Dr. David Harrington

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15 years, 22 days
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada

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I am a professor of chemistry at the University of Victoria, BC, Canada, where my research areas are electrochemistry and surface science. I have been a user of Maple since about 1992.

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These are replies submitted by dharr

The code you copy-pasted earlier worked OK as I already showed. Now you have edited that out of your question and are asking something else that is not clear. Suggest you upload your worksheet - use the large green uparrow to load the worksheet with the problem and someone can take a look at it.

@mehdi jafari I agree that it is not in general, but it is if tau=0 and sqrt(sigma1^2) = sigma1, i.e., if sigma1 is positive, which is an important case. 

Suggest you upload your worksheet using the large green up-arrow. Then it will be easier for someone to diagnose.

@arshl If you just want to solve the pde with the given initial and boundary conditions, then laplace/fourier is not the way I would go in this case and I do not know how to do this. Maple's pdsolve uses many methods that could help you solve it without transforms.

For the initial condition, you need to pay attention to @Rouben Rostamian's use of Heaviside, and then you can get the right result:

@arshl You have incorrectly factored out u from diff(u(x,t)*diff(u(x,t),x),x). But in any case, finding laplace/fourier transforms of products such as u(x,t)*diff(u(x,t),x,x) or (diff(u(x,t),x,x)^2 is problematic, and Maple is not succeeding here.

@Carl Love Yes, that was careless. I redid with Laplace, but last term might still have a problem. Not sure what assumptions are required.

@Carl Love Well, the PDF has all the information, though you have to recognize it as the normal distirbution. There are other things you can get (as on the Distribution help page), but as far as I can see, you can't back out the name of the distribution. I had the same hope as you, but it was dashed...

@one pound A lot depends on what you want to do. The sin-cos form with only positive frequencies is equivalent to the exp form with positive and negative frequencies. I usually work with the series with regular Maple commands such as integration to find the coefficients for a given periodic function.

Download Fourier.mw

@Carl Love Isn't the whole process called "changing variables" :)

@Nadeem_Malik Perhaps our responses got out of sequence, but you still haven't assigned the result of Infinitesimals to a variable and then used it for InfinitesimalGenerator. See worksheet Wsheet3.mw in my earlier response.

@Nadeem_Malik You didn't assign the infinitesimal output to the variable infies before using it. Here there is only one output so you don't need indexing. But this gives a new technical error, which I don't understand because I don't know much about Lie Algebra.


@Nadeem_Malik use variables or ctrl-L





u(x, t)*`will now be displayed as`*u


table( [(  ) = u(x, t) ] )


PDE := U[x,x]-U[t];

diff(diff(u(x, t), x), x)-(diff(u(x, t), t))


infies:=Infinitesimals(PDE);  #put the sequence of answers in a variable

[_xi[x](x, t, u) = 0, _xi[t](x, t, u) = 1, _eta[u](x, t, u) = 0], [_xi[x](x, t, u) = 1, _xi[t](x, t, u) = 0, _eta[u](x, t, u) = 0], [_xi[x](x, t, u) = 0, _xi[t](x, t, u) = 0, _eta[u](x, t, u) = u], [_xi[x](x, t, u) = (1/2)*x, _xi[t](x, t, u) = t, _eta[u](x, t, u) = 0], [_xi[x](x, t, u) = -2*t, _xi[t](x, t, u) = 0, _eta[u](x, t, u) = x*u], [_xi[x](x, t, u) = 0, _xi[t](x, t, u) = 0, _eta[u](x, t, u) = exp(_c[1]*t)*exp(_c[1]^(1/2)*x)], [_xi[x](x, t, u) = 0, _xi[t](x, t, u) = 0, _eta[u](x, t, u) = exp(_c[1]*t)/exp(_c[1]^(1/2)*x)], [_xi[x](x, t, u) = (1/2)*t*x, _xi[t](x, t, u) = (1/2)*t^2, _eta[u](x, t, u) = -(1/8)*x^2*u-(1/4)*u*t]


infies[-1];  #the last one

[_xi[x](x, t, u) = (1/2)*t*x, _xi[t](x, t, u) = (1/2)*t^2, _eta[u](x, t, u) = -(1/8)*x^2*u-(1/4)*u*t]


InfinitesimalGenerator(infies[-1],u(x,t)); # use it

proc (f) options operator, arrow; (1/2)*t*x*(diff(f, x))+(1/2)*t^2*(diff(f, t))+(-(1/8)*x^2*u-(1/4)*u*t)*(diff(f, u)) end proc


InfinitesimalGenerator((3)[-1],u(x,t)); # or use ctrl-L to get the equation number

proc (f) options operator, arrow; (1/2)*t*x*(diff(f, x))+(1/2)*t^2*(diff(f, t))+(-(1/8)*x^2*u-(1/4)*u*t)*(diff(f, u)) end proc




Download Infinitesimals2.mw


@ahmadtalaei  These cannot be converted to LegendreP because the first two arguments are not in the form [-a,a+1] for any a, as required for LegendreP. So these hypergeometric functions come from a differential equation that is not equivalent to the LegnedreP one.


sol[1] := hypergeom([3/4+(1/4)*sqrt(1+4*_c[1]), 3/4-(1/4)*sqrt(1+4*_c[1])], [1/2], cos(phi)^2)+hypergeom([5/4+(1/4)*sqrt(1+4*_c[1]), 5/4-(1/4)*sqrt(1+4*_c[1])], [3/2], cos(phi)^2)

hypergeom([3/4+(1/4)*(1+4*_c[1])^(1/2), 3/4-(1/4)*(1+4*_c[1])^(1/2)], [1/2], cos(phi)^2)+hypergeom([5/4+(1/4)*(1+4*_c[1])^(1/2), 5/4-(1/4)*(1+4*_c[1])^(1/2)], [3/2], cos(phi)^2)

X1 := op(op(1, sol[1]))[1]; Equate(%, [-a, a+1]); solve(%, allsolutions)

[3/4+(1/4)*(1+4*_c[1])^(1/2), 3/4-(1/4)*(1+4*_c[1])^(1/2)]

[3/4+(1/4)*(1+4*_c[1])^(1/2) = -a, 3/4-(1/4)*(1+4*_c[1])^(1/2) = a+1]

X2 := op(op(2, sol[1]))[1]; Equate(%, [-a, a+1]); solve(%, allsolutions)

[5/4+(1/4)*(1+4*_c[1])^(1/2), 5/4-(1/4)*(1+4*_c[1])^(1/2)]

[5/4+(1/4)*(1+4*_c[1])^(1/2) = -a, 5/4-(1/4)*(1+4*_c[1])^(1/2) = a+1]

FunctionAdvisor(definition, LegendreP)

[LegendreP(a, z) = hypergeom([-a, a+1], [1], 1/2-(1/2)*z), MathematicalFunctions:-`with no restrictions on `(a, z)], [LegendreP(a, b, z) = (z+1)^((1/2)*b)*hypergeom([-a, a+1], [1-b], 1/2-(1/2)*z)/((z-1)^((1/2)*b)*GAMMA(1-b)), (1-b)::(Not(nonposint))]



Download nosolutions.mw

@ahmadtalaei  Your last expression can be converted back to LegendreP in version 2017.3 using convert(expr,LegendreP)


You need to upload your worksheet using the green up-arrow. At the moment, I can't reproduce the solution of the differential equation shown in the link, and odeadvisor doesn't suggest Legendre. So maybe there is a typing error, which is why the others are suggesting you definitely need to be clear about what expression you want converted. In principle, convert(expr,StandardFunctions) with assumptions on n should do what you want.

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