salim-barzani

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1 years, 41 days

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These are questions asked by salim-barzani

is been a while i work on a test still i am study and there is a lot paper remain and is so important in PDEs, a lot paper explain in 2003 untill know and there is other way to find it too but i choose a easy one but is 2025 paper  which is explanation is so beeter than other paper, also some people write a package for take out this test with a second but maybe is not work for all i  search for that  but i didn't find it i will ask the question how we can find thus as shown in graph i did my train but need a little help while i am collect more information and style of solving 

Download p1.mw

In thus manuscript i got some reviewer comment which is asked to simplify this expresion and there is a lot of them maybe if i do by hand i  made a mistake becuase a lot of variable so how i can fix this issue and make thus square root are very simple as they demand

restart

B[2] := 0

0

(1)

K := sqrt(-(1/2)*sqrt(2)*sqrt(lambda*a[5]/a[4])+sqrt(-a[5]/(2*a[4]))*(B[1]*sqrt(-lambda)*sinh(xi*sqrt(-lambda))+B[2]*sqrt(-lambda)*cosh(xi*sqrt(-lambda)))/(B[1]*cosh(xi*sqrt(-lambda))+B[2]*sinh(xi*sqrt(-lambda))+mu/lambda)+sqrt(-(lambda^2*B[1]^2*a[5]-lambda^2*B[2]^2*a[5]-mu^2*a[5])/(2*lambda*a[4]))/(B[1]*cosh(xi*sqrt(-lambda))+B[2]*sinh(xi*sqrt(-lambda))+mu/lambda))*exp(I*(k*(xi+v*tau^alpha/alpha)+w*tau^alpha/alpha+gamma))

(1/2)*(-2*2^(1/2)*(lambda*a[5]/a[4])^(1/2)+2*(-2*a[5]/a[4])^(1/2)*B[1]*(-lambda)^(1/2)*sinh(xi*(-lambda)^(1/2))/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda)+2*(-2*(lambda^2*B[1]^2*a[5]-mu^2*a[5])/(lambda*a[4]))^(1/2)/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda))^(1/2)*exp(I*(k*(xi+v*tau^alpha/alpha)+w*tau^alpha/alpha+gamma))

(2)

simplify(K)

(1/2)*exp(I*((k*v+w)*tau^alpha+alpha*(k*xi+gamma))/alpha)*2^(3/4)*((lambda*(a[5]*(-lambda^2*B[1]^2+mu^2)/(lambda*a[4]))^(1/2)+(-B[1]*cosh(xi*(-lambda)^(1/2))*lambda-mu)*(lambda*a[5]/a[4])^(1/2)+sinh(xi*(-lambda)^(1/2))*lambda*(-a[5]/a[4])^(1/2)*(-lambda)^(1/2)*B[1])/(B[1]*cosh(xi*(-lambda)^(1/2))*lambda+mu))^(1/2)

(3)

subsindets(K, `&*`(rational, anything^(1/2)), proc (u) options operator, arrow; (u^2)^(1/2) end proc)

(1/2)*(-2*2^(1/2)*(lambda*a[5]/a[4])^(1/2)+2*(-2*a[5]/a[4])^(1/2)*B[1]*(-lambda)^(1/2)*sinh(xi*(-lambda)^(1/2))/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda)+2*(-2*(lambda^2*B[1]^2*a[5]-mu^2*a[5])/(lambda*a[4]))^(1/2)/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda))^(1/2)*exp(I*(k*(xi+v*tau^alpha/alpha)+w*tau^alpha/alpha+gamma))

(4)

latex(%)

\frac{\sqrt{-2 \sqrt{2}\, \sqrt{\frac{\lambda  a_{5}}{a_{4}}}+\frac{2 \sqrt{-\frac{2 a_{5}}{a_{4}}}\, B_{1} \sqrt{-\lambda}\, \sinh \left(\xi  \sqrt{-\lambda}\right)}{B_{1} \cosh \left(\xi  \sqrt{-\lambda}\right)+\frac{\mu}{\lambda}}+\frac{2 \sqrt{-\frac{2 \left(\lambda^{2} B_{1}^{2} a_{5}-\mu^{2} a_{5}\right)}{\lambda  a_{4}}}}{B_{1} \cosh \left(\xi  \sqrt{-\lambda}\right)+\frac{\mu}{\lambda}}}\, {\mathrm e}^{\mathrm{I} \left(k \left(\xi +\frac{v \,\tau^{\alpha}}{\alpha}\right)+\frac{w \,\tau^{\alpha}}{\alpha}+\gamma \right)}}{2}

 

KK := sqrt(-(1/2)*sqrt(2)*sqrt(lambda*a[5]/a[4])+sqrt(-a[5]/(2*a[4]))*(B[1]*sqrt(-lambda)*sinh(xi*sqrt(-lambda))+B[2]*sqrt(-lambda)*cosh(xi*sqrt(-lambda)))/(B[1]*cosh(xi*sqrt(-lambda))+B[2]*sinh(xi*sqrt(-lambda))+mu/lambda)+sqrt(-(lambda^2*B[1]^2*a[5]-lambda^2*B[2]^2*a[5]-mu^2*a[5])/(2*lambda*a[4]))/(B[1]*cosh(xi*sqrt(-lambda))+B[2]*sinh(xi*sqrt(-lambda))+mu/lambda))*exp(I*(k*(xi+v*tau^alpha/alpha)+w*tau^alpha/alpha+gamma))

(1/2)*(-2*2^(1/2)*(lambda*a[5]/a[4])^(1/2)+2*(-2*a[5]/a[4])^(1/2)*B[1]*(-lambda)^(1/2)*sinh(xi*(-lambda)^(1/2))/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda)+2*(-2*(lambda^2*B[1]^2*a[5]-mu^2*a[5])/(lambda*a[4]))^(1/2)/(B[1]*cosh(xi*(-lambda)^(1/2))+mu/lambda))^(1/2)*exp((k*(xi+v*tau^alpha/alpha)+w*tau^alpha/alpha+gamma)*I)

(5)

latex(KK)

\frac{\sqrt{-2 \sqrt{2}\, \sqrt{\frac{\lambda  a_{5}}{a_{4}}}+\frac{2 \sqrt{-\frac{2 a_{5}}{a_{4}}}\, B_{1} \sqrt{-\lambda}\, \sinh \left(\xi  \sqrt{-\lambda}\right)}{B_{1} \cosh \left(\xi  \sqrt{-\lambda}\right)+\frac{\mu}{\lambda}}+\frac{2 \sqrt{-\frac{2 \left(\lambda^{2} B_{1}^{2} a_{5}-\mu^{2} a_{5}\right)}{\lambda  a_{4}}}}{B_{1} \cosh \left(\xi  \sqrt{-\lambda}\right)+\frac{\mu}{\lambda}}}\, {\mathrm e}^{\mathrm{I} \left(k \left(\xi +\frac{v \,\tau^{\alpha}}{\alpha}\right)+\frac{w \,\tau^{\alpha}}{\alpha}+\gamma \right)}}{2}

 

NULL

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i  can determine the pdes by one variable which is work so good but in some of the pdes i have two function i can separate by hand but how i can do by maple?

Download linear.mw

i did a lot of trail to avoid for find my parameter in the last step of that i get this `[Length of output exceeds limit of 1000000]` and i don't know how to fix it i need to find that parameter but when i do substitution  is said this there is any way for hundle this situation 

help-parameter.mw

in here i did all steps but i know something is missing becuase in finding parameter all x and y and t must be remove but still one x is remain so something is wrong with my substitution for finding parameter where is mistake ?

n1.mw

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