|>> is continuous. Then, there exists a unique, continuous extension of > to >. <\corollary> :\\\> is continuous. Examples: <\enumerate> \=1/(2\)>. Let \\n> and >=\((n-\)/2)>. Then (C>\|x\|>)=C>\|x\|)>>. Why we care: u=\>. In Fourier space: <\eqnarray*> \|>||)>>>|>||)>\|\\|.>>|\>||)>|C>\|x\|.>>>> Prove (1) and (2) by testing against Gaussians. Consider constant coefficient linear PDE <\equation*> \u+\\|\1>c>\>u+\\|\2>c>\>u+\+\|\m>c>\>u=0. Here \\\\>, is the order of the equation, >\\>. Shorthand )u=0>. Here >,\,\>)> and =\> differentiation operators. <\equation*> P(D,\)=\+\P(D)+\+P(D). (D)=>polynomial in of order k>. <\equation*> P(D,\)u=\ for \>, 0> with\ <\eqnarray*> ||>>|u=\u>||>>|>||>>|u=\u>||>>>> at . <\equation*> P(D,\)u=0 with <\eqnarray*> ||>|u=\u>||>|>||>>|u=\u>||>>> at . (Initial conditions). Solution of General Problem from Standard Problem. First, suppose \0> and =f=\=f=0>. Consider the solution to a family of standard problems: <\eqnarray*> )U(x,t,s)>||(s\t)>>|U(x,t,s)>||(x,s)(t=s)>>|U(x,t,,s)>||(t=s,0\k\m-2)>>>> Consider <\equation*> u(x,t)=U(x,t,s)\s. This gives us <\eqnarray*> )u(x,t)>||P(D,\)U(x,t,s)\s+(\+\P(D)+\+P(D))U(x,t,t)>>|||(x,t)+0>>>> as desired. Similarly, getting rid of non-standard initial conditions involves consideration of <\eqnarray*> )>||>|||>>|u>||>>|>||>>|u>||>>>> Let > dentote the solution to the standard problem. Consider <\eqnarray*> ||>+(\+P(D))u>+(\+P(D)\+P(D))u>+\+(\+P(D)\+\+P(D))u>.>>>> Then <\eqnarray*> )u>||)u>+(\+P(D))P(D,\)u>+\>>|||>>> since )u>=0> for k\m-1>. We need to check the initial conditions: At , u>=0>, l\m-2>. Thus, all terms except the last one are 0. The last term is <\equation*> \+P(D)\+\+P(D)u>=\u>+m-2> >=f. Henceforth, only consider the standard problem <\eqnarray*> )>||>|u(x,0)>||(0\k\m-2),>>|u(x,0)>||>>> Solve by Fourier analysis: <\eqnarray*> (\,t)>||)>>e\>u(x,t)\x.>>>> Fourier transform of the above standard problem yields <\eqnarray*> ,\)>||>|(\,\)>||>|(\,0)>||(\)>>>> Fix > and suppose ,t)> denotes the solution > to the ODE <\equation*> P(i\,\)Z(\,t)=0 with initial conditions <\equation*> \Z(\,0)=0(0\k\m-2),\Z(\,0)=1. This is a constant coefficients ODE, an analytic solution for it exists for all . Clearly, by linearity <\equation*> (\,t)=Z(\,t)(\) and <\equation*> u(x,t)=)>>e\>Z(\,t)(\)\\. We want C> (``classical solution''). Problem: Need to show that ,t)> does not grow too fast (=faster than a polynomial) in >. Formally, <\eqnarray*> >\u(x,t)>||)>>e\>(i\)>\Z(\,t)(\)\\.>>>> Key estimate: For any 0>, there exists >, such that <\equation*> maxk\m>sup\\T>sup\\>\Z(\,t)\C(1+\|\\|) \; <\definition> The above standard problem is called if there exists a > solution for every \(\)>. <\theorem> The problem is hyperbolic iff c\\> such that ,\)\0> for all \\> and > with )\-c>. |gr-frame|>|gr-geometry||gr-edit-grid-aspect|||>|gr-line-arrows|none|gr-line-width|default|gr-fill-color|none||||>>>||>>||||>>>||>>|>|>|>|>|>|||||>||||>||||> for fixed >|>|=-c>|>||>||||>>||>|>||||>|||||>|||>>>|Nice cartoon.> <\proof> Typical solutions to ,\)Z=0> are of the form t>> with ,i\)=0>. We will only prove ``>'': We'll prove the estimate <\equation*> maxk\m>sup\\T>sup\\>Z(\,t)\C(1+\|\\|) assuming ,i\)\0> for )\-c>. Formula for ,t)>: <\equation*> Z(\,t)=>>t>|P(i\,i\)>\\. Claim: ,\)Z=0> (0>), Z=0> (k\m-2>, ), Z=1> (). <\equation*> \Z=>>)et>|P(i\,i\)>\\. Therefore <\eqnarray*> ,\)Z>||>>P(i\,i\)t>|P(i\,i\)>\\>>|||>>et>\\=0>>>> by Cauchy's Theorem. Suppose k\m-2>. Let >t>=1>. <\equation*> \Z=>>)|(i\)1+o\|>>\\. Suppose that > is the circle of radius 1> with center at 0. Then <\eqnarray*> Z\|>|>|>|R1+o>*\2\R=R1+o\0>>>> if m-2>. Thus, ||||Z=0>>>>>> for any > enclosing all roots. When , we have <\eqnarray*> Z>||i>>1+>|\>>>\\=1.>>>> Claim: Any root of ,i\)> staisfies <\equation*> \|\(\)\|\M(1+\|\\|). Estimate growth of roots: Suppose > solves ,i\)=0>. Then <\equation*> (i\)+(i\)P(i\)+\+P(i\)=0. Thus, <\equation*> -(i\)=(i\)P(i\)+(i\)P(i\)+\+P(i\). Observe that <\equation> \|P(i\)\|\C(1+\|\\|) for every , k\m>. Therefore, <\equation*> \|\\|\C\|\\|(1+\|\\|). this implies: <\equation*> \|\\|\(1+C)(1+\|\\|). Let <\equation*> \=\||1+\|\\|>. Then () implies <\equation*> \\C\\\\-1|\-1>(\\1). Cases: <\itemize> \1>>\|\1+\|\\|>>nothing to prove. \1>>\C\/(\-1)>>\1+C>>\|\(1+C)(1+\|\\|)>. Claim: <\equation*> \|\Z(\,t)\|\M*m*e(1+\|\\|). Here =bound from step 2, =order of )>, =constant in Gårding's criterion. |gr-frame|>|gr-geometry||gr-edit-grid-aspect|||>|gr-line-arrows|none|gr-fill-color|default||||>>>||>>||||>>||||>>>||>>||>|>|>|>|>|>|>|>|||>|||>|||>|||>|||>|>|>>>|Sketch.> Fix \\>. Let >=union of circles of unit radius abound each >. (wlog, no > on the boundary, else consider circles of radius >) <\equation*> P(i\,i\)=i(\-\(\)). On > we have -\(\)\|\1> for all >. Therefore ,i\)\|\1> on >. <\equation*> \Z(\,t)=>>)et>|P(i\,i\)>\\ Bound on t>\|> on >. we have )\-c-1> by Gårding's assumption. <\equation*> \|et>\|=e)t>\e. Thus, <\eqnarray*> Z(\,t)\|>|>|>sup\\>\|\\|em)|\>>>>>>||>|sup(\|\(\)\|+1)>>||>|M(1+\|\\|)+1>>>> since each (\)\M(1+\|\\|)>. <\equation*> \|\Z(\,t)\|\C*Mm*e(1+\|\\|). This implies that <\eqnarray*> >\u(x,t)\|>|>|)>>\|\Z(\,t)\|*\|\\|>(\)\\.>>||>|e|(2\)>>(1+\|\\|)\|\\|>\|\|(\)\\\\>>>> because \\>. <\theorem> Assume )> satisfies Gårding's criterion. Then there exist >> solutions for all \(\)>. For finite regularity, we only need check for \|\m>. We need <\equation*> (1+\|\\|)\|(\)\|\L(\). Need for every \0> <\equation*> (1+\|\\|)\|(\)\|\>|(1+\|\\|>)> or <\equation*> (\)\|\C>(1+\|\\|)>. =order of )>=regularity of solution, =space dimension. <\example> -\u=0>. )-(i\|\\|)=0>, =\\|\\|>> Growth estimate can't be improved. Is a hyperbolic equation hyperbolic in the sense that it is ``wavelike'' (meaning if has compact support, has compact support (in ) for each 0>. <\theorem> Suppose L(\)> with compact support. Then :\\\> is . <\proof> <\equation*> (\)=)>e\>g(x)\x. Formally differentiate once, then >> follows. \; <\theorem> Assume Gårding's criterion (restriction on roots). Then there is a >> solution to the standard problem for \(\)>. <\example> \; <\eqnarray*> )u>||-\u>>|,i\)>||+\|\\|>>>> The roots are =\\|\\|>, which satisfies (GC). <\example> Suppose ,i\)> is homogeneous <\equation*> P(i*s\,i*s*\)=sP(i\,i\) for every \>. (GC) holds>all roots are real--otherwise, we can scale them out as far as we need to. In general, we can write <\equation*> P(i\,i\)=p(i\,i\)+\+p(i\,i\), where > is homogeneous of degree . <\corollary> Suppose )> is hyperbolic. Then all roots of (i\,i\)> are real for every \\>. <\corollary> Suppose the roots of > are real and distinct for all \\>. Then is hyperbolic. (=order of ). <\proof> write =\\>, =\\>. where \|=1>, =\|\\|>. <\equation*> P(i\,i\)=0\p(i\,i\)+>p(i\,i\)+\+>p(i\,i\)=0. |gr-frame|>|gr-geometry||gr-grid||gr-edit-grid-aspect|||>|gr-edit-grid||gr-line-arrows|none||||>>>||>>||||>>>||>>||||||||>|||>|||>|||>|||>|)>|>|)>|>|>>|>>>|Illustrative Sketch. :-)> Use the Implicit Function Theorem to deduce that there exists \0> such that each >> perturbs (p)> for \\>. <\equation*> \|\>-\(p)\|\>. We want ),\)=0>. We know ,0)=0>. The distinctness is guaranteed by the derivative condition. <\definition> )> is called if all (\)> are real and distinct. Also say that (D,\)> is if roots are real and distinct. <\example> -\u=0> is strictly hyperbolic. <\example> () -\u+k*u=0> with \>. By Corollary , this equation is hyperbolic. <\theorem> Suppose (D,\)> is strictly hyperbolic. Suppose \(\)> and B(0,a)>. Then there exists a >> such that <\equation*> supp* u\B(0,a+c>t). >> is the largest wave speed. <\proof> (Main ingredients) <\itemize> Paley-Wiener Theorem: Suppose L(\)> and B(0,a)>. Then > extends to an entire function \\> and <\equation*> \|(\+i\)\|\|>|(2\)>e\|>. (Proof see below) Heuristic:\ <\itemize> Decay in >regularity of >. Regularity of >decay of >. Estimates of )> for complex +i\>. Use strict hyperbolicity to show <\equation*> Im(\)\c>(1+\|\\|) for all \\>. Plug into <\equation*> Z(i\,t)=>>t>|P(i\-\,\)>\\. Use <\equation*> u(x,t)=)>>e(\+i\)>Z(\+i\,\)g(\)\\. \; <\proof> (of Paley-Wiener) <\eqnarray*> (\+i\)\|>||)>e(\+i\)>g(x)\x>>||>|)>\|e(\+i\)>\|*\|g(x)\|\x>>||>|)>e\|>\|g(x)\|\x.>>>> <\initial> <\collection> <\references> <\collection> |1>> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > <\auxiliary> <\collection> <\associate|figure> > > > > |g=0>.|> > > > > |g> spreads in time.|> |f>. b) The sharp dropoff in |u(0,t)>.|> > > > > <\associate|toc> |math-font-series||Table of contents> |.>>>>|> |math-font-series||11-D Wave Equation> |.>>>>|> |1.1Boundary conditions |.>>>>|> > |1.2Method of Spherical Means |.>>>>|> > |1.3Wave equation in |\> |.>>>>|> > |1.4Method of spherical means |.>>>>|> > |1.5Hadamard's Method of Descent |.>>>>|> > |1.6Hadamard's Solution for all odd |n\3> |.>>>>|> > |math-font-series||2Distributions> |.>>>>|> |2.1The Schwartz Class |.>>>>|> > |2.2Fourier Transform |.>>>>|> > |2.2.1Basic Estimates |.>>>>|> > |2.2.2Symmetries and the Fourier Transform |.>>>>|> > |2.2.3Inversion Formula |.>>>>|> > |math-font-series||3More about the Wave Equation> |.>>>>|> |3.1Duhamel's Principle |.>>>>|> > |3.2Hyperbolicity and the Standard Problem |.>>>>|> >