Therefore, when we invert the transform, using the Laplace table:įor x ∞, we assume that u(x,t) 0. So we have an ODE in the variable x together with some boundary conditions. The boundary conditions become U(0,s) = U(l,s) = 1/s. Respect to t, the Laplace transform U(x,s) satisfies U(0,t) = u(l,t) = 1, u(x,0) = 1 + sin(πx/l)Īnd noting that the partials with respect to x commute with the transforms with Now, we can use the inverse Laplace Transform with respect to s to find Solving this as an ODE of variable x, U(x,s)=c(s)e-x + 1/s2 Partials with respect to “x” do not disappear) with boundary condition Taking the Laplace of the initial equation leaves Ux+ U=1/s2 (note that the
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