Question #204415

Q. Construct the solution of the heat equation using separation of variables method:

u_xx=4u_t , 0<x<40, t>0

u(0, t)=0, u(40, t)=0, t>0

u(x,0)= x, 0 less than or equal x less than or equal 20, u(x,0)=40-x, 20 less than or equal x less than or equal 40


1
Expert's answer
2021-06-09T16:17:46-0400

TT=XX=λ\frac{T'}{T}=\frac{X''}{X}=\lambda

0=u(0,t)=X(0)T(t)=X(0)=00=u(0,t)=X(0)T(t)=X(0)=0

0=ux(40,t)=X(40)T(t)=X(40)=00=u_x(40,t)=X'(40)T(t)=X'(40)=0

T(t)=λT(t)T'(t) = λT(t)

X(0)=X(40)=0.X(0) = X' (40) = 0.

X(40)=0=μC1cos(40μ)=0X'(40)=0=\mu C_1cos(40\mu )=0

λn=(n12)2λn =-(n-\frac{1}{2})^2

Xn(x)=sin(n12)xX_n(x)=sin(n- \frac{1}{2})^x

un(x,t)=Xn(x)Tn(t)=sin(n12)x)exp((n12)2t)u_n(x,t)=X_n(x)T_n(t)=sin(n-\frac{1}{2})x )exp(-(n-\frac{1}2)^2t)

un(x,t)=n1sin(n12)x)exp((n12)2t)u_n(x,t)=\sum_{n-1}^\infty sin(n-\frac{1}{2})x )exp(-(n-\frac{1}2)^2t)

f(x)=n1bnsin((n12)x)f(x)=\sum^\infty_{n-1}b_n sin((n-\frac{1}{2})x)

bn=2L0Lf(x)sin((n12)x)dx=040f(x)sin((n12)x)dxb_n=\frac{2}L \int_0^Lf(x) sin((n-\frac{1}2)x)dx=\int_0^{40} f(x)sin((n- \frac{1}2)x)dx

040sin((n12)x)sin((m12)x)dx=[200\int_0^{40} sin((n-\frac{1}2)x)sin((m-\frac{1}2)x)dx=[_{20}^0

bn=240040f(x)sin((n12)x)dx=640040sin(5x2)sin((n12)x)dxb_n=\frac{2}{40} \int _0^{40}f(x) sin ((n-\frac{1}2)x)dx=\frac{6}{40}\int_0^{40} sin(\frac{5x}2)sin((n- \frac{1}2)x)dx

bn=640040sin(100x)sin((n12)x)dx=[03b_n=\frac{6}{40} \int_0^{40} sin(100x)sin((n-\frac{1}2)x)dx=[_0^3

f(x)=n=1bnsin((n12)x)f(x)=\sum_{n=1}^\infty b_nsin((n-\frac{1}2)x)

u(x,t)=3sin(5x2)exp((12)t)u(x,t)=3 sin (\frac{5x}{2})exp (-(\frac{1}{2})t)



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